Principles and Applications of Electrica

G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Chapter 1 Instructor Notes Chapter 1 is introductory in nature, establishing some rationale for studying electrical engineering methods, even though the students' primary interest may lie in other areas. The material in this chapter should be included in every syllabus, and can typically be thoroughly covered in a single-day introductory lecture. Oftentimes, reading of this material is left up to the discretion of the student. 1.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Chapter 1 problem solutions treadmill _________________________________ 1.1 Miscellaneous A few examples are: lawn tools Bathroom power tools ventilation fan electric toothbrush _________________________________ hair dryer ________________________________ electric shaver 1.2 Several examples are listed below for each system: electric heater fan Kitchen microwave fan a) microwave turntable A ship Circuit Analysis mixer design of the ship's food processor electrical system blender Electromagnetics coffee grinder radar garbage disposal Solid-State Electronics ceiling fan radio electric clock sonar exhaust fan Electric Machines refrigerator compressor pump dish washer elevator Utility Room Electric Power Systems clothes washer lighting dryer generators air conditioner Digital Logic Circuits furnace blower elevator control pump Computer Systems Family Room navigation VCR drive Communication Systems cassette tape drive radio reel-to-reel tape drive telephone record turntable drive Electro-Optics computer fan Morse light bridge displays 1.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 1 Control Systems Instrumentation compass rudder speed indicator flaps Control Systems c) rudder Household Circuit Analysis HVAC design of the home's electrical system b) A Commercial Passenger Aircraft Electromagnetics Circuit Analysis Design of the plane's microwave oven electrical system stereo speakers Solid-State Electronics Electromagnetics radar television microwave oven stereo VCR Solid-State Electronics Electric Machines radio appliances Electric Machines turbines power tools fans fans Electric Power Systems Electric Power Systems lighting lighting HVAC HVAC receptacles Digital Logic Circuits Digital Logic Circuits seat belts Computer Systems clocks navigation timers Computer Systems Communication Systems radio microwave oven telephone programmable VCR Electro-Optics Communication Systems cockpit displays telephone Instrumentation compass CB radio air speed indicator television inclinometer radio Electro-Optics altimeter 1.3 G. Rizzoni, Principles and Applications of Electrical Engineering digital clocks Instrumentation electric meter Control Systems thermostat _________________________________ _________________________________ 1.3 Some examples are: a) HVAC lighting office equipment typewriter computer copy machine clock stapler shredder elevator b) conveyor punch press lighting ventilation drill press hoist lathe c) power saw drill lighting elevator pump compressor _________________________________ 1.4 Problem solutions, Chapter 1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Chapter 2 Instructor Notes Chapter 2 develops the foundations for the first part of the book. Coverage of the entire Chapter would be typical in an introductory course. The first four sections provide the basic definitions and cover Kirchoff’s Laws and the passive sign convention; the box Focus on Methodology: The Passive Sign Convention (p. 35) and two examples illustrate the latter topic. The sidebars Make The Connection: Mechanical Analog of Voltage Sources (p. 20) and Make The Connection: Hydraulic Analog of Current Sources (p. 22) present the concept of analogies between electrical and other physical domains; these analogies will continue through the first six chapters. Sections 2.5and 2.6 introduce the i-v characteristic and the resistance element. Tables 2.1 and 2.2 on p. 41 summarize the resistivity of common materials and standard resistor values; Table 2.3 on p. 44 provides the resistance of copper wire for various gauges. The sidebar Make The Connection: Electric Circuit Analog of Hydraulic Systems – Fluid Resistance (p. 40) continues the electric-hydraulic system analogy. Finally, Sections 2.7 and 2.8 introduce some basic but important concepts related to ideal and nonideal current sources, and measuring instruments. The Instructor will find that although the material in Chapter 2 is quite basic, it is possible to give an applied flavor to the subject matter by emphasizing a few selected topics in the examples presented in class. In particular, a lecture could be devoted to resistance devices, including the resistive displacement transducer of Focus on Measurements: Resistive throttle position sensor (pp. 52-54), the resistance strain gauges of Focus on Measurements: Resistance strain gauges (pp. 54-55), and Focus on Measurements: The Wheatstone bridge and force measurements (pp. 55-56). The instructor wishing to gain a more in-depth understanding of resistance strain gauges will find a detailed analysis in1. Early motivation for the application of circuit analysis to problems of practical interest to the nonelectrical engineer can be found in the Focus on Measurements: The Wheatstone bridge and force measurements. The Wheatstone bridge material can also serve as an introduction to a laboratory experiment on strain gauges and the measurement of force (see, for example2). Finally, the material on practical measuring instruments in Section2.8b can also motivate a number of useful examples. The homework problems include a variety of practical examples, with emphasis on instrumentation. Problem 2.36 illustrates analysis related to fuses; problems 2.44-47 are related to wire gauges; problem 2.52 discusses the thermistor; problems 2.54 and 2.55 discuss moving coil meters; problems 2.52 and 2.53 illustrate calculations related to temperature sensors; an problems 2.56-66 present a variety of problems related to practical measuring devices. It has been the author's experience that providing the students with an early introduction to practical applications of electrical engineering to their own disciplines can increase the interest level in the course significantly. Learning Objectives 1. Identify the principal elements of electrical circuits: nodes, loops, meshes, branches, and voltage and current sources. 2. Apply Kirchhoff’s Laws to simple electrical circuits and derive the basic circuit equations. 3. Apply the passive sign convention and compute power dissipated by circuit elements. 4. Apply the voltage and current divider laws to calculate unknown variables in simple series, parallel and series-parallel circuits. 5. Understand the rules for connecting electrical measuring instruments to electrical circuits for the measurement of voltage, current, and power. 1 E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990. 2 G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rd Edition, Kendall-Hunt, 1998. 2.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Section 2.1: Definitions Problem 2.1 Solution: Known quantities: Initial Coulombic potential energy, potential energy, Vi = 17kJ / C ; initial velocity, U i = 93M m ; final Coulombic s V f = 6kJ / C . Find: The change in velocity of the electron. Assumptions: ∆PE g << ∆PE c Analysis: Using the first law of thermodynamics, we obtain the final velocity of the electron: Qheat − W = ∆KE + ∆PE c + ∆PE g + ... Heat is not applicable to a single particle. W=0 since no external forces are applied. ∆KE = −∆PE c 1 me (U 2f − U i2 ) = −Qe (V f − Vi ) 2 2Q U 2f = U i2 − e (V f − Vi ) me ( ) m · 2 − 1.6 × 10 −19 C § = ¨ 93 M ¸ − (6kV − 17kV ) 9.11× 10 −37 g s¹ © 2 2 m2 15 m − 3 . 864 × 10 s2 s2 m U f = 6.917 × 10 7 s m m m U f − U i = 93 M − 69.17 M = 23.83 M . s s s = 8.649 × 1015 ________________________________________________________________________ Problem 2.2 Solution: Known quantities: MKSQ units. Find: Equivalent units of volt, ampere and ohm. 2.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: Joule J V= Coulomb C Coulomb C Current = Ampere = a= second s Volt Joule × second Resistance = Ohm = = Ampere Coulomb 2 Voltage = Volt = Conductance = Siemen or Mho = Ω= J ⋅s C2 Ampere C 2 = Volt J ⋅s ________________________________________________________________________ Problem 2.3 Solution: Known quantities: Battery nominal rate of 100 A-h. Find: a) Charge potentially derived from the battery b) Electrons contained in that charge. Assumptions: Battery fully charged. Analysis: a) C· s · § § 100 A × 1hr = ¨100 ¸(1hr )¨ 3600 ¸ s¹ hr ¹ © © = 360000 C b) charge on electron: − 1.602 × 10 −19 C no. of electrons = 360 × 103 C = 224.7 × 10 22 −19 1.602 × 10 C ________________________________________________________________________ Problem 2.4 Solution: Known quantities: Two-rate change charge cycle shown in Figure P2.4. 2.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: a) The charge transferred to the battery b) The energy transferred to the battery. Analysis: a) To find the charge delivered to the battery during the charge cycle, we examine the charge-current relationship: i= dq dt or dq = i ⋅ dt thus: t1 Q = ³ i (t )dt t0 5hrs 10 hrs Q = ³ 50mAdt + ³ 20mAdt 0 5hrs 18000 s 36000 s = ³ 0.05dt + ³ 0.02dt 0 18000 = 900 + 360 = 1260C b) To find the energy transferred to the battery, we examine the energy relationship p= dw dt or dw = p (t )dt t1 t1 t0 t0 w = ³ p (t )dt = ³ v(t )i (t )dt observing that the energy delivered to the battery is the integral of the power over the charge cycle. Thus, 18000 w= ³ 36000 0.05(1 + 0 = (0.05t + 0.25t 0.75t ) dt ) dt + ³ 0.02(1 + 18000 18000 18000 0.25 2 36000 0.75 2 18000 t ) 0 + (0.02t + t ) 18000 36000 36000 w = 1732.5 J ________________________________________________________________________ Problem 2.5 Solution: Known quantities: Rated voltage of the battery; rated capacity of the battery. Find: a) The rated chemical energy stored in the battery b) The total charge that can be supplied at the rated voltage. 2.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: a) ∆PE c ∆Q I= ∆Q ∆t Chemical energy = ∆PE c ∆V ≡ = ∆V ⋅ ∆Q = ∆V ⋅ (I ⋅ ∆t ) = 12 V 350 a hr 3600 s hr = 15.12 MJ . As the battery discharges, the voltage will decrease below the rated voltage. The remaining chemical energy stored in the battery is less useful or not useful. b) ∆Q is the total charge passing through the battery and gaining 12 J/C of electrical energy. ∆Q = I ⋅ ∆t = 350 a hr = 350 C s hr ⋅ 3600 s hr = 1.26 MC. ________________________________________________________________________ Problem 2.6 Solution: Known quantities: Resistance of external circuit. Find: a) Current supplied by an ideal voltage source b) Voltage supplied by an ideal current source. Assumptions: Ideal voltage and current sources. Analysis: a) An ideal voltage source produces a constant voltage at or below its rated current. Current is determined by the power required by the external circuit (modeled as R). I= Vs R P = Vs ⋅ I b) An ideal current source produces a constant current at or below its rated voltage. Voltage is determined by the power demanded by the external circuit (modeled as R). V = Is ⋅ R P = V ⋅ Is A real source will overheat and, perhaps, burn up if its rated power is exceeded. ________________________________________________________________________ 2.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Sections 2.2, 2.3: KCL, KVL Problem 2.7 Solution: Known quantities: Circuit shown in Figure P2.7 with currents I 0 = −2 A, I1 = −4 A, I S = 8 A, and voltage source VS = 12 V. Find: The unknown currents. Analysis: Applying KCL to node (a) and node (b): °I 2 = −(I 0 + I1 ) = 6 A I 0 + I1 + I 2 = 0 ® ® °̄I 3 = I 0 + I S + I1 = 2 A ¯ I 0 + I S + I1 − I 3 = 0 ________________________________________________________________________ 2.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Section 2.4: Sign Convention Focus on Methodology: Passive Sign Convention 1. 2. 3. 4. Choose an arbitrary direction of current flow. Label polarities of all active elements (current and voltage sources). Assign polarities to all passive elements (resistors and other loads); for passive elements, current always flows into the positive terminal. Compute the power dissipated by each element according to the following rule: If positive current flows into the positive terminal of an element, then the power dissipated is positive (i.e., the element absorbs power); if the current leaves the positive terminal of an element, then the power dissipated is negative (i.e., the element delivers power). Problem 2.8 Solution: Known quantities: Direction and magnitude of the current through the elements in Figure P2.8; voltage at the terminals. Find: Class of the components A and B. Analysis: The current enters the negative terminal of element B and leaves the positive terminal: its coulombic potential energy increases. Element B is a power source. It must be either a voltage source or a current source. The reverse is true for element A. The current loses energy as it flows through element A. Element A could be 1. a resistor or 2. a power source through which current is being forced to flows ‘backwards’. ________________________________________________________________________ Problem 2.9 Solution: Known quantities: Current absorbed by the heater; voltage at which the current is supplied; cost of the energy. Find: a) Power consumption b) Energy dissipated in 24 hr. c) Cost of the Energy Assumptions: The heater works for 24 hours continuously. Analysis: a) 2.7 G. Rizzoni, Principles and Applications of Electrical Engineering P = VI = 110 V (23 a ) = 2.53 K Problem solutions, Chapter 2 J a = 2.53 KW a s b) W = Pt = 2.53 K J s 24 hr 3600 = 218.6 MJ hr s c) Cost = ( Rate)W = 6 cents (2.53 KW )(24 hr ) = 364.3 cents = $3.64 KW hr ________________________________________________________________________ Problem 2.10 Solution: Known quantities: Current through elements A, B and C shown in Figure P2.10; voltage across elements A, B and C. Find: Which components are absorbing power, which are supplying power; verify the conservation of power. Analysis: A absorbs (35 V )(15 A) = 525 W B absorbs (15 V )(15 A) = 225 W C supplies (50 V )(15 A) = 750 W Total power supplied = PC = 750 W Total power absorbed = PB + PA = 225 W + 525 W = 750 W Total power supplied = Total power absorbed, so conservation of power is satisfied. ________________________________________________________________________ Problem 2.11 Solution: Known quantities: Vs = 12V ; internal resistance of the source, Rs = 5kΩ ; and resistance of the load, RL = 7kΩ . Circuit shown in Figure P2.11 with voltage source, Find: The terminal voltage of the source; the power supplied to the circuit, the efficiency of the circuit. Assumptions: Assume that the only loss is due to the internal resistance of the source. 2.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: KVL : − V S + I T RS + VT = 0 OL : VT = I T R L ∴ I T = − VS + VT = VT RL VT RS + VT = 0 RL VS 12 V = = 7V 5 KΩ RS 1+ 1+ 7 KΩ RL or VD : VT = VS R L 12 V 7 KΩ = = 7 V. R S + R L 5 KΩ + 7 KΩ VR2 VT2 (7 V ) PL = = = = 7 mW V RL RL K 7 a P P I 2R 7 KΩ RL = 2 T L2 = η = out = = 0.5833 or 58.33% . Pin PRS + PRL I T RS + I T RL 5 KΩ + 7 KΩ 2 ________________________________________________________________________ Problem 2.12 Solution: Known quantities: Circuit shown in Figure P2.12; Current through elements D and E; voltage across elements B, C and E. Find: a) Which components are absorbing power and which are supplying power b) Verify the conservation of power. Analysis: a) By KCL, the current through element B is 5 A, to the right. By KVL, v D = v E = 10 V (positive at the top) v A + 5 − 10 − 10 = 0 Therefore the voltage across element A is v A = 15 V (positive on top) A supplies (15 V )(5 A) = 75 W B supplies (5 V )(5 A) = 25 W C absorbs (10 V )(5 A) = 50 W D absorbs (10 V )(4 A) = 40 W E absorbs (10 V )(1 A) = 10 W b) Total power supplied = PA + PB = 75 W + 25 W = 100 W 2.9 G. Rizzoni, Principles and Applications of Electrical Engineering Total power absorbed Problem solutions, Chapter 2 = PC + PD + PE = 50 W + 40 W + 10 W = 100 W Total power supplied = Total power absorbed, so conservation of power is satisfied. ________________________________________________________________________ Problem 2.13 Solution: Known quantities: Headlights connected in parallel to a 24-V automotive battery; power absorbed by each headlight. Find: Resistance of each headlight; total resistance seen by the battery. Analysis: Headlight no. 1: P = v × i = 100 W = R= v2 or R v 2 576 = = 5.76Ω 100 100 Headlight no. 2: P = v × i = 75 W = R= v2 or R v 2 576 = = 7.68Ω 75 75 The total resistance is given by the parallel combination: 1 RTOTAL = 1 1 or RTOTAL = 3.29 Ω + 5.76Ω 7.68Ω ________________________________________________________________________ Problem 2.14 Solution: Known quantities: Headlights and 24-V automotive battery of problem 2.13 with 2 15-W taillights added in parallel; power absorbed by each headlight; power absorbed by each taillight. Find: Equivalent resistance seen by the battery. Analysis: The resistance corresponding to a 75-W headlight is: R 75W = v 2 576 = = 7.68Ω 75 75 For each 15-W tail light we compute the resistance: 2.10 G. Rizzoni, Principles and Applications of Electrical Engineering R 15W = Problem solutions, Chapter 2 v 2 576 = = 38.4Ω 15 15 Therefore, the total resistance is computed as: 1 RTOTAL = 1 1 1 1 or RTOTAL = 3.2 Ω + + + 7.68Ω 7.68Ω 38.4Ω 38.4Ω ________________________________________________________________________ Problem 2.15 Solution: Known quantities: Circuit shown in Figure P2.15 with voltage source, Vs = 20V ; and resistor, Ro = 5Ω . Find: The power absorbed by variable resistor R (ranging from 0 to 20 Ω). Analysis: The current flowing clockwise in the series circuit is: i= 20 5+ R The voltage across the variable resistor R, positive on the left, is: v R = Ri = 20 R R+5 2 § 20 · Therefore, PR = v R i = ¨ ¸ R ©5+ R ¹ ________________________________________________________________________ 2.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.16 Solution: Known quantities: Circuit shown in Figure P2.16 with source voltage, Vs = 12V ; internal resistance of the source, Rs = 0.3Ω . Current, I T = 0, 5, 10, 20, 30 A. Find: a) b) c) d) The power supplied by the ideal source as a function of current The power dissipated by the nonideal source as a function of current The power supplied by the source to the circuit Plot the terminal voltage and power supplied to the circuit as a function of current Assumptions: There are no other losses except that on Rs. Analysis: a) Ps = power supplied by the source = VS I S = VS I T . b) Rs = equivalent resistance for internal losses Ploss = I T2 RS c) VT = voltage at the battery terminals: VD : VT = VS − RS I T P0 = power supplied to the circuit ( RL in this case) = I T VT . Conservation of energy: PS = Ploss + P0 . I T ( A) PS (W ) Ploss (W ) VT (V ) P0 (W ) 0 2 5 10 20 30 0 30 60 120 240 360 0 1.875 7.5 30 120 270 0 11.4 10.5 9 6 3 0 28.13 52.5 90 120 90 2.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Terminal Voltage 12 11 10 9 t V (V) 8 7 6 5 4 3 0 5 10 15 It (A) 20 25 30 400 350 300 250 P(W) Pi=Ps 200 150 P0 100 50 0 0 5 10 15 It(a) 20 25 Note that the power supplied to the circuit is maximum when RL = P0 120 Va V = = 30 m = 30 mΩ 2 2 a I T (20 a ) RS = Ploss 120 Va = = 30 mΩ I T2 (20 a ) 2 30 I T = 20a . R L = RS ________________________________________________________________________ 2.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.17 Solution: Known quantities: Circuit shown in Figure P2.17 if the power delivered by the source is 40 mW; the voltage v = v1 /4; and R1 = 8kΩ, R2 = 10kΩ, R3 = 12kΩ Find: The resistance R, the current i and the two voltages v and v1 Analysis: P = v ⋅ i = 40 mW (eq. 1) v v1 = R2 ⋅ i = 10000 ⋅ i = (eq. 2) 4 From eq.1 and eq.2, we obtain: i = 1.0 mA and v = 40 V. Applying KVL for the loop: − v + 8000i + 10000i + Ri + 12000i = 0 or, 0.001R = 10 Therefore, R = 10kΩ and v1 = 10V . ________________________________________________________________________ Problem 2.18 Solution: Known quantities: Rated power; rated optical power; operating life; rated operating voltage; open-circuit resistance of the filament. Find: a) The resistance of the filament in operation b) The efficiency of the bulb. Analysis: a) PR 60 Va = = 521.7 ma VR 115 V 115 V V VR = = = 220.4 Ω OL: R = I I 521.7 ma P = VI ∴ I = b) Efficiency is defined as the ratio of the useful power dissipated by or supplied by the load to the total power supplied by the source. In this case, the useful power supplied by the load is the optical power. From any handbook containing equivalent units: 680 lumens=1 W 2.14 G. Rizzoni, Principles and Applications of Electrical Engineering Po ,out = Optical Power Out = 820 lum η= efficiency = Po,out PR = Problem solutions, Chapter 2 W = 1.206 W 680 lum 1.206 W = 0.02009 = 2.009 % . 60 W ________________________________________________________________________ Problem 2.19 Solution: Known quantities: Rated power; rated voltage of a light bulb. Find: The power dissipated by a series of three light bulbs connected to the nominal voltage. Assumptions: The resistance of each bulb doesn’t vary when connected in series. Analysis: When connected in series, the voltage of the source will divide equally across the three bulbs. The across each bulb will be 1/3 what it was when the bulbs were connected individually across the source. Power dissipated in a resistance is a function of the voltage squared, so the power dissipated in each bulb when connected in series will be 1/9 what it was when the bulbs were connected individually, or 11.11 W: Ohm’s Law: P = IVB = I 2 RB = VB2 RB VB = VS = 110 V V B2 (110 V ) RB = = = 121 Ω 100 Va P 2 Connected in series and assuming the resistance of each bulb remains the same as when connected individually: − VS + VB1 + VB 2 + VB 3 = 0 OL: − VS + IRB 3 + IRB 2 + IR B1 = 0 110 V VS = = 303ma I= RB1 + RB 2 + RB 3 121 + 121 + 121 V a V· 1 2§ PB1 = I 2 RB1 = (303 ma ) ¨121 ¸ = 11.11 W = 100 W . a¹ 9 © KVL: ________________________________________________________________________ Problem 2.20 Solution: Known quantities: Rated power and rated voltage of the two light bulbs. 2.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: The power dissipated by the series of the two light bulbs. Assumptions: The resistance of each bulb doesn’t vary when connected in series. Analysis: When connected in series, the voltage of the source will divide equally across the three bulbs. The across each bulb will be 1/3 what it was when the bulbs were connected individually across the source. Power dissipated in a resistance is a function of the voltage squared, so the power dissipated in each bulb when connected in series will be 1/9 what it was when the bulbs were connected individually, or 11.11 W: Ohm’s Law: P = IVB = I 2 RB = VB2 RB VB = VS = 110 V VB2 (110 V ) R60 = = = 201.7 Ω P60 60 Va 2 V2 (110 V ) = 121 Ω = B = 100 Va P100 2 R100 When connected in series and assuming the resistance of each bulb remains the same as when connected individually: − VS + VB 60 + VB100 = 0 OL: − VS + IR B 60 + IR B100 = 0 VS 110 V I= = = 340.9 ma V RB 60 + RB100 201.7 + 121 a V· 2§ PB 60 = I 2 RB 60 = (340.9 ma ) ¨ 201.7 ¸ = 23.44 W a¹ © V· 2§ PB100 = I 2 RB100 = (340.9 ma ) ¨121 ¸ = 14.06 W a¹ © KVL: Notes: 1.It’s strange but it’s true that a 60 W bulb connected in series with a 100 W bulb will dissipate more power than the 100 W bulb. 2. If the power dissipated by the filament in a bulb decreases, the temperature at which the filament operates and therefore its resistance will decrease. This made the assumption about the resistance necessary. ________________________________________________________________________ Problem 2.21 Solution: Known quantities: Schematic of the circuit shown in Figure P2.21. Find: The resistor values, including the power rating, necessary to achieve the indicated voltages for: a) V = 30V , R1 = 10kΩ, vout = 10V 2.16 G. Rizzoni, Principles and Applications of Electrical Engineering b) Problem solutions, Chapter 2 V = 12V , R1 = 140kΩ, vout = 8.5V Assumptions: Resistors are available in 1 8 - 1 4 - 1 2 -, and 1-W ratings. Analysis: (a) § R2 · R2 ¸¸ ⋅V = vout = ¨¨ ⋅ (30) = 10 R 2 + 10000 © R2 + R1 ¹ R 2 (30 − 10) = 10 ⋅10 ⋅10 3 R 2 = 5 kΩ 2 § 30 · P2 = I R = ¨ ¸ ⋅ (5000) = 20 mW © 15000 ¹ 1 PR2 = W 8 P1 = I 2 R1 = 40 mW 1 PR1 = W . 8 2 (b) § R2 · § 140 · ¸¸ ⋅ V = 12 ⋅ ¨¨ ¸¸ = 8.5 vout = ¨¨ © R2 + R1 ¹ © 140 + R1 ¹ R1 = 57Ω V 12 V I= = = 61 ma P1 = I 2 R1 = 212.1 mW R1 + R2 57 Ω + 140 Ω P2 = I 2 R2 = 520.9 mW 1 PR1 = W 4 PR2 = 1 W ________________________________________________________________________ Problem 2.22 Solution: Known quantities: Schematic of the circuit shown in Figure P2.22 with resistances, voltages, Ro = 1.6kΩ, R2 = 4.3kΩ ; and V = 110V , vout = 64.3V . Find: The resistor values, including the power rating, necessary to achieve the indicated voltages. 2.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Assumptions: Resistors are available in 1 8 - ¼- ½-, and 1-W ratings. Analysis: vout = · § R2 4300 ¸¸ = 64.3 V = 110 ⋅ ¨¨ R0 + R1 + R2 © 1600 + R1 + 4300 ¹ R1 = 1.45kΩ V 110 V I= = = 15 ma R0 + R1 + R2 1600 Ω + 1450 Ω + 4300 Ω 1 P0 = I 2 R0 = 360 mW PR0 = W 2 1 P1 = I 2 R1 = 326.25 mW PR1 = W 2 2 P2 = I R2 = 967.5 mW PR 2 = 1 W ________________________________________________________________________ Problem 2.23 Solution: Known quantities: Schematic of the circuit shown in Figure P2.23 with source voltage, Ro = 8Ω, R1 = 10Ω, R2 = 2Ω . Find: a) b) c) d) e) v = 24V ; and resistances, The equivalent resistance seen by the source The current i The power delivered by the source The voltages v1 and v 2 The minimum power rating required for R1 Analysis: a) The equivalent resistance seen by the source is Req = R0 + R1 + R2 = 8 + 10 + 2 = 20Ω b) Applying KVL: V − Req i = 0 , therefore i = V 24V = = 1.2A Req 20Ω c) Psource = Vi = 24V ⋅1.2 A = 28.8 W d) Applying Ohm's law: v1 = R1i = 10Ω ⋅1.2 A = 12 V , and v2 = R2i = 2Ω ⋅1.2 A = 2.4 V e) 2.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 P1 = R1i 2 = 10Ω ⋅ (1.2 A) = 14.4 W , therefore the minimum power rating for R1 is 16 W. 2 ________________________________________________________________________ Problem 2.24 Solution: Known quantities: Schematic of the circuit shown in Figure P2.24 with resistors, R1 = 25Ω, R2 = 10Ω, R3 = 5Ω, R4 = 7Ω . Find: a) The currents i1 and i 2 b) The power delivered by the 3-A current source and the 12-V voltage source c) The total power dissipated by the circuit. Analysis: a) KCL at node 1 requires that: v1 - 12 V v1 + -3A=0 R2 R3 v1 we have (4 + R3 )R2 = 18 V v1 = 3 R2 + R3 Solving for Therefore, v1 18 = − = − 1 .8 A R2 10 12 − v1 6 i2 = = − = − 1 .2 A 5 R3 i1 = − b) The power delivered by the 3-A source is: P3-A = (v3-A)(3) Thus, we can compute the voltage across the 3-A source as v3-A = 3R1 + v1 = 3 ⋅ 25 + 18 = 93 V Thus, P3-A = (93)(3) = 279 W. Similarly, the power supplied by the 12-V source is: P12-V = (12)(I12-V) We have I12-V = 12 + i2 = 514.3 mA, thus: R4 P12-V = (12)(I12-V) = 6.17 W c) Since the power dissipated equals the total power supplied: 2.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Pdiss = P3-A + P12-V =279 + 6.17 = 285.17 W ________________________________________________________________________ Problem 2.25 Solution: Known quantities: Schematic of the circuit shown in Figure P2.25. Find: The power delivered by the dependent source. Analysis: 24V 24 A =2A = (7 + 5)Ω 12 isource = 0.5i 2 = 0.5 ⋅ (4 ) = 2 A i= The voltage across the dependent source (+ ref. taken at the top) can be found by KVL: − vD + (2 A)(15Ω) + 24V = 0 vD = 54 V Therefore, the power delivered by the dependent source is PD = vD isource = 54 ⋅ 2 = 108 W. ________________________________________________________________________ Problem 2.26 Solution: Known quantities: Schematic of the circuit in Figure P2.26. Find: If V1 = 12.0V , R1 = 0.15Ω, RL = 2.55Ω , the load current and the power dissipated by the load b) If a second battery is connected in parallel with battery 1with V2 = 12.0V , R2 = 0.28Ω , determine the variations in the load current and in the power dissipated by the load due to the parallel connection with a second battery. a) Analysis: a) IL = V1 12 12 = = = 4.44 A R1 + RL 0.15 + 2.55 2.7 PLoad = I L2 RL = 50.4 W. b) with another source in the circuit we must find the new power dissipated by the load. To do so, we write KVL twice using mesh currents to obtain 2 equations in 2 unknowns: 2.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 I 2 R2 − V1 + V2 − I 1 R1 = 0 0.28 ⋅ I 2 − 0.15 ⋅ I 1 = 0 ® ® ¯(I1 + I 2 )RL + I 2 R2 = V2 ¯2.55 ⋅ (I1 + I 2 ) + 0.28 ⋅ I 2 = 12 Solving the above equations gives us: I1 = 2.95 A, I 2 = 1.58 A I L = I1 + I 2 = 4.53 A PLoad = I L2 RL = 52.33 W This is an increase of 1%. ________________________________________________________________________ Problem 2.27 Solution: Known quantities: Open-circuit voltage at the terminals of the power source is 50.8 V; voltage drop with a 10-W load attached is to 49 V. Find: a) The voltage and the internal resistance of the source b) The voltage at its terminals with a 15-Ω load resistor attached c) The current that can be derived from the source under short-circuit conditions. Analysis: (a) (49V ) 2 = 10W RL = 240Ω RL vs = 50.8V , the open-circuit voltage RL 240 vS = 49 50.8 = 49 RS + RL RS + 240 (240)(50.8) RS = − 240 = 8.82Ω 49 (b) v= RL 15 vS = 50.8 = 32.0V 8.82 + 15 RS + RL (c) iCC ( RL = 0) = vS 50.8 = = 5.76 A RS 8.82 ________________________________________________________________________ Problem 2.28 Solution: Known quantities: Voltage of the heater, maximum and minimum power dissipation; number of coils, schematics of the configurations. 2.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: a) The resistance of each coil b) The power dissipation of each of the other two possible arrangements. Analysis: P = 2000 W. Therefore, (220)2 + R2 (a) For the parallel connection, (220) 2 2000 = R1 1 · 2§ 1 ¸¸ = (220 ) ¨¨ + R R 1 2 ¹ © or, 1 1 5 + = . R1 R2 121 For the series connection, (220) P = 300 W. Therefore, 2 300 = R1 + R2 or, 1 3 = . R1 + R2 484 Solving, we find that R1 = 131.6Ω and R2 = 29.7Ω . (b) the power dissipated by R1 alone is: PR1 2 ( 220 ) = R1 = 368W and the power dissipated by R2 alone is PR2 = (220)2 R2 = 1631W . ________________________________________________________________________ 2.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Section 2.5, 2.6: Resistance and Ohm’s Law Problem 2.29 Solution: Known quantities: Diameter of the cylindrical substrate; length of the substrate; conductivity of the carbon. Find: The thickness of the carbon film required for a resistance R of 33 KΩ. Assumptions: Assume the thickness of the film to be much smaller than the radius Neglect the end surface of the cylinder. Analysis: d d ≅ σ ⋅ A σ ⋅ 2πa ⋅ ∆t 9 ⋅ 10 −3 m d ∆t = = S R ⋅ 2πa ⋅ σ 33 ⋅ 10 3 Ω ⋅ 2.9 ⋅ 10 6 ⋅ 2π ⋅ 1 ⋅ 10 −3 m m R= ________________________________________________________________________ Problem 2.30 Solution: Known quantities: The constants A and k; the open-circuit resistance. Find: The rated current at which the fuse blows, showing that this happens at: I= 1 . AkR0 Assumptions: Here the resistance of the fuse is given by: R = R0 [1 + A(T − T0 )] where T0 , room temperature, is assumed to be 25°C. We assume that: T − T0 = kP where P is the power dissipated by the resistor (fuse). Analysis: R = R0 (1 + A ⋅ ∆T ) = R0 (1 + AkP) = R0 (1 + AkI 2 R) R − R0 AkI 2 R = R0 2.23 G. Rizzoni, Principles and Applications of Electrical Engineering R= R0 1 − R0 AkI 2 →∞ when Problem solutions, Chapter 2 I − R0 AkI 2 → 0 m °C V − = (0.7 0.35 0.11 ) 2 =6.09 a. °C Va a AkR0 1 1 I= ________________________________________________________________________ Problem 2.31 Solution: Known quantities: Vs = 10V and resistors, R1 = 20Ω, R2 = 40Ω, R3 = 10Ω, R4 = R5 = R6 = 15Ω . Circuit shown in Figure P2.31 with voltage source, Find: The current in the 15-Ω resistors. Analysis: Since the 3 resistors must have equal currents, I15Ω = 1 ⋅I 3 and, I= Therefore, VS 10 10 = = = 303 mA R 1 + R2 || R3 + R4 || R5 || R6 20 + 8 + 5 33 I15Ω = 10 = 101 mA 99 ________________________________________________________________________ Problem 2.32 Solution: Known quantities: Schematic of the circuit in Figure P2.7 with currents I 0 = −2 A, I1 = −4 A, I S = 8 A, voltage source VS = 12 V, and resistance R 0 = 2 Ω . Find: The unknown resistances R 1 , R 2 , R 3 , R 4 and R 5 . Assumption: In order to solve the problem we need to make further assumptions on the value of the resistors. For example, we may assume that R4 = 2 1 R 1 and R 2 = R 1 . 3 3 Analysis: We can express each current in terms of the adjacent node voltages: 2.24 G. Rizzoni, Principles and Applications of Electrical Engineering I0 = I1 = Problem solutions, Chapter 2 v a − vb v −v = a b = −2 2 R0 + R4 2 + R1 3 v a − vb = −4 R1 I2 = va v = a =6 R2 1 R1 3 I3 = vb =2 R3 IS = VS − vb 12 − vb = =8 R5 R5 Solving the system we obtain: va = 3 V , vb = 9 V , R 1 = 1.5 Ω , R 2 = 0.5 Ω , R 3 = 4.5 Ω , R 4 = 1 Ω and R 5 = 0.375 Ω . ________________________________________________________________________ Problem 2.33 Solution: Known quantities: R 1 = 2Ω , R 2 = 5Ω , R 3 = 4Ω , R 4 = 1Ω , R 5 = 3Ω , voltage source VS = 54 V, and current I 2 = 4 A. Schematic of the circuit in Figure P2.7 with resistors Find: The unknown currents I 0 , I1 , I 3 , I S and the resistor R 0 . Analysis: We can express each current in terms of the adjacent node voltages: I0 = v a − vb R0 + R4 I1 = v a − vb R1 I2 = va = 4 va = 4 ⋅ 5 = 20 V R2 2.25 G. Rizzoni, Principles and Applications of Electrical Engineering I3 = vb R3 IS = VS − vb R5 Problem solutions, Chapter 2 Applying KCL to node (a) and (b) : 20 − vb 20 − vb ° R +1 + 2 + 4 = 0 I 0 + I1 + I 2 = 0 ° 0 ® ® ¯ I 0 + I S + I1 − I 3 = 0 ° 20 − vb + 54 − vb + 20 − vb − vb = 0 °¯ R 0 + 1 3 2 4 Solving the system we obtain: vb = 24 V and R 0 = 1 Ω . ________________________________________________________________________ Problem 2.34 Solution: Known quantities: NOTE: Typo in Problem Statement for units of R3 Schematic of the circuit shown in Figure P2.34 with resistors and voltage source R0 = 2Ω, R1 = 1Ω, R2 = 4 / 3Ω, R3 = 6Ω VS = 12 V. Find: a) The mesh currents ia , ib , ic b) The current through each resistor. Analysis: Applying KVL to mesh (a), mesh (b) and mesh (c): 2ia + (ia − ib ) = 0 ia R 0 + (ia − ib )R 1 = 0 ° 4 ° ° ( ) ( ) − − + − = i i R i R i i R 0 ® a b 1 b 2 ®(ia − ib ) − ib + 6(ic − ib ) = 0 c b 3 3 ° °V = (i − i )R c b 3 ¯ S i i 6 − = ( ) °¯ c b 12 Solving the system we obtain: I R 0 ° = i 2 A a °I R 1 ° ®ib = 6 A ® °I R 2 °i = 8 A ¯c °I ¯ R3 = ia = 2 A (positive in the direction of ia ) = ib − ia = 4 A (positive in the direction of ib ) = ib = 6 A (positive in the direction of ib ) = ic − ib = 2 A (positive in the direction of ic ) ________________________________________________________________________ 2.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.35 Solution: Known quantities: NOTE: Typo in Problem Statement for units of R3 Schematic of the circuit shown in Figure P2.34 with resistors and voltage source R0 = 2Ω, R1 = 2Ω, R2 = 5Ω, R3 = 4Ω VS = 24 V. Find: a) The mesh currents ia , ib , ic b) The current through each resistor. Analysis: Applying KVL to mesh (a), mesh (b) and mesh (c): ia R 0 + (ia − ib )R 1 = 0 2ia + 2(ia − ib ) = 0 ° ° ®(ia − ib )R 1 − ib R 2 + (ic − ib )R 3 = 0 ®2(ia − ib ) − 5ib + 4(ic − ib ) = 0 °V = (i − i )R °4(i − i ) = 24 c b 3 ¯ S ¯ c b Solving the system we obtain: VR 0 ° ia = 2 A ° °VR1 ®ib = 4 A ® °i = 10 A °VR 2 ¯c °V ¯ R3 = R 0 ia = 4 V (⊕ up) = R 1 (ib − ia ) = 4 V (⊕ down) = R 2 ib = 20 V (⊕ up) = R 3 (ic − ib ) = 24 V (⊕ up) ________________________________________________________________________ Problem 2.36 Solution: Known quantities: NOTE: Typo in Problem Statement for units of R3 Schematic of the circuit shown in Figure P2.34 with resistors and of the current source R0 = 1Ω, R1 = 3Ω, R2 = 2Ω, R3 = 4Ω I S = 12 A. Find: The voltage across each resistance. Analysis: Applying KVL to mesh (a), mesh (b) and mesh (c): ia R 0 + (ia − ib )R 1 = 0 ia + 3(ia − ib ) = 0 ° ° ®(ia − ib )R 1 − ib R 2 + (ic − ib )R 3 = 0 ®3(ia − ib ) − 2ib − 4ib + 48 = 0 °i = I °i = 12 A ¯c S ¯c 2.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Solving the system we obtain: 16 VR 0 = R 0 ia = 5.33 V °ia = 3 A ° ° 64 ° °VR1 = R 1 (ib − ia ) = 5.33 V A = i ®b ® 9 ° °VR 2 = R 2 ib = 14.22 V °ic = 12 A °V = R (i − i ) = 19.55 V 3 c b ¯ R3 ° ¯ (⊕ up) (⊕ down) (⊕ up) (⊕ up) ________________________________________________________________________ Problem 2.37 Solution: Known quantities: Schematic of voltage divider network shown of Figure P2.37. Find: a) The worst-case output voltages for b) The worst-case output voltages for ± 10 percent tolerance ± 5 percent tolerance Analysis: a) 10% worst case: low voltage R2 = 4500 Ω, R1 = 5500 Ω vOUT ,MIN = 4500 5 = 2.25V 4500 + 5500 10% worst case: high voltage R2 = 5500 Ω, R1 = 4500 Ω vOUT ,MAX = 5500 5 = 2.75V 4500 + 5500 b) 5% worst case: low voltage R2 = 4750 Ω, R1 = 5250 Ω vOUT ,MIN = 4750 5 = 2.375V 4750 + 5250 5% worst case: high voltage R2 = 5250 Ω, R1 = 4750 Ω vOUT ,MAX = 5250 5 = 2.625V 5250 + 4750 ________________________________________________________________________ 2.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.38 Solution: Known quantities: Schematic of the circuit shown in figure P2.38 with resistances, R0 = 4Ω, R1 = 12Ω, R2 = 8Ω, R3 = 2Ω, R4 = 16Ω, R5 = 5Ω . Find: The equivalent resistance of the circuit. Analysis: Starting from the right side, we combine the two resistors in series: Then, we can combine the two parallel resistors, namely the 5 Ω resistor and 10 Ω resistor: 1 R parallel = ( ) 1 1 10 + Ω −1 R parallel = Ω 5 10 3 Then, we can combine the two resistors in series, namely the 3.33 Ω and the 16 Ω resistor: Then, we can combine the two parallel resistors, namely the 12 Ω resistor and 19.33 Ω resistor: 1 R parallel = ( ) 1 1 + Ω −1 R parallel = 7.4 Ω 12 19.33 2.29 G. Rizzoni, Principles and Applications of Electrical Engineering Therefore, Problem solutions, Chapter 2 Req = 4 + 7.4 = 11.4 Ω. _______________________________________________________________________ Problem 2.39 Solution: Known quantities: = 12V ; and resistances, R0 = 4Ω, R1 = 2Ω, R2 = 50Ω, R3 = 8Ω, R4 = 10Ω, R5 = 12Ω, R6 = 6Ω . Schematic of the circuit shown in Figure P2.39 with source voltage, Vs Find: The equivalent resistance of the circuit seen by the source; the current i through the resistance R2 . Analysis: Starting from the right side, we can combine the three parallel resistors, namely the 10 Ω resistor, the 12 Ω resistor and the 6 Ω resistor: 1 R parallel = ( ) 1 1 1 −1 20 + + Ω R parallel = Ω 10 12 6 7 Then, we can combine the two resistors in series, namely the 8 Ω and the 2.86 Ω resistor: Then, we can combine the three parallel resistors, namely the 4 Ω resistor, the 50 Ω resistor and the 10.86 Ω resistor: 2.30 G. Rizzoni, Principles and Applications of Electrical Engineering 1 R parallel Therefore, = Problem solutions, Chapter 2 ( ) 1 1 1 + + Ω −1 R parallel = 2.76 Ω 4 50 10.86 Req = 2 + 2.76 = 4.76 Ω. Looking at the following equivalent circuit: We can apply KVL and KCL to the above circuit: I 0 = 12.5i VS − 2 I1 − 4 I 0 = 0 ° I = 6 − 25i ° ° 1 ® i = 107 mA ® I1 = I 0 + i + I eq = I i 17 . 48 eq ° ° ¯4 I 0 = 50i = 2.86 I eq °i = I1 − I 0 − I eq ¯ ________________________________________________________________________ Problem 2.40 Solution: Known quantities: Schematic of the circuit shown in Figure P2.40 with source voltage, Vs = 50V ; resistances, R1 = 20Ω, R2 = 5Ω, R3 = 2Ω, R4 = 8Ω, R5 = 8Ω, R6 = 30Ω ; and power absorbed by the 20-Ω resistor. Find: The resistance R . Analysis: Starting from the right side, we can replace resistors Ri (i=2..6) with a single equivalent resistors: Req = R2 + (R3 + (R4 || R5 )) || R6 = 10 Ω 2.31 G. Rizzoni, Principles and Applications of Electrical Engineering The same voltage appears across both Problem solutions, Chapter 2 R1 and Req and, therefore, these element are in parallel. Applying the voltage divider rule: VR1 = R1 || Req R + R1 || Req VS = 1000 3R + 20 The power absorbed by the 20-Ω resistor is: (V ) = 2 P20Ω R1 R1 2 1 § 1000 · 50000 = 20 R = 10 Ω = ¨ ¸ = 20 © 3R + 20 ¹ (3R + 20)2 ________________________________________________________________________ Problem 2.41 Solution: Known quantities: Schematic of the circuit shown in Figure P2.41. Find: The equivalent resistance Req of the infinite network of resistors. Analysis: We can see the infinite network of resistors as the equivalent to the circuit in the picture: R Req Req R R Therefore, Req = R + (R || Req ) + R = 2 R + RReq R + Req ( ) Req = 1+ 3 R ________________________________________________________________________ 2.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.42 Solution: Known quantities: Vs = 110V ; and resistances, R1 = 90Ω, R2 = 50Ω, R3 = 40Ω, R4 = 20Ω, R5 = 30Ω, R6 = 10Ω, R7 = 60Ω, R8 = 80Ω . Schematic of the circuit shown in Figure P2.42 with source voltage, Find: a) The equivalent resistance of the circuit seen by the source. b) The current through and the power absorbed by the resistance 90-Ω resistance. Analysis: a) Starting from the right side, we can combine the two parallel resistors, namely the 20 Ω resistor and the 30 Ω resistor: 1 R parallel = ( ) 1 1 + Ω −1 R parallel = 12 Ω 20 30 Then we can combine the two parallel resistors in the bottom, namely the 60 Ω resistor and the 80 Ω, and the two resistor in series: 1 R parallel = ( ) 1 1 + Ω −1 R parallel = 34.3 Ω 60 80 Then we can combine the two parallel resistors on the right, namely the 40 Ω resistor and the 22 Ω: 1 R parallel = ( ) 1 1 + Ω −1 R parallel = 14.2 Ω 40 22 2.33 G. Rizzoni, Principles and Applications of Electrical Engineering Therefore, Problem solutions, Chapter 2 ( ) 1 1 1 = + Ω −1 Req = 47 Ω . Req 90 (50 + 14.2 + 34.3) b) The current through and the power absorbed by the 90-Ω resistor are: I 90 Ω = P90 Ω = VS 110 = = 1.22 A R1 90 (VS )2 R1 = 110 2 = 134.4 W 90 ________________________________________________________________________ Problem 2.43 Solution: Known quantities: Schematic of the circuit shown in Figure P2.43. Find: The equivalent resistance at terminals a,b in the case that terminals c,d are a) open b) shorted; the same for terminals c,d with respect to terminals a,b. Analysis: With terminals c-d open, Req= (360 + 540) (180 + 540)Ω = 400Ω , with terminals c-d shorted, Req= with terminals a-b open, Req= (360 180) + (540 540)Ω = 390Ω , (540 + 540) (360 + 180)Ω = 360Ω , with terminals a-b shorted, Req= (360 540) + (180 540 )Ω = 351Ω . ________________________________________________________________________ Problem 2.44 Solution: Known quantities: Layout of the site shown in Figure P2.44; characteristics of the cables; rated voltage of the generator; range of voltages and currents absorbed by the engine at full load. Find: The minimum AWG gauge conductors which must be used in a rubber insulated cable. 2.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: The cable must meet two requirements: 1. The conductor current rating must be greater than the rated current of the motor at full load. This requires AWG #14. 2. The voltage drop due to the cable resistance must not reduce the motor voltage below its minimum rated voltage at full load. KVL : − VG + V RC1 + VM − Min + VRC 2 = 0 − VG + I M − FL RC1 + VM − Min + I M − FL RC 2 = 0 RC1 + RC 2 = VG − VM − Min = I M − FL 110 V − 105 V = 703.9 mΩ 7.103 A 1 [703.9 mΩ] RC1 RC 2 2 Ω RMax = = = = 2.346 m 150 m d d m = Therefore, AWG #8 or larger wire must be used. ________________________________________________________________________ Problem 2.45 Solution: Known quantities: Layout of the building shown in Figure P2.45; characteristics of the cables; rated voltage of the generator; total electrical load in the building. Find: The minimum AWG gauge conductors which must be used in a rubber insulated cable. Analysis: The cable must meet two requirements: 1. The conductor current rating must be greater than the rated current of the motor at full load. This requires AWG #4. 2. The voltage drop due to the cable resistance must not reduce the motor voltage below its minimum rated voltage at full load. KVL : − VS + VRC1 + VL − Min + VRC 2 = 0 − VS + I L − FL RC1 + VL − Min + I L − FL RC 2 = 0 RC1 + RC 2 = VS − VL − Min = I L − FL 450 V − 446 V = 77.6 mΩ 51.57 A 1 [77.6 mΩ] RC1 RC 2 2 Ω RMax = = = = 0.4565 m 85 m d d m = Therefore, AWG #0 or larger wire must be used. ________________________________________________________________________ 2.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.46 Solution: Known quantities: Layout of the site shown in Figure P2.46; characteristics of the cables; rated voltage of the generator; electrical characteristics of the engine. Find: The maximum length of a rubber insulated cable with AWG #14 which can be used to connect the motor and the generator. Analysis: The voltage drop due to the cable resistance must not reduce the motor voltage below its minimum rated voltage at full load. KVL : − VG + VRC1 + VM − Min + VRC 2 = 0 − VG + I M − FL RC1 + VM − Min + I M − FL RC 2 = 0 RC1 + RC 2 = VG − VM − Min = I M − FL 110 V − 105 V = 703.9 mΩ 7.103 A 1 [703.9 mΩ] RC1 RC 2 2 = = d Max = = 42.48 m Ω Rrated Rrated 8.285 m m = ________________________________________________________________________ Problem 2.47 Solution: Known quantities: Layout of the building shown in Figure P2.47; characteristics of the cables; rated voltage of the generator; total electrical load in the building. Find: The maximum length of a rubber insulated cable with AWG #4 which can be used to connect the source to the load. Analysis: The voltage drop due to the cable resistance must not reduce the motor voltage below its minimum rated voltage at full load. 2.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 KVL : − VS + VRC1 + VL − Min + VRC 2 = 0 − VS + I L − FL RC1 + VL − Min + I L − FL RC 2 = 0 RC1 + RC 2 = VS − VL − Min = I L − FL 450 V − 446 V = 77.6 mΩ 51.57 A 1 [77.6 mΩ] RC1 RC 2 2 = = d Max = = 47.59 m Ω Rrated Rrated 0.8153 m m = ________________________________________________________________________ Problem 2.48 Solution: Known quantities: Schematic of the circuit shown in Figure P2.48 with resistances, R1 = 2.2kΩ, R2 = 18kΩ , R3 = 220kΩ, R4 = 3.3kΩ . Find: The equivalent resistance between A and B. Analysis: Shorting nodes C and D creates a single node to which all four resistors are connected. Req1 = R1 R3 = R1 R3 [2.2 KΩ][4.7 KΩ] = 1.499 KΩ = R1 + R3 2 .2 + 4 .7 K Ω Req 2 = R2 R4 = R2 R 4 [18 KΩ][3.3 KΩ] = 2.789 KΩ = R2 + R 4 18 + 3.3 KΩ Req = Req1 + Req 2 = 1.499 + 2.789 KΩ = 4.288 KΩ ________________________________________________________________________ Problem 2.49 Solution: Known quantities: Schematic of the circuit shown in Figure P2.49 with source voltage, R1 = 11kΩ, R2 = 220kΩ, R3 = 6.8kΩ, R4 = 0.22mΩ Find: The voltage between nodes A and B. 2.37 Vs = 12V ; and resistances, G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Analysis: The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4 are connected in series. SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THIS WILL HAVE TO BE A WILD GUESS AT THIS POINT. Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. The actual polarities are not difficult to determine. Do so. VD : VR 3 = VD : VR 4 = [12 V ][6.8 kΩ] = 4.584 V VS R3 = R1 + R3 11 + 6.8 kΩ [ ] [12 V ] 0.22 ×10−6 kΩ = 1.20 ×10−8 V ≈ 0 VS R4 = R2 + R4 220 + 0.22 × 10−6 kΩ ( ) KVL : − VR 3 + VAB + VR 4 = 0 ∴VAB = VR 3 − VR 4 = 4.584V The voltage is negative indicating that the polarity of V AB is opposite of that specified. A solution is not complete unless the assumed positive direction of a current or assumed positive polarity of a voltage IS SPECIFIED ON THE CIRCUIT. ________________________________________________________________________ Problem 2.50 Solution: Known quantities: Schematic of the circuit shown in Figure P2.49 with source voltage, Vs = 5V ; and resistances, R1 = 2.2kΩ, R2 = 18kΩ, R3 = 4.7kΩ, R4 = 3.3kΩ Find: The voltage between nodes A and B. Analysis: The same current flows through R1 and R3. Therefore, they are connected in series. Similarly, R2 and R4 are connected in series. SPECIFY THE ASSUMED POLARITY OF THE VOLTAGE BETWEEN NODES A AND B. THIS WILL HAVE TO BE A WILD GUESS AT THIS POINT. Specify the polarities of the voltage across R3 and R4 which will be determined using voltage division. The actual polarities are not difficult to determine. Do so. VD : VR 3 = VD : VR 4 = VS R3 [5 V ][4.7 KΩ] = 3.406 V = R1 + R3 2.2 KΩ + 4.7 KΩ VS R4 [5 V ][3.3 KΩ] = 0.917 V = R2 + R4 18 KΩ + 3.3 KΩ KVL : − VR 3 + V AB + VR 4 = 0 V AB = VR 3 − VR 4 = 2.489 V A solution is not complete unless the assumed positive direction of a current or assumed positive polarity of a voltage IS SPECIFIED ON THE CIRCUIT. ________________________________________________________________________ 2.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.51 Solution: Known quantities: Schematic of the circuit shown in Figure P2.51 with source voltage, Vs = 12V ; and resistances, R1 = 1.7mΩ, R2 = 3kΩ, R3 = 10kΩ Find: The voltage across the resistance R3 . Analysis: The same voltage appears across both R2 and R3 and, therefore, these element are in parallel. Applying the voltage divider rule: VR3 = Note that since R2 || R3 2.3kΩ VS = 12 V = 11.999991 V (⊕ down) R1 + R2 || R3 1.7 mΩ + 2.3kΩ R1 << R2 || R3 , then VR3 ≅ VS . ________________________________________________________________________ 2.39 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Sections 2.7, 2.8: Practical Sources and Measuring Devices Problem 2.52 Solution: Known quantities: Parameters R0 = 300 Ω (resistance at temperature T0 = 298 K), and β = −0.01 K -1 , value of the second resistor. Find: a) Plot Rth (T ) versus T in the range 350 T 750 [°K] b) The equivalent resistance of the parallel connection with the 250-Ω resistor; plot Req (T ) versus T in the range 350 T 750 [°K] for this case on the same plot as part a. Assumptions: Rth (T ) = R0 e − β (T −T0 ) . Analysis: a) Rth (T ) = 300e b) −0.01⋅(T − 298 ) Req (T ) = Rth (T ) || 250Ω = 1500 e −0.01 (T − 298 ) 5 + 6 e −0.01 (T − 298 ) The two plots are shown below. In the above plot, the solid line is for the thermistor alone; the dashed line is for the thermistor-resistor combination. ________________________________________________________________________ 2.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Problem 2.53 Solution: Known quantities: A potentiometer shown in the circuit of Figure P.253 with The value of the resistance of the resistor Rm , the total length xT and voltage source v S . Find: vout ( x) . Plot vout / v S versus x / xT . b) The distance x when vout = 5 V . c) Assuming the resistance Rm becomes infinite, repeat parts a and b a) The expression for Assumptions: vout 1 = v S 1 ( x xT ) + (RP Rm )(1 − x xT ) RP ( x) = 200e x Analysis: a) vout ( x ) = 10 x = 500 x 2 x 1 ( x 0.02 ) + 200e 100 (1 − x 0.02 ) 1 − 10 e x + 5 ⋅10 3 e x x 2 ( In the above plot, the solid line is for b) ) Rm = 100 Ω ; the dashed line is for Rm → ∞ . x(vout = 5 V ) = 14.18 cm c) Now Rm → ∞ . 2.41 G. Rizzoni, Principles and Applications of Electrical Engineering vout ( x ) = Problem solutions, Chapter 2 10 → 500 x 1 (x 0.02 ) + 200e x ∞ (1 − x 0.02 ) ( ) x(vout = 5 V ) = 10 cm ________________________________________________________________________ Problem 2.54 Solution: Known quantities: Meter resistance of the coil; meter current for full scale deflection; max measurable pressure. Find: a) The circuit required to indicate the pressure measured by a sensor b) The value of each component of the circuit; the linear range c) The maximum pressure that can accurately be measured. Assumptions: Sensor characteristics follow what is shown in Figure P2.54 Analysis: a) A series resistor to drop excess voltage is required. b) At full scale, meter: I mˆ FS = 10µA rmˆ = 200Ω 0.L. : Vmˆ FS = I mˆ FS rmˆ = 2 mV . at full scale, sensor (from characteristics): PFS = 100 kPa VTFS = 9.5 mV KVL : − VTFS + VRFS + Vmˆ FS = 0 VRFS = VTFS − Vmˆ FS = 9.5 mV − 2 mV = 7.5 mV I RFS = ITFS = I mˆ FS = 10µA VRFS 7.5 mV = = 750 Ω . Ohm law: R = I RFS 10 µA c) from sensor characteristic: 30 kPa –110 kPa. ________________________________________________________________________ Problem 2.55 Solution: Known quantities: Meter resistance of the coil; meter current for full scale deflection; max measurable pressure. Find: a) Redesign the circuit to meet the specification. b) The value of each component of the circuit. c) The linear range of the system. 2.42 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Assumptions: Sensor characteristics follow what is shown in Figure P2.55 Analysis: a) A series resistor to drop excess voltage is required. b) At full scale, meter: I mˆ FS = 50µA rmˆ = 1.8kΩ 0.L. : Vmˆ FS = I mˆ FS rmˆ = 90 mV . at full scale, sensor (from characteristics): VTFS = 9.5 V KVL : − VTFS + VRFS + Vmˆ FS = 0 VRFS = VTFS − Vmˆ FS = 9.5 V − 90 mV = 9.41 V I RFS = ITFS = I mˆ FS = 50µA VRFS 9.41 V = = 188.2 kΩ . Ohm law: R = I RFS 50 µA c) from sensor characteristic: 20 kPa –110 kPa. ________________________________________________________________________ Problem 2.56 Solution: Known quantities: Meter resistance of the coil; meter voltage for full scale deflection; max measurable temperature. Find: a) The circuit required to meet the specifications of the new sensor. b) The value of each component of the circuit. c) The linear range of the system. Assumptions: Sensor characteristics follow what is shown in Figure P2.56 Analysis: a) A parallel resistor is required to shunt (bypass) the excess current. b) At full scale, meter: Vmˆ FS = 250 mV rmˆ = 2.5 kΩ 0.L. : I mˆ FS = Vmˆ FS = 100 µA. rmˆ at full scale, sensor (from characteristics): TFS = 400 °C I TFS = 8.5 mA KCL : − I TFS + I RFS + I mˆ FS = 0 2.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 I RFS = ITFS − I mˆ FS = 8.5 mA − 100 µA = 8.4 mA VRFS = VTFS = Vmˆ FS = 250 mV VRFS 250 mV = = 29.76 Ω . Ohm law: R = I RFS 8.4 mA c) from sensor characteristic: 220 °C –410 °C. ________________________________________________________________________ Problem 2.57 Solution: Known quantities: Meter resistance of the coil; meter voltage at full scale; max measurable temperature. Find: a) The circuit required to meet the specifications of the new sensor. b) The value of each component of the circuit c) The linear range of the system. Assumptions: Sensor characteristics follow what is shown in Figure P2.57 Analysis: a) A parallel resistor is required to shunt (bypass) the excess current. b) At full scale, meter: Vmˆ FS = 250 mV rmˆ = 2.5 kΩ 0.L. : I mˆ FS = Vmˆ FS = 100 µA. rmˆ at full scale, sensor (from characteristics): TFS = 400 °C I TFS = 8.5 mA KCL : − I TFS + I RFS + I mˆ FS = 0 I RFS = ITFS − I mˆ FS = 8.5 mA − 100 µA = 8.4 mA VRFS = VTFS = Vmˆ FS = 250 mV VRFS 250 mV = = 29.76 Ω . Ohm law: R = I RFS 8.4 mA c) from sensor characteristic: 220 °C –410 °C. ________________________________________________________________________ Problem 2.58 Solution: Known quantities: Schematic of the circuit shown in Figure P2.58; voltage at terminals with switch open and closed for fresh battery; same voltages for the same battery after 1 year. 2.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: The internal resistance of the battery in each case. Analysis: a) § 10 · ¸V Vout = ¨ © 10 + rB ¹ oc §V · § 2.28 · − 1¸ rB = 10¨¨ oc − 1¸¸ = 10¨ © 2.27 ¹ © Vout ¹ = 0.044Ω b) §V · § 2.2 · − 1¸ rB = 10¨¨ oc − 1¸¸ = 10¨ © 0.31 ¹ © Vout ¹ = 60.97Ω ________________________________________________________________________ Problem 2.59 Solution: Known quantities: Ammeter shown in Figure P2.59; Current for full-scale deflection; desired full scale values. Find: Value of the resistors required for the given full scale ranges. Analysis: We desire R1, R2, R3 such that Ia = 30 µA for I = 10 mA, 100 mA, and 1 A, respectively. We use conductances to simplify the arithmetic: Ga = 1 Ra G1,2,3 = = 1 S 1000 1 R 1,2,3 By the current divider rule: Ia = Ga Ga + Gx I or: § I· 1 1 § Ia · ¨ ¸ G x = Ga ¨ ¸ − Ga or = G x Ga © I − I a ¹ © Ia ¹ § I · Rx = Ra ¨ a ¸ . © I − Ia ¹ 2.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 We can construct the following table: x I Rx (Approx.) 1 10-2 A 3Ω 2 10-1 A 0.3 Ω 3 100 A 0.03 Ω ________________________________________________________________________ Problem 2.60 Solution: Known quantities: Schematic of the circuit shown in Figure P2.60; for part b: value of R p and current displayed on the ammeter. Find: The current i; the internal resistance of the meter. Assumptions: ra << 50 kΩ Analysis: ra << 50 kΩ V 12 i≈ s = = 240 µA R s 50000 a) Assuming that b) With the same assumption as in part a) imeter = 150⋅(10)-6 = Rp ra + R p i or: 150⋅(10)-6 = Therefore, 15 240 ⋅ 10 -6 . ra + 15 ra = 9 Ω. ________________________________________________________________________ Problem 2.61 Solution: Known quantities: Voltage read at the meter; schematic of the circuit shown in Figure P2.61 with source voltage, Vs = 12V and source resistance, Rs = 25kΩ . 2.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Find: The internal resistance of the voltmeter. Analysis: Using the voltage divider rule: V = 11.81 = rm (12) rm + R s Therefore, rm = 1.55 MΩ. ________________________________________________________________________ Problem 2.62 Solution: Known quantities: Circuit shown in Figure P2.61 with source voltage, Vs = 24V ; and ratios between R s and rm . Find: The meter reads in the various cases. Analysis: By voltage division: V= rm (24) rm + R s Rs 0.2 rm V 20 V 0.4 rm 17.14 V 0.6 rm 15 V 1.2 rm 10.91 V 4 rm 4.8 V 6 rm 3.43 V 10 rm 2.18 V For a voltmeter, we always desire rm >> R s . ________________________________________________________________________ Problem 2.63 Solution: Known quantities: Schematic of the circuit shown in Figure P2.63, values of the components. Find: The voltage across R 4 with and without the voltmeter for the following values: 2.47 G. Rizzoni, Principles and Applications of Electrical Engineering R4 b) R4 c) R4 d) R4 a) Problem solutions, Chapter 2 = 100Ω = 1kΩ = 10kΩ = 100kΩ . Assumptions: The voltmeter behavior is modeled as that of an ideal voltmeter in parallel with a 120- kΩ resistor. Analysis: We develop first an expression for VR 4 in terms of R 4 . Next, using current division: § · RS ¸¸ ° I R1 = I S ¨¨ ° © RS + R1 + R2 || (R3 + R4 ) ¹ ® · R2 ° I = I §¨ ¸¸ R1 ¨ ° R4 © R2 + R3 + R4 ¹ ¯ Therefore, · § · § RS R2 ¸¸ ⋅ ¨¨ ¸¸ I R4 = I S ¨¨ © RS + R1 + R2 || (R3 + R4 ) ¹ © R2 + R3 + R4 ¹ VR4 = I R4 R4 § · § · RS R4 R2 ¸¸ ⋅ ¨¨ ¸¸ = I S ¨¨ ( ) || + + + + + R R R R R R R R 1 2 3 4 ¹ © 2 3 4 ¹ © S 66000 ⋅ R4 = R4 + 2.1352 ⋅10 6 Without the voltmeter: a) VR4 = 3.08 V b) VR4 = 30.47 V c) VR4 = 269.91 V d) VR4 = 1260.7 V. Now we must find the voltage drop across R 4 with a 120-kΩ resistor across R 4 . This is the voltage that the voltmeter will read. · § · § RS R2 ¸¸ ⋅ ¨¨ ¸¸ I R4 = I S ¨¨ © RS + R1 + R2 || (R3 + ( R4 || 120kΩ ) ¹ © R2 + R3 + ( R4 || 120kΩ) ¹ VR4 = I R4 R4 § · § · RS R4 R2 ¸¸ ⋅ ¨¨ ¸¸ = I S I S ¨¨ © RS + R1 + R2 || (R3 + ( R4 || 120kΩ ) ¹ © R2 + R3 + ( R4 || 120kΩ) ¹ (120000 + R4 ) ⋅ R4 = 82.5 7319 R4 + 320.28 ⋅10 6 2.48 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 With the voltmeter: a) VR4 = 3.08 V b) VR4 = 30.47 V c) VR4 = 272.57 V d) VR4 = 1724.99 V. ________________________________________________________________________ Problem 2.64 Solution: Known quantities: Schematic of the circuit shown in Figure P2.64, value of the components. Find: The current through R5 b) R5 c) R5 d) R5 a) R 5 both with and without the ammeter, for the following values of the resistor R 5 : = 1kΩ = 100Ω = 10Ω = 1Ω . Analysis: First we should find an expression for the current through R 5 in terms of R 5 and the meter resistance, R m . By the voltage divider rule we have: VR 3 = R 3 || (R 4 + R 5 + R m )VS R 3 || (R 4 + R 5 + R m ) + R 2 + (R 1 || R S ) and I R3 = VR 3 R4 + R5 + Rm Therefore, I R3 = R 3 || (R 4 + R 5 + R m )VS 1 ⋅ R 3 || (R 4 + R 5 + R m ) + R 2 + (R 1 || R S ) R 4 + R 5 + R m 5904 = 208350 + 373 ⋅ (R m + R S ) Using the above equation will give us the following table: 2.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 with meter in without meter circuit in circuit a 10.15 mA 9.99 mA b 24.03 mA 23.15 mA c 27.84 mA 26.67 mA d 28.29 mA 27.08 mA ________________________________________________________________________ Problem 2.65 Solution: Known quantities: Schematic of the circuit and geometry of the beam shown in Figure P2.65, characteristics of the material, reads on the bridge. Find: The force applied on the beam. Assumptions: Gage Factor for Strain gauge is 2 Analysis: R1 and R2 are in series; R3 and R4 are in series. VS R2 VS ( R0 − ∆R) V ( R − ∆R) = = S 0 Voltage Division: V R2 = R1 + R2 R0 + ∆R + R0 − ∆R 2 R0 V S R4 VS ( R0 − ∆R) V ( R + ∆R) = = S 0 Voltage Division: V R4 = R3 + R4 R0 − ∆R + R0 + ∆R 2 R0 KVL: − V R2 − V BA + V R4 = 0 V (2 )(6 )LF VS ( R0 + ∆R) VS ( R0 − ∆R) VS 2∆R − = = VS GFε = S wh 2Y 2 R0 2 R0 2 R0 N 0.050 V (0.025 m)(0.100 m) 2 69 × 109 2 2 V wh y m = 19.97 kN . F = BA = Vs 12 L 12 V (12) 0.3 m VBA = VR4 − VR2 = ________________________________________________________________________ Problem 2.66 Solution: Known quantities: Schematic of the circuit and geometry of the beam shown in Figure P2.65, characteristics of the material, reads on the bridge. Find: The force applied on the beam. 2.50 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 2 Assumptions: Gage Factor for Strain gauge is 2 Analysis: R1 and R2 are in series; R3 and R4 are in series. VS R2 VS ( R0 − ∆R) V ( R − ∆R) = = S 0 VD: V R2 = R1 + R2 R0 + ∆R + R0 − ∆R 2 R0 V S R4 VS ( R0 − ∆R) V ( R + ∆R) = = S 0 VD: V R4 = R3 + R4 R0 − ∆R + R0 + ∆R 2 R0 KVL: − V R2 − V BA + V R4 = 0 VS ( R0 + ∆R) VS ( R0 − ∆R) VS 2∆R V (2 )(6 )LF − = = VS GFε = S 2 R0 2 R0 2 R0 wh 2 y N VBA (0.03 m)(0.07 m) 2 200 × 109 2 2 V wh y m = F = 1.3 × 106 N = BA Vs 12 L 24V (12)1.7m VBA = VR4 − VR2 = VBA = 1.3 × 106 N × 24V (12)1.7m (0.03 m)(0.07 m) 2 200 × 109 N m2 = 21.6V ________________________________________________________________________ 2.51 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Chapter 3 Instructor Notes Chapter 3 presents the principal topics in the analysis of resistive (DC) circuits. The presentation of node voltage and mesh current analysis is supported by several solved examples and drill exercises, with emphasis placed on developing consistent solution methods, and on reinforcing the use of a systematic approach. The aim of this style of presentation, which is perhaps more detailed than usual in a textbook written for a non-majors audience, is to develop good habits early on, with the hope that the orderly approach presented in Chapter 3 will facilitate the discussion of AC and transient analysis in Chapters 4 and 5. Make The Connection sidebars (pp. 75-77) introduce analogies between electrical and thermal circuit elements. These analogies are encountered again in Chapter 5. A brief discussion of the principle of superposition precedes the discussion of Thèvenin and Norton equivalent circuits. Again, the presentation is rich in examples and drill exercises, because the concept of equivalent circuits will be heavily exploited in the analysis of AC and transient circuits in later chapters. The Focus on Methodology boxes (p.76 – Node Analysis; p. 86 – Mesh Analysis; pp. 103, 107, 111 – Equivalent Circuits) provide the student with a systematic approach to the solution of all basic network analysis problems. After a brief discussion of maximum power transfer, the chapter closes with a section on nonlinear circuit elements and load-line analysis. This section can be easily skipped in a survey course, and may be picked up later, in conjunction with Chapter 9, if the instructor wishes to devote some attention to load-line analysis of diode circuits. Finally, those instructors who are used to introducing the op-amp as a circuit element, will find that sections 8.1 and 8.2 can be covered together with Chapter 3, and that a good complement of homework problems and exercises devoted to the analysis of the op-amp as a circuit element is provided in Chapter 8. The homework problems present a graded variety of circuit problems. Since the aim of this chapter is to teach solution techniques, there are relatively few problems devoted to applications. We should call the instructor's attention to the following end-of-chapter problems: 3.8 and 3.19 on the Wheatstone bridge; 3.21, 3.22, 3.23, on three-wire residential distribution service; 3.24, 3.25, 3.26 on AC three-phase electrical distribution systems; 3.28-3.31 on fuses; 3.62-66 on various nonlinear resistance devices. Learning Objectives 1. Compute the solution of circuits containing linear resistors and independent and dependent sources using node analysis. 2. Compute the solution of circuits containing linear resistors and independent and dependent sources using mesh analysis. 3. Apply the principle of superposition to linear circuits containing independent sources. 4. Compute Thévenin and Norton equivalent circuits for networks containing linear resistors and independent and dependent sources. 5. Use equivalent circuits ideas to compute the maximum power transfer between a source and a load. 6. Use the concept of equivalent circuit to determine voltage, current and power for nonlinear loads using load-line analysis and analytical methods. 3.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Sections 3.1, 3.2, 3.3, 3.4: Nodal and Mesh Analysis I cus on Methodology: Node Voltage Analysis Method 1. 2. 3. Select a reference node(usually ground). This node usually has most elements tied to it. All other nodes will be referenced to this node. Define the remaining n–1 node voltages as the independent or dependent variables. Each of the m voltage sources in the circuit will be associated with a dependent variable. If a node is not connected to a voltage source, then its voltage is treated as an independent variable. Apply KCL at each node labeled as an independent variable, expressing each current in terms of the adjacent node voltages. Focus on Methodology: Mesh Current Analysis Method 1. 2. 3. 4. Define each mesh current consistently. Unknown mesh currents will be always defined in the clockwise direction; known mesh currents (i.e., when a current source is present) will always be defined in the direction of the current source. In a circuit with n meshes and m current sources, n–m independent equations will result. The unknown mesh currents are the n–m independent variables. Apply KVL to each mesh containing an unknown mesh current, expressing each voltage in terms of one or more mesh currents.. Solve the linear system of n–m unknowns. Problem 3.1 Solution: Known quantities: Circuit shown in Figure P2.1 with mesh currents: I1 = 5 A, I2 = 3 A, I3 = 7 A. Find: The branch currents through: a) R1, b) R2, c) R3. Analysis: a) Assume a direction for the current through node A: KCL: R1 (e.g., from node A to node B). Then summing currents at − I1 + I R1 + I 3 = 0 I R1 = I1 − I 3 = −2 A This can also be done by inspection noting that the assumed direction of the current through direction of I1 are the same. R1 and the b) Assume a direction for the current through R2 (e.g., from node B to node A). Then summing currents at node B: KCL: I 2 + I R2 − I3 = 0 3.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 I R2 = I 3 − I 2 = 4 A This can also be done by inspection noting that the assumed direction of the current through direction of I3 are the same. c) Only one mesh current flows through direction, then: R2 and the R3. If the current through R3 is assumed to flow in the same I R1 = I 3 = 7 A . ______________________________________________________________________________________ Problem 3.2 Solution: Known quantities: Circuit shown in Figure P3.1 with source and node voltages: VS 1 = VS 2 = 110 V , V A = 103 V , VB = −107 V . Find: The voltage across each of the five resistors. Analysis: Assume a polarity for the voltages across R1 and R2 (e.g., from ground to node A, and from node B to ground). R1 is connected between node A and ground; therefore, the voltage across R1 is equal to this node voltage. R2 is connected between node B and ground; therefore, the voltage across R2 is equal to the negative of this voltage. VR1 = V A = 103 V, VR 2 = −VB = +107 V The two node voltages are with respect to the ground which is given. Assume a polarity for the voltage across R3 (e.g., from node B to node A). Then: KVL: V A + VR 3 + VB = 0 VR 3 = V A − VB = 210 V Assume polarities for the voltages across R4 and R5 (e.g., from node A to ground , and from ground to node B): KVL: − VS 1 + V R 4 + V A = 0 V R 4 = VS 1 − V A = 7 V KVL: − VS 2 − V B − V R 5 = 0 VR 5 = −VS 2 − VB = −3 V ______________________________________________________________________________________ Problem 3.3 Solution: Known quantities: Circuit shown in Figure P3.3 with known source currents and resistances, R1 3.3 = 3Ω, R2 = 1Ω, R3 = 6Ω . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Find: The currents I1, I2 using node voltage analysis. Analysis: At node 1: §1 · v1 ⋅ ¨ + 1¸ + v2 ⋅ (− 1) = 1 ©3 ¹ At node 2: § 1· v1 ⋅ (− 1) + v 2 ⋅ ¨1 + ¸ = −2 © 6¹ Solving, we find that: v1 = −1.5 V v2 = −3 V Then, v1 = − 0 .5 A 3 v i 2 = 2 = − 0 .5 A 6 i1 = ______________________________________________________________________________________ Problem 3.4 Solution: Known quantities: Circuit shown in Figure P3.3 with known source currents and resistances, R1 Find: The currents I1, I2 using mesh analysis. Analysis: At mesh (a): ia = 1 A At mesh (b): 3 (ib − ia ) + ib + 6 (ib − ic ) = 0 At mesh (c): ic = 2 A Solving, we find that: ib = 1.5 A Then, i1 = (ia − ib ) = −0.5 A i2 = (ib − ic ) = −0.5 A 3.4 = 3Ω, R2 = 1Ω, R3 = 6Ω . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 ______________________________________________________________________________________ Problem 3.5 Solution: Known quantities: Circuit shown in Figure P3.5 with resistance values, current and voltage source values. Find: The current, i, through the voltage source using node voltage analysis. Analysis: At node 1: v −v v1 v −v + 1 2 + 1 3 =0 200 5 100 At node 2: v2 − v1 + i + 0 .2 = 0 5 At node 3: −i + v3 − v1 v3 + =0 100 50 For the voltage source we have: v3 − v2 = 50 V Solving the system, we obtain: v1 = −45.53 V , v2 = −48.69 V , v3 = 1.31 V and, finally, i = 491 mA . ______________________________________________________________________________________ Problem 3.6 Solution: Known quantities: The current source value, the voltage source value and the resistance values for the circuit shown in Figure P3.6. Find: The three node voltages indicated in Figure P3.6 using node voltage analysis. Analysis: 50 Ω 75 Ω V1 V2 10V + V3 i 0.2 A 200 Ω 25 Ω 3.5 100 Ω G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 At node 1: v1 v −v + 1 2 = 0 .2 A 200 75 At node 2: v2 − v1 v2 v2 − v3 + + +i = 0 75 25 50 At node 3: −i + v3 − v2 v3 + =0 50 100 For the voltage source we have: v3 + 10 = v2 Solving the system, we obtain: v1 = 14.24 V , v2 = 4.58V , v3 = −5.42 V and, finally, i = −254 mA . ______________________________________________________________________________________ Problem 3.7 Solution: Known quantities: The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7. Find: The current, i, drawn from the independent voltage source using node voltage analysis. Analysis: At node 1: v1 − 3 v1 v1 − v2 + + =0 0.5 0.5 0.25 At node 2: v2 − v1 v + 2 + 0 .5 = 0 0.25 0.75 Solving the system, we obtain: v1 = 1.125 V , v2 = 0.75 V Therefore, i = 3 − v1 = 3.75 A . 0 .5 ______________________________________________________________________________________ Problem 3.8 Solution: Known quantities: The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8. Find: The voltage at nodes a and b, Va and Vb , and their difference, Va − Vb using node voltage analysis. 3.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: Using nodal analysis at the two nodes a and b, we write the equations Vb − 15 Vb + =0 18 20 Va − 15 Va + =0 36 20 Rearranging the equations, 38Vb − 300 = 0 14Va − 75 = 0 Solving for the two unknowns, Va = 5.36 V and Vb = 7.89 V Therefore, Va − Vb = −2.54 V ______________________________________________________________________________________ Problem 3.9 Solution: Known quantities: The voltage source value, 15 V, and the four resistance values, indicated in Figure P3.8. Find: The voltage at nodes a and b, Va and Vb , and their difference, Va − Vb using mesh analysis. Analysis: Using mesh analysis at the two meshes a and b, we write the equations 36 (ia − ib ) + 20 (ia − ib ) = 15 18ib + 20ib + 20 (ib − ia ) + 36 (ib − ia ) = 0 Rearranging the equations, 15 + ib 56 94ib − 56ia = 0 ia = Solving for the two unknowns, ia = 662 mA and ib = 395 mA Therefore, Va = 20 (ib − ia ) = 5.36 V , Vb = 20ib = 7.89 V and Va − Vb = −2.54 V . ______________________________________________________________________________________ Problem 3.10 Solution: Known quantities: Circuit of Figure P3.10 with voltage source, VS , current source, IS, and all resistances. 3.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Find: a. The node equations required to determine the node voltages. b. The matrix solution for each node voltage in terms of the known parameters. Analysis: a) Specify the nodes (e.g., A on the upper left corner of the circuit in Figure P3.10, and B on the right corner). Choose one node as the reference or ground node. If possible, ground one of the sources in the circuit. Note that this is possible here. When using KCL, assume all unknown current flow out of the node. The direction of the current supplied by the current source is specified and must flow into node A. KCL: − IS + Va − VS Va − Vb + =0 R2 R1 § 1 § 1 · V 1 · + ¸¸ + Vb ¨¨ − ¸¸ = I S + S Va ¨¨ R2 © R2 R1 ¹ © R1 ¹ KCL: Vb − Va Vb − VS Vb − 0 + + =0 R1 R3 R4 § 1 § 1 · 1 1 · VS ¸¸ = + Va ¨¨ − ¸¸ + Vb ¨¨ + © R1 ¹ © R1 R3 R4 ¹ R3 b) Matrix solution: IS + Va = VS R2 VS R3 1 1 + R1 R2 1 − R1 1 R1 § V ·§ 1 1 1 1 1 1 · § V ·§ 1 · ¨¨ I S + S ¸¸¨¨ + + + + ¸¸ − ¨¨ S ¸¸¨¨ − ¸¸ R2 ¹© R1 R3 R4 ¹ © R3 ¹© R1 ¹ R1 R3 R4 = © 1 §1 1 ·§ 1 1 1 · § 1 ·§ 1 · − ¨¨ + ¸¸¨¨ + + ¸¸ − ¨¨ − ¸¸¨¨ − ¸¸ R1 © R1 R2 ¹© R1 R3 R4 ¹ © R1 ¹© R1 ¹ 1 1 1 + + R1 R3 R4 − 1 1 V + IS + S R1 R2 R2 §1 V · VS 1 ·§ V · § 1 ·§ 1 ¨¨ + ¸¸¨¨ S ¸¸ − ¨¨ − ¸¸¨¨ I S + S ¸¸ − R2 ¹ R1 R3 © R1 R2 ¹© R3 ¹ © R1 ¹© = Vb = 1 1 1 §1 1 ·§ 1 1 1 · § 1 ·§ 1 · + − ¨¨ + ¸¸¨¨ + + ¸¸ − ¨¨ − ¸¸¨¨ − ¸¸ R1 R2 R1 © R1 R2 ¹© R1 R3 R4 ¹ © R1 ¹© R1 ¹ 1 1 1 1 − + + R1 R1 R3 R4 Notes: 1. The denominators are the same for both solutions. 2. The main diagonal of a matrix is the one that goes to the right and down. 3. The denominator matrix is the "conductance" matrix and has certain properties: a) The elements on the main diagonal [ i(row) = j(column) ] include all the conductance connected to node i = j. 3.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 b) The off-diagonal elements are all negative. c) The off-diagonal elements are all symmetric, i.e., the i j-th element = j i-th element. This is true only because there are no controlled (dependent) sources in this circuit. d) The off-diagonal elements include all the conductance connected between node i [row] and node j [column]. ______________________________________________________________________________________ Problem 3.11 Solution: Known quantities: Circuit shown in Figure P3.11 VS1 = VS 2 = 110 V R1 = 500 mΩ R2 = 167 mΩ R3 = 700 mΩ R4 = 200 mΩ R5 = 333 mΩ Find: a. The most efficient way to solve for the voltage across R3. Prove your case. b. The voltage across R3. Analysis: a) There are 3 meshes and 3 mesh currents requiring the solution of 3 simultaneous equations. Only one of these mesh currents is required to determine, using Ohm's Law, the voltage across R3. In the terminal (or node) between the two voltage sources is made the ground (or reference) node, then three node voltages are known (the ground or reference voltage and the two source voltages). This leaves only two unknown node voltages (the voltages across R1, VR1, and across R2, VR2). Both these voltages are required to determine, using KVL, the voltage across R3, VR3. A difficult choice. Choose node analysis due to the smaller number of unknowns. Specify the nodes. Choose one node as the ground node. In KCL, assume unknown currents flow out. b) KCL: KCL: VR1 − VS1 VR1 − 0 VR1 − VR 2 + + =0 R4 R1 R3 VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1 + + =0 R5 R2 R3 § 1 § 1 · V 1 1 · ¸¸ + VR 2 ¨¨ − ¸¸ = S1 + VR1 ¨¨ + © R1 R3 R4 ¹ © R3 ¹ R4 § 1 · § 1 V 1 1 · ¸¸ = − S 2 + + VR1 ¨¨ − ¸¸ + VR 2 ¨¨ R5 © R3 ¹ © R5 R2 R3 ¹ 3.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 1 1 1 1 1 1 + + = + + = 8.43 Ω-1 −3 −3 −3 R1 R3 R4 500 ⋅10 700 ⋅10 200 ⋅10 1 1 1 1 1 1 + + = + + = 10.42 Ω-1 −3 −3 −3 R5 R2 R3 333 ⋅10 167 ⋅10 700 ⋅10 1 1 = = 1.43 Ω-1 R3 700 ⋅10 −3 VS 1 VS 2 110 110 = = 550 A = = 330 A −3 R4 200 ⋅10 R5 333 ⋅10 −3 − 1.43 550 VR1 = − 330 10.42 (5731) − (472) = 61.30 V = 8.43 − 1.43 (87.84 ) − (2.04 ) − 1.43 10.42 8.429 550 − 1.429 − 330 (− 2782 ) − (− 786 ) = = −23.26 V 85.790 85.80 − VR1 + VR 3 + VR 2 = 0 VR 3 = VR1 − VR 2 = 84.59 V VR1 = KVL: ______________________________________________________________________________________ Problem 3.12 Solution: Known quantities: Circuit shown in Figure P3.12 VS 2 = kT k = 10 V/°C VS1 = 24 V RS = R1 = 12 kΩ R2 = 3 kΩ R3 = 10 kΩ . R4 = 24 kΩ VR 3 = −2.524 V The voltage across R3, which is given, indicates the temperature. Find: The temperature, T. Analysis: Specify nodes (A between R1 and R3, C between R3 and R2) and polarities of voltages (VA from ground to A, Vc from ground to C, and VR3 from C to A). When using KCL, assume unknown currents flow out. KVL: − V A + VR 3 + VC = 0 VC = V A − VR 3 Now write KCL at node C, substitute for VC, solve for VA: KCL: VC − VS1 VC − VA VC + + =0 R2 R3 R4 3.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 § 1 V A VS 1 1 1 · ¸¸ = 0 − + (V A − VR 3 )¨¨ + + R3 R2 © R2 R3 R4 ¹ § 1 VS 1 1 1 · 24 1 1 · § 1 ¸¸ + VR 3 ¨¨ + + + (− 2.524)¨ + + ¸ 3 3 3 R 24 ⋅103 ¹ © R2 R3 R4 ¹ = 3 ⋅10 © 3 ⋅10 10 ⋅10 = 18.14 V VA = 2 1 1 1 1 + + R2 R4 3 ⋅103 24 ⋅10 3 − VC = V A − VR 3 = 18.14 − (− 2.524) = 20.66 V Now write KCL at node A and solve for VS2 and T: KCL: VA − VS1 VA − VS 2 VA − VC + + =0 R1 RS R3 R R VS 2 = V A + S (V A − VS1 ) + S (V A − VC ) = R1 R3 12 ⋅10 3 12 ⋅10 3 ( ) (18.14 − 20.66) = 9.26 V 18 . 14 − 24 + 12 ⋅10 3 10 ⋅10 3 V 9.26 T = S2 = = 0.926 °C 10 k = 18.14 + ______________________________________________________________________________________ Problem 3.13 Solution: Known quantities: Circuit shown in Figure P3.13 VS = 5 V R2 = 1.8 kΩ AV = 70 R1 = 2.2 kΩ R3 = 6.8 kΩ R4 = 220 Ω Find: The voltage across R4 using KCL and node voltage analysis. Analysis: A node analysis is not a method of choice because the dependent source is [1] a voltage source and [2] a floating source. Both factors cause difficulties in a node analysis. A ground is specified. There are three unknown node voltages, one of which is the voltage across R4. The dependent source will introduce two additional unknowns, the current through the source and the controlling voltage (across R1) that is not a node voltage. Therefore 5 equations are required: 3.11 G. Rizzoni, Principles and Applications of Electrical Engineering [1] KCL V1 − VS V1 − V3 V1 − V2 + + =0 R1 R3 R2 [2] KCL V2 − V1 − I CS = 0 R2 [3] KCL V3 − V1 V + I CS + 3 = 0 R3 R4 [4] KVL [5] KVL − VS + VR1 + V1 = 0 Problem solutions, Chapter 3 VR1 = VS − V1 V2 = V3 + AV VR1 = V3 + AV (VS − V1 ) Substitute using Equation [5] into Equations [1], [2] and [3] and eliminate V2 (because it only appears − V3 − AV VR1 + V2 = 0 twice in these equations). Collect terms: §1 § 1 V V A 1 1 AV · 1 · ¸¸ + V3 ¨¨ − − ¸¸ + I CS (0) = S + S V V1 ¨¨ + + + R1 R2 © R1 R3 R2 R2 ¹ © R3 R2 ¹ § 1 AV · § 1 · V A ¸¸ + V3 ¨¨ ¸¸ + I CS (− 1) = − S V V1 ¨¨ − − R2 © R2 R2 ¹ © R2 ¹ § 1 · § 1 1 · V1 ¨¨ − ¸¸ + V3 ¨¨ + ¸¸ + I CS (+ 1) = 0 © R3 ¹ © R3 R4 ¹ 1 1 1 1 = = 555.6 ⋅10 -6 Ω-1 = = 147.1 ⋅10 -6 Ω-1 3 R2 1.8 ⋅10 R3 6.8 ⋅10 3 1 1 1 1 + = + = 702.6 ⋅10 -6 Ω-1 3 R3 R2 6.8 ⋅10 1.8 ⋅10 3 1 1 1 1 1 AV 1 + 70 + = + = 4.69 ⋅10 -3 Ω-1 + = = 39.44 ⋅10 -3 Ω-1 3 3 3 R3 R4 6.8 ⋅10 R2 R2 1.8 ⋅10 0.22 ⋅10 1 1 1 AV 1 1 1 + 70 + + + = + + = 40.05 ⋅10 -3 Ω-1 3 3 3 R1 R3 R2 R2 2.2 ⋅10 6.8 ⋅10 1.8 ⋅10 VS AV (5)(70 ) = = 194.4 mA R2 1.8 ⋅ 103 VS VS AV 5 (5)(70) + = + = 196.7 mA 3 R1 R2 2.2 ⋅ 10 1.8 ⋅10 3 Solving, we have: VR 4 = V3 = 5.1 mV Note: 1. This solution was not difficult in terms of theory, but was terribly long and arithmetically cumbersome. This was because the wrong method was used. There are only 2 mesh currents in the circuit; the sources were voltage sources; therefore, a mesh analysis is the method of choice. 2. In general, a node analysis will have fewer unknowns (because one node is the ground or reference node) and will, in such cases, be preferable. ______________________________________________________________________________________ 3.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.14 Solution: Known quantities: The values of the resistors and of the voltage sources (see Figure P3.14). Find: The voltage across the 10 Ω resistor in the circuit of Figure P3.14 using mesh current analysis. Analysis: For mesh (a): ia (50 + 20 + 20) − ib (20) − ic (20) = 12 For mesh (b): − ia (20 ) + ib (20 + 10) − ic (10) + 5 = 0 For mesh (c): − ia (20) − ib (10) + ic (20 + 10 + 15) = 0 Solving, ia = 127.5 mA ib = −67.8 mA and ic = 41.6 mA vR4 = 10 (ib − ic ) = 10 (− 0.109 ) = −1.09 V . ______________________________________________________________________________________ Problem 3.15 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.15. Find: The voltage across the current source using mesh current analysis. Analysis: For mesh (a): ia (20 + 30) + ib (− 30) = 3 For meshes (b) and (c): ia (− 30) + ib (10 + 30) + ic (30 + 20) = 0 For the current source: ic − ib = 0.5 Solving, 3.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 ia = −133 mA , ib = −322 mA and ic = 178 mA . Therefore, v = ic (30 + 20) = 8.89 V . ______________________________________________________________________________________ Problem 3.16 Solution: Known quantities: The values of the resistors and of the voltage source in the circuit of Figure P3.16. Find: The current i through the resistance R4 mesh current analysis. Analysis: For mesh (a): ia (50 + 1200) + ib (− 1200) = 5.6 For meshes (b) and (c): ia (− 1200) + ib (1200 + 330) + ic (440) = 0 For the current source: ic − ib = 0.2v x = 0.2 (1200 (ia − ib )) = 240 (ia − ib ) Solving, ia = 136 mA , ib = 137 mA and ic = −106 mA . Therefore, i = ic = −106 mA . ______________________________________________________________________________________ Problem 3.17 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5. Find: The current through the voltage source using mesh current analysis. Analysis: For mesh (a): ia (100 + 5) + ib (− 5) + 50 = 0 For the current source: ib − ic = 0.2 3.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 For meshes (b) and (c): − ia (5) + ib (200 + 5) + ic (50) = 50 Solving, ia = −465 mA , ib = 226 mA and ic = 26 mA . Therefore, i = ic − ia = 491 mA . ______________________________________________________________________________________ Problem 3.18 Solution: Known quantities: The values of the resistors and of the current source in the circuit of Figure P3.6. Find: The current through the voltage source in the circuit of Figure P3.6 using mesh current analysis. Analysis: 50 Ω 75 Ω 10V ia + i 0.2 A 200 Ω ib 25 Ω 100 Ω ic For mesh (a): ia (100) + 10 = 0 For mesh (b): ib (200 + 75 + 25) + ic (− 25) + 0.2 (− 200) = 0 For mesh (c): ib (− 25) + ic (50 + 25) = 10 Solving, ia = −100 mA , ib = 148 mA and ic = 183 mA . Therefore, i = ic − ia = 283 mA ______________________________________________________________________________________ Problem 3.19 Solution: Known quantities: The values of the resistors in the circuit of Figure P3.19. 3.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Find: The current in the circuit of Figure P3.19 using mesh current analysis. Analysis: 1/2 Ω 1Ω i2 I i i1 1/5 Ω i3 1/4 Ω 1/3 Ω Since I is unknown, the problem will be solved in terms of this current. For mesh #1, it is obvious that: i1 = I For mesh #2: § 1 1· § 1· i1 (− 1) + i2 ¨1 + + ¸ + i3 ¨ − ¸ = 0 © 2 5¹ © 5¹ For mesh #3: § 1· § 1· §1 1 1· i1 ¨ − ¸ + i2 ¨ − ¸ + i3 ¨ + + ¸ = 0 © 4¹ © 5¹ © 4 3 5¹ Solving, i2 = 0.645 I i3 = 0.483I i = i3 − i2 Then, and i = 0.483I − 0.645 I = −0.163I ______________________________________________________________________________________ Problem 3.20 Solution: Known quantities: The values of the resistors of the circuit in Figure P3.20. Find: The voltage gain, AV = v2 , in the circuit of Figure P3.20 v1 using mesh current analysis. Analysis: Note that v= i1 − i2 2 For mesh #1: 3.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 § 1· § 1· i1 ¨1 + ¸ + i2 ¨ − ¸ + i3 (0 ) = v1 © 2¹ © 2¹ For mesh #2: § 1· §1 1 1· § 1· i1 ¨ − ¸ + i2 ¨ + + ¸ + i3 ¨ − ¸ = 2v © 4¹ ©2 4 4¹ © 2¹ For mesh #3: or i1 (− 1.5) + i2 (2) + i3 (− 0.25) = 0 §1 1· § 1· i1 (0) + i2 ¨ − ¸ + i3 ¨ + ¸ = −2v ©4 4¹ © 4¹ or i1 (1) + i2 (− 1.25) + i3 (0.5) = 0 i3 = −0.16v1 Solving, 1 from which v2 = i3 = −0.04v1 4 v and AV = 2 = −0.04 v1 ______________________________________________________________________________________ Problem 3.21 Solution: Known quantities: Circuit in Figure P3.21 and the values of the voltage sources, VS1 = VS 2 = 450 V , and the values of the 5 resistors: R1 = 8 Ω R2 = 5 Ω R4 = R5 = 0.25 Ω R3 = 32 Ω Find: The voltages across R1, R2 and R3 using KCL and node analysis. Analysis: Choose a ground/reference node. The node common to the two voltage sources is the best choice. Specify polarity of voltages and direction of the currents. KCL: KCL: VR1 − VS 1 VR1 − 0 VR1 − VR 2 + + =0 R4 R1 R3 VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1 + + =0 R5 R2 R3 Collect terms in terms of the unknown node voltages: § 1 § 1 · V 1 1 · VR1 ¨¨ + + ¸¸ + VR 2 ¨¨ − ¸¸ = S1 © R4 R1 R3 ¹ © R3 ¹ R4 § 1 · § 1 V 1 1 · + ¸¸ = − S 2 VR1 ¨¨ − ¸¸ + VR 2 ¨¨ + R5 © R3 ¹ © R5 R2 R3 ¹ 3.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Evaluate the coefficients of the unknown node voltages: VS1 VS 2 450 = = = 1.8 kA R4 R5 0.25 1 1 = = 0.03125 Ω−1 R3 32 1 1 1 1 1 1 + + = + + = 4.14 Ω−1 R1 R4 R3 0.25 8 32 1 1 1 1 1 1 + + = + + = 4.23 Ω−1 R5 R2 R3 0.25 5 32 VR1 = 1800 − 31.25 ⋅ 10 −3 − 1800 4.23 4.16 − 31.25 ⋅10 −3 − 31.25 ⋅ 10 −3 4.23 4.156 1800 −3 − 31.25 ⋅10 − 1800 = −422.2 V 17.59 − VR1 + VR 3 + VR 2 = 0 VR 2 = KVL: = 429.5 V VR 3 = VR1 − VR 2 = 852.0 V ______________________________________________________________________________________ Problem 3.22 Solution: Known quantities: Circuit in Figure P3.22 with the values of the voltage sources, VS1 = VS 2 = 115 V , and the values of the 5 resistors: R1 = R2 = 5 Ω R3 = 10 Ω R4 = R5 = 200 mΩ Find: The new voltages across R1, R2 and R3, in case F1 "blows" or opens using KCL and node analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. KCL: KCL: VR1 − 0 VR1 − VR 2 + =0 R1 R3 VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1 + + =0 R5 R2 R3 0+ Collect terms in unknown node voltages: 3.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 § 1 § 1 · 1 · VR1 ¨¨ + ¸¸ + VR 2 ¨¨ − ¸¸ = 0 © R1 R3 ¹ © R3 ¹ § 1 · § 1 V 1 1 · + ¸¸ = − S 2 VR1 ¨¨ − ¸¸ + VR 2 ¨¨ + R5 © R3 ¹ © R5 R2 R3 ¹ 1 1 = = 0.1 Ω−1 R3 10 1 1 + = 0.3 Ω−1 R1 R3 VS 2 115 = = 575 A R5 200 ⋅ 10 -3 0 − 0 .1 VR1 = − 575 5.3 (0) − (57.5) = −36.39 V = 0.3 − 0.1 (1.59 ) − (0.01) − 0.1 5.3 0 .3 VR 2 = 1 1 1 + + = 5.3 Ω−1 R5 R2 R3 0 − −0.1 − 575 (− 172.5) − (0 ) = = −109.2 V 1.58 1.58 − VR1 + VR 3 + VR 2 = 0 KVL: VR 3 = VR1 − VR 2 = 72.81 V − VS1 + VR 4 + VF + VR1 = 0 KVL: VR 4 = I1 R4 = 0 VF = 115 − 0 − (− 36.39) = 151.4 V Note the voltages are strongly dependent on the loads (R1, R2 and R3) connected at the time the fuse blows. With other loads, the result will be quite different. ______________________________________________________________________________________ Problem 3.23 Solution: Known quantities: Circuit in Figure P3.22 and the values of the voltage sources, VS1 = VS 2 = 120 V , and the values of the 5 resistors: R1 = R2 = 2 Ω R3 = 8 Ω R4 = R5 = 250 mΩ Find: The voltages across R1, R2, R3, and F1 in case F1 "blows" or opens using KCL and node analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. KCL: 0+ VR1 − 0 VR1 − VR 2 + =0 R1 R3 3.19 G. Rizzoni, Principles and Applications of Electrical Engineering KCL: Problem solutions, Chapter 3 VR 2 − (− VS 2 ) VR 2 − 0 VR 2 − VR1 + + =0 R5 R2 R3 § 1 § 1 · 1 · VR1 ¨¨ + ¸¸ + VR 2 ¨¨ − ¸¸ = 0 © R1 R3 ¹ © R3 ¹ § 1 · § 1 V 1 1 · + ¸¸ = − S 2 VR1 ¨¨ − ¸¸ + VR 2 ¨¨ + R5 © R3 ¹ © R5 R2 R3 ¹ 1 1 = = 0.125 Ω−1 R3 8 1 1 + = 0.625 Ω−1 R1 R3 VS 2 120 = = 480 A R5 250 ⋅ 10 -3 0 − 0.125 VR1 = 4.625 (0 ) − (60) = −20.87 V = − 0.125 (2.89 ) − (0.016 ) − 0.125 4.625 − 480 0.625 0.625 VR 2 = KVL: KVL: 1 1 1 + + = 4.625 Ω−1 R5 R2 R3 0 − 0.125 − 480 (− 300 ) − (0 ) = = −104.35 V 2.87 2.87 − VR1 + VR 3 + VR 2 = 0 VR 3 = VR1 − VR 2 = 83.48 V − VS1 + VR 4 + VF + VR1 = 0 VR 4 = I1 R4 = 0 VF = 120 − 0 − (− 20.87 ) = 140.9 V ______________________________________________________________________________________ Problem 3.24 Solution: Known quantities: The values of the voltage sources, VS1 = VS 2 = VS 3 = 170 V , and the values of the 6 resistors in the circuit of Figure P3.24: RW 1 = RW 2 = RW 3 = 0.7 Ω R1 = 1.9 Ω R2 = 2.3 Ω R3 = 11 Ω Find: a. The number of unknown node voltages and mesh currents. b. Unknown node voltages. Analysis: If the node common to the three sources is chosen as the ground/reference node, and the series resistances are combined into single equivalent resistances, there is only one unknown node voltage. On the other 3.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 hand, there are two unknown mesh currents. A node analysis is the method of choice! Specify polarity of voltages and direction of currents. Req1 = RW 1 + R1 = 2.6 Ω Req 2 = RW 2 + R2 = 3.0 Ω Req3 = RW 3 + R3 = 11.7 Ω KCL: VN − VS1 VN − (− VS 2 ) VN − VS 3 + + =0 Req1 Req 2 Req 3 VS 1 VS 2 VS 3 170 170 170 − + − + Req1 Req 2 Req 3 VN = = 2.6 3.0 11.7 = 28.94 V 1 1 1 1 1 1 + + + + Req1 Req 2 Req 3 2.6 3.0 11.7 KVL: − VS1 + I1 RW 1 + I1 R1 + V N = 0 I1 = VS1 − VN 170 − 28.94 = = 54.26 A RW 1 + R1 2.6 ______________________________________________________________________________________ Problem 3.25 Solution: Known quantities: The values of the voltage sources, VS 1 = VS 2 = VS 3 = 170 V , the common node voltage, VN = 28.94 V ,and the values of the 6 resistors in the circuit of Figure P3.24: RW 1 = RW 2 = RW 3 = 0.7 Ω R1 = 1.9 Ω R2 = 2.3 Ω R3 = 11 Ω Find: The current through and voltage across R1. Analysis: KVL: − VS1 + I1 RW 1 + I1 R1 + V N = 0 I1 = VS1 − VN 170 − 28.94 = = 54.26 A RW 1 + R1 2.6 = I1 R1 = (54.26)(1.9) = 103.1 V OL: VR1 ______________________________________________________________________________________ Problem 3.26 Solution: Known quantities: The values of the voltage sources, VS1 = VS 2 = VS 3 = 170 V , and the values of the 6 resistors in the circuit of Figure P3.24: 3.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 RW 1 = RW 2 = RW 3 = 0.7 Ω R1 = 1.9 Ω R2 = 2.3 Ω R3 = 11 Ω Find: The mesh (or loop) equations and any additional equation required to determine the current through R1 in the circuit shown in Figure P3.24. Analysis: KVL: − VS1 + I1RW 1 + I1R1 + (I1 − I 2 )R2 + (I1 − I 2 )RW 2 − VS 2 = 0 KVL: VS 2 + (I 2 − I1 )RW 2 + (I 2 − I1 )R2 + I 2 R3 + I 2 RW 3 + VS 3 = 0 I1 (R1 + RW 1 + R2 + RW 2 ) + I 2 (− R2 − RW 2 ) = VS1 + VS 2 I1 (− R2 − RW 2 ) + I 2 (R2 + RW 2 + R3 + RW 3 ) = −VS 2 − VS 3 I R1 = I 1 = (VS1 + VS 2 ) − (R2 + RW 2 ) − (VS 2 + VS 3 ) (R2 + RW 2 + R3 + RW 3 ) (R1 + RW 1 + R2 + RW 2 ) − (R2 + RW 2 ) (R2 + RW 2 + R3 + RW 3 ) − (R2 + RW 2 ) ______________________________________________________________________________________ Problem 3.27 Solution: Known quantities: The values of the voltage sources, VS1 = 90 V, VS 2 = VS 3 = 110 V , and the values of the 6 resistors in the circuit of Figure P3.24: RW 1 = RW 2 = RW 3 = 1.3 Ω R1 = 7.9 Ω R2 = R3 = 3.7 Ω Find: The branch currents, using KVL and loop analysis. Analysis: Three equations are required. Voltages will be summed around the 2 loops that are meshes, and KCL at the common node between the resistances. Assume directions of the branch currents and the associated polarities of the voltages. After like terms are collected: KVL: KVL: KCL: − VS1 + I1RW 1 + I1R1 + (I1 − I 2 )R2 + (I1 − I 2 )RW 2 − VS 2 = 0 VS 2 + (I 2 − I1 )RW 2 + (I 2 − I1 )R2 + I 2 R3 + I 2 RW 3 − VS 3 = 0 I 3 = I1 − I 2 Plugging in the given parameters results in the following system of equations: 14.2 I1 − 5.0 I 2 = 200 5.0 I1 − 10.0 I 2 = 0 I 3 = I1 − I 2 Solving the system of equations gives: I1 = 17.09 A I 2 = 8.55 A 3.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 I 3 = 8.55 A Hence, the assumed polarity of the second and third branch currents is actually reversed. ______________________________________________________________________________________ Problem 3.28 Solution: Known quantities: The values of the voltage sources, Figure P3.22: R1 VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of 5 R2 R3 10 R4 R5 200 m Find: The voltages across R1, R2 and R3, under normal conditions, i.e., no blown fuses using KVL and a mesh analysis. Analysis: KVL: I1 R1 R4 R1I3 I2 R2 R2 I3 Vs2 R5 -I1R1 I2 R2 Vs1 R1 R2 R3 I3 0 Rearranging the above equations: 5.2 I1 − 5I 3 = 115 5.2 I 2 − 5I 3 = 115 5I1 + 5I 2 − 20 I 3 = 0 I1 = I 2 I1 = 2 I 3 −5 115 0 115 5.2 − 5 0 5 − 20 − 11960 = = 42.6 A I1 = I 2 = −5 5.2 0 − 280.8 0 5.2 − 5 5 5 − 20 I 3 = 21.3 A VR1 = R1 (I1 − I 3 ) = 106.5 V VR 2 = R2 (I 3 − I 2 ) = −106.5 V VR 3 = R3 I 3 = 213 V ______________________________________________________________________________________ 3.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.29 Solution: Known quantities: The values of the voltage sources, VS1 = VS 2 = 110 V , and the values of the 5 resistors in the circuit of Figure P3.22: R1 = 100 Ω R2 = 22 Ω R3 = 70 Ω R4 = R5 = 13 Ω Find: The voltage across R1 using KVL and mesh analysis. Analysis: KVL: I1 (R1 + R4 ) − R1 I 3 = VS1 I 2 (R2 + R5 ) − R2 I 3 = VS 2 -I1 R1 − I 2 R2 + (R1 + R2 + R3 )I 3 = 0 Rearranging the above equations: 113I1 − 100 I 3 = 110 35 I 2 − 22 I 3 = 110 100 I1 + 22 I 2 − 192 I 3 = 0 110 110 0 I1 = 113 0 100 0 35 22 0 35 22 − 100 − 22 − 192 − 935660 = = 2.64 A − 100 − 354668 − 22 − 192 I 3 = 1.88 A VR1 = R1 (I 1 − I 3 ) = 75.89 V ______________________________________________________________________________________ Problem 3.30 Solution: Known quantities: The values of the voltage sources, VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of Figure P3.22: R1 = 5 Ω R2 = 5 Ω R3 = 10 Ω R4 = R5 = 0.2 Ω Find: The voltage across R1 using KVL and mesh analysis. 3.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. KVL: KVL: I1 = 0 − VS 2 + (I 2 − I 3 )R2 + I 2 R5 = 0 I 3 R1 + I 3 R3 + (I 3 − I 2 )R2 = 0 I 2 (R2 + R5 ) + I 3 (− R2 ) = VS 2 I 2 (− R2 ) + I 3 (R1 + R2 + R3 ) = 0 R2 + R5 = 5.2 Ω R1 + R2 + R3 = 20 Ω 115 − 5 0 20 (2300) − (0) I2 = = = 29.11 A 5.2 − 5 (104) − (25) − 5 20 5.2 115 −5 0 (0) − (− 575) = 7.28 A I3 = = 5.2 − 5 79 − 5 20 VR1 = I R1 R1 = − I 3 R1 = −36.39 V VR 2 = I R 2 R2 = (I 3 − I 2 )R2 = −109.16 V VR 3 = I R 3 R3 = I 3 R3 = 72.78 V KVL: − VS1 + I1 R4 + VF + VR1 = 0 VF = 151.39 V ______________________________________________________________________________________ Problem 3.31 Solution: Known quantities: The values of the voltage sources, VS1 = VS 2 = 115 V , and the values of the 5 resistors in the circuit of Figure P3.22: R1 = 4 Ω R2 = 7.5 Ω R3 = 12.5 Ω R4 = R5 = 1 Ω Find: The voltages across R1, R2, R3, and across the open fuse using KVL and mesh analysis. Analysis: Specify polarity of voltages. The ground is already specified. The current through the fuse F1 is zero. I1 = 0 3.25 G. Rizzoni, Principles and Applications of Electrical Engineering KVL: Problem solutions, Chapter 3 I 2 (R2 + R5 ) − R2 I 3 = VS 2 − I 2 R2 + (R1 + R2 + R3 )I 3 = 0 Rearranging the above equations: 8.5 I 2 − 7.5 I 3 = 115 7.5 I 2 − 24 I 3 = 0 115 − 7.5 0 − 24 − 2760 I2 = = = 18.68 A 8.5 − 7.5 − 147.75 7.5 − 24 I 3 = 5.84 A VR1 = − I 3 R1 = −23.35 V VR 2 = R2 (I 3 − I 2 ) = −96.3 V VR 3 = R3 I 3 = 73 V VF = VS1 − VR1 = 138.35 V ______________________________________________________________________________________ 3.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Section 3.5: Superposition Problem 3.32 Solution: Known quantities: The values of the voltage sources, VS1 = 110 V , VS 2 = 90 V and the values of the 3 resistors in the circuit of Figure P3.32: R1 = 560 Ω R2 = 3.5 kΩ R3 = 810 Ω Find: The current through R1 due only to the source VS2. Analysis: Suppress VS1. Redraw the circuit. Specify polarity of VR1. Choose ground. KCL: OL: − VR1−2 − 0 − V R1−2 − 0 − VR1− 2 − (− VS 2 ) + + =0 R1 R2 R3 VS 2 90 R3 810 VR1−2 = = = 33.61 V 1 1 1 1 1 1 + + + + R1 R2 R3 560 3500 810 V 33.61 = 60.02 mA I R1−2 = R1−2 = 560 R1 ______________________________________________________________________________________ Problem 3.33 Solution: Known quantities: The values of the current source, of the voltage source and of the resistors in the circuit of Figure P3.33: I B = 12 A RB = 1 Ω VG = 12 V RG = 0.3 Ω R = 0.23 Ω Find: The voltage across R1 using superposition. Analysis: Specify a ground node and the polarity of the voltage across R. Suppress the voltage source by replacing it with a short circuit. Redraw the circuit. KCL: − IB + VR − I VR − I VR − I + + =0 RB RG R 3.27 G. Rizzoni, Principles and Applications of Electrical Engineering VR − I = Problem solutions, Chapter 3 IB 12 = = 1.38 V 1 1 1 1 1 1 + + + + RB RG R 1 0.3 0.23 Suppress the current source by replacing it with an open circuit. KCL: VR −V VR −V − VG VR −V + + =0 RB RG R VG 12 RG 0 .3 = = 4.61 V VR −V = 1 1 1 1 1 1 + + + + RB RG R 1 0.3 0.23 VR = VR − I + VR −V = 5.99 V Note: Superposition essentially doubles the work required to solve this problem. The voltage across R can easily be determined using a single KCL. ______________________________________________________________________________________ Problem 3.34 Solution: Known quantities: The values of the voltage sources and of the resistors in the circuit of Figure P3.34: VS1 = VS 2 = 12 V R1 = R2 = R3 = 1 kΩ Find: The voltage across R2 using superposition. Analysis: Specify the polarity of the voltage across R2 . Suppress the voltage source VS1 by replacing it with a short circuit. Redraw the circuit. 1 1 kΩ = 0.5 kΩ 2 R2 (12)(1000) = 8 V = R2 + Req 1000 + 500 Req = R1 R3 = V R 2 − 2 = VS 2 Suppress the voltage source VS 2 by replacing it with a short circuit. Redraw the circuit. 1 1 k Ω = 0 .5 k Ω 2 Req (12 V )(0.5 kΩ) = −4 V = −VS1 = R1 + Req 1 kΩ + 0.5 kΩ Req = R2 R3 = VR 2−1 VR 2 = VR 2−1 + V R 2−2 = −4 V + 8 V = 4 V 3.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Note: Although superposition is necessary to solve some circuits, it is a very inefficient and very cumbersome way to solve a circuit. This method should, if at all possible, be avoided. It must be used when the sources in a circuit are AC sources with different frequencies, or where some sources are DC and others are AC. ______________________________________________________________________________________ Problem 3.35 Solution: Known quantities: The values of the voltage sources and of the resistors in the circuit of Figure P3.35: VS1 = VS 2 = 450 V R1 = 7 Ω R2 = 5 Ω Find: The component of the current through R3 = 10 Ω R4 = R5 = 1 Ω R3 that is due to VS2, using superposition. Analysis: Suppress VS1 by replacing it with a short circuit. Redraw the circuit. A solution using equivalent resistances R1 and R4 are in parallel: RR (7 )(1) = 0.875 Ω R14 = 1 4 = R1 + R4 7 + 1 R14 is in series with R3 : R143 = R14 + R3 = 0.875 + 10 = 10.875 Ω looks reasonable. Req = R5 + (R2 R143 ) = R5 + R2 R143 (5)(10.875) = 4.425 Ω = 1+ R2 + R143 5 + 10.875 VS 2 450 = = 101.695 A Req 4.425 OL: IS = CD: I R 3−2 = I S R2 (101.695)(5) = 32.03 A = 5 + 10.875 R2 + R143 ______________________________________________________________________________________ Problem 3.36 Solution: Known quantities: The values of the voltage sources and of the resistors in the circuit of Figure P3.24: VS1 = VS 2 = VS 3 = 170 V RW 1 = RW 2 = RW 3 = 0.7 Ω R1 = 1.9 Ω Find: The current through R2 = 2.3 Ω R3 = 11 Ω R1 , using superposition. 3.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: Req1 = RW 1 + R1 = 2.6 Ω Req 2 = RW 2 + R2 = 3 Ω Req 3 = RW 3 + R3 = 11.7 Ω Specify the direction of I1. Suppress VS2 and VS3. Redraw circuit. Req = Req1 + I I −1 = Req 2 Req 3 Req 2 + Req 3 = 4.99 Ω VS 1 = 34.08 A Req Suppress VS1 and VS3. Redraw circuit. KCL: VA − (− VS 2 ) VA V + + A =0 Req 2 Req1 Req3 VS 2 = −70.54 V VA = − Req 2 Req 2 1+ + Req1 Req 3 I I −2 = − VA = 27.13 A Req1 Suppress VS1 and VS2. Redraw circuit. KCL: V A − (− VS 3 ) V A − 0 V A − 0 + + =0 Req 3 Req1 Req 2 VS 3 VA = − 1+ I I −3 = − Req 3 Req1 + Req3 = 18.09 V Req 2 VA = −6.96 A Req1 I = I I −1 + I I − 2 + I I −3 = 54.25 A Note: Superposition should be used only for special conditions, as stated in the solution to Problem 3.34. In the problem above a better method is: a. mesh analysis using KVL (2 unknowns) b. node analysis using KCL (1 unknown but current must be obtained using OL). ________________________________________________________________________ Problem 3.37 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.5. Find: The current through the voltage source using superpoistion. Analysis: (1) Suppress voltage source V. Redraw the circuit. 3.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 For mesh (a): ia (100 + 5) + ib (− 5) = 0 For the current source: ib − ic = 0.2 For meshes (b) and (c): − ia (5) + ib (200 + 5) + ic (50) = 0 Solving, ia = 2 mA , ib = 39 mA and ic = −161 mA . Therefore, i1 = ic − ia = −163 mA . (2) Suppress current source I. Redraw the circuit. For mesh (a): ia (100 + 5) + ib (− 5) + 50 = 0 For mesh (b): − ia (5) + ib (200 + 5 + 50) = 50 Solving, ia = −467mA and ib = 187 mA . Therefore, i2 = ib − ia = 654 mA . Using the principle of superposition, i = i1 + i2 = 491 mA ________________________________________________________________________ Problem 3.38 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.6. Find: The current through the voltage source using superposition. Analysis: (1) Suppress voltage source V. Redraw the circuit. 50 Ω 75 Ω ia i1 0.2 A 200 Ω ib 25 Ω 3.31 ic 100 Ω G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 For mesh (a): ia = 0 For mesh (b): ib (200 + 75 + 25) + ic (− 25) − 40 = 0 For mesh (c): ib (− 25) + ic (25 + 100) = 0 Solving, ib = 136 mA and ic = 27 mA . Therefore, i1 = ic = 27 mA . (2) Suppress current source I. Redraw the circuit. 50 Ω 75 Ω 10V ia + i2 200 Ω ib 25 Ω ic 100 Ω For mesh (a): ia (50) − 10 = 0 For mesh (b): ib (200 + 75 + 25) + ic (− 25) = 0 For mesh (c): ib (− 25) + ic (25 + 100) = −10 Solving, ia = 200 mA , ib = −6.8 mA and ic = −81 mA . Therefore, i2 = ic − ia = −281 mA . Using the principle of superposition, i = i1 + i2 = −254 mA ________________________________________________________________________ 3.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.39 Solution: Known quantities: The voltage source value, 3 V, and the five resistance values, indicated in Figure P3.7. Find: The current, i, drawn from the independent voltage source using superposition. Analysis: (1) Suppress voltage source V. Redraw the circuit. At node 1: v1 v v −v + 1 + 1 2 =0 0.5 0.5 0.25 At node 2: v2 − v1 v + 2 + 0 .5 = 0 0.25 0.75 Solving the system, we obtain: v1 = −0.075 V , v2 Therefore, i1 = − = −0.15 V v1 = 150 mA . 0 .5 (2) Suppress current source I. Redraw the circuit. At node 1: v1 − 3 v1 v1 + + =0 0.5 0.5 (0.25 + 0.5 + 0.25) Solving, v1 = 1.2 V 3 − v1 Therefore, i1 = = 3 .6 A . 0 .5 Using the principle of superposition, i = i1 + i2 = 3.75 A ________________________________________________________________________ Problem 3.40 Solution: Known quantities: Circuit in Figure P3.12 VS 2 = kT k = 10 V/°C VS1 = 24 V RS = R1 = 12 kΩ R2 = 3 kΩ R3 = 10 kΩ R4 = 24 kΩ . VR 3 = −2.524 V 3.33 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 The voltage across R3, which is given, indicates the temperature. Find: The temperature, T using superposition. Analysis: (1) Suppress voltage source VS 2 . Redraw the circuit. For mesh (a): ia (24k ) + ib (− 12k ) + ic (− 12k ) = 24 For mesh (b): ia (− 12k ) + ib (46k ) + ic (− 10k ) = 0 For mesh (c): ia (− 12k ) + ib (− 10k ) + ic (25k ) = 0 Solving, ia = 2.08 mA , ib = 0.83 mA and ic = 1.33 mA . Therefore, VR 3, S 2 = 10000(ib − ic ) = −5 V . (2) Suppress voltage source VS1 . Redraw the circuit. For mesh (a): ia (24k ) + ib (− 12k ) + ic (− 12k ) + 10T = 0 For mesh (b): ia (− 12k ) + ib (46k ) + ic (− 10k ) = 10T For mesh (c): ia (− 12k ) + ib (− 10k ) + ic (25k ) = 0 Solving, ia = −0.52T mA , ib = 0.029T mA and ic = −0.2381T mA . Therefore, VR 3, S1 = 10000(ib − ic ) = 2.671T V . Using the principle of superposition, VR 3 = VR 3,S 2 + VR 3,S 1 = −5 + 2.671T = −2.524 V Therefore, T = 0.926 °C. ________________________________________________________________________ Problem 3.41 Solution: Known quantities: The values of the resistors and of the voltage sources (see Figure P3.14). Find: The voltage across the 10 Ω resistor in the circuit of Figure P3.14 using superposition. 3.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: (1) Suppress voltage source VS1 . Redraw the circuit. For mesh (a): ia (50 + 20 + 20) − ib (20) − ic (20 ) = 0 For mesh (b): − ia (20 ) + ib (20 + 10) − ic (10) + 5 = 0 For mesh (c): − ia (20) − ib (10) + ic (20 + 10 + 15) = 0 Solving, ia = −73.8 mA , ib = −245 mA and ic = −87.2 mA . Therefore, V10Ω ,S 1 = 10(ib − ic ) = −1.578 V . (2) Suppress voltage source VS 2 . Redraw the circuit. For mesh (a): ia (50 + 20 + 20) − ib (20) − ic (20) = 12 For mesh (b): − ia (20) + ib (20 + 10) − ic (10) = 0 For mesh (c): − ia (20) − ib (10) + ic (20 + 10 + 15) = 0 Solving, ia = 201 mA , ib = 177 mA and ic = 129 mA . Therefore, V10Ω , S1 = 10(ib − ic ) = 0.48 V . Using the principle of superposition, V10Ω = V10 Ω ,S 2 + V10Ω , S1 = −1.09 V . ________________________________________________________________________ Problem 3.42 Solution: Known quantities: The values of the resistors, of the voltage source and of the current source in the circuit of Figure P3.15. Find: The voltage across the current source using superposition. Analysis: (1) Suppress voltage source. Redraw the circuit. 3.35 G. Rizzoni, Principles and Applications of Electrical Engineering For mesh (a): ia (20 + 30) + ib (− 30 ) = 0 For meshes (b) and (c): ia (− 30) + ib (10 + 30) + ic (30 + 20) = 0 For the current source: ic − ib = 0.5 Solving, ia = −208 mA , ib = −347 mA and ic = 153 mA . Therefore, vV = ic (30 + 20) = 7.65 V . (2) Suppress current source. Redraw the circuit. For mesh (a): ia (20 + 30) + ib (− 30) = 3 For mesh (b): ia (− 30) + ib (90) = 0 Solving, ia = 75 mA and ib = 25 mA . Therefore, v I = ib (30 + 20) = 1.25 V . Using the principle of superposition, v = vV + v I = 8.9 V . 3.36 Problem solutions, Chapter 3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Section 3.6: Equivalent circuits Focus on Methodology: Computation of equivalent resistance of a one-port network that does not contain dependent sources 1. 2. 3. Remove the load, leaving the load terminals open circuited. Zero all independent voltage and current sources Compute the total resistance between load terminals, with the load removed. This resistance is equivalent to that which would be encountered by a current source connected to the circuit in place of the load. Focus on Methodology: Computing the Thevenin voltage 1. 2. 3. 4. Remove the load, leaving the load terminals open circuited. Define the open-circuit voltage vOC across the open load terminals. Apply any preferred method (e.g.: nodal analysis) to solve for vOC. The Thevenin voltage is vT = vOC. Focus on Methodology: Computing the Norton current 1. 2. 3. 4. Replace the load with a short circuit. Define the short-circuit current iSC to be the Norton equivalent current. Apply any preferred method (e.g.: nodal analysis) to solve for iSC. The Norton current is iN = iSC. Problem 3.43 Solution: Known quantities: The schematic of the circuit (see Figure P3.1). Find: The Thévenin equivalent resistance seen by resistor Norton (short-circuit) current when R3 , the Thévenin (open-circuit) voltage and the R3 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = R1 || R4 + R2 || R5 = RR R1 R4 + 2 5 R1 + R4 R2 + R5 (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 v1 − VS1 + =0 R1 R4 For node #2: 3.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 v 2 v 2 + VS 2 + =0 R2 R5 Solving the system, v1 = R1 VS 1 R1 + R4 v2 = − R2 VS 2 R2 + R5 Therefore, vOC = v1 − v2 = R2 R1 VS 2 VS 1 + R2 + R5 R1 + R4 (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (R1 + R4 ) − R1ic = VS1 For mesh (b): ib (R2 + R5 ) − R2 ic = VS 2 For mesh (c): − R1ia − R2ib + ic (R1 + R2 ) = 0 Solving the system, (R1 R2 + R1 R5 + R2 R5 )VS1 + R1 R2VS 2 R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 ) R R V + (R1 R2 + R1 R4 + R2 R4 )VS 2 ib = 1 2 S1 R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 ) R (R + R5 )VS1 + R2 (R1 + R4 )VS 2 ic = 1 2 R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 ) ia = Therefore, R1 (R2 + R5 )VS 1 + R2 (R1 + R4 )VS 2 R1 R4 (R2 + R5 ) + R2 R5 (R1 + R4 ) ________________________________________________________________________ iSC = ic = Problem 3.44 Solution: Known quantities: The schematic of the circuit (see Figure P3.6). Find: The Thévenin equivalent resistance seen by resistor Norton (short-circuit) current when R5 , the Thévenin (open-circuit) voltage and the R5 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. 3.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 50 Ω 75 Ω 200 Ω 25 Ω RT RT = 25 Ω || (75 Ω + 200 Ω ) = 22.92 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. 50 Ω 75 Ω V1 0.2 A 200 Ω V2 25 Ω For node #1: v1 v1 − v2 + = 0 .2 200 75 For node #2: v2 − v1 v2 v2 − v3 + + + i10V = 0 75 25 50 For node #3: v3 − v2 = i10V 50 For the voltage source: v3 + 10 = v2 Solving the system, v1 = 13.33 V , v2 = 3.33 V and v3 = −6.67 V . Therefore, vOC = v3 = −6.67 V . (3) Replace the load with a short circuit. Redraw the circuit. 3.39 10V + V3 + voc _ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 50 Ω 75 Ω 10V ia + 0.2 A 200 Ω ib 25 Ω ic isc For mesh (a): ia (50) = 10 For mesh (b): ib (300) − ic (25) = 40 For mesh (c): ib (25) − ic (25) = 10 Solving the system, ia = 200 mA , ib = 109 mA and ic = −291 mA . Therefore, iSC = ic = −291 mA . ________________________________________________________________________ Problem 3.45 Solution: Known quantities: The schematic of the circuit (see Figure P3.7). Find: The Thévenin equivalent resistance seen by resistor R5 , the Thévenin (open-circuit) voltage and the Norton (short-circuit) current when R5 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = 0.5 Ω + 0.25 Ω + (0.5 Ω || 0.5 Ω ) = 1 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 − 3 v1 v1 − v2 + + =0 0.5 0.5 0.25 For node #2: v2 − v1 + 0 .5 = 0 0.25 Solving the system, 3.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 v1 = 1.375 V and v2 = 1.25 V . Therefore, vOC = v2 = 1.25 V . (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (0.5 + 0.5) − ib (0.5) = 3 For meshes (b) and (c): − ia (0.5) + ib (0.5 + 0.25) + ic (0.5) = 0 For the current source: ib − ic = 0.5 Solving the system, ia = 3.875 A , ib = 1.75 A and ic = 1.25 A . Therefore, iSC = ic = 1.25 A . ________________________________________________________________________ Problem 3.46 Solution: Known quantities: The schematic of the circuit (see Figure P3.12). Find: The Thévenin equivalent resistance seen by resistor R3 , the Thévenin (open-circuit) voltage and the Norton (short-circuit) current when R3 is the load. Assumption: As in P3.12, we assume T = 0.926 °C, so that VS 2 = 9.26 V . Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = 12 kΩ || 12 kΩ + 3 kΩ || 24 kΩ = 8.67 kΩ (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 − 24 v1 − 9.26 + =0 12000 12000 For node #2: v2 − 24 v2 + =0 3000 24000 Solving the system, v1 = 16.63 V and v2 = 21.33 V . Therefore, vOC = v1 − v2 = −4.7 V . 3.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (24k ) − ib (12k ) − ic (12k ) = 24 − 9.26 For mesh (b): − ia (12k ) + ib (36k ) = 9.26 For mesh (c): − ia (12k ) + ic (15k ) = 0 Solving the system, ia = 1.71 mA , ib = 0.83 mA and ic = 1.37 mA . Therefore, iSC = ib − ic = −0.54 mA . ________________________________________________________________________ Problem 3.47 Solution: Known quantities: The schematic of the circuit (see Figure P3.14). Find: The Thévenin equivalent resistance seen by resistor R4 , the Thévenin (open-circuit) voltage and the Norton (shortcircuit) current when R4 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = R2 || (R3 + (R1 || R5 )) = 20 Ω || (20 Ω + (50 Ω || 15 Ω )) = 12.24 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 − 12 v1 − v2 + + i5V = 0 50 20 For node #2: v2 − v1 v2 + =0 20 20 For node #3: v3 − i5V = 0 15 For the 5-V voltage source: v1 − v3 = 5 Solving the system, v1 = 5.14 V , v2 = 2.57 V , v1 = 0.13 V and i5V = 8.95 mA . 3.42 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Therefore, vOC = v2 − v3 = 2.44 V . (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (90) − ib (20) − ic (20) = 12 For mesh (b): − ia (20 ) + ib (20 ) + 5 = 0 For mesh (c): − ia (20) + ic (35) = 0 Solving the system, ia = 119.5 mA , ib = −130.5 mA and ic = 68.3 mA . Therefore, iSC = ic − ib = 198.8 mA . ________________________________________________________________________ Problem 3.48 Solution: Known quantities: The schematic of the circuit (see Figure P3.15). Find: The Thévenin equivalent resistance seen by resistor R5 , the Thévenin (open-circuit) voltage and the Norton (shortcircuit) current when R5 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = 30 Ω + 10 Ω + (20 Ω || 30 Ω ) = 52 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 − 3 v1 v1 − v2 + + =0 20 30 10 For node #2: v2 − v1 = 0 .5 10 Solving the system, v1 = 7.8 V and v2 = 12.8 V . Therefore, vOC = v2 = 12.8 V . (3) Replace the load with a short circuit. Redraw the circuit. 3.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 For mesh (a): ia (20 + 30) − ib (30) = 3 For meshes (b) and (c): − ia (30 ) + ib (30 + 10) + ic (30 ) = 0 For the current source: ic − ib = 0.5 Solving the system, ia = −92 mA , ib = −254 mA and ic = 246 mA . Therefore, iSC = ic = 246 mA . ________________________________________________________________________ Problem 3.49 Solution: Known quantities: The schematic of the circuit (see Figure P3.33). Find: The Thévenin equivalent resistance seen by resistor R , the Thévenin (opencircuit) voltage and the Norton (short-circuit) current when R is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = 1 Ω || 0.3 Ω = 0.23 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 v1 − 12 + = 12 1 0 .3 Solving, v1 = 12 V . Therefore, vOC = v1 = 12 V . (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (1 + 0.3) − ib (0.3) − 12(1) + 12 = 0 For mesh (b): − ia (0.3) + ib (0.3) = 12 Solving the system, ia = 12 A and ib = 52 A . Therefore, 3.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 iSC = ib = 52 A . ________________________________________________________________________ Problem 3.50 Solution: Known quantities: The schematic of the circuit (see Figure P3.35). Find: The Thévenin equivalent resistance seen by resistor R3 , the Thévenin (open-circuit) voltage and the Norton (short-circuit) current when R3 is the load. Analysis: (1) Remove the load, leaving the load terminals open circuited, and the voltage sources. Redraw the circuit. RT = 1 Ω || 7 Ω + 1 Ω || 5 Ω = 1.71 Ω (2) Remove the load, leaving the load terminals open circuited. Redraw the circuit. For node #1: v1 − 450 v1 + =0 1 7 For node #2: v2 + 450 v2 + =0 1 5 Solving the system, v1 = 393.75 V and v2 = −375 V . Therefore, vOC = v1 − v2 = 768.75 V . (3) Replace the load with a short circuit. Redraw the circuit. For mesh (a): ia (1 + 7 ) − ic (7 ) = 450 For mesh (b): ib (5 + 1) − ic (5) = 450 For mesh (c): − ia (7 ) − ib (5) + ic (7 + 5) = 0 Solving the system, ia = 450 A , ib = 450 A and ic = 450 A . Therefore, iSC = ic = 450 A . ________________________________________________________________________ 3.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.51 Solution: Known quantities: The values of the voltage source, VS = 110 V , and the values of the 4 resistors in the circuit of Figure P3.51: R1 = R2 = 930 mΩ R3 = 100 mΩ RS = 19 mΩ Find: The change in the voltage across the total load, when the customer connects the third load R3 in parallel with the other two loads. Analysis: Choose a ground. If the node at the bottom is chosen as ground (which grounds one terminal of the ideal source), the only unknown node voltage is the required voltage. Specify directions of the currents and polarities of voltages. Without R3: I S + I1 + I 2 KCL: =0 OL: VRS VR1 VR 2 + + =0 RS R1 R2 VOi − VS VOi − 0 VOi − 0 + + =0 RS R1 R2 VS RS RS 110 = = 105.7 V VOi = 1 1 1 RS 1.04086 + + RS R1 R2 With R3: IS KCL: + I1 + I 2 + I 3 = 0 OL: VRS VR1 VR 2 VR 3 + + + =0 RS R1 R2 R3 VOf − VS VOf − 0 VOf − 0 VOf − 0 + + =0 RS R1 R2 R3 VS RS RS 110 = = 89.37 V VOf = 1 1 1 1 RS 1.04086 + 0.19 + + + RS R1 R2 R3 Therefore, the voltage decreased by: ∆ Vo = VOf − VOi = −16.33 V + Notes: 3.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 1. "Load" to an EE usually means current rather than resistance. 2. Additional load reduces the voltage supplied to the customer because of the additional voltage dropped across the losses in the distribution system. ________________________________________________________________________ Problem 3.52 Solution: Known quantities: The values of the voltage source, VS = 450 V , and the values of the 4 resistors in the circuit of Figure P3.52 R1 = R2 = 1.3 Ω R3 = 500 mΩ RS = 19 mΩ Find: The change in the voltage across the total load, when the customer connects the third load with the other two loads. Analysis: See Solution to Problem 3.40 for a detailed mathematical analysis. R3 in parallel ∆ Vo = VOf − VOi = −15.6 V ________________________________________________________________________ Problem 3.53 Solution: Known quantities: The circuit shown in Figure P3.53, the values of the terminal voltage, VT , before and after the application of the load, respectively VT = 20 V and VT = 18 V , and the value of the load resistor RL = 2.7 kΩ . Find: The internal resistance and the voltage of the ideal source. Analysis: KVL: − VS + I T RS + VT = 0 VS = VT = 20 V If IT = 0: If VT = 18 OL: V: IT = VT = 6.67 mA RL and RS = VS − VT = 300 Ω IT Note that RS is an equivalent resistance, representing the various internal losses of the source and is not physically a separate component. VS is the voltage generated by some internal process. The source voltage can be measured directly by reducing the current supplied by the source to zero, i.e., no-load or open-circuit conditions. The source resistance cannot be directly measured; however, it can be determined, as was done above, using the interaction of the source with an external load. ________________________________________________________________________ 3.47 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.54 Solution: Known quantities: The values of the voltage source, VCC = 20 V , and the values of the 2 resistors in the circuit of Figure P3.54: R1 = 1.3 MΩ R2 = 220 kΩ Find: The Thévenin equivalent circuit with respect to the port shown in Figure P3.54. Analysis: The Thévenin equivalent voltage is the open circuit voltage [with I = 0] between the terminals of the port. Specify the polarity of the voltage. KCL: VTH − VCC VTH + +I =0 R1 R2 VTH = VCC R1 1 1 + R1 R2 = 2.895 V Note that, since I = 0, R1 and R2 are effectively in series, using a VD relation would be easier. Suppress the ideal, independent voltage source, by shorting it. Determine the equivalent resistance with respect to the terminals of the port. REQ = R1 R2 = (1.3 ⋅10 )(220 ⋅10 ) = 188.2 kΩ 6 3 1.3 ⋅10 6 + 220 ⋅10 3 ______________________________________________________________________________________ Problem 3.55 Solution: Known quantities: The values of the battery voltage, VB = 11 V , the value of the generator voltage, VG = 12 V , and the values of the 3 resistors in the circuit of Figure P3.55: RB = 0.7 Ω RG = 0.3 Ω RL = 7.2 Ω Find: a. The Thévenin equivalent of the circuit to the right of the terminal pair of port x-x'. b. The terminal voltage of the battery, i.e., the voltage between x and x'. 3.48 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: a. Specify the polarity of the voltage: VD: VTH = VG RL (12)(7.2) = 11.52 V = RG + RL 0.3 + 7.2 Suppress generator source: R EQ = RL RG = b. (7.2)(0.3) = 288 mΩ 0 .3 + 7 .2 Specify the polarity of the terminal voltage. Choose a ground. KCL: VT − VB VT − VTH + =0 RB REQ VB VTH + RB REQ = 11.37 V VT = 1 1 + RB REQ ______________________________________________________________________________________ Problem 3.56 Solution: The values of the battery voltage, VB = 11 V , the value of the generator voltage, VG = 12 V , and the values of the 3 resistors in the circuit of Figure P3.56: RB = 0.7 Ω RG = 0.3 Ω RL = 7.2 Ω Find: a. The Thévenin equivalent of the circuit to the left of the terminal pair of port y-y'. b. The terminal voltage of the generator, i.e., the voltage between y and y'. Analysis: a. Specify the polarity of the Thévenin equivalent voltage: VD: VTH = VB R L (11)(7.2) = 10.03 V = R B + R L 0 .7 + 7 .2 Suppress generator source: RT = R L RB = b. (7.2)(0.7 ) = 638 mΩ 0 .7 + 7 .2 Specify the polarity of the terminal voltage. Choose a ground. KCL: VT − VG VT − vT + =0 RG RT VG vT + RG RT VT = = 11.37 V 1 1 + RG RT 3.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Section 3.7: Maximum power transfer Problem 3.57 Solution: Known quantities: The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57: VTH = 12 V Req = 8 Ω Assumptions: Assume the conditions for maximum power transfer exist. Find: a. The value of RL . b. c. The power developed in RL . The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the source. Analysis: a. For maximum power transfer: R L = Req = 8 Ω b. VRL = VD: PRL VTH RL (12)(8) = 6 V = 8+8 Req + RL V 2 RL (6 )2 = = = 4.5 W RL 8 P0 PRL I S2 RL RL c. η = = = 2 = = 0.5 = 50% 2 PS PRe q + PRL I S Req + I S RL Req + RL ________________________________________________________________________ Problem 3.58 Solution: Known quantities: The values of the voltage and of the resistor in the equivalent circuit of Figure P3.57: VTH = 300 V Req = 600 Ω Assumptions: Assume the conditions for maximum power transfer exist. Find: a. The value of RL . b. c. The power developed in RL . The efficiency of the circuit, that is the ratio of power absorbed by the load to power supplied by the source. Analysis: a. For maximum power transfer: 3.50 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 R L = R eq = 600 Ω b. VRL = VD: PRL VTH RL (35)(600) = 17.5 V = Req + RL 600 + 600 V 2 RL (17.5)2 = = = 510.4 mW RL 600 P0 PRL I 2R RL = = 2 S L2 = = 0.5 = 50% PS PRe q + PRL I S Req + I S RL Req + RL ________________________________________________________________________ c. η= Problem 3.59 Solution: Known quantities: The values of the voltage source, VS source, = 12 V , and of the resistance representing the internal losses of the RS = 0.3 Ω , in the circuit of Figure P3.59. Find: a. Plot the power dissipated in the load as a function of the load resistance. What can you conclude from your plot? b. Prove, analytically, that your conclusion is valid in Power dissipated in the load all cases. 120 Analysis: − VS = IRS + IR = 0 b. VS RS + R R [Ω] I [A] 0 40 0.1 30 0.3 20 0.9 10 2.1 5 KVL: I= PR = I R = 2 VS2 R (R + RS ) 2 PR = I 2 R 80 Power [W] a. 100 PR [W] 0.0 90.0 120.0 90.0 52.5 60 40 20 0 = VS2 R (R + RS ) −2 dPR −2 −3 = VS2 (1)(R + RS ) + VS2 (R )(− 2 )(R + RS ) (1) = 0 dR (R + RS )1 − 2 R = 0 R = RS 3.51 0 0.5 1 1.5 Load Resistance [Ohm] 2 2.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Section 3.8: Nonlinear circuit elements Problem 3.60 Solution: Known quantities: The two nonlinear resistors, in the circuit of Figure P3.60, are characterized by: ia = 2va3 ib = vb3 + 10vb Find: The node voltage equations in terms of v1 and v2. Analysis: At node 1, At node 2, But v1 + ia = 1 v1 + 2va3 = 1 1 ib − ia = 26 vb3 + 10vb − 2va3 = 26 va = v1 − v2 and vb = v2 . Therefore, the node equations are v1 + 2(v1 − v 2 ) = 1 3 and v23 + 10v2 − 2(v1 − v2 ) = 26 ________________________________________________________________________ 3 Problem 3.61 Solution: Known quantities: The characteristic curve I-V shown in Figure P3.61, and the values of the voltage, VT resistance, RT = 15 V , and of the = 200 Ω , in the circuit of Figure P3.61. Find: a. The operating point of the element that has the characteristic curve shown in Figure P3.61. b. The incremental resistance of the nonlinear element at the operating point of part a. c. If VT were increased to 20 V, find the new operating point and the new incremental resistance. Analysis: a. KVL: − 15 + 200 I + V = 0 − 15 + 200(0.0025V 2 ) + V = 0 Solving for V and I, I = 52.2 mA V = 4.57 V or − 6.57 V The second voltage value is physically impossible. Rinc = 10(0.0522)−0.5 = 43.8 Ω c. I = 73 mA V = 5.40 V Rinc = 37 Ω ________________________________________________________________________ b. 3.52 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.62 Solution: Known quantities: The characteristic curve the resistance, R I-V shown in Figure P3.62, and the values of the voltage, VS = 450 V , and of = 9 Ω , in the circuit of Figure P3.62. Find: The current through and the voltage across the nonlinear device. Analysis: The I-V characteristic for the nonlinear device is given. Plot the circuit I-V characteristic, i.e., the DC load line. KVL: − VS + I D R + V D = 0 VS − VD 450 − VD = = R 9 if VD = 450 V =0A ID = = 50 A VD = 0 if The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them with a straight line gives the DC load line. The solution for V and I is at the intersection of the device and circuit characteristics: I DQ ≈ 26 A V DQ ≈ 210 V . ________________________________________________________________________ Problem 3.63 Solution: Known quantities: The I-V characteristic shown in Figure P3.63, and the values of the voltage, VS resistance, R = VT = 1.5 V , and of the = Req = 60 Ω , in the circuit of Figure P3.63. Find: The current through and the voltage across the nonlinear device. Analysis: The solution is at the intersection of the device and circuit characteristics. The device I-V characteristic is given. Determine and plot the circuit I-V characteristic. KVL: − VS + I D R + VD = 0 VS − VD 1.5 V − VD = = R 60 Ω =0A if VD = 1.5 V ID = = 25 mA if VD = 0 The DC load line [circuit characteristic] is linear. Plotting the two intercepts above and connecting them with a straight line gives the DC load line. The solution is at the intersection of the device and circuit characteristics, or "Quiescent", or "Q", or "DC operating" point: I DQ ≈ 12 mA VDQ ≈ 0.77 V . ________________________________________________________________________ 3.53 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Problem 3.64 Solution: Known quantities: The I-V characteristic shown in Figure P3.64 as a function of pressure. VS = VT = 2.5 V R = Req = 125 Ω Find: The DC load line, the voltage across the device as a function of pressure, and the current through the nonlinear device when p = 30 psig. Analysis: Circuit characteristic [DC load line]: KVL: − VS + I D R + VD = 0 VS − VD 2.5 V − VD = = R 125 Ω =0A if VD = 2.5 V ID = = 20 mA if VD = 0 The circuit characteristic is a linear relation. Plot the two intercepts and connect with a straight line to plot the DC load line. Solutions are at the intersections of the circuit with the device characteristics, i.e.: p [psig] VD [V] 10 20 25 30 40 2.14 1.43 1.18 0.91 0.60 The plot is nonlinear. At p = 30 psig: VD = 1.08 V I D = 12.5 mA . ________________________________________________________________________ Problem 3.65 Solution: Known quantities: The I-V characteristic of the nonlinear device in the circuit shown in Figure P3.65: I D = I 0e vD VT I 0 = 10 −15 A VT = 26 mV VS = VT = 1.5 V R = Rea = 60 Ω Find: An expression for the DC load line. The voltage across and current through the nonlinear device. 3.54 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 3 Analysis: Circuit characteristic [DC load line]: KVL: − VS + I D R + VD = 0 VS − VD 1.5 − VD = R 60 §I · § I · VD = VT ln¨¨ D ¸¸ = 0.026 ⋅ ln¨ D−15 ¸ © 10 ¹ © I0 ¹ [1] ID = [2] Iterative procedure: Initially guess V D = 750 mV . Note this voltage must be between zero and the value of the source voltage. Then: a. Use Equation [1] to compute a new ID. b. Use Equation [2] to compute a new VD. c. Iterate, i.e., go step a. and repeat. VD [mV] ID [mA] 750 12.5 784.1 11.93 782.9 11.95 782.9 11.95 … … I DQ ≈ 11.95 mA VDQ ≈ 782.9 mV . ________________________________________________________________________ Problem 3.66 Solution: Known quantities: The I-V characteristic shown in Figure P3.66 as a function of pressure. VS = VT = 2.5 V R = Req = 125 Ω Find: The DC load line, and the current through the nonlinear device when p = 40 psig. Analysis: Circuit characteristic [DC load line]: KVL: − VS + I D R + VD = 0 VS − VD 2.5 V − VD = = 125 Ω R =0 A if VD = 2.5 V ID = = 20 mA if VD = 0 The circuit characteristic is a linear relation that can be plotted by plotting the two intercepts and connecting them with a straight line. Solutions are at the intersections of the circuit and device characteristics. At p = 40 psig: VD = 0.60 V I D = 15.2 mA ________________________________________________________________________ 3.55 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Chapter 4 Instructor Notes The chapter starts by developing the dynamic equations for energy storage elements. The analogy between electrical and hydraulic circuits (Make The Connection: Fluid (hydraulic) Capacitance, p. 138, Make The Connection: Fluid (hydraulic)Iinertance, p. 150, Table 4.2) is introduced early to permit a connection with ideas that may already be familiar to the student from a course in fluid mechanics, such as mechanical, civil, chemical and aerospace engineers are likely to have already encountered. A Focus on Measurements boxes: Capacitive displacement transducer and microphones, pp. 147-148, permits approaching the subject of capacitance in a pragmatic fashion, if so desired. The instructor wishing to gain a more in-depth understanding of such transducers will find a detailed analysis in1. Next, signal sources are introduced, with special emphasis on sinusoids. The material in this section can also accompany a laboratory experiment on signal sources. The emphasis placed on sinusoidal signals is motivated by the desire to justify the concepts of phasors and impedance, which are introduced next. The author has found that presenting the impedance concept early on is an efficient way of using the (invariably too short) semester or quarter. The chapter is designed to permit a straightforward extension of the resistive circuit analysis concepts developed in Chapter 3 to the case of dynamic circuits excited by sinusoids. The ideas of nodal and mesh analysis, and of equivalent circuits, can thus be reinforced at this stage. The treatment of AC circuit analysis methods is reinforced by the usual examples and drill exercises, designed to avoid unnecessarily complicated circuits. Two Focus on Methodology boxes (pp. 165 and 180) provide the student with a systematic approach to the solution of basic AC analysis problems using phasor and impedance concepts. The capacitive displacement transducer example is picked up again in Focus on Measurements: Capacitive displacement transducer (pp.175-177) to illustrate the use of impedances in a bridge circuit. This type of circuit is very common in mechanical measurements, and is likely to be encountered at some later time by some of the students. The homework problems in this chapter are mostly exercises aimed at mastery of the techniques. Learning Objectives 1. Compute currents, voltages and energy stored in capacitors and inductors. 2. Calculate the average and root-mean-square value of an arbitrary (periodic) signal. 3. Write the differential equation(s) for circuits containing inductors and capacitors. 4. Convert time-domain sinusoidal voltages and currents to phasor notation, and vice-versa, and represent circuits using impedances. 5. Apply the circuit analysis methods of Chapter 3 to AC circuits in phasor form. 1 E. O. Doebelin, Measurement Systems – Application and Design, 4th Edition, McGraw-Hill, New York, 1990. 4.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Section 4.1: Energy Storage Elements Problem 4.1 Solution: Known quantities: Inductance value, L = 0.5 H ; the current through the inductor as a function of time. Find: The voltage across the inductor, (Eq. 4.9), as a function of time. Assumptions: iL (t ≤ 0) = 0 Analysis: Using the differential relationship for the inductor, we may obtain the voltage by differentiating the current: v L (t ) = L di L (t ) di (t ) ª π ·º § = 0.5 L = 0.5 × «− 377 × 2 sin¨ 377t + ¸» dt dt 6 ¹¼ © ¬ π 5π · · § § = 377 sin¨ 377t + − π ¸ = 377 sin¨ 377t − ¸ V 6 6 ¹ © ¹ © ______________________________________________________________________________________ Problem 4.2 Solution: Known quantities: Capacitance value C = 100 µF ; capacitor terminal voltage as a function of time. Find: The current through the capacitor as a function of time for each case: a) vc (t ) = 40 cos(20t − π / 2)V b) vc (t ) = 20 sin(100t )V vc (t ) = −60 sin(80t + π / 6)V d) vc (t ) = 30 cos(100t + π / 4)V . c) Assumptions: The capacitor is initially discharged: vC (t = 0) = 0 Analysis: Using the defining differential relationship for the capacitor, (Eq. 4.4), we may obtain the current by differentiating the voltage: iC (t ) = C dvC (t ) dv (t ) dv (t ) = 100 × 10 −6 C = 10 − 4 C dt dt dt a) 4.2 G. Rizzoni, Principles and Applications of Electrical Engineering b) Problem solutions, Chapter 4 ª π ·º π· § § iC (t ) = 10 − 4 «− 20 × 40 sin¨ 20t − ¸ » = −0.08 sin¨ 20t − ¸ 2 ¹¼ 2¹ © © ¬ π π· § · § = 0.08 sin¨ 20t − + π ¸ = 0.08 sin¨ 20t + ¸ A 2 2¹ © ¹ © iC (t ) = 10−4 [100 × 20 cos100t ] = 0.2 cos100t A c) d) ª π ·º π· § § iC (t ) = 10 − 4 « − 80 × 60 cos¨ 80t + ¸» = −0.48 cos¨ 80t + ¸ 6 ¹¼ 6¹ © © ¬ π 5π · § · § = 0.48 cos¨ 80t + − π ¸ = 0.48 cos¨ 80t − ¸ A 6 6 ¹ © ¹ © ª π ·º π· § § iC (t ) = 10 − 4 «− 100 × 30 sin¨100t + ¸» = −0.3 sin¨100t + ¸ 4 ¹¼ 4¹ © © ¬ π 3π · § · § = 0.3 sin¨100t + − π ¸ = 0.3 sin¨100t − ¸ A 4 4 ¹ © ¹ © ______________________________________________________________________________________ Problem 4.3 Solution: Known quantities: Inductance value, L = 250 mH ; the current through the inductor, as a function of time. Find: The voltage across the inductor as a function of time for each case a) iL (t ) = 5 sin 25tA iL (t ) = −10 cos 50tA c) iL (t ) = 25 cos(100t + π / 3) A d) iL (t ) = 20 sin(10t − π / 12) A . b) Assumptions: iL (t ≤ 0) = 0 Analysis: Using the differential relationship for the inductor, (Eq. 4.9), we may obtain the voltage by differentiating the current: v L (t ) = L a) b) di L (t ) di (t ) di (t ) = 250 × 10 −3 L = 0.25 L dt dt dt v L (t ) = 0.25[25 × 5 cos 25t ] = 31.25 cos 25t V v L (t ) = 0.25[− 50 × (− 10 sin 50t )] = 125 sin 50t V c) 4.3 G. Rizzoni, Principles and Applications of Electrical Engineering d) Problem solutions, Chapter 4 ª π ·º π· § § v L (t ) = 0.25«− 100 × 25 sin¨100t + ¸» = −625 sin¨100t + ¸ 3 ¹¼ 3¹ © © ¬ 2π · π § · § = 625 sin¨100t + − π ¸ = 625 sin¨100t − ¸ V 3 3 ¹ © ¹ © ª π ·º π · § § v L (t ) = 0.25«10 × 20 cos¨10t − ¸» = 50 cos¨10t − ¸ V 12 ¹¼ 12 ¹ © © ¬ ______________________________________________________________________________________ Problem 4.4 Solution: Known quantities: Inductance value; resistance value; the current through the circuit shown in Figure P4.4 as a function of time. Find: The energy stored in the inductor as a function of time. Analysis: The magnetic energy stored in an inductor may be found from, (Eq. 4.16): wL (t ) = For − ∞ < t < 0 , 1 1 2 Li (t ) = (2 ) i 2 (t ) = i 2 (t ) 2 2 wL (t ) = 0 For 0 ≤ t < 10 s wL (t ) = t 2 J For 10 s ≤ t < +∞ wL (t ) = 100 J ______________________________________________________________________________________ Problem 4.5 Solution: Known quantities: Inductance value; resistance value; the current through the circuit in Figure P4.4 as a function of time. Find: The energy delivered by the source as a function of time. Analysis: The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy stored in the inductor: wS (t ) = wR (t ) + wL (t ) = Ri 2 (t ) + 1 2 Li (t ) 2 4.4 G. Rizzoni, Principles and Applications of Electrical Engineering = (1) i 2 (t ) + For − ∞ < t < 0 , Problem solutions, Chapter 4 1 (2)i 2 (t ) = 2i 2 (t ) 2 wS (t ) = 0 For 0 ≤ t < 10 s wS (t ) = 2t 2 J For 10 s ≤ t < +∞ wS (t ) = 200 J ______________________________________________________________________________________ Problem 4.6 Solution: Known quantities: Inductance value; resistance value; the current through the circuit shown in Figure P4.4 as a function of time. Find: The energy stored in the inductor and the energy delivered by the source as a function of time. Analysis: The magnetic energy stored in an inductor may be found from, (Eq. 4.16): wL (t ) = 1 1 2 Li (t ) = (2 ) i 2 (t ) = i 2 (t ) 2 2 −∞ < t < 0, wL (t ) = 0 For 0 ≤ t < 10 s For wL (t ) = t 2 J For 10 ≤ t < 20 s wL (t ) = (20 − t ) = 400 − 40t + t 2 J For 20 s ≤ t < +∞ wL (t ) = 0 J 2 The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy stored in the inductor: wS (t ) = wR (t ) + wL (t ) = Ri 2 (t ) + = (1) i 2 (t ) + −∞ < t < 0, wL (t ) = 0 For 0 ≤ t < 10 s 1 2 Li (t ) 2 1 (2)i 2 (t ) = 2i 2 (t ) 2 For wL (t ) = 2t 2 J 4.5 G. Rizzoni, Principles and Applications of Electrical Engineering For Problem solutions, Chapter 4 10 ≤ t < 20 s wL (t ) = 2 (20 − t ) = 800 − 80t + 2t 2 J For 20 s ≤ t < +∞ wL (t ) = 0 J 2 ______________________________________________________________________________________ Problem 4.7 Solution: Known quantities: Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of time. Find: The energy stored in the capacitor as a function of time. Analysis: The energy stored in a capacitor may be found from: 1 1 2 2 2 Cv (t ) = (0.1)v(t ) = 0.05v(t ) 2 2 wC (t ) = −∞ <t < 0 wC (t ) = 0 For 0 ≤ t < 10 s wC (t ) = 0.05 t 2 J For 10 s ≤ t < +∞ 2 wC (t ) = 0.05(10 ) = 5 J For ______________________________________________________________________________________ Problem 4.8 Solution: Known quantities: Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of time. Find: The energy delivered by the source as a function of time. Analysis: The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy stored in the capacitor: wS (t ) = wR (t ) + wC (t ) = = For − ∞ < t < 0 , v 2 (t ) 1 2 + Cv (t ) R 2 1 1 2 v (t ) + (0.1)v 2 (t ) = 0.55i 2 (t ) 2 2 wS (t ) = 0 For 0 ≤ t < 10 s 4.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 wS (t ) = 0.55t 2 J For 10 s ≤ t < +∞ wS (t ) = 55 J ______________________________________________________________________________________ Problem 4.9 Solution: Solution: Known quantities: Capacitance value; resistance value; the voltage applied to the circuit shown in Figure P4.7 as a function of time. Find: The energy stored in the capacitor and the energy delivered by the source as a function of time. Analysis: The energy stored in a capacitor may be found from: wC (t ) = For − ∞ < t < 0 1 1 2 2 2 Cv (t ) = (0.1)v(t ) = 0.05v(t ) 2 2 wC (t ) = 0 For 0 ≤ t < 10 s wC (t ) = 0.05 t 2 J For 10 ≤ t < 20 s 2 wC (t ) = 0.05 (20 − t ) = 20 − 2t + 0.05t 2 J For 10 s ≤ t < +∞ wC (t ) = 0 The energy delivered by the source is the sum of the energy absorbed by the resistance and the energy stored in the capacitor: wS (t ) = wR (t ) + wC (t ) = = For − ∞ < t < 0 v 2 (t ) 1 2 + Cv (t ) R 2 1 1 2 v (t ) + (0.1)v 2 (t ) = 0.55i 2 (t ) 2 2 wC (t ) = 0 For 0 ≤ t < 10 s wC (t ) = 0.55 t 2 J For 10 ≤ t < 20 s 2 wC (t ) = 0.55 (20 − t ) = 220 − 22t + 0.55t 2 J For 10 s ≤ t < +∞ wC (t ) = 0 ______________________________________________________________________________________ 4.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.10 Solution: Known quantities: Capacitance, resistance and inductance values; the source voltage vS = 6 V applied to the circuit shown in Figure P4.10. Find: The energy stored in each capacitor and inductor. Analysis: Under steady-state conditions, all the currents are constant, no current can flow through the capacitors, and the voltage across any inductor is equal to zero. v2 F = v4 Ω 6 − v4 Ω v4 Ω v4Ω = + v4Ω = 3.43 V 2 4 8 1 1 2 w2 F = C2 F v22F = (2 F )(3.43 V ) = 11.76 J 2 2 v1F = v 2 H = 0 i2 H = v4 Ω = 0.43 A 8 w1F = 1 1 2 C1F v12F = (1 F )(0 ) = 0 2 2 1 1 2 L2 H i22H = (2 H )(0.43 A ) = 0.18 J 2 2 1 1 2 = C3 F v32F = (3 F )(3.43 V ) = 17.65 J 2 2 w2 H = v3 F = v4Ω = 3.43 V w3 F ______________________________________________________________________________________ Problem 4.11 Solution: Known quantities: Capacitance, resistance and inductance values; the voltage v A = 12 V applied to the circuit shown in Figure P4.11. Find: The energy stored in each capacitor and inductor. Analysis: Under steady-state conditions, all the currents are constant, no current can flow across the capacitors, and the voltage across any inductor is equal to zero. The voltage for the 1-F capacitor is equal to the 12-Volt input. Since the voltage is the same on either end of the 3-Ω resistor in parallel with the 2-F capacitor, there is no voltage drop through either component. Finally, since there is no voltage drop through the 3-Ω resistor in parallel with the 2-F capacitor, there is no current flow through the resistor, and the current through the 1-H inductor is equal to the current through the 2-H inductor. Therefore, 4.8 G. Rizzoni, Principles and Applications of Electrical Engineering v1F = v A = 12 V w1F = 12 V =2A 6Ω Problem solutions, Chapter 4 1 1 2 C1F v12F = (1 F )(12 V ) = 72 J 2 2 1 1 2 L1H i12H = (1 H )(2 A ) = 2 J 2 2 1 1 2 i2 H = i1H = 2 A w2 H = L2 H i22H = (2 H )(2 A ) = 4 J 2 2 1 1 2 v2 F = 0 V w2 F = C2 F v22F = (2 F )(0 V ) = 0 J 2 2 i1H = w1H = ______________________________________________________________________________________ Problem 4.12 Solution: Known quantities: Capacitance value C = 80 µF ; the voltage applied to the capacitor as a function of time as shown in Figure P4.12. Find: The current through the capacitor as a function of time. Analysis: Since the voltage waveform is piecewise continuous, the derivative must be evaluated over each continuous segment. For 0 < t < 5 ms vC (t ) = mvC t + q vC where: m vC = [− 10 V] − [+ 20 V] = −6 V ms [5 ms] − [0] q vC = +20 V dvC (t ) d V· § =C mvC t + q vC = CmvC = (80 µF )¨ − 6 ¸ = −480 mA dt dt ms ¹ © For 5 ms < t < 10 ms vC (t ) = −10 V [ iC = C iC = C For ] dvC (t ) d = C [− 10 V ] = 0 dt dt t > 10 ms vC (t ) = 0 iC = C dvC (t ) d = C [0] = 0 dt dt A capacitor is fabricated from two conducting plates separated by a dielectric constant. Dielectrics are also isolators; therefore, current cannot really flow through a capacitor. Positive charge, however, entering one plate exerts a repulsive force on and forces positive carriers to exit the other plate. Current then appears to 4.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 flow through the capacitor. Such currents are called electric displacement currents. ______________________________________________________________________________________ Problem 4.13 Solution: Known quantities: Inductance value, L = 35 mH ; the voltage applied to the inductor as shown in Figure P4.12; the initial condition for the current i L (0 ) = 0 . Find: The current across the inductor as a function of time. Analysis: Since the voltage waveform is piecewise continuous, integration must be performed over each continuous segment. Where not indicated t is supposed to be expressed in seconds. For 0 < t ≤ 5 ms v L (t ) = mvL t + q vL where: mvL = [− 10 V] − [+ 20 V] = −6 V ms [5 ms] − [0] q vL = +20 V ( t ) 1 t 1 t 1 ª1 º i L = i L (0) + ³ v L (τ )dτ = 0 + ³ mvL τ + qv L dτ = « mvL τ 2 + qvL τ » = L 0 L 0 L ¬2 ¼0 1 1 V 2 § · = mvL t 2 + q vL t = ⋅ t + 20 V ⋅ t ¸ = − 85.71 ⋅ 10 3 t 2 + 571.4t A ¨− 6 2L 2 ⋅ 35 mH © ms ¹ ( ) ( ( ) ) i L (t = 5 ms = 0.005 s ) = − 85.71 ⋅ 10 3 ⋅ (0.005) + 571.4 ⋅ 0.005 A = 714.3 mA For 5 ms < t ≤ 10 ms v L ( t ) = cvL = −10 V 2 ( ) [ ] 1 t 1 t 1 v L (τ )dτ = i L (0.005) + ³ cvL dτ = i L (0.005) + cvL τ ³ 0 005 0 005 . . L L L 1 = 714.3 mA − ⋅ (− 10 V ) ⋅ (t − 0.005 s ) = (2.143 − 285.7t ) A 35 mH i L = i L (0.005) + i L (t = 10 ms = 0.01 s ) = (2.143 − 285.7(0.01)) A = −713.5 mA For t > 10 ms v L ( t ) = cvL = 0 1 t 1 t 1 t v L (τ )dτ = i L (0.01) + ³ (0 )dτ = i L (0.01) + [0]0.005 = ³ L 0.01 L 0.01 L = i L (0.01) = −713.5 mA i L = i L (0.01) + 4.10 t 0.005 = G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 ______________________________________________________________________________________ Problem 4.14 Solution: Known quantities: Inductance value L = 0.75 mH ; the voltage applied to the inductor as a function of time as shown in Figure P4.14. Find: The current through the inductor at the time t = 15 µs . Assumptions: iL (t ≤ 0) = 0 Analysis: Since the voltage waveform is a piecewise continuous function of time, integration must be performed over each continuous segment. Where not indicated, t is expressed in seconds. For 0 < t ≤ 5 µs v L (t ) = mvL t + q vL where: mvL = [3.5 V] − [0 V] = 0.7 V µs [5 µs ] − [0] q vL = 0 V For t > 5 µs v L ( t ) = cvL = −1.9 V Therefore: ( ) ( ) 1 15 µs 1 5 µs 1 15 µs v L (τ )dτ = iL (0 ) + ³ mvLτ dτ + ³ c v dτ = ³ 0 − ∞ L L L 5 µs L V 5 µs 0 .7 1 ª1 1 1 º 15 µs ms ⋅ (5 µs )2 − 0 + = iL (0 ) + « mvLτ 2 » + cvLτ 5 µs = 0 + ⋅ L ¬2 L 0.75 mH 2 ¼0 iL (t = 15 µs ) = [ ] + ( 1 ⋅ (− 1.9 V ) ⋅ (15 µs − 5 µs ) = −13.67 mA 0.75 mH ______________________________________________________________________________________ 4.11 ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.15 Solution: Known quantities: Capacitance value C v Peak = 680 nF ; the periodic voltage applied to the capacitor as shown in Figure P4.15: = 20 V , T = 40 µs . Find: The waveform and the plot for the current through the capacitor as a function of time. Analysis: Since the voltage waveform is not a continuous function of time, differentiation can be performed only over each continuous segment. In the discontinuity points the derivative of the voltage will assume an infinite value, the sign depending on the sign of the step. Where not indicated, t is expressed in seconds. For each period 0 < t < T , T consider only the first period: For 0 < t < T < t < 2T , the behavior of the capacitor will be the same; thus, we vC (t ) = mvC t + q vC where: m vC = [v Peak ] − [0] = 0.5 V [T ] − [0] µs q vC = 0 V dvC (t ) d nAs ·§ V· § =C mvC t = CmvC = ¨ 480 ¸ ¨ 0 .5 ¸ = 340 mA dt dt V ¹© µs ¹ © For t = T , 2T , dv (t ) iC = C C = C [− ∞] dt iC = C [ ] Figure P 4.15 shows the current waveform. Note: For the voltage across the capacitor to decrease instantaneously to zero at t = T , 2T , , the charge on the plates of the capacitor should be instantaneously discharged. This requires an infinite current which is not physically possible. If this were a practical waveform, the slope at t = T , 2T , , would be finite, not infinite. A large negative spike of current over a finite period of time would result instead of the infinite spike over zero time. These large spike of current (or voltage) degrade the performance of many circuits. ______________________________________________________________________________________ Problem 4.16 Solution: Known quantities: Inductance value, L = 16 µH ; the voltage applied to the inductor as a function of time as shown in Figure P4.16; the initial condition for the current i L (0 ) = 0 . 4.12 G. Rizzoni, Principles and Applications of Electrical Engineering Find: The current through the inductor at Problem solutions, Chapter 4 t = 30 µs . Analysis: Since the voltage waveform is piecewise continuous, the integration can be performed over each continuous segment. Where not indicated t is supposed to be expressed in seconds. i L (t = 30 µs ) = 1 30 µs 1 20 µ s 1 30 µ s v L (τ )dτ = i L (0) + ³ v L (τ )dτ + ³ v L (τ )dτ = ³ L −∞ L 0 L 20 µ s ( 20 µ s ) 1 ª3 Vº 1 1 V 30 µ s 3 = i L (0) + « τ 3 2 » + [1.2τ nV ]20 µ s = 0 + ⋅ 1 2 ⋅ (20 µs ) − 0 + L ¬3 L 16 µH s ¼0 s 1 + ⋅ (1.2 nV ) ⋅ (30 µs − 20 µs ) = 1.250 nA 16 µH ______________________________________________________________________________________ Problem 4.17 Solution: Known quantities: Resistance value R = 7 Ω ; inductance value L = across the components as shown in Figure P4.17. Find: The current through each component. Assumptions: 7 mH ; capacitance value C = 0.5 µF ; the voltage i R (t ≤ 0) = i L (t ≤ 0) = iC (t ≤ 0 ) = 0 Analysis: Since the voltage waveform is piecewise continuous, integration and differentiation can only be performed over each continuous segment. Where not indicated, t is expressed in seconds. t ≤ 0: i R (t ) = i L (t ) = iC (t ) = 0 For 0 < t < 5 ms : v(t ) = mv t + qv For where: mv = [15 V] − [0] = 3 V [5 ms] − [0] ms q vC = 0 V v(t ) mv ⋅ t i R (t ) = = = R R 3 V ⋅t ms = 428.6 ⋅ t A 7Ω t 1 t 1 t 1 ª1 1 º i L (t ) = ³ v(τ )dτ = ³ mv ⋅ τ dτ = « mv ⋅ τ 2 » = mv ⋅ t 2 = 0 0 L L L ¬2 ¼ 0 2L = 1 V 2 ⋅3 ⋅ t = 214.3 ⋅ 10 3 ⋅ t 2 A 2 ⋅ 7 mH ms 4.13 G. Rizzoni, Principles and Applications of Electrical Engineering iC (t ) = C For t = 5 ms : Problem solutions, Chapter 4 dvC (t ) d µAs ·§ V · § = C [mv t ] = Cmv = ¨ 0.5 ¸¨ 3 ¸ = 1.5 mA dt dt V ¹© ms ¹ © ( ) (5 ⋅ 10 ) = 214.3 ⋅10 (5 ⋅ 10 ) = 1.5 mA i R 5 ⋅ 10 −3 = 428.6 ⋅ t A = 428.6 ⋅ 5 ⋅ 10 −3 A = 2.143 A ( ) 2 −3 3 iL ⋅ t 2 A = 214.3 ⋅ 10 3 ⋅ 5 ⋅ 10 −3 A = 5.357 A −3 iC For 5 ms < t < 10 ms : v(t ) = c v = 15 V v(t ) cv 15 V i R (t ) = = = = 2.143 A R R 7Ω 1 t 1 t i L (t ) = i L 5 ⋅ 10 −3 + ³ −3 v(τ )dτ = i L 5 ⋅ 10 −3 + ³ −3 cv dτ = L 5⋅10 L 5⋅10 1 1 t = i L 5 ⋅ 10 −3 + [cv ⋅ τ ]5⋅10 −3 = i L 5 ⋅ 10 −3 + ⋅ cv ⋅ t − 5 ⋅ 10 −3 = L L 1 = 5.357 A + ⋅ 15 V ⋅ t − 5 ⋅ 10 −3 s = −5.357 + 2.143 ⋅ 10 3 ⋅ t A 7 mH dv (t ) d iC (t ) = C C = C [cv ] = 0 dt dt For t = 10 ms : i R (0.01) = 2.143 A i L (0.01) = −5.357 + 2.143 ⋅ 10 3 ⋅ t A = −5.357 + 2.143 ⋅ 10 3 ⋅ 0.01 A = 16.07 A iC (0.01) = 0 For t > 10 ms : v(t ) = 0 v(t ) 0 i R (t ) = = =0 R R 1 t 1 t i L (t ) = i L (10 ⋅ 10 −3 ) + ³ − 3 v(τ )dτ = i L (10 ⋅ 10 −3 ) + ³ − 3 0 dτ = L 10⋅10 L 10⋅10 1 t = i L (10 ⋅ 10 −3 ) + [0]10⋅10− 3 = i L (10 ⋅ 10 −3 ) = 16.07 A L dvC (t ) d iC (t ) = C = C [0] = 0 dt dt ( ) ( ) ( ( ) ) ( ) 4.14 ( ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 ______________________________________________________________________________________ Problem 4.18 Solution: Known quantities: The voltage across and the current through an ideal capacitor as shown in Figure P4.18. Find: The capacitance of the capacitor. Analysis: Considering the period: − 2.5 µs < t < +2.5 µs : ic = C dvc ∆v = C c , since the voltage has a linear waveform. dt ∆t Substituting: 12 A = C [+ 10 V] − [− 10 V] 5 µs C = 12 A ⋅ 5 µs = 3 µF 20 V ______________________________________________________________________________________ Problem 4.19 Solution: Known quantities: The voltage across and the current through an ideal inductor as shown in Figure P4.19. 4.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Find: The inductance of the inductor. Analysis: vL = L di L ∆i = L L , since the current has a linear waveform. dt ∆t Substituting: 2V=L [2 A] − [1 A ] 10 ms − 5 ms L = 2 V⋅ 5 ms = 10 mH 1 A ______________________________________________________________________________________ Problem 4.20 Solution: Known quantities: The voltage across and the current through an ideal capacitor as shown in Figure P4.20. Find: The capacitance of the capacitor. Analysis: Considering the period: 0 < t < 5 ms : ic = C dvc ∆v = C c , since the voltage has a linear waveform. dt ∆t Substituting: 1.5 mA = C [15 V] − [0] 5 ms C = 1.5 mA ⋅ 5 ms = 0.5 µF 15 V ______________________________________________________________________________________ Problem 4.21 Solution: Known quantities: The voltage across and the current through an ideal capacitor as shown in Figure P4.21. Find: The capacitance of the capacitor. Analysis: Considering the period: 0 < t < 5 ms : ic = C dvc ∆v = C c , since the voltage has a linear waveform. dt ∆t Substituting: 3 mA = C [7 V] − [0] 5 ms C = 3 mA ⋅ 5 ms = 2.14 µF 7 V ______________________________________________________________________________________ 4.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Section 4.2: Time-Dependent Signals Problem 4.22 Solution: Known quantities: The signal x(t ) = 2 cos(ωt ) + 2.5 . Find: The average and rms value of the signal. Analysis: The average value is: v(t ) = ϖ 2π = 1 2π 2π ª 2ωπ º ω 2π 2π º « » 1 ª ω + 2.5t ω ( ) 2cos ( t)dt + 2 . 5 dt sin ( ) ω ϖ t = − ³ 0 0 » «³ » 2π «¬ ¼ 0 «¬ 0 »¼ [sin(0) − sin(2π )] + 2.5 = 1 [0 − 0] + 2.5 = 2.5 2π The rms value is: x rms ω = 2π 2π ω ω 2 ³0 (2cos (ωt) + 2.5) dt = 2π ª ω « = ⋅ 4⋅ 2π « «¬ = ω 2π 2π ω ³ [cos(ωt )] 0 2 2π ω ³ [4 ⋅ [cos(ωt )] 2 ] + 10 ⋅ cos(ωt ) + 6.25 dt = 0 2π ω 2π ω º » dt + 10 ⋅ ³ sin (ωt )dt + 6.25 ⋅ ³ dt » = 0 0 »¼ 2π º ª 1 2π ⋅ «4 ⋅ ⋅ + 10 ⋅ 0 + 6.25 ⋅ » = 8.25 = 2.87 ω¼ ¬ 2 ω Note: The integral of a sinusoid over an integer number of period is identically zero. This is a useful and important result. ______________________________________________________________________________________ Problem 4.23 Solution: Known quantities: The sinusoidal voltage v(t ) of 110 V rms shown in Figure P4.23. Find: The average and rms voltage. Analysis: The rms value of a sinusoidal is equal to 0.707 times the peak value: V peak = 110 2 The average value is: 4.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 θ 2π º 1 ª ( ) + t dt 110 2 sin 110 2 sin (t )dt » = «³ ³ 2π ¬ 0 2π −θ ¼ θ 2π 1 ª º= = − 110 2 cos(t ) − 110 2 cos(t ) « »¼ 0 2 π θ − 2π ¬ v(t ) = =− The rms value is: vrms 50 2 [cos(θ ) − 1 + cos(2π ) − cos(2π − θ )] = 0 π ° 1 =® °̄ 2π ( ) ( ) 12 2π §θ ·½ 2 2 ¨ ³ 110 2 sin (t ) dt + ³ 110 2 sin (t ) dt ¸ °¾ ¨ ¸° 2π −θ ©0 ¹¿ = 12 2π θ °12100 § 1 · ½° ¸¾ ¨ (− cos(t )sin (t ) + t ) + 1 (− cos(t )sin (t ) + t ) =® ¸ 2 °̄ π ¨© 2 0 2π −θ ¹ ° ¿ = 12 6050 (− cos(θ )sin (θ ) + θ + 2π − (− cos(2π − θ )sin (2π − θ ) + 2π − θ ))½¾ =® ¯ π ¿ 12 6050 =® (2θ )½¾ ¯ π ¿ = 110 = θ π ______________________________________________________________________________________ Problem 4.24 Solution: Known quantities: The sinusoidal voltage v(t ) of 110 V rms shown in Figure P4.23. Find: The angle θ that correspond to delivering exactly one-half of the total available power in the waveform to a resistive load. Analysis: From vrms = 110 θ π 2 vrms = 110 2 , we obtain: θ 110 2 = π 2 θ= π 2 ______________________________________________________________________________________ Problem 4.25 Solution: Known quantities: The signal v(t ) shown in Figure P4.25. Find: The ratio between average and rms value of the signal. 4.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Analysis: The average value is: v = 0.004 1 ª 0.002 ( − 9 )dt + ³ (1)dt º = 250(− 0.018 + 0.002 ) = −4 V ³ »¼ 0.002 0.004 «¬ 0 The rms value is: 8 0.004 1 ª 0.002 2 2 ( − 9) dt + ³ (1) dt º = 250 ⋅ [81 ⋅ 0.002 + 0.004 − 0.002] = 6.40 V ³ »¼ 0.002 0.004 «¬ 0 v rms = Therefor, v vrms =− 4 = −0.625 6.40 ______________________________________________________________________________________ Problem 4.26 Solution: Known quantities: The signal i (t ) shown in Figure P4.26. Find: The power dissipated by a 1-Ω resistor. Analysis: The rms value is: ( ) 2 1 p 2 10 ⋅ sin ( t ) dt = p ³0 irms = 1 p 100 ⋅ sin 4 (t )dt = ³ 0 p 1 3p ⋅ 100 ⋅ = 6.12 A p 8 Therefore, the power dissipated by a 1-Ω resistor is: 2 P1Ω = Rirms = (1)(6.12) W = 37.5 W 2 ______________________________________________________________________________________ Problem 4.27 Solution: Known quantities: The signal x(t ) shown in Figure P4.27. Find: The average and rms value of the signal. Analysis: The average value is: V = 1 t0 +τ t Vm dt = Vm ³ T t0 T where t0 is the left-hand side of the pulse. The rms value is: 4.19 G. Rizzoni, Principles and Applications of Electrical Engineering Vrms = Problem solutions, Chapter 4 t 1 t0 +τ 2 Vm dt = Vm ³ T t0 T Therefore, V Vrms = t T ______________________________________________________________________________________ Problem 4.28 Solution: Known quantities: The signal i (t ) shown in Figure P4.28. Find: The rms value of the signal. Analysis: The rms value is: irms 3T º ª T4 2 2 2 T 4 1« §8 · § 8 · §8 · » = ¨ t ¸ dt + ³ ¨ − t + 4 ¸ dt + ³ ¨ t − 8 ¸ dt » = T « ³0 © T ¹ T ¹ ¹ » 3T © T T© « 4 4 ¼ ¬ 3T º ª T4 2 2 T 4 1 « § 64 2 · § 64 2 64 · § 64 2 128 · » t ¸ dt + ³ ¨ 2 t − t + 16 ¸ dt + ³ ¨ 2 t − t + 64 ¸ dt » = = ¨ T « ³0 © T 2 ¹ T T ¹ ¹ » 3T © T T ©T « 4 4 ¼ ¬ = 1 ª1 1 64 º T + 9T − 18T + 12T − T + 2T − 4T + T − 64T + 64T − 9T + 36T − 48T » = « T ¬3 3 3 ¼ = 1 T 2 ª4 º T = = 1.15 A «3 » 3 ¬ ¼ ______________________________________________________________________________________ Problem 4.29 Solution: Known quantities: The signal v(t ) . Find: The rms value of the signal. Analysis: The rms value is: vrms 1 = 2π 2π ³ (v(t )) d (ωt ) 2 0 4.20 G. Rizzoni, Principles and Applications of Electrical Engineering v 2 rms 1 = 2π 2π = 1 2π 2π = 1 2π § 2 · 1 § 2 · V2 V2 ¨¨ VDC [ωt ]02π + 0 + 0 [ωt ]02π + 0 ¸¸ = ¨¨VDC (2π − 0 ) + 0 + 0 (2π − 0 ) + 0 ¸¸ 2 2 © ¹ 2π © ¹ ³ (V + V0 cos(ωt )) 2 DC 0 1 d (ωt ) = 2π 2π Problem solutions, Chapter 4 ³ (V 2 DC ) + 2VDCV0 cos(ωt ) + V02 cos 2 (ωt ) d (ωt ) = 0 § 2 · V02 V02 ¨ ( ) + + + V V V cos t cos 2 (ωt )¸¸ d (ωt ) = 2 ω ³0 ¨© DC DC 0 2 2 ¹ 2 v rms = VDC + V02 = 2 (50 V )2 + 1 (70.7 V )2 2 = 70.7 V Note: 1. T = period in units of time and ωt = period in angular units, i.e., 2π radians. Considering ωt as a single variable is useful when dealing with sinusoids. 2. A sinusoid integrated over one or more whole periods gives 0 which is very useful. ______________________________________________________________________________________ 4.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Section 4.4: Phasor Solution of Circuits with Sinusoidal Excitation Focus on Methodology: Phasors 1. Any sinusoidal signal may be mathematically represented in one of two ways: a time domain form: v(t) A cos( t ) , and a frequency domain form: 2. 3. Ae j j t . Note the jω in the notation V( j ) , indicating the e dependence of the phasor. In the remainder of this chapter, bold uppercase quantities indicate phasor voltages and currents. A phasor is a complex number, expressed in polar form, consisting of a magnitude equal to the peak amplitude of the sinusoidal signal and a phase angle equalto the phase shift of the sinusoidal signal referenced to a cosine signal. when one is using phaso notation, it is important tonote the specific frequency w of the sinusoidal signal,since this is not explicitly apparent in the phasor expression. V( j ) A Problem 4.30 Solution: Known quantities: The current through and the voltage across a component. Find: a) Whether the component is a resistor, capacitor, inductor b) The value of the component in ohms, farads, or henrys. Analysis: a) The current and the voltage can be expressed in phasor form: I = 17∠ − 15 o mA , V = 3.5∠75 o V Z= 3.5∠75 o V V = = 205.9∠90 o Ω = 0 + j ⋅ 205.9 Ω o I 17∠ − 15 mA The impedance has a positive imaginary or reactive component and a positive angle of 90 degree indicating that this is an inductor (see Fig. 4.39). b) Z L = j ⋅ X L = j ⋅ ωL = j ⋅ 205.9 Ω L = 205.9 Ω Vs = 327.7 m = 327.7 mH rad A 628.3 s ______________________________________________________________________________________ Problem 4.31 Solution: Known quantities: The waveform of a signal shown in Figure P4.31. Find: The sinusoidal description of the signal. Analysis: From the graph of Figure P4.31: 4.22 G. Rizzoni, Principles and Applications of Electrical Engineering φ =+ π 180 o 2π = 60 o , V0 = 170 V , ω = 2πf = 3 π T ( Problem solutions, Chapter 4 … not given. ) v r (t ) = V0 cos(ωt + φ ) = 170 cos ωt + 60 o V Phasor form: V = V0 ∠φ = 170∠60 o V = 170 V ⋅ e j 60 o ______________________________________________________________________________________ Problem 4.32 Solution: Known quantities: The waveform of a signal shown in Figure P4.32. Find: The sinusoidal description of the signal. Analysis: From graph: 3π 180 o 2π rad = −135 o , I 0 = 8 mA , ω = 2πf = = 1571 4 π T s rad § · ⋅ t − 135 o ¸ mA i (t ) = I 0 cos(ωt + φ ) = 8 cos¨1571 s © ¹ φ =− Phasor form: I = I 0 ∠φ = 8∠ − 135 o mA = 8 mA ⋅ e − j 135 o ______________________________________________________________________________________ Problem 4.33 Solution: Known quantities: The waveform of a signal shown in Figure P4.33. Find: The sinusoidal description of the signal. Analysis: From graph: 3π 180 o 2π rad = −135 o , I 0 = 8 mA , ω = 2πf = = 1571 4 π T s rad § · i (t ) = I 0 cos(ωt + φ ) = 8 cos¨1571 ⋅ t − 135 o ¸ mA s © ¹ φ =− Phasor form: I = I 0 ∠φ = 8∠ − 135 o mA = 8 mA ⋅ e − j 135 o ______________________________________________________________________________________ 4.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.34 Solution: Known quantities: The current through ( ) i (t ) = I 0 cos ωt + 45o , I 0 = 3 mA, ω = 6.283 v(t ) = V0 cos(ωt ), V0 = 700 mV , ω = 6.283 rad , and the voltage across s rad an electrical component. s Find: a) Whether the component is inductive or capacitive. b) The waveform of the instantaneous power p (t ) as a function of ωt over the range 0 < ωt < 2π . c) The average power dissipated as heat in the component. d) The same as b. and c. with the phase of the current equal to zero. Analysis: a) Phasor notation: I = 3∠45o mA , V = 700∠0 o mV Z= V 700∠0o mV = = 233.3∠ − 45o Ω = 165.0 − j165.0 Ω I 3∠90o mA The component is inductive because it is lagging. b) ( ) ( ( ) ( )) 1 p(t ) = v(t )i (t ) = V0 I 0 cos ωt + 45o cos(ωt ) = V0 I 0 cos 2ωt + 45o + cos 45o = 2 1 = (700 mV )(3 mA ) cos 2ωt + 45o + 0.707 = 1050 cos 2ωt + 45o + 742.4 µW 2 ( ( ) ) ( ( c) P= 1 ωT ωT ³ p(t )dωt = 0 1 ωT ωT ³ (1050 µW cos(2ωt + 45 ) + 742.4)dωt = 0 [ ( )] 1 (1050 µW ) 1 sin 2ωt + 45o 02π + 1 742.4(ωt ) 02π = ωT ωT 2 1 = (1050 µW ) 1 sin 765o − sin (45°) + 742.4 = ωT 2 1 1 (1050 µW ) [0 − 0] + 742.4 = 742.4µW = 2π 2 = [ ( d) o ] ) ( ( )) 1 1 p(t ) = v(t )i (t ) = V0 I 0 cos(ωt ) cos(ωt ) = V0 I 0 cos(2ωt ) + cos 0o = 2 2 1 = (700 mV )(3 mA )(cos(2ωt ) + 1) = 1050(cos(2ωt ) + 1) µW 2 4.24 ) ) G. Rizzoni, Principles and Applications of Electrical Engineering 1 P= ωT ωt ³ 0 1 p(t )dωt = ωT Problem solutions, Chapter 4 ωt ³ (1050 µW(cos(2ωt ) + 1))dωt = 0 = 1 (1050 µW )§¨ 1 [sin (2ωt )]02π + [ωt ]02π ·¸ = ωT ©2 ¹ = 1 (1050 µW )§¨ 1 (0 − 0) + (2π − 0)·¸ = 1050 µW 2π ©2 ¹ ______________________________________________________________________________________ Problem 4.35 Solution: Known quantities: The values of the impedance, R1 = 2.3 kΩ , R2 = 1.1 kΩ , L = 190 mH , C = 55 nF and the o voltage applied to the circuit shown in Figure P4.35, v s (t ) = 7 cos (3000t + 30 ) V . Find: The equivalent impedance of the circuit. Analysis: rad · § X L = ωL = ¨ 3 k ¸(190 mH ) = 0.57 kΩ Z L = + j ⋅ X L = + j ⋅ 0.57 kΩ s ¹ © 1 1 = 6.061 kΩ Z C = − j ⋅ X C = − j ⋅ 6.061 kΩ = rad · ωC § ¨3 k ¸(55 nF) s ¹ © = Z R1 + Z L = R1 + jX L = 2.3 + j ⋅ 0.57 kΩ = 2.37∠13.92 o kΩ XC = Z eq1 Z eq 2 = Z R1 + Z C = R1 − jX C = 1.1 − j ⋅ 6.061 kΩ = 6.16∠ − 79.71o kΩ Z eq = Z eq1 ⋅ Z eq 2 Z eq1 + Z eq 2 = (2.37∠13.92 )( ) kΩ ⋅ 6.16∠ − 79.71o kΩ = (2.3 + j ⋅ 0.57 kΩ ) + (1.1 − j ⋅ 6.061 kΩ ) o 14.60∠ − 65.79o kΩ 2 14.60∠ − 65.79o kΩ 2 = = = 2.261∠ − 7.56o kΩ o 3.4 − j ⋅ 5.491 kΩ 6.458∠ − 58.23 kΩ ______________________________________________________________________________________ Problem 4.36 Solution: Known quantities: The values of the impedance, R1 = 3.3 kΩ , R2 = 22 kΩ , L = 1.90 H , C = 6.8 nF and the voltage o applied to the circuit shown in Figure P4.35, v s (t ) = 636 cos (3000t + 15 ) V . Find: The equivalent impedance of the circuit. 4.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Analysis: rad · § X L = ωL = ¨ 3 k ¸(1.90 H ) = 5.7 kΩ Z L = + j ⋅ X L = + j ⋅ 5.7 kΩ s ¹ © 1 1 = 49.02 kΩ Z C = − j ⋅ X C = − j ⋅ 49.02 kΩ = rad · ωC § ¸(6.8 nF) ¨3 k s ¹ © = Z R1 + Z L = R1 + jX L = 3.3 + j ⋅ 5.7 kΩ = 6.59∠59.93o kΩ XC = Z eq1 Z eq 2 = Z R1 + Z C = R1 − jX C = 22 − j ⋅ 49.02 kΩ = 53.73∠ − 65.83o kΩ Z eq = Z eq1 ⋅ Z eq 2 Z eq1 + Z eq 2 = (6.59∠59.93 )( ) kΩ ⋅ 53.73∠ − 65.83o kΩ = (3.3 + j ⋅ 5.7 kΩ ) + (22 − j ⋅ 49.02 kΩ ) o 354.08∠ − 5.9o kΩ 2 354.08∠ − 5.9o kΩ 2 = = = 7.05∠53.81o kΩ o 25.3 − j ⋅ 43.32 kΩ 50.17∠ − 59.71 kΩ ______________________________________________________________________________________ Problem 4.37 Solution: Known quantities: The current in the circuit, ( ) i s (t ) = I 0 cos ωt + 30 o , I 0 = 13 mA , ω = 1000 the capacitance present in the circuit shown in Figure P4.37 C = 0.5 µF . rad , and the value of s Find: a) The phasor notation for the source current. b) The impedance of the capacitor. c) The voltage across the capacitor, showing all the passages and using phasor notation only. Analysis: a) Phasor notation: I s = I 0 ∠φ = 13∠30 o mA b) Z C = − jX C − j c) ( 1 1 = 0 − j 2 kΩ = 2∠ − 90 o kΩ =−j rad · ωC § ¨1000 ¸(0.5 µF ) s ¹ © )( ) VC = I s ⋅ Z C = 13∠30 o mA ⋅ 2∠ − 90 o kΩ = 26∠ − 60 o V ( ) vC (t ) = 26 cos 1000t − 60 o V Note that conversion from phasor notation to time notation or vice versa can be done at any time. ______________________________________________________________________________________ 4.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.38 Solution: Known quantities: The values of two currents in the circuit shown in Figure P4.38: ( ) i2 (t ) = 50 cos ωt + 53.13o mA , ω = 377 Find: The current ( ) i1 (t ) = 141.4 cos ωt + 135 o mA , rad . s i3 (t ) . Analysis: A solution using trigonometric identities is possible but inefficient, cumbersome, and takes a lot of time. Phasors are better! Note that one current is described with a sine and the other with a cosine function. When using phasors, all currents and voltages must be described with either sine functions or cosine functions. Which does not matter, but it is a good idea to adopt one and use it consistently. Therefore, first converts to cosines. KCL: − i1 (t ) + i2 (t ) + i3 (t ) = 0 + i3 (t ) = i1 (t ) − i 2 (t ) ( ) ( ) = 141.4 cos(ωt + 135 ) mA − 50 cos(ωt − 53.13 − 90 ) mA i3 (t ) = 141.4 cos ωt + 135 o mA − 50 sin ωt − 53.13o mA = o o o I 3 = 141.4 mA ∠135 o − 50 mA ∠ − 143.13o = = (− 99.98 + j ⋅ 99.98) mA − (− 40.00 − j ⋅ 30.00) mA = = (− 59.98 + j ⋅ 129.98) mA = 143.2 mA ∠114.8 o ( ) i3 (t ) = 143.2 cos ωt + 114.8 o mA If sine functions were used, the result in phasor notation would differ in phase by 90 degrees. ______________________________________________________________________________________ Problem 4.39 Solution: Known quantities: Z 1 = 5.9∠7 o kΩ , Z 2 = 2.3∠0 o Ω , Z 3 = 17∠11o Ω and the voltages applied to the circuit shown in Figure P4.39, v s1 (t ) = v s 2 (t ) = 170 cos (377t ) V . The values of the impedance, Find: The current through Z3 . Analysis: Vs1 = Vs 2 = 170∠0 o V = (170 + j 0 ) V KVL: − Vs1 − Vs 2 + I 3 Z 3 = 0 I3 = Vs1 + Vs 2 170∠0 o V + 170∠0 o V 340∠0 o V = = = 20∠ − 11o A Z3 17∠11o Ω 17∠11o Ω 4.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 rad § · i3 (t ) = 20 cos¨ 377 ⋅ t − 11o ¸ A s © ¹ Note also: KVL: − Vs1 + I 1 Z 1 = 0 I 1 = Vs1 Vs 2 , − Vs 2 − I 2 Z 2 = 0 I 2 = − Z1 Z2 ______________________________________________________________________________________ Problem 4.40 Solution: Known quantities: The values of the impedance in the circuit shown in Figure P4.40, Z s = (13000 + jω 3) Ω , R = 120 Ω , L = 19 mH , C = 220 pF . Find: The frequency such that the current I i and the voltage V0 are in phase. Analysis: Z s is not a factor in this solution. Only R, L, and C will determine if the voltage across this combination is in phase with the current through it. If the voltage and current are in phase, then, the equivalent impedance must have an "imaginary" or reactive part which is zero! V0 = Z eq ∠0 o = Req + jX eq , X eq (ω ) = 0 Ii (Z + Z L ) ⋅ Z C = (R + jX L ) ⋅ (− jX C ) = X L X C − jRX C R − j ( X L − X C ) = Z eq = R Z R + Z L + ZC R + jX L − jX C R + j( X L − X C ) R − j( X L − X C ) Z eq = X L X C R − RX C ( X L − X C )] − j [R 2 X C + X L X C ( X L − X C )] [ = 2 R 2 + (X L − X C ) At the resonant frequency the reactive component of this impedance must equal zero: R2 X C + X L X C (X L − X C ) X eq (ω ) = = 0 R 2 + X L (X L − X C ) = 0 2 2 R + (X L − X C ) 1 · L § 2 2 2 R 2 + ωL¨ ωL − ¸=0 ω L = −R ωC ¹ C © ω= 1 R2 − 2 = LC L = 293.3 G (120 Ω)2 (19 mH )(220 pF) (19 mH )2 1 − = rad rad rad − 39.89 M = 489.1 k s s s Notes: 1. To separate the equivalent impedance into real (resistive) and "imaginary" (reactive) components, the denominator had to be "rationalized". This was done by multiplying numerator and denominator by the complex conjugate of the denominator, and multiplying term by term. Remember that j = −1 , etc. 2. The term with R had a negligible effect on the resonant frequency in this case. If R is sufficiently large, however, it will significantly affect the answer. ______________________________________________________________________________________ 2 4.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.41 Solution: Known quantities: The circuit shown in Figure P4.40 Find: The frequency such that the current I i and the voltage V0 are in phase. Analysis: If the voltage and current are in phase, then, the equivalent impedance must have an "imaginary" or reactive part which is zero! V0 = Z eq ∠0 o = Req + jX eq , X eq (ω ) = 0 Ii (Z + Z L ) ⋅ Z C = (R + jX L ) ⋅ (− jX C ) = X L X C − jRX C R − j ( X L − X C ) = Z eq = R Z R + Z L + ZC R + jX L − jX C R + j( X L − X C ) R − j( X L − X C ) Z eq = = [X L X C R − RX C ( X L − X C )] − j[R 2 X C + X L X C ( X L − X C )] 2 R 2 + (X L − X C ) At the resonant frequency the reactive component of this impedance must equal zero: X eq (ω ) = R2 X C + X L X C (X L − X C ) = 0 R 2 + X L (X L − X C ) = 0 2 2 R + (X L − X C ) L 1 · § 2 2 2 R 2 + ωL¨ ωL − ¸=0 ω L = −R C ω C © ¹ ω= 1 R2 − 2 LC L ______________________________________________________________________________________ Problem 4.42 Solution: Known quantities: The values of the impedance, Rs = 50 Ω , Rc = 40 Ω , L = 20 µH , C = 1.25 nF , and the voltage rad o applied to the circuit shown in Figure P4.42, v s (t ) = V0 cos (ωt + 0 ), V0 = 10 V , ω = 6 M . s Find: The current supplied by the source. Analysis: Assume clockwise currents: rad · § o X L = ωL = ¨ 6 M ¸(20 µH ) = 1203 Ω Z L = 0 + j120 Ω = 120∠90 Ω s ¹ © XC = 1 1 = = 133.3 Ω Z C = 0 − j133.3 Ω = 133.3∠ − 90 o Ω rad · ωC § ¨6 M ¸(1.25 nF) s ¹ © 4.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Z Rc = 40 − j Ω = 40∠0 o Ω , Z Rs = 50 − j Ω = 50∠0 o Ω Equivalent impedances: Z eq1 = Z Rc + Z L = 40 + j120 Ω = 126.5∠71.56 o Ω Z eq = Z Rs + Z C ⋅ Z eq1 Z C + Z eq1 = 50 + j 0 Ω + OL: Is = (133.3∠ − 90 Ω)⋅ (126.5∠71.56 Ω) = = 50 + j 0 Ω + o o 133.3∠ − 90 o Ω + 126.5∠71.56 o Ω 16.87∠ − 18.44 o kΩ2 = 50∠0 o Ω + 400∠0 o Ω = 450∠0 o Ω o 42.161∠ − 18.44 Ω Vs 10∠0 o V = = 22.22∠0 o mA i s (t ) = 22.22 cos ωt + 0 o mA Z eq 450∠0 o Ω ( ) Note: The equivalent impedance of the parallel combination is purely resistive; therefore, the frequency given is the resonant frequency of this network. ______________________________________________________________________________________ Problem 4.43 Solution: Known quantities: The values of the impedance and the voltage applied to the circuit shown in Figure P4.43. Find: The current in the circuit. Analysis: Assume clockwise currents: rad o , VS = 12∠0 V s 1 ZC = = − j Ω , Z L = jωL = j 9 Ω Z total = 3 + j 9 − j = 3 + j8 Ω jω C 12 I= = 0.4932 − j1.3151 A = 1.4045∠ − 69.44 o A , 3 + j8 i(t ) = 1.4 cos ωt − 69.4 o A ω =3 ( ) ______________________________________________________________________________________ Problem 4.44 Solution: Known quantities: The values of the impedance and the current source shown in Figure P4.44. Find: The voltage. Analysis: Assume clockwise currents: ω=2 rad 1 o , I S = 10∠0 A , Z L = jωL = j 6 Ω , Z C = = − j 1.5 Ω jωC s 4.30 G. Rizzoni, Principles and Applications of Electrical Engineering Z eq = Problem solutions, Chapter 4 1 1 1 = = = 0.9231 − j1.3846 Ω 1 1 1 1 1 2 0.33 + j 0.5 + + −j +j R Z L ZC 3 6 3 V = I S Z eq = 10 A ⋅ (0.9231 − j1.3846) Ω = 9.231 − j13.846 a *10 V = 16.641∠ − 56.31o V ______________________________________________________________________________________ Problem 4.45 Solution: Known quantities: The values of the impedance and the current source for the circuit shown in Figure P4.45. Find: The current I1. Analysis: Specifying the positive directions of the currents as in figure P4.45: Z eq = 1 1 § 1 · ¸ +¨ 2 ¨© − j 4 ¸¹ = 1.79∠26.56 o Ω ( ) ( ) VS = I S Z eq = 10∠ − 22.5o A ⋅ 1.79∠26.56 o Ω = 17.9∠4.06 o V I1 = VS = 8.95∠4.06 o A R ______________________________________________________________________________________ Problem 4.46 Solution: Known quantities: The values of the impedance and the voltage source for circuit shown in Figure P4.46. Find: The voltage V2. Analysis: Specifying the positive directions as in figure P4.46: Z L = jωL = j12 Ω V2 = R12Ω R6Ω 6Ω 150∠0 o Ω V = 6.93∠ − 33.7 o V V= 25∠0 o V = (12 + j12 + 6) Ω + Z L + R6Ω 18 + j12 Ω ______________________________________________________________________________________ Problem 4.47 Solution: Known quantities: The values of the impedance and the current source of circuit shown in Figure P4.44. Find: The value of ω for which the current through the resistor is maximum. 4.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Analysis: Assume clockwise currents: I S = 10∠0 o A , Z L = jωL = j 3ω Ω , Z C = 1 3 =−j Ω jωC ω 1 1 10ω 3 R IS = 10∠0 o = IR = ω 1 1 1 1 1 ω + j ω 2 −1 + + −j +j R Z L ZC 3 3ω 3 ( ) The maximum of I R is obtained for ω = 1. , therefore iR (t ) = 10 cos(t ). ______________________________________________________________________________________ Problem 4.48 Solution: Known quantities: The values of the impedance and the current source for circuit shown in Figure P4.48. Find: The current through the resistor. Analysis: Specifying the positive directions as in figure P4.48: By current division: IR = − 1 R 1 1 + R ZC ⋅ Is = − 1 1 1 − jωRC ⋅ Is = − ⋅ Is = − ⋅ Is = 2 R 1+jωRC ( ) 1 + RC ω 1+ jωC = −(0.0247 − j 0.1552) A ⋅ 1∠0 o A = 157 ⋅ 10 −3 ∠99.04 o A i R (t ) = 157 cos (200πt + 99.04 o ) mA ______________________________________________________________________________________ 4.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.49 Solution: Known quantities: The values of the reactance X L = 1 kΩ , X C = 10 kΩ , and the current source I = 10∠45o mA for circuit shown in Figure P4.49. Find: The voltage vout . Analysis: Specifying the positive directions of the currents as in figure P4.49: Vout = Z eq I = (Z L + Z C )I = (0 + jX L + 0 − jX C )I = ( j1 kΩ − j10 kΩ ) ⋅10∠45o mA = = (− j 9 kΩ ) ⋅10∠45o mA = 9∠ − 90o kΩ ⋅10∠45o mA = 90∠ − 45o V vout = 90 cos(ωt − 45°) V ______________________________________________________________________________________ Problem 4.50 Solution: Known quantities: The circuit shown in Figure P4.50, the values of the resistance, inductance, L = 1 4 H , and the frequency ω = 4 Find: The impedance Analysis: R = 2 Ω , capacitance, C = 1 8 F , rad . s Z. 1 1 1 1 Z L = jω L = j 4 Ω = j Ω , Z C = =−j =−j = − j2 Ω jω C ωC 4 4 ⋅ (1 8) ( j 2) (− 1 − j ) 1 1 j2 = j+ = j+ = j+ Z = Z L + ZC R = Z L + 1 1 1 1 (− 1 + j ) (− 1 − j ) −1+ j + + ZC R − j2 2 = j+ j 2(− 1 − j ) = j − j +1 = 1Ω 1+1 ______________________________________________________________________________________ Problem 4.51 Solution: Known quantities: Circuit shown in Figure P4.51, the values of the resistance, inductance, L = 4 5 H , and the frequency ω = 5 rad . s 4.33 R = 3 Ω , capacitance, C = 1 10 F , G. Rizzoni, Principles and Applications of Electrical Engineering Find: The admittance Analysis: Problem solutions, Chapter 4 Y. 1 1 4 = = − j 2 Ω , Z L = j5 ⋅ = j 4 Ω jωC j 5 ⋅ (1 10 ) 5 1 1 1 1 1 1 1 = = + = + Y= = 1 Z Z C (R + Z L ) ZC R + Z L − j2 3 + j4 1 1 + ZC R + Z L ZC = (1) (3 − j 4) = j 1 + 3 − j 4 1 = j + 2 (3 + j 4 ) (3 − j 4 ) 2 9 + 16 1 3 − j4 1 3 4 = j + = j + −j = 0.12 + j 0.34 S 2 25 2 25 25 ______________________________________________________________________________________ 4.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Section 4.5: AC Circuit Analysis methods Focus on Methodology: AC Circuit Analysis 1. 2. 3. 4. 5. Identify the sinusoidal source(s) and note the excitation frequency. Convert the source(s) tophasor form. Represent each circuit element by ots impedance. Splve the resulting phasor circuit, using appropriate circuit analysis tools. Convert the (phasor-form) answer to its time-domain equivalent, using equation 4.50. Problem 4.52 Solution: Known quantities: Circuit shown in Figure P4.52, the values of the resistance, R = 9 Ω , capacitance, C = 1 18 F , π· § inductance, L1 = 3 H , L2 = 3 H , L3 = 3 H , and the voltage source v s (t ) = 36 cos¨ 3t − ¸ V . 3¹ © Find: The voltage across the capacitance tehcniques. Analysis: v using phasor rad , Vs = 36∠ − 60° V s = jωL2 = j 3 ⋅ 3 = j 9 Ω ω =3 Z L2 1 1 = = − j6 Ω jωC j 3 ⋅ (1 18) Z L3 = jωL3 = j 3 ⋅ 3 = j 9 Ω ZC = Z eq = ( 1 Z L3 Z L2 + Z C ) = 1 1 1 + Z L3 Z L2 + Z C ( = ) 1 1 1 + j9 j3 = j9 = 2.25∠90 o Ω 4 Z T = Z R + Z L1 + Z eq = 9 + j 3 ⋅ 3 + j 2.25 = 9 + j11.25 = 14.407∠51.34 o Ω I= VS 36∠ − 60o V = = 2.499∠ − 111.34o A Z T 14.407∠51.340 Ω Veq = IZ eq = (2.499∠ − 111.34o )(2.25∠90o ) = 5.623∠ − 21.34o V V= − j6 ZC 5.623∠ − 21.34o = 11.25∠158.66o V Veq = j3 Z L2 + Z C ( ( ) ) v = 11.25 cos 3t − 158.66o V ______________________________________________________________________________________ 4.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.53 Solution: Known quantities: Circuit shown in Figure P4.53, the values of the resistance, R = 5 Ω , capacitance, C = 1 2 F , inductance, L1 = 0.5 H , L2 = 1 H , L2 = 10 H , and the current source i s (t ) = 6 cos(2t ) A . Find: The current through the inductance iL2 . Analysis: rad , Z L2 = jωL2 = j 2 Ω s 1 ZC = = − j Ω , Z L3 = jωL3 = j 20 Ω jω C Z L3 + Z C j 20 − j I= IS = 6∠0 o Z L3 + Z C + R + Z L2 ( j 20 − j ) + (5 + j 2) ω=2 ( ) ( ) j19 6∠0 o = 5.28∠13.4 o A 5 + j 21 i = 5.28 cos 2t + 13.4 o A = ( ) ______________________________________________________________________________________ Problem 4.54 Solution: Known quantities: Circuit shown in Figure P4.54 the values of the resistance, inductance, RS = RL = 500 Ω , R = 1 kΩ and the L = 10 mH . Find: a) The Thèvenin equivalent circuit if the voltage applied to the circuit is vS (t ) = 10 cos(1,000t ) . b) The Thèvenin equivalent circuit if the voltage applied to the circuit is vS (t ) = 10 cos(1,000 ,000t ) . Analysis: a) Z L = jωL = j1000 rad ⋅ 10 mH = j10 Ω , s The equivalent impedance is: ZT = ZL ⋅ R ( j10)1000 + 500 = 500 + j103 = 500.1 + j9.999 Ω + RS = (Z L + R ) 100 + j j10 + 1000 The equivalent Thèvenin voltage is: VT = VS = 10∠0 o V 6 rad ⋅ 10 mH = j10 4 Ω , b) Z L = jωL = j10 s The equivalent impedance is: 4.36 G. Rizzoni, Principles and Applications of Electrical Engineering ZT = Problem solutions, Chapter 4 ZL ⋅ R ( j10 4 )1000 + 500 = 500 + j10 4 = 1490.1 + j99.01 Ω + RS = (Z L + R ) 1 + j10 j10 4 + 1000 The equivalent Thèvenin voltage is: VT = VS = 10∠0 o V ______________________________________________________________________________________ Problem 4.55 Solution: Circuit shown in Figure P4.55 the values of the impedance, the voltage source vin (t ) = 12 cos(10t ) V . L = 0.1 H , capacitance, C = 100 µF , and Find: The Thèvenin equivalent of the circuit as seen by the load resistor RL . Analysis: 1 = − j1000 Ω rad ⋅ 100 µF j10 s rad Z L = jωL = j10 ⋅ 0.1 H = j1 Ω s ZC = 1 = jω C The equivalent impedance is: ZT = Z L Z C = Z L ⋅ ZC j (− j1000 ) 1000 = = = 1.001∠90 o Ω = j1.001 Ω Z L + ZC j − j1000 − j 999 The Thèvenin voltage is: VT = ZC − j1000 1000 Vin = ⋅ 12∠0 o = ⋅ 12∠0 o = 12.012∠0 o V Z L + ZC j − j1000 999 ______________________________________________________________________________________ Problem 4.56 Solution: Known quantities: Circuit shown in Figure P4.56 the values of the resistance, R1 = 4 Ω , R2 = 4 Ω , capacitance, C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V . Find: The current in the circuit i L (t ) using phasor techniques. Analysis: VS (t ) = 2∠0 o V 1 1 = = − j2 Ω ZC = 1 jωC j2 4 Z L = jω L = j 2 ⋅ 2 = j 4 Ω Applying the voltage divider rule: 4.37 G. Rizzoni, Principles and Applications of Electrical Engineering (Z L || (Z C + Z 2 )) V S Z1 + (Z L || (Z C + Z 2 )) VL = = Problem solutions, Chapter 4 4∠36.8° 2∠0o = 1.05∠18.4° V 4∠0° + 4∠36.8° Therefore, the current is: IL = VL 1.05∠18.4° = = 0.2635∠ − 71.6o A ZL 4∠90° ( ) iL (t ) = 0.2635 cos 2t − 71.6o A ______________________________________________________________________________________ Problem 4.57 Solution: Known quantities: Circuit shown in Figure P4.57, the values of the resistance, R1 = 75 Ω , R2 = 100 Ω , capacitance, C = 1 µF , inductance, L = 0.5 H , and the voltage source v s (t ) = 15 cos(1,500t ) V . Find: The currents in the circuit i1 (t ) and i2 (t ) . Analysis: In the phasor domain: ZC = -j 2000 =−j = − j 666.7 Ω , Z L = j (1500)(0.5) = j 750 Ω -6 3 1500( 1 × 10 ) By applying KVL in the first loop, we have VS = R1 I 1 + Z C (I1 − I 2 ) By applying KVL in the second loop, we have 0 = (Z C )(I 2 − I 1 ) + (Z L + R2 )I 2 That is: 2000 · 2000 § o °15∠0 = ¨ 75 − j 3 ¸ I1 + j 3 I 2 ° © ¹ ® °0 = j 2000 I + §¨100 + j 250 ·¸ I 1 2 °¯ 3 3 ¹ © By solving above equations, we have I1 = 3.8 ⋅10 −3 ∠46.6 o A I 2 = 19.6 ⋅10 −3 ∠ − 83.2 o A i1 (t ) = 3.8 cos 1,500t + 46.6 o mA i2 (t ) = 19.6 cos 1,500t − 83.2 o mA ( ( ) ) ______________________________________________________________________________________ Problem 4.58 Solution: Known quantities: Circuit shown in Figure P4.58, the values of the resistance, R1 = 40 Ω , R2 = 10 Ω , capacitance, C = 500 µF , inductance, L = 0.2 H , and the current source is (t ) = 40 cos(100t ) A. 4.38 G. Rizzoni, Principles and Applications of Electrical Engineering Find: The voltages in the circuit Problem solutions, Chapter 4 v1 (t ) and v 2 (t ) . Analysis: ZC = 1 -j = = -j20 Ω , Z L = jωL = j100 ⋅ 0.2 = j20 Ω jωC 100 ⋅ 500 ⋅ 10 - 6 Applying KCL at node 1, we have: IS = §1 V1 V1 − V2 1 · 1 j · j § 1 ¸¸V1 − V2 40∠0o = ¨ + ¸V1 − V2 I S = ¨¨ + + R1 ZC ZC 20 © 40 20 ¹ © R1 Z C ¹ Applying KCL at node 2, we have § 1 V1 − V2 V2 V2 V 1 1 = + 1 = ¨¨ + + ZC R2 Z L Z C © R2 Z L Z C Therefore: · V 1 1 1 ¸¸V2 j 1 = §¨ − j + j ·¸V2 20 © 10 20 20 ¹ ¹ j · j § 1 o j · j § 1 o °40∠0 = ¨ 40 + 20 ¸V1 − 20 V2 © ¹ °40∠0 = ¨ + ¸(− j 2V2 ) − V2 ° ® 20 © 40 20 ¹ ® °V = − j 2V ° j V1 = §¨ 1 ·¸V 2 ¯ 1 °¯ 20 © 10 ¹ 2 j 1 j j· §1 o °40∠0 = − V2 + V2 − V2 = ¨ − ¸V2 20 10 20 © 10 10 ¹ ® °V = − j 2V ¯ 1 2 40∠0o = 282.84∠45o V, V1 = − j 2V2 = 565.68∠ − 45o V j· §1 ¨ − ¸ © 10 10 ¹ v2 (t ) = 282.84 cos 100t + 45o V, v1 (t ) = 568.68 cos 100t − 45o V V2 = ( ) ( ) ______________________________________________________________________________________ Problem 4.59 Solution: Known quantities: The circuit called Wheatstone bridge shown in Figure P4.59. a) The balanced status for the bridge: v ab = 0 . R1 = 100 Ω , R2 = 1 Ω , the capacitance, C 3 = 4.7 µF , the inductance, L3 = 0.098 H , that are necessary to balance the bridge: v ab = 0 , and the voltage applied to the bridge, v s = 24 sin(2 ,000t ) V . b) The values of the resistance, Find: a) The unknown reactance X 4 in terms of the circuit elements. b) The value of the unknown reactance X 4 . c) The source frequency that should be avoided in this circuit. Analysis: a) Assuming a balanced circuit, we have v ab = 0 , that is, v a 4.39 = vb G. Rizzoni, Principles and Applications of Electrical Engineering From the voltage divider: jX L3 Problem solutions, Chapter 4 R2 jX 4 = - jX C3 + R2 R1 + jX 4 R2 jX 4 = j R1 + jX 4 jωL3 + R2 ωC 3 Inverting both sides and equating imaginary parts: b) § 1 · R1R2 ¸¸ X 4 X 4 = R1R2 = ¨¨ − ωL3 + ωC3 ¹ § 1 · © ¨¨ − ωL3 ¸¸ © ωC3 ¹ X4 = 100 ⋅ 1 1 § · - 2000 ⋅ 0.098 ¸ ¨ −6 © 2000 ⋅ 4.7 ⋅ 10 ¹ = −1.116 Ω Negative reactance implies that the component is a capacitor. 1 1 = 1.116 Ω C = = 448 µF ωC ω ⋅ 1.116 c) If the reactances of L3 and C3 cancel, the bridge can not measure X4. Thus, the condition to be avoided is: ωL 3 − 1 1 = 0 L3C 3 = 2 ω = ωC 3 ω 1 L3 C 3 = 1 0.098 ⋅ 4.7 ⋅ 10 − 6 = 1473 rad s f = 234.5 Hz ______________________________________________________________________________________ Problem 4.60 Solution: Known quantities: Circuit shown in Figure P4.56, the values of the resistance, R1 = 4 Ω , R2 = 4 Ω , capacitance, C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V . Find: The Thévenin impedance seen by resistor R2 . Analysis: Z T = (R1 Z L ) + (Z C ) = (4 j 4 ) + (-j 2 ) = j 2(1-j )-j 2 = 2 + j 2 + (-j 2) = 2 Ω ______________________________________________________________________________________ Problem 4.61 Solution: Known quantities: Circuit shown in Figure P4.58, the values of the resistance, R1 = 10 Ω , R2 = 40 Ω , capacitance, C = 500 µF , inductance, L = 0.2 H , and the current source i s (t ) = 40 cos(100t ) A . Find: The Thévenin voltage seen by inductance L . Analysis: The Thévenin equivalent voltage source is the open-circuit voltage at the load terminals: 4.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 VT = R2 I 2 = 40I 2 From the current division, we have I2 = R1 10 IS = 40∠0 o = 7.43∠21.8 o A (R2 + Z C ) + R1 (40-j 20) + 10 VT = R2 I 2 = 40 ⋅ 7.43∠21.8 o = 297∠21.8 o V vT (t ) = 297 cos 100t + 21.8 o V ( ) ______________________________________________________________________________________ Problem 4.62 Solution: Known quantities: Circuit shown in Figure P4.62, the values of the impedance, the voltage source R = 8 Ω , Z C = − j8 Ω , Z L = j8 Ω , and Vs = 5∠ − 30 V . o Find: The Thévenin equivalent circuit seen from the terminals a-b. Analysis: The Thévenin equivalent circuit is given by: § 8 + j8 · ¸¸5∠ − 30° = (1 + j )5∠ − 30 o = 7.07∠15 V VTH = ¨¨ + − 8 j 8 j 8 © ¹ (8 + j8)(− j8) = (8 − j8) = 8 2∠ − 45o Ω Z TH = 8 + j8 − j 8 ______________________________________________________________________________________ Problem 4.63 Solution: Known quantities: Circuit shown in Figure P4.56, the values of the resistance, R1 = 4 Ω , R2 = 4 Ω , capacitance, C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V . Find: The Thévenin equivalent voltage seen by the resistor R2 . Analysis: The Thévenin equivalent circuit is given by: j4 2∠0 o = (1 + j ) = 2∠45 o = 1.414∠45 o V 4+j 4 vT (t ) = 1.414 cos 2t + 45 o V VT = ( ) ______________________________________________________________________________________ 4.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 Problem 4.64 Solution: Known quantities: Circuit shown in Figure P4.56, the values of the resistance, R1 = 4 Ω , R2 = 4 Ω , capacitance, C = 1 4 F , inductance, L = 2 H , and the voltage source v s (t ) = 2 cos(2t ) V . Find: The Norton equivalent circuit seen by the resistor Analysis: From the result of Problem 4.60, we have IN = R2 . Z T = 2 Ω . From the current divider: j4 I = 2I j 4-j 2 and j4 − j2 = (− j 2)( j 4) = − j 4 j2 The current is: I= 2∠0 o 2 = ∠45 o = 0.353∠45 o A 4-j 4 4 Therefore: I N = 2I = 0.707∠45 o A ______________________________________________________________________________________ Problem 4.65 Solution: Known quantities: Circuit shown in Figure P4.66. Find: The equations required to solve for the loop currents in the circuit in: a. Integral-differential form; b. Phasor form. Analysis: KVL: KVL: 1 t (i1 − i2 )dt + (i1 − i2 )R1 = 0 C ³0 t (i2 − i1 )R1 − vc (0 ) + 1 ³0 (i2 − i1 )dt + L di2 + i2 R2 = 0 dt C − v S + i1 RS + vc (0 ) + Note: The initial voltage across the capacitor must, in general, be considered. It is modeled as an ideal voltage source in series with the capacitor. KVL: − VS + I 1 RS + (I 1 − I 2 )Z C + (I 1 − I 2 )R1 = 0 KVL: (I 2 − I 1 )R1 + (I 2 − I1 )Z C + +I 2 Z L + I 2 R2 =0 Note: 1. The i-v characteristics of the inductor and capacitor, i.e. the integral and derivative, have been replaced here by the impedance. 2. This form of the equation is applicable only when the waveforms of the currents and voltages are sinusoids! 4.42 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 ______________________________________________________________________________________ Problem 4.66 Solution: Known quantities: Circuit shown in Figure P4.65. Find: The node equations required to solve for all currents and voltages in the circuit. Analysis: I1 = I 2 + I 3 VS − I1RS = I 2 (Z C + R1 ) + I 3 (Z L + R2 ) ______________________________________________________________________________________ Problem 4.67 Solution: Known quantities: The voltages at the nodes of the circuit shown in Figure P4.67, Va = 450∠0 o V , Vb = 440∠30 o V , Vc = 420∠ − 200 o V , Vbc = 779.5∠5.621o V , Vcd = 153.9∠68.93 o V , Vba = 230.6∠107.4 o V , and the voltage sources, v s1 = 450 cos(ωt ) V , v s 2 = 450 cos(ωt ) V . Find: The new values of Vb and Vbc , if the ground is moved from Node e to Node d. Analysis: A node voltage is defined as the voltage between a node and the ground node. If the ground node is changed, then all node voltages in the circuit will change. With the ground at Node d: Vb = Vbd = Vbe + Ved = Vbe + Vs 2 = 440∠30 o V + 450∠0 o V = = (381.1 + j 220.0 ) V + (450 + j 0) V = 831.1 + j 220.0 V = 859.6∠14.83o V The voltages between any two nodes in a circuit do not depend on which is the ground node; therefore, the voltage between Node b and Node c remains the same when the ground is moved from Node e to Node d: Vbc = 779.5∠5.621o V ______________________________________________________________________________________ Problem 4.68 Solution: Known quantities: Circuit shown in Figure P4.68, the values of the resistance, RL = 120 Ω , the capacitance, C = 12.5 µF , and the inductance, L = 60 mH , and the voltage source, ( ) vi = 4 cos 1,000t + 30 o V . Find: The new value of V0 . Analysis: The circuit has 3 unknown mesh currents but only 1 unknown node voltage. 4.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 1 1 =−j = − j80 Ω = 80∠ − 90 o Ω rad · ωC § ¨1,000 k ¸(12.5 µF ) s ¹ © rad · § o Z L = jX L = jωL = j ¨1,000 k ¸(60 mH ) = j 60 Ω = 60∠90 Ω s © ¹ o Reference phasor: Vi = 4∠30 V Z C = − jX C = − j KCL: V0 − 0 V0 − 0 V0 − Vi + + =0 Z RL ZC ZL Vi Vi ZL 4∠30 o V = V0 = = = ZL ZL 1 1 1 60∠90 o Ω 60∠90 o Ω + + + +1 +1 + Z RL Z C Z L Z RL Z C 120∠0 o Ω 80∠ − 90 o Ω = 4∠30 o V 4∠30 o V = = 0.5∠90 o + 0.75∠180 o + 1 (0 + j 0.5) + (− 0.75 + j 0 ) + (1 + j 0 ) = 4∠30 o V 4∠30 o V = = 7.155∠ − 33.43 o V o 0.25 + j 0.5 0.559∠63.43 ( ) v0 (t ) = 7.155 cos ωt − 33.43o V ______________________________________________________________________________________ Problem 4.69 Solution: Known quantities: Circuit shown in Figure P4.69, the mesh currents and node voltages, ( ) ( ) ( ( ) ) i1 (t ) = 3.127 cos ωt − 47.28 o A , i2 (t ) = 3.914 cos ωt − 102.0 o A , i3 (t ) = 1.900 cos(ωt + 37.50 o ) A , v1 (t ) = 130.0 cos ωt + 10.08 o V , rad v 2 (t ) = 130.0 cos ωt − 25.00 o V , where ω = 377.0 . s Find: One of the following: Analysis: KCL: − I1 L1 , C 2 , R3 , L3 . + I Z1 + I 3 = 0 I Z1 = I1 − I 3 = (2.121 − j 2.297 ) A − (1.507 + j1.157 ) A = 3.508∠ − 79.92 o A V1 130∠10.08 o V = = 37.05∠90 o Ω = ωL1∠90 o I Z1 3.508∠ − 79.92 o A OL: Z1 = KCL: 37.05 Ω = 98.29 mH rad 377 s I 2 + I Z2 − I 3 = 0 L1 = 4.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 4 I Z 2 = I 3 − I 2 = (1.507 − j1.157 ) A − (− 0.8138 − j 3.828) A = 5.499∠65.03o A OL: V2 130∠24.97 o V 1 Z2 = = = 23.64∠ − 90 o Ω = ∠ − 90 o o I Z 2 5.499∠65.03 A ωC 2 1 = 112.2 µF rad · § ¨ 377 ¸(23.64 Ω) s ¹ © V2 − V1 + VZ3 = 0 C2 = KVL: VZ 3 = V1 − V2 = (128.0 + j 22.75) V − (117.8 − j 54.88) V = 78.29∠82.56 o V OL: VZ 3 78.29∠82.56 o V Z3 = = = 4.21∠45.06 o Ω = 29.11 + j 29.17 Ω = R3 + jωL3 o I3 1.9∠37.5 A 29.17 Ω R3 = 29.11 Ω, R3 = = 77.37 mH rad 377 s ______________________________________________________________________________________ 4.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Chapter 5 Instructor Notes Chapter 5 has been reorganized in response to a number of suggestions forwarded by users of the third edition of this book. The material is now divided into three background sections (overview of transient analysis, writing differential equations, and DC steady-state solution) and two major sections: first and second order transients. A few examples have been added, and all previous examples have been reorganized to follow the methodology outlined in the text. In the section on first order transients, a Focus on Methodology: First Order Transient Response (p. 215) clearly outlines the methodology that is followed in the analysis of first order circuits; this methodology is then motivated and explained, and is applied to eight examples, including four examples focusing on engineering applications (5.8 - Charging a camera flash; 5.9 and 5.11 dc motor transients; and 5.12, transient response of supercapacitor bank). The analogy between electrical and thermal systems that was introduced in Chapter 3 is now extended to energy storage elements and transient response (Make The Connection: Thermal Capacitance, p. 204; Make The Connection: Thermal System Dynamics, p. 205; Make The Connection: First-Order Thermal System, p. 218-219;); similarly, the analogy between hydraulic and electrical circuits, begun in Chapter 2 and continued in Chapter 4, is continued here (Make The Connection: Hydraulic Tank, pp. 214-215). The box Focus on Measurements: Coaxial Cable Pulse Response (pp. 230-232) illustrates an important transient analysis computation (this problem was suggested many years ago by a Nuclear Engineering colleague). The section on second order transients summarizes the analysis of second order circuits in the boxes Focus on Methodology: Roots of Second-Order System (p. 240) and Focus on Methodology: Second Order Transient Response (pp. 244-245). These boxes clearly outline the methodology that is followed in the analysis of second order circuits; the motivation and explanations in this section are accompanied by five very detailed examples in which the methodology is applied step by step. The last of these examples takes a look at an automotive ignition circuit (with many thanks to my friend John Auzins, formerly of Delco Electronics, for suggesting a simple but realistic circuit). The analogy between electrical and mechanical systems is explored in Make The Connection: Automotive Suspension, pp. 239-240 and pp. 245-246. The homework problems are divided into four sections, and contain a variety of problems ranging from very basic to the fairly advanced. The focus is on mastering the solution methods illustrated in the chapter text and examples. Learning Objectives 1. Write differential equations for circuits containing inductors and capacitors. 2. Determine the DC steady state solution of circuits containing inductors and capacitors. 3. Write the differential equation of first order circuits in standard form and determine the complete solution of first order circuits excited by switched DC sources. 4. Write the differential equation of second order circuits in standard form and determine the complete solution of second order circuits excited by switched DC sources. 5. Understand analogies between electrical circuits and hydraulic, thermal and mechanical systems. 5.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Section 5.2: Writing Differential Equations for Circuits Containing Inductors and Capacitors Problem 5.1 Solution: Known quantities: L = 0.9 mH , Vs = 12 V , R1 = 6 k Ω, R2 = 6 kΩ, R3 = 3 kΩ. Find: The differential equation for t > 0 (switch open) for the circuit of P5.21. Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. Note that the top node voltage is the inductor voltage, v L . vL v + iL + L = 0 R1 + R2 R3 Next, use the definition of inductor voltage to eliminate the variable v L from the nodal equation: L diL L diL + iL + =0 R1 + R2 dt R3 dt (R1 + R2 )R3 i = 0 di L + L dt L(R1 + R2 + R3 ) Substituting numerical values, we obtain the following differential equation: di L + 2.67 ⋅10 6 iL = 0 dt ______________________________________________________________________________________ Problem 5.2 Solution: Known quantities: V1 = 12 V , C = 0.5 µF , R1 = 0.68 k Ω, R2 = 1.8 kΩ. Find: The differential equation for t > 0 (switch closed) for the circuit of P5.23. Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. Note that the top node voltage is the capacitor voltage, vC . iC + vC vC − V1 + =0 R2 R1 Next, use the definition of capacitor current to eliminate the variable C dvC R1 + R2 V vC = 1 + dt R1 R2 R1 iC from the nodal equation: dvC R1 + R2 V vC = 1 + dt C (R1 R2 ) CR1 Substituting numerical values, we obtain the following differential equation: 5.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 dvC + (4052 )vC − 35292 = 0 dt ______________________________________________________________________________________ Problem 5.3 Solution: Known quantities: V1 = 12 V , R1 = 0.68 k Ω, R2 = 2.2 kΩ, R3 = 1.8 kΩ, C = 0.47 µF . Find: The differential equation for t > 0 (switch closed) for the circuit of P5.27. Analysis: Apply KCL at the two node (nodal analysis) to write the circuit equation. Note that the node #1 voltage is the capacitor voltage, vC . For node #1: iC + vC − v2 =0 R2 For node #2: v2 − vC v2 v2 − V1 + + =0 R2 R3 R1 Solving the system: R3 R R + R1 R3 + R2 R3 V1 − 1 2 iC R1 + R3 R1 + R3 R3 RR v2 = V1 − 1 3 iC R1 + R3 R1 + R3 vC = Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation: C (R1 R2 + R1 R3 + R2 R3 ) dvC R3 V1 + vC = R1 + R3 dt R1 + R3 dvC R1 + R3 R3 vC = V1 + dt C (R1 R2 + R1 R3 + R2 R3 ) C (R1 R2 + R1 R3 + R2 R3 ) Substituting numerical values, we obtain the following differential equation: dvC + (790 )vC − 6876 = 0 dt ______________________________________________________________________________________ Problem 5.4 Solution: Known quantities: VS 2 = 13 V , L = 170 mH , R2 = 4.3 kΩ, R3 = 29 kΩ. Find: The differential equation for t > 0 (switch open) for the circuit of P5.29. 5.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Analysis: Applying KVL we obtain: (R2 + R3 )iL + vL + VS 2 = 0 Next, using the definition of inductor voltage to eliminate the variable v L from the nodal equation: (R2 + R3 )iL + L diL + VS 2 = 0 dt V di L (R2 + R3 ) + i + S2 = 0 dt L L L Substituting numerical values, we obtain the following differential equation: di L + 1.96 ⋅ 10 5 i L + 76.5 = 0 dt ______________________________________________________________________________________ Problem 5.5 Solution: Known quantities: I 0 = 17 mA, C = 0.55 µF , R1 = 7 k Ω, R2 = 3.3 kΩ. Find: The differential equation for t > 0 for the circuit of P5.32. Analysis: Using the definition of capacitor current: C dvC = I0 dt dvC I 0 = C dt Substituting numerical values, we obtain the following differential equation: dvC − 30909 = 0 dt ______________________________________________________________________________________ Problem 5.6 Solution: Known quantities: VS 1 = VS 2 = 11V , C = 70 nF , R1 = 14 k Ω , R2 = 13 kΩ , R3 = 14 kΩ . Find: The differential equation for t > 0 (switch closed) for the circuit of P5.34. Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. v1 − VS 2 v + iC + 1 = 0 R1 R3 Note that the node voltage v1 is equal to: v1 = R2 iC + vC Substitute the node voltage v1 in the first equation: 5.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 · §R R §1 1 · V ¨¨ + ¸¸vC + ¨¨ 2 + 2 + 1¸¸iC − S 2 = 0 R1 © R1 R3 ¹ © R1 R3 ¹ Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation: · dv V §R R §1 1 · ¨¨ + ¸¸vC + ¨¨ 2 + 2 + 1¸¸C C − S 2 = 0 R1 © R1 R3 ¹ dt © R1 R3 ¹ Substituting numerical values, we obtain the following differential equation: dvC + 714.3vC − 3929 = 0 dt ______________________________________________________________________________________ Problem 5.7 Solution: Known quantities: VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: The differential equation for t > 0 (switch closed) for the circuit of P5.41. Analysis: Apply KCL at the two node (nodal analysis) to write the circuit equation. Note that the node #1 voltage is equal to the two capacitor voltages, vC1 = vC 2 = vC . For node #1: iC1 + iC 2 + vC − v 2 =0 R2 For node #2: v 2 − vC v 2 v 2 + + − IS = 0 R2 R3 R4 Solving the system: v2 = R3 R4 (iC1 + iC 2 − I S ) R3 + R4 § RR · RR vC = −¨¨ R2 + 3 4 ¸¸(iC1 + iC 2 ) + 3 4 I S R3 + R4 R3 + R4 ¹ © Next, use the definition of capacitor current to eliminate the variables iCi from the nodal equation: § RR · dv RR vC + ¨¨ R2 + 3 4 ¸¸(C1 + C 2 ) C − 3 4 I S = 0 R3 + R4 ¹ dt R3 + R4 © Substituting numerical values, we obtain the following differential equation: dvC 1 1 + vC − = 0 dt 48 6 ______________________________________________________________________________________ 5.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.8 Solution: Known quantities: C = 1 µF , RS = 15 k Ω , R3 = 30 kΩ . Find: The differential equation for t Assume: Assume that > 0 (switch closed) for the circuit of P5.47. VS = 9 V, R1 =10kΩ and R2 = 20kΩ . Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. 1. Before the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation. § 1 V 1 1 1 · ¨¨ + + + ¸¸vC + iC − S = 0 RS © RS R1 R2 R3 ¹ Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation: vC − VS vC v v + + iC + C + C = 0 RS R1 R2 R3 § 1 dv V 1 1 1 · ¨¨ + + + ¸¸vC + C C − S = 0 dt RS © RS R1 R2 R3 ¹ Substituting numerical value, we obtain the following differential equation: dvC + 250vC − 600 = 0 dt 2. After the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation. § 1 V vC − VS vC 1· + + iC = 0 ¨¨ + ¸¸vC + iC − S = 0 RS RS R1 © RS R1 ¹ Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation: § 1 dv V 1· ¨¨ + ¸¸vC + C C − S = 0 dt RS © RS R1 ¹ Substituting numerical values, we obtain the following differential equation: dvC 500 vC − 600 = 0 + dt 3 ______________________________________________________________________________________ Problem 5.9 Solution: Known quantities: Values of the voltage source, of the inductance and of the resistors. Find: The differential equation for t > 0 (switch open) for the circuit of P5.49. Analysis: Apply KCL at the top node (nodal analysis) to write the circuit equation. 5.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 v1 − 100 v1 + + iL = 0 0.3v1 + iL − 10 = 0 10 5 Note that the node voltage v1 is equal to: v1 = 2.5iL + v L Substitute the node voltage v1 in the first equation: (1.75)iL + (0.3)vL − 10 = 0 Next, use the definition of inductor voltage to eliminate the variable (0.3)(0.1) di L + (1.75)i L − 10 = 0 dt v L from the nodal equation: di L + (58.33)iL − 333 = 0 dt ______________________________________________________________________________________ Problem 5.10 Solution: Known quantities: I S = 5 A, L1 = 1H , L2 = 5H , R = 10kΩ . Find: The differential equation for t > 0 (switch open) for the circuit of P5.52. Analysis: Applying KCL at the top node (nodal analysis) to write the circuit equation. v1 + iL = 0 R Note that the node voltage v1 is equal to: v1 = v L1 + v L 2 Substitute the node voltage v1 in the first equation: v L1 + v L 2 + iL − I S = 0 R − IS + Next, use the definition of inductor voltage to eliminate the variable (L1 + L2 ) diL + i R dt L v L from the nodal equation: − IS = 0 Substituting numerical value, we obtain the following differential equation: di L 5000 25000 iL − + =0 dt 3 3 ______________________________________________________________________________________ 5.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Section 5.3: DC Steady-State Solution of Circuits Containing Inductors and Capacitors - Initial and Final Conditions Problem 5.11 Solution: Known quantities: L = 0.9 mH ,Vs = 12 V , R1 = 6 k Ω, R2 = 6 kΩ, R3 = 3 kΩ. Find: The initial and final conditions for the circuit of P5.21. Analysis: Before opening, the switch has been closed for a long time. Thus we have a steady-state condition, and we treat the inductor as a short circuit. The voltages across the resistances R1 and R3 is equal to zero, since they are in parallel to the short circuit, so all the current flow through the resistor i L ( 0) = R2 : VS = 2 mA. R2 After the switch has been opened for a long time, we have again a steady-state condition, and we treat the inductor as a short circuit. When the switch is open, the voltage source is not connected to the circuit. Thus, i L (∞ ) = 0 A. ______________________________________________________________________________________ Problem 5.12 Solution: Known quantities: V1 = 12 V , C = 0.5 µF , R1 = 0.68 k Ω, R2 = 1.8 kΩ. Find: The initial and final conditions for the circuit of P5.23. Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the circuit. Thus, vC (0) = 0 V. After the switch has been closed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R2 : vC (∞ ) = R2 1800 V1 = 12 = 8.71 V. R1 + R2 1800 + 680 ______________________________________________________________________________________ 5.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.13 Solution: Known quantities: V1 = 12 V , R1 = 0.68 k Ω, R2 = 2.2 kΩ, R3 = 1.8 kΩ, C = 0.47 µF . Find: The initial and final conditions for the circuit of P5.27. Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. When the switch is open, the voltage source is not connected to the circuit. Thus, vC (0) = 0 V. After the switch has been closed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. Since the current flowing through the resistance R2 is equal to zero, the voltage across the capacitor is equal to the voltage across the resistance vC (∞ ) = R3 : R3 1800 V1 = 12 = 8.71 V. R1 + R3 1800 + 680 ______________________________________________________________________________________ Problem 5.14 Solution: Known quantities: VS1 = VS 2 = 13 V , L = 170 mH , R2 = 4.3 kΩ, R3 = 29 kΩ. Find: The initial and final conditions for the circuit of P5.29. Analysis: In a steady-state condition we can treat the inductor as a short circuit. Before the switch changes, applying the KVL we obtain: VS 1 − VS 2 = (R1 + R2 )iL (0) iL (0 ) = VS1 − VS 2 = 0 mA. R1 + R2 After the switch has changed for a long time, we have again a steady-state condition, and we treat the inductor as a short circuit. Thus, applying the KVL we have: iL (∞ ) = − VS 2 = −0.39 mA. R2 + R3 ______________________________________________________________________________________ 5.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.15 Solution: Known quantities: I 0 = 17 mA, C = 0.55 µF , R1 = 7 k Ω, R2 = 3.3 kΩ. Find: The initial and final conditions for the circuit of P5.32. Analysis: Before the switch changes, the capacitor is not connected to the circuit, so we don’t have any information about its initial voltage. After the switch has changed, the current source and the capacitor will be in series so the current to the capacitor will be constant at I 0 . Therefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently, the voltage across the capacitor will rise at a constant rate, without reaching any equilibrium state. ______________________________________________________________________________________ Problem 5.16 Solution: Known quantities: VS 1 = 17 V , VS 2 = 11V , C = 70 nF , R1 = 14 k Ω , R2 = 13 kΩ , R3 = 14 kΩ . Find: The initial and final conditions for the circuit of P5.34. Analysis: In a steady-state condition we can treat the capacitor as an open circuit. Before the switch changes, applying the KVL we have that the voltage across the capacitor is equal to the source voltage VS 1 : vC (0) = VS 1 = 17 V. After the switch has changed for a long time, we have again a steady-state condition, and we treat the capacitor as an open circuit. Since the current flowing through the resistance R2 is equal to zero, the voltage across the capacitor is equal to the voltage across the resistance vC (∞ ) = R3 : R3 14000 VS 2 = 11 = 5.5 V. R1 + R3 14000 + 14000 ______________________________________________________________________________________ Problem 5.17 Solution: Known quantities: VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: The initial and final conditions for the circuit of P5.41. Analysis: The switch S1 is always open and the switch S 2 closes at t = 0 . Before closing, the switch S 2 has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitors as open circuits. When the switch is open, the current source is not connected to the circuit. Thus, 5.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 vC1 (0) = 0 V, vC 2 (0) = 0 V. After the switch S 2 has been closed for a long time, we have again a steady-state condition, and we treat the capacitors as open circuits. The voltages across the capacitors are both equal to the voltage across the resistance R3 : vC1 (∞ ) = vC 2 (∞ ) = (R3 || R4 )I S = 6⋅3 4 = 8 V. 6+3 ______________________________________________________________________________________ Problem 5.18 Solution: Known quantities: C = 1 µF , RS = 15 k Ω , R3 = 30 kΩ . Find: The initial and final conditions for the circuit of P5.47. Assume: Assume that VS = 9 V, R1 =10kΩ and R2 = 20kΩ . Analysis: Before opening, the switch has been closed for a long time. Thus, we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistances R1 , R2 , and R3 . Thus, vC (0) = R1 || (R2 || R3 ) 10k || 12k VS = 9 = 2.4 V. RS + R1 || (R2 || R3 ) 15k + 10k || 12k After opening, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R1 . Thus, vC (∞ ) = R1 10000 VS = 9 = 3.6 V. R1 + RS 10000 + 15000 _____________________________________________________________________________________ Problem 5.19 Solution: Known quantities: Values of the voltage source, of the inductance and of the resistors. Find: The initial and final conditions for the circuit of P5.49. Analysis: Before the switch changes, apply KCL at the top node (nodal analysis) to write the following circuit equation. v1 − 100 v1 v1 + + =0 1000 5 2 .5 v1 = 100 = 0.165 V. 601 5.11 G. Rizzoni, Principles and Applications of Electrical Engineering i L ( 0) = Problem solutions, Chapter 5 v1 40 = = 66 mA. 2.5 601 After the switch has changed, apply KCL at the top node (nodal analysis) to write the following circuit equation. v1 − 100 v1 v1 + + =0 10 5 2 .5 i L (∞ ) = v1 = 100 = 14.285 V. 7 v1 40 = = 5.714 A. 2 .5 7 ______________________________________________________________________________________ Problem 5.20 Solution: Known quantities: I S = 5 A, L1 = 1H , L2 = 5H , R = 10kΩ . Find: The initial and final conditions for the circuit of P5.52. Analysis: Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the inductors as short circuits. The values of the two resistors are equal so the current flowing through the inductors is: i L ( 0) = IS = 2.5 A. 2 After the switch has been closed for a long time, we have again a steady-state condition, and we treat the inductors as short circuits. In this case the resistors are short-circuited and so all the current is flowing through the inductors. i L (∞) = I S = 5 A. ______________________________________________________________________________________ 5.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Section 5.4: Transient Response of First-Order Circuits Focus on Methodology First-order transient response 1. 2. 3. 4. 5. Solve for the steady-state response of the circuit before the switch changes state (t = 0-), and after the transient has died out (t → ∞). We shall generally refer to these responses as x(0-) and x(∞). Identify the initial condition for the circuit, x(0+), using continuity of capacitor voltages and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), as illustrated in Section 5.4. Write the differential equation of the circuit for t = 0+, that is, immediately after the switch has changed position. The variable x(t) in the differential equation will be either a capacitor voltage, vC(t), or an inductor current, iL(t). It is helpful at this time to reduce the circuit to Thévenin or Norton equivalent form, with the energy storage element (capacitor or inductor) treated as the load for the Thévenin (Norton) equivalent circuit. Reduce this equation to standard form (Equation 5.8). Solve for the time constant of the circuit: τ = RTC for capacitive circuits, τ = L/RT for inductive circuits. Write the complete solution for the circuit in the form: x(t) x( ) x(0) x( ) e t/ Problem 5.21 Solution: Known quantities: Circuit shown in Figure P5.21, L = 0.9 mH ,Vs = 12 V , R1 = 6 k Ω, R2 = 6 kΩ, R3 = 3 kΩ. Find: If the steady-state conditions exist just before the switch was opened. Assumptions: iL = 1.70 mA before the switch is opened at t = 0 . Analysis: Determine the steady state current through the inductor at t < 0 . If this current is equal to the current specified, steady-state conditions did exit; otherwise, opening the switch interrupted a transient in progress. To determine the steady-state current before the switch was opened, replace the inductor with an equivalent DC short-circuit and compute the steady-state current through the short circuit. At steady state, the inductor is modeled as a short circuit: ( ) VR 3 0 − = 0 ( ) iR 3 0 − = 0 Thus, the short-circuit current through the inductor is simply the current through by applying Kirchoff’s Laws and Ohm’s Law. Apply KVL: ( ) − Vs + iR 3 0 − R2 = 0 Apply KCL: ( ) ( ) ( ) iR 2 0 − = ( ) Vs 12 = = 2 mH R2 6 × 10 3 ( ) ( ) − iR 2 0 − + i L 0 − + i R 3 0 − = 0, i L 0 − = i R 2 0 − = 2 mH 5.13 R2 which can be found G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 The actual steady state current through the inductor is larger than the current specified. Therefore, the circuit is not in a steady state condition just before the switch is opened. ______________________________________________________________________________________ Problem 5.22 Solution: Known quantities: Circuit shown in Figure P5.22, VS1 = 35V , VS 2 = 130 V , C = 11 µF , R1 = 17 k Ω, R2 = 7 kΩ, R3 = 23 kΩ. Find: At t = 0 + the initial current through R3 just after the switch is changed. Assumptions: None. Analysis: To solve this problem, find the steady state voltage across the capacitor before the switch is thrown. Since the voltage across a capacitor cannot change instantaneously, this voltage will also be the capacitor voltage immediately after the switch is thrown. At that instant, the capacitor may be viewed as a DC voltage source. − At t = 0 : Determine the voltage across the capacitor. At steady state, the capacitor is modeled as an open circuit: ( ) ( ) iR1 0 − = iR 2 0 − = 0 Apply KVL: ( ) VS1 + 0 − VC 0 − + 0 − VS 2 = 0 ( ) VC 0 − = VS1 − VS 2 = −95V At t = 0+ : VC 0 + = VC 0 − = −95V iR2 Apply KVL: ( ) ( ) (0 ) = i (0 ) + + R3 ( ) ( ) (0 ) = ( ) VS 2 − i R 3 0 + R2 + VC 0 + − i R 3 0 + R3 = 0 ( ) iR3 0 + = VS 2 + VC R2 + R3 + 130 − 95 = 1.167 mA 7 × 10 3 + 23 × 10 3 ______________________________________________________________________________________ Problem 5.23 Solution: Known quantities: Circuit shown in Figure P5.23, V1 = 12 V , C = 0.5 µF , R1 = 0.68 k Ω, R2 = 1.8 kΩ. Find: The current through the capacitor just before and just after the switch is closed. Assumptions: The circuit is in steady-state conditions for t < 0 . 5.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Analysis: − At t = 0 , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. These are the steady state conditions, i.e.: iC (0 − ) = 0 At VC (0 − ) = 0 t = 0 + , the switch is closed and the transient starts. Continuity requires: VC (0+ ) = VC (0 − ) = 0 At this instant, treat the capacitor as a DC voltage source of strength zero i.e. a short-circuit. Therefore, all of the voltage V1 is across the resistor R1 and the resulting current through R1 is the current into the capacitor. Apply KCL: (Sum of the currents out of the top node) ( ) ( ) VC 0 + − 0 VC 0 + − V1 + =0 iC 0 + R2 R1 ( ) + ( ) iC 0 + = V1 12 = = 17.65 mA R1 0.68 × 10 3 The CURRENT through the capacitor is NOT continuous but changes from 0 to 17.65 mA when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY. ______________________________________________________________________________________ Problem 5.24 Solution: Known quantities: Circuit shown in Figure P5.23, V1 = 12 V , C = 150 µF , R1 = 400 m Ω, R2 = 2.2 kΩ. Find: The current through the capacitor just before and just after the switch is closed. Assumptions: The circuit is in steady-state conditions for t < 0 . Analysis: − At t = 0 , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. These are the steady state conditions, i.e.: iC (0 − ) = 0 At VC (0 − ) = 0 t = 0 + , the switch is closed and the transient starts. Continuity requires: VC (0+ ) = VC (0 − ) = 0 At this instant, treat the capacitor as a DC voltage source of strength zero i.e. a short-circuit. Therefore, all of the voltage capacitor. V1 is across the resistor R1 and the resulting current through R1 is the current into the Apply KCL: (Sum of the currents out of the top node) 5.15 G. Rizzoni, Principles and Applications of Electrical Engineering ( ) Problem solutions, Chapter 5 ( ) VC 0 + − 0 VC 0 + − V1 + =0 iC 0 + R2 R1 ( ) + ( ) iC 0 + = V1 12 = = 30 A R1 400 × 10 −3 The CURRENT through the capacitor is NOT continuous but changes from 0 to 30 A when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY. ______________________________________________________________________________________ Problem 5.25 Solution: Known quantities: Circuit shown in Figure P5.21, Find: The voltage across Vs = 12 V , L = 0.9 mH , R1 = 6 k Ω, R2 = 6 kΩ, R3 = 3 kΩ. R3 just after the switch is open. Assumptions: iL = 1.70 mA before the switch is opened at t = 0 . Analysis: When the switch is opened the voltage source is disconnected from the circuit and plays no role. Since the current through the inductor cannot change instantaneously the current through the inductor at t = 0+ is also 1.70 mA. At this instant, treat the inductor as a DC current source and solve for the voltage across R3 by current division or KCL and Ohm’s Law. Specify the polarity of the voltage across iL (0 ) = iL (0 ) = 1.7 mA Req = R1 + R2 = 12 kΩ + − R3 . Apply KCL: (Sum of the currents out of the top node) VR 3 (0 + ) V (0 + ) + i L (0 + ) + R 3 =0 Req R3 VR 3 (0 + ) = − iL (0 + ) 1.7 ×10 −3 = −4.080V = 1 1 1 1 + + Req R3 12 ×10 3 3 ×10 3 _____________________________________________________________________________________ Problem 5.26 Solution: Known quantities: Circuit shown in Figure P5.26, V1 = 12 V , Rs = 0.7 Ω, R1 = 22 kΩ, L = 100 mH . Find: The voltage through the inductor just before and just after the switch is changed. Assumptions: The circuit is in steady-state conditions for t < 0 . 5.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Analysis: In steady-state the inductor acts like a short-circuit so it has no voltage across it for t < 0. However, its current is non-zero and is equal to the current out of the source VS and through RS. At the instant the switch is changed the current through the inductor is unchanged since the current through an inductor cannot change instantaneously. Also notice that after the switch is changed the current through R1 is always equal to the inductor current and the voltage across R1 is always equal to the inductor voltage. Thus, at t = 0+ the voltage across the inductor must be non-zero. That’s fine since the voltage across an inductor can change instantaneously (or relatively so.) Assume a polarity for the voltage across the inductor. t = 0 − : Steady state conditions exist. The inductor can be modeled as a short circuit with: VL (0 − ) = 0 Apply KVL; − V S + i L (0 − ) RS + V L (0 − ) = 0 i L (0 − ) = At VS 12 = = 17.14 A RS 0.7 t = 0 + , the transient commences. Continuity requires: iL (0+ ) = iL (0 − ) Apply KVL: iL (0+ )R1 + VL (0 + ) = 0 VL (0 + ) = − iL (0+ )R1 = −17.14 × 22 × 10 3 = −337.1kV ______________________________________________________________________________________ Problem 5.27 Solution: Known quantities: Circuit shown in Figure P5.27, V1 = 12 V , R1 = 0.68 k Ω, R2 = 2.2 kΩ, R3 = 1.8 kΩ, C = 0.47 µF . Find: The current through the capacitor at t = 0 + , just after the switch is closed. Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: For t < 0, the switch is open and no power source is connected to the left half of the circuit. In steady state, by definition, the voltage across the capacitor and the current out of it must be constant. However, without a power source to replenish the energy dissipated by the resistors, that constant must be zero. Otherwise, current would flow out of the capacitor, its voltage would drop as it lost charge, and the energy of that charge would be dissipated by the resistors. This process would continue until no net charge remained on the capacitor and its voltage was zero. At steady state, then, the voltage across the capacitor is zero. At t = 0+, the voltage across the capacitor is still zero since the voltage across a capacitor cannot change instantaneously. At that instant, the capacitor can be treated as a voltage source of strength zero (i.e. a short-circuit.) However, the current through the capacitor can change instantaneously (or relatively so) from 0 to a new value. In this problem it will change as the switch is closed because the voltage source V1 will drive current through R1and the parallel combination of R2 and R3. The current through R2 is the capacitor current. 5.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 VC (0+ ) = VC (0 − ) = 0 Apply KCL: ( ) ( ) ( ) VR 3 0 + − 0 VR 3 0 + VR 3 0 + − V1 + + =0 R2 R3 R1 ( ) VR 3 0 + V1 R1 V1 12 = = = = 7.114V 1 1 1 R1 R1 0.68 0.68 + + 1+ + + 1+ R1 R2 R3 2 .2 1 .8 R2 R3 Recall that the voltage across the capacitor (Volts = Joules/Coulomb) represents the energy stored in the electric field between the plates of the capacitor. The electric field is due to the amount of charge stored in the capacitor and it is not possible to instantaneously remove charge from the capacitor’s plates. Therefore, the voltage across the capacitor cannot change instantaneously when the circuit is switched. However, the rate at which charge is removed from the plates of the capacitor (i.e. the capacitor current) can change instantaneously (or relatively so) when the circuit is switched. Note also that these conditions hold only at the instant t = 0+. For t > 0+, the capacitor is gaining charge, all voltages and currents exponentially approach their final or steady state values. Apply KVL: − VC (0 + ) + i C (0 + )R2 + VR 3 (0 + ) = 0 iC (0 + ) = VR 3 (0 + ) − VC (0 + ) 7.114 = = 3.234 mA R2 2.2 × 10 3 ______________________________________________________________________________________ Problem 5.28 Solution: Known quantities: NOTE: Typo in problem statement: should read “At t < 0, …” Circuit shown in Figure P5.22, VS 1 = 35 V , VS 2 = 130 V , C = 11 µF , R1 = 17 k Ω, R2 = 7 KΩ, R3 = 23 kΩ. Find: The time constant of the circuit for t > 0. Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single capacitor system is ReqC where Req is the Thévenin equivalent resistance as seen by the capacitor, i.e., with respect to the port or terminals of the capacitor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the capacitor to the other through the network?” Then: 5.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Req = R2 + R3 = 7 × 10 3 + 23 × 10 3 = 30 kΩ τ = ReqC = 30 × 10 3 × 11 × 10 −6 = 330.0 ms ______________________________________________________________________________________ Problem 5.29 Solution: Known quantities: Circuit shown in Figure P5.29, VS 1 = 13 V , VS 2 = 13 V , L = 170 mH , R1 = 2.7 k Ω, R2 = 4.3 kΩ, R3 = 29 kΩ. Find: The time constant of the circuit for t > 0. Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single inductor system is L/Req where Req is the Thévenin equivalent resistance as seen by the inductor, i.e., with respect to the port or terminals of the inductor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the inductor to the other through the network?” Then: Req = R2 + R3 = 4.3 × 10 3 + 29 × 10 3 = 33.30 kΩ τ= L 170 × 10 −3 = = 5.105 µs Req 33.30 × 10 3 ______________________________________________________________________________________ Problem 5.30 Solution: Known quantities: Circuit shown in Figure P5.27, V1 = 12 V , C = 0.47 µF , R1 = 680 Ω, R2 = 2.2 kΩ, R3 = 1.8 kΩ. Find: The time constant of the circuit for t > 0. Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single capacitor system is ReqC where Req is the Thévenin equivalent resistance as seen by the capacitor, i.e., with respect to the port or terminals of the capacitor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the capacitor to the other through the network?” Then: Req = R2 + R3 R1 1.8 × 10 3 × 0.68 × 10 3 = 2.2 × 10 3 + = 2.694 kΩ R3 + R1 1.8 × 10 3 + 0.68 × 10 3 τ = ReqC = 2.694 × 10 3 × 0.47 × 10 −6 = 1.266 ms ______________________________________________________________________________________ 5.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.31 Solution: Known quantities: Circuit shown in Figure P5.21, VS = 12 V , L = 0.9 mH , R1 = 6 k Ω, R2 = 6 kΩ, R3 = 3 kΩ. Find: The time constant of the circuit for t > 0. Assumptions: The current through the inductor is iL = 1.70 mA before the switch is opened at t = 0. Analysis: For t > 0, the transient is in progress. The time constant is a relative measure of the rate at which the voltages and currents are changing in the transient phase. The time constant for a single inductor system is L/Req where Req is the Thévenin equivalent resistance as seen by the inductor, i.e., with respect to the port or terminals of the inductor. To calculate Req turn off (set to zero) the ideal independent voltage source and ask the question “What is the net equivalent resistance encountered in going from one terminal of the inductor to the other through the network?” Then: Req = τ= R3 ( R1 + R2 ) 3 × 10 3 (6 × 10 3 + 6 × 10 3 ) = = 2.400 kΩ R3 + ( R1 + R2 ) 3 × 10 3 + (6 × 10 3 + 6 × 10 3 ) L 0.9 × 10 −3 = = 0.3750 µs Req 2.4 × 10 3 ______________________________________________________________________________________ Problem 5.32 Solution: Known quantities: Circuit shown in Figure P5.32, Find: The voltage Vc (0 − ) = −7 V , I 0 = 17 mA, C = 0.55 µF , R1 = 7 k Ω, R2 = 3.3 kΩ. Vc (t ) across the capacitor for t > 0. Assumptions: Before the switch is thrown the voltage across the capacitor is –7 V. Analysis: The current source and the capacitor will be in series so the current to the capacitor will be constant at I0. Therefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently, the voltage across the capacitor will rise at a constant rate. The integral form of the capacitor i-V relationship best expresses this accumulation process. The continuity of the voltage across the capacitor requires: 5.20 G. Rizzoni, Principles and Applications of Electrical Engineering ( ) Problem solutions, Chapter 5 ( ) VC 0+ = VC 0 − = −7 V iC (t ) = I 0 = 17 mA t 1 t 1 0 i t dt i t dt ( ) = ( ( ) + C C ³0 I 0 dt ) C ³−∞ C ³−∞ I t I = VC 0+ + 0 ³ dt = VC 0+ + 0 t |t0 C 0 C −3 17 × 10 = −7 + t = −7 + 30.91 × 10 3 t −6 0.55 × 10 VC (t ) = ( ) ( ) ______________________________________________________________________________________ Problem 5.33 Solution: Known quantities: Circuit shown in Figure P5.29, VS1 = 23 V ,VS 2 = 20V , L = 23 mH , R1 = 0.7 Ω, R2 = 13 kΩ, R3 = 330 kΩ. Find: The current iR 3 (t ) through resistor R3 for t > 0. Assumptions: The circuit is in steady-state conditions for t < inductor) so an assumed solution is of the form 0 . It is a resistive circuit with one storage element (e.g. i R 3 (t ) = I SS + ( I 0 − I SS )e −t / τ Analysis: The approach here is to first find the initial condition at t = 0+ for the inductor and use it to determine the initial condition on the current through resistor R3. Second, find the final steady-state condition of the circuit for t > 0. To do so, simply apply DC circuit analysis to solve for the current through resistor R3 i.e. replace the inductor with a short-circuit. Finally, solve for the time constant of the circuit for t > 0. Each of these three results is needed to construct the complete transient solution. − At t = 0 : Assume steady state conditions exist. At steady state, the inductor is modeled as a short-circuit: Apply KVL: − VS 2 + iL (0 − ) R1 + iL (0 − ) R2 + VS 1 = 0 i L (0 − ) = VS 2 − VS 1 20 − 23 = = −0.22 A R1 + R2 0.7 + 13 This current is flowing in the direction from the inductor to the switch. + Find I0 at t = 0 : Continuity of the current through the inductor requires that: i L (0 + ) = i L (0 − ) = −0.22 A I 0 = i R 3 (0 + ) = i L (0 + ) = −0.22 A Find ISS at t = infinity: Assume that enough time has elapsed for steady state conditions to return. In steady state the inductor is modeled as a short circuit; therefore, the voltage across the inductor is zero. The result is a simple series 5.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 connection of resistors R2 and R3. The current across R3 is found directly from Ohm’s Law in this case. I SS = i R3 = VS 2 20 V = ≅ 58 mA R2 + R3 343 Ω Find τ for t > 0 : To find the time constant t one first needs to determine the Thevenin equivalent resistance RTH across the terminals of the inductor. To do so, set all independent ideal sources to zero and determine the equivalent resistance "seen" by the inductor, i.e. with respect to the port or terminals of the inductor: Req = R2 + R3 = 13 + 330 × 10 3 = 330.0 kΩ τ= L 23 × 10 −3 = = 69.70 ns Req 330.0 × 10 3 The complete response can now be written using the solution for transient voltages and currents in firstorder circuits: i R 3 (t ) = i R 3 (∞) + (i R 3 (0 + ) − i R 3 (∞))e = 0.058 + (−0.22 − 0.058)e = 0.058 − 0.28 e − t 69.70×10 − 9 − t −t τ 69.70×10 − 9 A ______________________________________________________________________________________ Problem 5.34 Solution: Known quantities: Circuit shown in Figure P5.34, VS1 = 17 V , VS 2 = 11V , R1 = 14 k Ω, R2 = 13 kΩ, R3 = 14 kΩ, C = 70 nF . Find: a) V (t ) for t > 0. b) The time for V (t ) to change by 98% of its total change in voltage after the switch is operated. Assumptions: The circuit is in steady-state conditions for t < 0 . It is a resistive circuit with one storage element (e.g. capacitor) so an assumed solution is of the form VR 3 (t ) = VSS + (V0 − VSS )e −t / τ Analysis: In general, the approach in problems such as this one is to first find the initial condition at t = 0+ for the capacitor and use it to determine the initial condition on the voltage across resistor R3. Second, find the final steady-state condition of the circuit for t > 0. To do so, simply apply DC circuit analysis to solve for the voltage across the resistor R3 (i.e. replace the capacitor with an open-circuit and solve.) Finally, solve for the time constant of the circuit for t > 0 by finding the Thevenin equivalent resistance RTH across the terminals of the capacitor. Each of these three results is needed to construct the complete transient solution. a) At t = 0− : Steady state conditions are specified. At steady state, the capacitor is modeled as an open circuit: iC ( 0 − ) = 0 Apply KVL: 5.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 VS1 = iC (0 − ) R2 + VC (0 − ) VC (0 − ) = VS1 = 17V At t = 0+ : The voltage across the capacitor remains the same: VC (0 + ) = VC (0 − ) = 17 V Apply KCL: V (0+ ) − (VS 2 ) V (0+ ) − VC (0 + ) V (0+ ) − 0 + + =0 R1 R2 R3 VC (0+ ) VS 2 + R2 R1 + V (0 ) = 1 1 1 + + R1 R2 R3 17 11 + 3 13 × 10 14 × 103 = = 9.525V 1 1 1 + + 14 × 103 13 × 103 14 × 103 At t > 0: Determine the equivalent resistance as "seen" by the capacitor, ie, with respect to the port or terminals of the capacitor. Suppress the independent ideal voltage source: Req = R2 + (R1 R3 ) = R2 + R1 R3 14 × 10 3 × 14 × 10 3 = 13 × 10 3 + R1 + R3 14 × 10 3 + 14 × 10 3 = 20.00 kΩ τ = Req C = 20 × 10 3 × 70 × 10 −9 = 1.400 ms At t = infinity: Steady state is again established. At steady state the capacitor is once again modeled as an open circuit: iC (∞) = 0 Since the current through the capacitor branch is zero in steady state the voltage across R3 may be found quickly by voltage division V (∞) = = VS 2 R3 R3 + R1 11× 14 × 103 = 5.500V 14 × 103 + 14 × 103 The complete response for t > 0 is then: 5.23 G. Rizzoni, Principles and Applications of Electrical Engineering V (t ) = V (∞) + (V (0 + ) − V (∞))e = 5.5 + (9.525 − (5.5))e = 5.5 + 4.025e −t 1.4×10−3 −t −t Problem solutions, Chapter 5 τ 1.4×10−3 V b) The time required for V(t) to reach 98% of its final value is found from 0.98 = V (t ) − V (0 + ) V (∞) − V (0 + ) or ∆V = V (∞) − V (0 + ) = 5.5 − (9.525) = −4.025V V (t1 ) = V (0 + ) + 0.98∆V = 9.525 + 0.98 × (−4.025) − t1 −3 = 5.5805 = 5.5 + 4.025e 1.4×10 5.5805 − 5.5 t1 = −1.4 × 10 −3 ln( ) = 5.477 ms 4.025 ______________________________________________________________________________________ Problem 5.35 Solution: Known quantities: Circuit shown in Figure P5.35, Find: The value of VG = 12 V , RG = 0.37 Ω, R = 1.7 kΩ. L and R1 . Assumptions: The voltage across the spark plug gap VR just after the switch is changed is 23 kV and the voltage will change exponentially with a time constant τ = 13 ms . Analysis: − At t = 0 : Assume steady state conditions exist. At steady state the inductor is modeled as a short circuit: VL (0 − ) = 0 The current through the inductor at this point is given directly by Ohm’s Law: i L (0 − ) = At VG RG + R1 t = 0+ : Continuity of the current through the inductor requires that: 5.24 G. Rizzoni, Principles and Applications of Electrical Engineering i L (0 + ) = i L (0 − ) = VG RG + R1 V R ( 0 + ) = −i L ( 0 + ) R = − R1 = − Problem solutions, Chapter 5 VG R RG + R1 VG R − RG V R (0 + ) =− 12 × 1.7 × 10 3 − 0.37 = 0.5170 Ω − 23 × 10 3 Note that the voltage across the gap VR was written as –23 kV since the current from the inductor flows opposite to the polarity shown for VR; that is, the actual polarity of the voltage across R is opposite that shown. At t > 0 : Determine the Thevenin equivalent resistance as "seen" by the inductor, ie, with respect to the port or terminals of the inductor: Req = R1 + R τ= L L = Req R1 + R L = τ ( R1 + R) = 13 × 10 −3 × (0.5170 + 1.7 × 10 3 ) = 22.11 H ______________________________________________________________________________________ Problem 5.36 Solution: Known quantities: Circuit shown in Figure P5.36, when iL ≥ +2 mA , the relay functions. VS = 12 V , L = 10.9 mH , R1 = 3.1 kΩ. Find: R2 so that the relay functions at t = 2.3 s . Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: In this problem the current through the inductor is clearly zero before the switch is thrown. The task is determine the value of the resistance R2 such that the current through the inductor will need 2.3 seconds to rise to 2 mA. Once again, we must find the complete transient solution, this time for the current through the inductor. Assume a solution of the form i L (t ) = i∞ + (i0 − i∞ )e −t / τ − At t = 0 : The current through the inductor is zero since no source is connected. i L (0 − ) = 0 At t = 0+ : 5.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 i 0 = i L (0 + ) = i L ( 0 − ) = 0 At t > 0: Determine the Thevenin equivalent resistance as "seen" by the inductor, i.e., with respect to the port or terminals of the inductor: RTH = (R1 R2 ) = R1 R2 R1 + R2 And so τ= At L(R1 + R2 ) L = RTH R1 R2 t = infinity: Steady state is again established. At steady state the inductor is again modeled as a short circuit. Thus, the current through R2 is zero and the current through the inductor is given by i ∞ = i L (∞ ) = VS 12 = = 3.87 mA R1 3.1 × 10 3 Plug in the above quantities to the complete solution and set the current through the inductor equal to 2 mA and the time equal to 2.3 s. 2 × 10 −3 = 3.87 × 10 −3 (1 − e −2.3 / τ ) or 2 º − 2.3 ª = ln «1 − τ ¬ 3.87 »¼ or τ ≅ 3.16 seconds Solving for R2: 3.16 = L( R1 + R2 ) R1 R2 or R2 = LR1 ≅ 3.4 mΩ 3.16 R1 − L ______________________________________________________________________________________ Problem 5.37 Solution: Known quantities: Circuit shown in Figure P5.37, V1 = 12 V , R1 = 400 m Ω, R2 = 2.2 kΩ, C = 150 µF . Find: The current through the capacitor just before and just after the switch is closed. Assumptions: The circuit is in steady-state conditions for t < 0 . 5.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Analysis: − At t = 0 , assume steady state conditions exist. If charge is stored on the plates of the capacitor, then there will be energy stored in the electric field of the capacitor and a voltage across the capacitor. This will cause a current flow through R2 which will dissipate energy until no energy is stored in the capacitor at which time the current ceases and the voltage across the capacitor is zero. Thus, in this circuit, and others like it that contain no connected sources prior to the switch being thrown, the steady state condition is zero currents and zero voltages. iC (0 − ) = 0 At VC (0 − ) = 0 t = 0 + , the switch is closed; the transient starts. Continuity requires: VC (0+ ) = VC (0 − ) = 0 Since the voltage across the capacitor is zero at this instant, the voltage across the resistor R2 is also zero at this instant which implies that no current passes through the resistor at t = 0+. Therefore, at t = 0+, the current out of the source must be the current into the capacitor. ( ) iC 0 + = V1 12 = = 30 A R1 0.4 The CURRENT through the capacitor is NOT continuous but changes from 0 to 30 A when the switch is closed. The voltage across the capacitor is continuous because the stored energy CANNOT CHANGE INSTANTANEOUSLY. ______________________________________________________________________________________ Problem 5.38 Solution: Known quantities: Circuit shown in Figure P5.38, VS = 12 V , RS = 0.24 Ω, R1 = 33 kΩ, L = 100 mH . Find: The voltage across the inductor before and just after the switch is changed. Assumptions: The circuit is in steady-state conditions for t < 0 . Analysis: The solution to this problem can be visualized by noting that before the switch is thrown the circuit is presumed to be in a DC steady state. Therefore, before the switch is thrown the inductor may be modeled as a short circuit such that the voltage across it is zero. Moreover, the current through the inductor before the switch is thrown is simply VS/RS. Immediately after the switch is thrown the current through the inductor must be this same value since the current through an inductor cannot change instanteneously. However, after the switch is thrown this current must also be the current drawn through R1 since that resistor and the inductor are then in series. Finally, the voltage across the inductor after the switch is thrown is always equal to the voltage across R1, which is simply the product of iL and R1. Thus, the voltage across the inductor immediately after the switch is thrown must be iL(t=0+) times R1. In quantitative terms, at t = 0-, v L (0 − ) = 0 V 12 i L (0 − ) = S = = 50 A RS 0.24 At t = 0+, 5.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 i L (0 + ) = i L (0 − ) = 50 A and v L (0 + ) = i L (0 + ) R1 = (50 A)(33 kΩ) = 1.65 MV This high side of this voltage is located at the ground terminal due to the direction of current flow through R1. ANSWER: 0V , − 1.65 MV ______________________________________________________________________________________ Problem 5.39 Solution: Known quantities: Circuit shown in Figure P5.27, V1 = 12 V , C = 150 µF , R1 = 4 M Ω, R2 = 80 MΩ , R3 = 6 MΩ . Find: The time constant of the circuit for t > 0. Assumptions: The circuit is in steady-state conditions for t < 0. Analysis: + At t = 0 , just after the switch is closed, a transient starts. Since this is a first order circuit (a single independent capacitance), the transient will be exponential with some time constant. That time constant is the product of the capacitance of the capacitor and the Thevenin equivalent resistance seen across the terminals of the capacitor. To find the Thevenin equivalent resistance, suppress the independent, ideal voltage source which is equivalent to replacing it with a short circuit. Then with respect to the terminals of the capacitor: RTH = R2 + ( R1 R3 ) = 80 + (4 6) = 82.4 MΩ The time constant is simply τ = RTH C = (82.4 MΩ)(150 µF ) = 12360 sec = 206 minutes = 3.43 hours + Notice that this value is quite independent of the magnitude of the voltage source. ANSWER: 2.36 ks = 206.0 min = 3.433 hr ______________________________________________________________________________________ Problem 5.40 Solution: Known quantities: Circuit shown in Figure P5.21, VS = 12 V , L = 100 mH , R1 = 400 Ω, R2 = 400 Ω , R3 = 600 Ω . Find: The time constant of the circuit for t > 0. Assumptions: iL = 1.70 mA just before the switch is opened at t = 0 . Analysis: + At t = 0 , just after the switch is opened, a transient starts. Since this is a first order circuit (a single independent capacitance), the transient will be exponential with some time constant. That time constant is 5.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 the product of the inductance of the inductor and the Thevenin equivalent resistance seen across the terminals of the inductor. In this problem, the Thevenin equivalent resistance is particularly easy to find since there are no sources connected. With respect to the terminals of the inductor: RTH = ( R1 + R2 ) R3 = 800 600 ≅ 343 Ω The time constant is simply τ= L 0 .1 = ≅ 292 ns RTH 343 ANSWER: 291.7 ns ______________________________________________________________________________________ Problem 5.41 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: a) The capacitor voltage VC (t ) at t b) The time constant τ for t ≥ 0 . = 0+ . VC (t ) and sketch the function. d) Find VC (t ) for each of the following values of t : 0 ,τ , 2τ , 5τ ,10τ . c) The expression for Assumptions: Switch S1 is always open and switch S2 closes at t = 0 . Analysis: a) Without any power sources connected the steady state voltages are zero due to relentless dissipation of energy in the resistors. VC (0 − ) = VC (0 + ) = 0 V When the initial condition on a transient is zero, the general solution for the transient simplifies to ( VC (t ) = V (∞) 1 − e −t / τ ) b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the current source (i.e. replacing it with an open circuit) and computing R2 + R3||R4. RTH = 4 + (3 6) = 6 Ω τ = RTH C = (6)(8) = 48 s c) The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by current division VC (∞) = 6Ω (4 A)(3 Ω) = 8V 3Ω + 6 Ω 5.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Plug in to the generalized solution given above to find ( VC (t ) = 8 1 − e − t / 48 ) VC (t ) = 0 d) , t≥0 , t≤0 VC (0) = 0V ; VC (τ ) = 5.06V ; VC ( 2τ ) = 6.9 V ; VC (5τ ) = 7.95V ; VC (10τ ) = 8.0V ______________________________________________________________________________________ Problem 5.42 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: a) The capacitor voltage VC (t ) at t b) The time constant τ for t ≥ 0 . = 0+ . VC (t ) and sketch the function. d) Find VC (t ) for each of the following values of t : 0 ,τ , 2τ , 5τ ,10τ . c) The expression for Assumptions: Switch S1 has been open for a long time and closes at long time and opens at t = 0. t = 0 ; conversely, switch S2 has been closed for a Analysis: a) The capacitor voltage immediately after the switch S1 is closed and the switch S 2 is opened is equal to that when the switches were the first opened and the second closed respectively. Since the second switch was closed for a long time one can assume that the capacitor voltage had reached a long-term steady state value. This value is found by replacing both capacitors with DC open circuits and solving for the voltage across R3. This voltage is found readily by current division VC (0 + ) = VC (0 − ) = 6Ω (4 A)(3 Ω) = 8V 3Ω + 6 Ω b) The Thevenin equivalent resistance seen by the parallel capacitors is R1 || (R2 + R3 ) . 1 1 −1 70 ) × ( 4 + 4) = τ = RTH C = ( + s 5 4+3 3 c) The generalized solution for the transient is VC (t ) = V (∞) + [V (0 + ) − V (∞)]e − t / τ The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by voltage division. Thus, VC (∞) = R2 + R3 4 Ω + 3Ω VS = (20V ) = 35 V 5Ω + 4 Ω + 3 Ω R1 + R2 + R3 3 5.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Plug in to the generalized solution given above to find [ ] 35 § 35 · 35 11 −3t / 70 − e + ¨ 8 − ¸e −3t / 70 = 3 © 3¹ 3 3 , t≤0 VC (t ) = V (∞) + V (0 + ) − V (∞) e −t / τ = VC (t ) = 8 d) VC (0) = 8 V ; VC (τ ) = 10.318 V ; VC (2τ ) = 11.170 V ; VC (5τ ) = 11.642 V ; , t≥0 VC (10τ ) = 11.666 V ______________________________________________________________________________________ Problem 5.43 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: a) The capacitor voltage VC (t ) at b) The expression for t = 0+ . VC (t ) and sketch the function. Assumptions: Switch S 2 is always open; switch S1 has been closed for a long time, and opens at t = 0 . At t = t1 = 3τ , switch S1 closes again. Analysis: The approach here is to find the transient solution in the interval 0 < t < 3 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S1 closes again. a) S1 has been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by voltage division VC (0 + ) = VC (0 − ) = 7 × 20 = 11.67 V 12 b) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S1 closes at t = 3 seconds. To do so it is first necessary to find the complete transient solution for when the switch is open (i.e. as if the switch never closed again.) The long-term steady state capacitor voltage when the switch is held open is zero. The time constant is simply RTH CEQ = 56 seconds. Thus, the complete transient solution for the first 3 seconds is VC (t ) = V (0 + )e − t / 56 = 11.67e − t / 56 , 0≤t ≤3 At t = 3 seconds, the capacitor voltage is VC (t = 3 − ) = 11.67e −3 / 56 = 11.06 At t=3 seconds the switch S1 closes again. Continuity of voltage across the capacitors still holds so VC (t = 3 + ) = VC (t = 3 − ) = 11.06 5.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 With the switch closed the long-term steady state capacitor voltage is the same as that found in part a. VC (∞) = 7 × 20 = 11.67 V 12 The new time constant is found after suppressing the independent voltage source (i.e. replacing it with a short circuit) and finding the new Thevenin equivalent resistance seen by the capacitors. RTH = (4 + 3) 5 = 2.92 Ω and τ = R TH C = (2.92)(4 + 4) = 23.3 seconds Finally, the transient solution for t > 3 is VC (t ) = 11.67 + [11.06 − 11.67 ]e − t / 23.3 = 11.67 − 0.61e − (t −3) / 23.3 , t ≥3 Notice the use of the shifted time scale (t-3) in the exponent. ______________________________________________________________________________________ Problem 5.44 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: a) The capacitor voltage VC (t ) at t b) The time constant τ for t ≥ 0 . = 0+ . VC (t ) and sketch the function. d) Find VC (t ) for each of the following values of t : 0 ,τ , 2τ , 5τ ,10τ . c) The expression for Assumptions: Both switches S1 and S2 close at t = 0 . Analysis: a) Without any power sources connected the steady state voltages are zero due to the complete dissipation of all circuit energy by the resistors. VC (0 − ) = VC (0 + ) = 0 V When the initial condition on a transient is zero, the general solution for the transient simplifies to VC (t ) = V (∞)(1 − e − t / τ ) b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current source with an open circuit and the voltage source with a short circuit) and computing R1||(R2 + R3||R4). [ ] RTH = 5 (4 + (3 6)) = [5 6] = 30 ≅ 2.73 Ω 11 5.32 G. Rizzoni, Principles and Applications of Electrical Engineering τ = RTH C = c) Problem solutions, Chapter 5 240 30 8= ≅ 21.8 s 11 11 At this point only the long-term steady state capacitor voltage is needed to write down the complete transient solution. In DC steady state the capacitors can be modeled as open circuits. Furthermore, R3||R4 can be replaced with an equivalent resistance. This resistance is in parallel with the independent current source and the two can be replaced with a Thevenin source transformation of an appropriate voltage source in series with the same resistance. Once this replacement is made it is a simple matter of voltage division to determine the capacitor voltage. R3 R4 = 3 6 = 2 Ω The source transformation results in an 8V voltage source in series with this resistance. Then, by voltage division, VC (∞) = 8 + 160 4+2 (20 − 8) = ≅ 14.55 V 11 4+2+5 Now plug in to the general form of the transient solution to find [ VC (t ) ≅ 14.5 1 − e −11t / 240 d) ] , t≥0 VC (0) = 0V ; VC (τ ) = 9.17 V ; VC ( 2τ ) = 12.5V ; VC (5τ ) = 14.4 V ; VC (10τ ) = 14.5V ______________________________________________________________________________________ Problem 5.45 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: + a) The capacitor voltage VC (t ) at t = 0 . b) The time constant τ for 0 ≤ t ≤ 48s . VC (t ) valid for 0 ≤ t ≤ 48s . d) The time constant τ for t > 48s . e) The expression for VC (t ) valid for t > 48s . f) Plot VC (t ) for all time. c) The expression for Assumptions: Switch S1 opens at t = 0 ; switch S2 opens at t = 48s . Analysis: The approach here is to find the transient solution in the interval 0 < t < 48 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S2 opens. 5.33 G. Rizzoni, Principles and Applications of Electrical Engineering a) Problem solutions, Chapter 5 S1 and S 2 have been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by node analysis and voltage division VS − VC (0− ) VC (0− ) R2 VS (R2 + R3 + R4 ) − = VC (0 ) = (R1 + R2 + R3 + R4 ) R1 R2 (R2 + R3 + R4 ) 20(4 + 3 + 6) 260 VC (0+ ) = VC (0− ) = V ≅ 14.4V = (5 + 4 + 3 + 6) 18 b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current source with an open circuit) and computing (R2 + R3||R4). RTH = R2 + (R3 || R4 ) = 4 + (3 || 6) = 4 + 2 = 6Ω τ = RTH (C1 + C 2 ) = 6 ⋅ 8 = 48 s c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S2 opens at t = 48 s. To do so it is first necessary to find the complete transient solution for when only the switch S1 is open (i.e. as if the switch S2 never opens.). The generalized solution for the transient is VC (t ) = V (∞) + [V (0 + ) − V (∞)]e − t / τ The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R3. This voltage is found readily by current division. Thus, VC (∞) = 6Ω (4 A)(3 Ω) = 8V 3Ω + 6 Ω Plug in to the generalized solution given above to find [ ] VC (t ) = V (∞) + V (0 + ) − V (∞) e − t / τ = 8 + (14.4 − 8)e − t / 48 = 8 + 6.4e − t / 48 , 0 ≤ t ≤ 48 At t = 48 s, the capacitor voltage is VC (t = 48− ) = 8 + 6.4e −1 = 10.35V Continuity of voltage across the capacitors still holds so VC (48+ ) = VC (48− ) = 10.35V d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by the 8 F capacitance is found by computing (R2 + R3). RTH = R2 + R3 = 4 + 3 = 7Ω τ = RTH (C1 + C 2 ) = 7 ⋅ 8 = 56 s e) The generalized solution for the transient is + VC (t ) = V (∞) + [V (t 0 ) − V (∞)]e − (t −t0 ) / τ The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus, VC (∞) = 0V Plug in to the generalized solution given above to find VC (t ) = V (48+ )e −t / τ = 10.35e − (t −48) / 56 f) The plot of , t > 48 VC (t ) for all time is shown in the following figure. 5.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Capacitor Voltage VC(t) [V] 15 10 5 0 0 20 40 60 80 100 Time [sec] 120 140 160 180 200 ______________________________________________________________________________________ Problem 5.46 Solution: Known quantities: Circuit shown in Figure P5.41, VS = 20 V , R1 = 5 Ω, R2 = 4 Ω , R3 = 3 Ω, R4 = 6 Ω, C 1 = 4 F ,C2 = 4 F , I S = 4 A. Find: + a) The capacitor voltage VC (t ) at t = 0 . b) The time constant τ for 0 ≤ t ≤ 96s . VC (t ) valid for 0 ≤ t ≤ 96s . d) The time constant τ for t > 96 s . e) The expression for VC (t ) valid for t > 96 s . f) Plot VC (t ) for all time. c) The expression for Assumptions: Switch S1 opens at t = 96 s ; switch S2 opens at t = 0 . 5.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Analysis: The approach here is to find the transient solution in the interval 0 < t < 96 seconds and use that solution to determine the initial condition (the capacitor voltage) for the new transient after the switch S1 opens. a) S1 and S 2 have been closed for a long time and in DC steady state the capacitors can be replaced with open circuits. Thus, by node analysis and voltage division VS − VC (0− ) VC (0− ) R2 VS (R2 + R3 + R4 ) − = VC (0 ) = (R1 + R2 + R3 + R4 ) R1 R2 (R2 + R3 + R4 ) 20(4 + 3 + 6) 260 VC (0+ ) = VC (0− ) = V ≅ 14.4V = (5 + 4 + 3 + 6) 18 b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current source with an open circuit) and computing RTH = R1 (R2 + R3 ) = 5 (4 + 3) = 2.92Ω R1 (R2 + R3 ) . τ = RTH (C1 + C2 ) = 2.92 ⋅ 8 = 23.3 s c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage when switch S1 opens at t = 96 s. To do so it is first necessary to find the complete transient solution for when only the switch S2 is open (i.e. as if the switch S1 never opens.). The generalized solution for the transient is VC (t ) = V (∞) + [V (0 + ) − V (∞)]e −t / τ The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across division. Thus, VC (∞) = R2 and R3 . This voltage is found readily by voltage 4 Ω + 3Ω (20V ) = 11.67 V 5 Ω + 4 Ω + 3Ω Plug in to the generalized solution given above to find [ ] VC (t ) = V (∞) + V (0+ ) − V (∞) e −t / τ = 11.67 + (14.4 − 11.67 )e −t / 23.3 VC (t ) = 11.67 + 2.73e −t / 23.3 , 0 ≤ t ≤ 96 At t = 96 s, the capacitor voltage is VC (t = 96 − ) = 11.67 + 2.73e −96 / 23.3 = 11.71V Continuity of voltage across the capacitors still holds so VC (96 + ) = VC (96 − ) = 11.71V d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by the 8 F capacitance is found by computing (R2 + R3). RTH = R2 + R3 = 4 + 3 = 7Ω τ = RTH (C1 + C 2 ) = 7 ⋅ 8 = 56 s e) The generalized solution for the transient is + VC (t ) = V (∞) + [V (t 0 ) − V (∞)]e − (t −t0 ) / τ The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus, 5.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 VC (∞) = 0V Plug in to the generalized solution given above to find VC (t ) = V (96 + )e −t / τ = 11.71e − (t −96) / 56 f) The plot of , t > 96 VC (t ) for all time is shown in the following figure. Capacitor Voltage VC(t) [V] 15 10 5 0 0 50 100 150 200 250 Time [sec] ______________________________________________________________________________________ Problem 5.47 Solution: Known quantities: RS = 15 k Ω , τ = 1.5 ms, Find: The value of resistors τ ' = 10 ms, R3 = 30 kΩ , C = 1 µF . R1 and R2 . Assumptions: None. Analysis: Before the switch opens: Req = RS // R1 // R2 // R3 τ = Req C = 1.5 ms After the switch opens: 5.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Req' = RS // R1 τ ' = Req' C = 10 ms Solving the system of equations we have, R1 = RS τ ' 15000 ⋅ 0.01 = = 30kΩ RS C − τ ' 15000 ⋅10 −6 − 0.01 −1 −1 §C 1 § 10 −6 1 1 · 1 1 1 · ¸¸ = 1875Ω R2 = ¨¨ − − − ¸¸ = ¨¨ − − − © 0.0015 15000 30000 30000 ¹ © τ RS R1 R3 ¹ ______________________________________________________________________________________ Problem 5.48 Solution: Known quantities: Circuit shown in Figure P5.47, VS = 100V , RS = 4 k Ω, R1 = 2 k Ω, R2 = R3 = 6 kΩ, C = 1 µF . Find: The value of the voltage across the capacitor after t = 2.666 ms. Assumptions: None. Analysis: Before opening, the switch has been closed for a long time. Thus we have a steady-state condition, and we treat the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R1. Thus, VC (t 0− ) = VC (t 0+ ) = RS || R1 || R2 || R3 300 VS = V ≅ 23.077V RS 13 After the switch opens, the time constant of the circuit is Req = RS // R1 = τ = Req C = 4000 Ω 3 1 ms ≅ 1.3ms 750 the generalized solution for the transient is + VC (t ) = V (∞) + [V (t 0 ) − V (∞)]e − (t −t0 ) / τ The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and solving for the voltage across R1. This voltage is found readily by voltage division. Thus, VC (∞) = R1 2000 Ω 20 VS = 20V = V ≅ 6.67V RS + R1 4000 Ω + 2000 Ω 3 Plug in to the generalized solution given above to find + VC (t ) = V (∞) + [V (t 0 ) − V (∞)]e −( t −t0 ) / τ = 20 § 300 20 · −750 (t −t0 ) 20 640 −750( t −t0 ) +¨ − ¸e = + e 3 © 13 3 ¹ 3 39 Finally, 5.38 G. Rizzoni, Principles and Applications of Electrical Engineering VC (t 0 + 2.666ms) = Problem solutions, Chapter 5 20 640 −750 ( 0.002666 ) + e = 8.888V 3 39 ______________________________________________________________________________________ Problem 5.49 Solution: Known quantities: As described in Figure P5.49. Find: The time at which the current through the inductor is equal to 5 A, and the expression for iL (t ) for t ≥ 0 . Assumptions: None. Analysis: At t < 0 : Using the current divider rule: iL (0 − ) = ( 100 5 )( ) = 66.5 mA 1000 + 5 // 2.5 5 + 2.5 At t > 0 : Using the current divider rule: iL (∞ ) = ( 100 5 )( ) = 5.71 A 10 + 5 // 2.5 5 + 2.5 To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor: Req = 10 // 5 + 2.5 = 5.83 Ω τ= L 0.1 = = 17.1 ms Req 5.83 Finally, we can write the solution: iL (t ) = iL (∞ ) − ( iL (∞ ) − iL (0 ))e = 5.71 − (5.71 − 0.0665)e = 5.71 − 5.64e − t 17.1×10 − 3 −t − t τ 17.1×10 − 3 A Solving the equation we have, −3 iL (tˆ ) = 5.71 − 5.64e 17.1×10 = 5 tˆ = 35.437 ms The plot of iL (t ) for all time is shown in the following figure. − tˆ 5.39 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 ______________________________________________________________________________________ Problem 5.50 Solution: Known quantities: As described in Figure P5.49. Find: The expression for iL (t ) for 0 ≤ t ≤ 5ms . The maximum voltage between the contacts during the 5-ms duration of the switch. Assumptions: The mechanical switching action requires 5 ms. Analysis: a) At t < 0 : Using the current divider rule: iL (0 − ) = ( 100 5 )( ) = 66.5 mA 1000 + 5 // 2.5 5 + 2.5 For 0 ≤ t ≤ 5ms : The long term steady state inductor current after the switch has been opened is zero since no independent source is connected to the circuit and all the initial energy in the circuit is eventually dissipated by the resistors. Thus, iL (∞ ) = 0 A To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor: Req = 5 + 2.5 = 7.5 Ω τ= L 0.1 = = 13.33 ms Req 7.5 Finally, we can write the solution: iL (t ) = iL (0 )e −t τ = (0.0665)e − t 13.33×10 − 3 A (0 ≤ t ≤ 5ms ) b) The voltage between the contacts during the 5-ms duration of the switching is equal to: 5.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Vcont (t ) = VS − V5Ω = 100 − (VL + 2.5iL ) where, VL (t ) = L Therefore, −t di L (t ) (0.1)(0.0665) − t 13.33×10−3 13.33×10− 3 =− e = ( − 0 . 5 ) e V −3 dt 13.33 × 10 Vcont (t ) = 100 − [(2.5)(0.0665) − 0.5]e − t 13.33×10 − 3 = 100 + (0.33)e − t 13.33×10 − 3 V Thus, the maximum voltage between the contacts during the 5-ms duration of the switch is: MAX Vcont = Vcont (t = 0) = 100.33V ______________________________________________________________________________________ Problem 5.51 Solution: Known quantities: As described in Figure P5.51. The switch closes when the voltage across the capacitor voltage reaches vMC ; The switch opens when the voltage across the capacitor voltage reaches vMO = 1V . The period of the capacitor voltage waveform is 200 ms. Find: The voltage vMC . Assumptions: The initial capacitor voltage is 1V and the switch has just opened. Analysis: With the switch open: VC (∞ ) = 10V τ = RC = 0.15 s VC (t ) = VC (∞ ) − [VC (∞ ) − VC (0 )]e = 10 − (10 − 1)e = 10 − 9e −t 0.15 −t −t τ 0.15 V Now we must determine the time when VC (t ) = v M . Using the expression for the capacitor voltage: C v = 10 − 9e C M −t 0 0.15 e − t0 0.15 10 − v MC = 9 § 10 − v MC t 0 = −0.15 ln¨¨ © 9 · ¸¸ ¹ With the switch closed, the capacitor sees the Thevenin equivalent defined by: 10 × 10 ≈ 1× 10 − 2 V 10 + 10000 Req = 10kΩ 10Ω ≅10 Ω Veq = VC (∞) = (voltage division) τ = Req C = 0.15 ms The initial value of this part of the transient is for the capacitor voltage: vMC at t = t0. With these values we can write the expression 5.41 G. Rizzoni, Principles and Applications of Electrical Engineering VC (t ) = VC (∞ ) − [VC (∞ ) − VC (t 0 )]e ( ) = 0.01 + v MC − 0.01 e − ( t −t 0 ) − ( t −t 0 ) 0.15×10 −3 Problem solutions, Chapter 5 τ V The end of one full cycle of the waveform across the 10Ω resistor occurs when the second transient reaches vMO = 1V . If we call the time at which this event occurs t1, then: VC (t ) = 1V at t = t1 = 200ms and so ( ) 1 = 0.01 + v MC − 0.01 e − ( t1 −t 0 ) 0.15×10 −3 Graphically, the solution is the intersection between the following function: vMC = 10 − 9e −t 0 (blue line) (1 − 0.01) vMC = 0.01 + e which correspond to 0.15 −( 0.2 −t0 ) 0.15×10 (red line) −3 vMC = 7.627 V and t 0 ≅ 0.1997 ms. ______________________________________________________________________________________ Problem 5.52 Solution: Known quantities: As describes in Figure P5.52. At Find: a) iL (t ) for t = 0 , the switch closes. t ≥ 0. 5.42 G. Rizzoni, Principles and Applications of Electrical Engineering b) Problem solutions, Chapter 5 VL1 (t ) for t ≥ 0 . Assumptions: iL (0) = 0 A . Analysis: a) In the long-term DC steady state after the switch is closed the inductors may be modeled as short circuits and so all of the current from the source will travel through the inductors. iL (∞ ) = 5 A With the current source suppressed (treated as an open circuit) the Thevenin equivalent resistance seen by the inductors in series and the associated time constant are Req = 10 kΩ Leq = L1 + L2 = 6 H τ= Leq Req = 0.6 ms iL (t ) = iL (∞) ª1 − e «¬ −t τ º »¼ − t −3 º ª = 5«1 − e 0.6×10 » A ¬ ¼ b) The voltage across either of the inductors is derived directly from the differential relationship between current and voltage for an inductor. di L (t ) dt − t d −3 = (1)(5) (1 − e 0.6×10 ) dt − t 1 −3 · § = 5¨ e 0.6×10 ¸ −3 © 0.6 × 10 ¹ VL1 (t ) = L1 = 8.333e − t 0.6×10 − 3 kV ______________________________________________________________________________________ Problem 5.53 Solution: Known quantities: As describes in Figure P5.52. At t = 0 , the switch closes. Find: The voltage across the 10-kΩ resistor in parallel with the switch for t ≥ 0. Assumptions: None. Analysis: When the switch closes at t = 0, the 10-kΩ resistor is in parallel with a short circuit, so its voltage is equal to zero for all time (t ≥0). ______________________________________________________________________________________ 5.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Section 5.5: Transient Response of Second-Order Circuits Focus on Methodology – roots of second order systems Case 1: Real and distinct roots. This case occurs when ζ>1, since the term under the square root is positive in this case, and the roots are: overdamped response. 2 s1,2 n n 1. This leads to an Case 2: Real and repeated roots. This case holds when ζ=1, since the term under the square root is zero in this case, and s1,2 n n . This leads to a critically damped response. Case 3: Complex conjugate roots. This case holds when ζ<1, since the term under the square root is negative in this case, and response. s1,2 j n n 1 2 . This leads to an underdamped Focus on Methodology Second-order transient response 1. Solve for the steady-state response of the circuit before the switch changes state (t = 0-), and after the transient has died out (t → ∞). We shall generally refer to these responses as x(0-) and x(∞). 2. Identify the initial conditions for the circuit, x (0 ), and x (0 ) using continuity of capacitor voltages and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), and circuit analysis. This will be illustrated by examples. Write the differential equation of the circuit for t = 0+, that is, immediately after the switch has changed position. The variable x(t) in the differential equation will be either a capacitor voltage, vC(t), or an inductor current, iL(t). Reduce this equation to standard form (Equation 5.9, or 5.48). Solve for the parameters of the second-order circuit: ωn and ζ . Write the complete solution for the circuit in one of the three forms given below, as appropriate: 3. 4. 5. + + 1t n Overdamped case (ζ > 1): 2 x(t) x N (t) xF (t) n 1e n 2 2e 1t n x( ) t 0 Critically damped case (ζ = 1): x(t) x N (t) xF (t) n 1e t 2 te n t j n x( ) t 0 Underdamped case (ζ = 1): x(t) 6. x N (t) xF (t) 1e n j n 1 2 t 2e n Apply the initial conditions to solve for the constants α1 and α2. 5.44 1 2 t x( ) t 0 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.54 Solution: Known quantities: Circuit shown in Figure P5.54, VS 1 = 15 V , VS 2 = 9 V , RS1 = 130 Ω, RS 2 = 290 Ω, R1 = 1.1 kΩ, R2 = 700 Ω, L = 17 mH , C = 0.35 µF . Find: The voltage across the capacitor and the current through the inductor and RS 2 as t approaches infinity. Assumptions: The circuit is in DC steady-state conditions for t < 0 . Analysis: The conditions that exist at t < 0 have no effect on the long-term DC steady state conditions at t → ∞ . In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 9V source travels through the inductor in this case. i L (∞) = VS 2 9 = = 31.03 mA RS 2 290 Of course, this current is also the current traveling through the 290 Ω resistor. iRS 2 (∞) = i L (∞) = 31.03 mA And since the voltage across the inductor in the long-term DC steady state is zero (short circuit) 0 + VC (∞) + i R2 (∞)R2 = 0 VC (∞) = 0 __________________________________________________________________________ Problem 5.55 Solution: Known quantities: Circuit shown in Figure P5.54, VS 1 = 12 V , VS 2 = 12 V , RS1 = 50 Ω, RS 2 = 50 Ω, R1 = 2.2 kΩ, R2 = 600 Ω, L = 7.8 mH , C = 68 µF . Find: The voltage across the capacitor and the current through the inductor as Assumptions: The circuit is in DC steady-state conditions for t approaches infinity. t < 0. Analysis: The conditions that exist at t < 0 have no effect on the long-term DC steady state conditions at t → ∞ . In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 12V source travels through the inductor in this case. i L (∞ ) = VS 2 12 = = 240 mA RS 2 50 Of course, this current is also the current traveling through the 290 Ω resistor. i RS 2 (∞) = i L (∞) = 240 mA And since the voltage across the inductor in the long-term DC steady state is zero (short circuit) 5.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 0 + VC (∞) + i R2 (∞)R2 = 0 VC (∞) = 0 ______________________________________________________________________________________ Problem 5.56 Solution: Known quantities: Circuit shown in Figure P5.56, VS = 170 V , RS = 7 kΩ, R1 = 2.3 k Ω, R2 = 7 KΩ, L = 30 mH , C = 130 µF . Find: The current through the inductor and the voltage across the capacitor and R1 at steady state. Assumptions: None. Analysis: As t → ∞ , the circuit will return to DC steady state conditions. In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the current through R2 is zero and thus the voltage across the capacitor must be equal to the voltage across R1. Furthermore, the current through the inductor and R1 is simply VS/(RS + R1). iC ( ∞ ) = 0 i L (∞) = i R1 (∞) = VS 170 = ≅ 18.3 mA RS + R1 7000 + 2300 VR1 (∞) = i L (∞) R1 = 18.28 × 10 −3 × 2.3 × 10 3 = 42.04V and VC (∞) = VR1 (∞) = 42.04V ______________________________________________________________________________________ Problem 5.57 Solution: Known quantities: Circuit shown in Figure P5.57, VS = 12 V , C = 130 µF , R1 = 2.3 k Ω, R2 = 7 KΩ, L = 30 mH . Find: The current through the inductor and the voltage across the capacitor and R1 at steady state. Assumptions: None. Analysis: As t → ∞ , the circuit will return to DC steady state conditions (practically after about 5 time constants.) In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the voltage across the capacitor must be equal to the voltage across R2. Furthermore, the current through the inductor and R2 is simply VS/(R1 + R2). iC (∞) = 0 V L (∞ ) = 0 i L (∞ ) = i R 2 (∞ ) = VS 12 = = 1.29 mA R1 + R2 9.3 × 10 3 VR1 (∞) = iS (∞) R1 = 1.29 × 10 −3 × 2.3 × 10 3 = 2.968V 5.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 And by observation VC (∞) = VR2 (∞) = (1.29 × 10 −3 )(7 × 10 3 ) = 9.03V All answers are positive indicating that the directions of the currents and polarities of the voltages assumed initially are correct. (You did do that, didn't you?) ______________________________________________________________________________________ Problem 5.58 Solution: Known quantities: Circuit shown in Figure P5.58, VS = 12 V , C = 0.5 µF , R1 = 31 k Ω, R2 = 22 KΩ, L = 0.9 mH . Find: The current through the inductor and the voltage across the capacitor at steady state. Assumptions: None. Analysis: As t → ∞ , the circuit will return to DC steady state conditions (practically after about 5 time constants). In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the voltage across the capacitor must be equal to the voltage across R2. Furthermore, the current through the inductor and R2 is simply VS/(R1 + R2). iC (∞) = 0 V L (∞ ) = 0 i L (∞ ) = i R 2 (∞ ) = VS 12 = ≅ 226 µA R1 + R2 (31 + 22) ×10 3 VR2 (∞) = iR2 (∞) R2 = (226 × 10 −3 )(22 × 10 3 ) ≅ 4.98 V And by observation VC (∞) = VR2 (∞) = 4.98 V Theoretically, when the switch is OPENED, the current through the inductor must continue to flow, at least momentarily. However, the inductor is in series with an OPEN switch through which current CANNOT flow. What the theory does not predict is that a very large voltage is developed across the gap and this causes an arc with a current (kind of like a teeny, weeny lightning bolt). The energy stored in the magnetic field of the inductor is rapidly dissipated in the arc. The same effect will be important later when discussing transistors as switches. ______________________________________________________________________________________ Problem 5.59 Solution: Known quantities: Circuit shown in Figure P5.59, 5.47 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 VS = 12 V , C = 3300 µF , R1 = 9.1 kΩ, R2 = 4.3 kΩ, R1 = 4.3 kΩ, L = 16 mH , . Find: The initial voltage across R2 just after the switch is changed. Assumptions: At t < 0 the circuit is at steady state and the voltage across the capacitor is + 7V . Analysis: It is important to remember that only the values of the capacitor voltage and the inductor current are guaranteed continuity from immediately before the switch is thrown to immediately afterward. Therefore, to determine the initial voltage across R2 it is necessary to first determine the initial voltage across the capacitor and the initial current through the inductor. Assume that before the switch was thrown DC steady state conditions existed. In DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. The initial voltage across the capacitor is given as +7V. The initial current through the inductor is equal to the current through R3, which is given by Ohm’s Law. i L (0 + ) = i L (0 − ) = and VS 12 = = 2.791 mA R3 4.3 × 10 3 VC (0 + ) = VC (0 − ) = 7 V Apply KCL: VR 2 (0 + ) − VC (0 + ) VR 2 (0 + ) + + i L (0 + ) = 0 R1 R2 VC (0 + ) − i L (0 + ) (V (0 + ) − i L (0 + ) R1 ) R2 R1 V R 2 (0 + ) = = C 1 1 R2 + R1 + R1 R2 = (7 − 2.791 × 10 −3 × 9.1 × 10 3 ) × 4.3 × 10 3 = −5.93V 3 4.3 × 10 + 9.1× 10 3 One could also solve for VR2 by superposition. V R 2 (0 + ) = R2 RR (7V ) − 1 2 (2.8mA) = −5.93 V R1 + R2 R1 + R2 ______________________________________________________________________________________ Problem 5.60 Solution: Known quantities: Circuit shown in Figure P5.60, 5.48 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 VS 1 = 15 V , VS 2 = 9 V , RS1 = 130 Ω, RS 2 = 290 Ω, R1 = 1.1 kΩ, R2 = 700 Ω, L = 17 mH , C = 0.35 µF . Find: The current through and the voltage across the inductor and the capacitor and the current through RS 2 at + t=0 . Assumptions: The circuit is in DC steady-state conditions for t < 0. Analysis: Since this was not done in the specifications above, you must note on the circuit the assumed polarities of voltages and directions of currents. − At t = 0 : Assume that steady state conditions exist. At steady state the inductor is modeled as a short circuit and the capacitor as an open circuit. Choose a ground. Note that because the inductor is modeled as a short circuit, there is no voltage drop from the top node to the bottom node and so before the switch there is no current through R1. ( ) V L 0− = 0 ( ) iC 0 − = 0 Apply KCL: 0 − VS 1 0 − VS 2 0 + i L (0 − ) + + 0 + =0 RS 1 R1 RS 2 i L (0 − ) = = Apply KVL: VS 1 VS 2 + = RS 1 RS 2 15 9 + = 146.4 mA 130 290 ( ) ( ) (0 )= 0 VC 0− + iC 0 − R2 = 0 VC At − t = 0+ : iL (0 + ) = iL (0 − ) = 146.4 mA VC (0 + ) = VC (0 − ) = 0 Apply KVL: − VL (0 + ) + 0 + iC (0 + )R2 = 0 V L (0 + ) iC (0 ) = R2 + Apply KCL: 5.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 V L (0 + ) V L ( 0 + ) V L (0 + ) − VS 2 + + =0 i L (0 ) + R1 R2 RS 2 + VS 2 − i L (0 + ) R V − i (0 + ) RS 2 = S2 L V L (0 + ) = S 2 1 1 1 RS 2 RS 2 + + + +1 R1 R2 RS 2 R1 R2 9 − 146.4 × 10 −3 × 0.29 × 10 3 = −19.94V 0.29 0.29 + +1 1.1 0.7 V (0 + ) − 19.94 iC (0 + ) = L = = −28.49 mA R2 0.7 × 10 3 = Apply KVL again: − VL (0 + ) + iRS 2 (0 + )RS 2 + VS 2 = 0 iRS 2 (0 + ) = VL (0 + ) − VS 2 − 19.94 − 9 = −99.79 mA = RS 2 0.29 × 10 3 ______________________________________________________________________________________ Problem 5.61 Solution: Known quantities: Circuit shown in Figure P5.60, VS 1 = 12 V , VS 2 = 12 V , RS1 = 50 Ω, RS 2 = 50 Ω, R1 = 2.2 kΩ, R2 = 600 Ω, L = 7.8 mH , C = 68 µF . Find: The voltage across the capacitor and the current through the inductor as Assumptions: The circuit is in DC steady-state conditions for t approaches infinity. t < 0. Analysis: The conditions that exist at t < 0 have no effect on steady state conditions as t → ∞ . In the long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches and the current through them are zero. In other words all of the current produced by the 12V source travels through the inductor in this case. Apply KCL; i L (∞) + i R1 (∞) + i R 2 (∞) + 0 − VS 2 =0 RS 2 i R1 (∞) = i R 2 (∞) = 0 V 12 i L (∞ ) = S 2 = = 240 mA RS 2 50 Apply KVL: 5.50 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 0 + VC (∞) + i R2 (∞)R2 = 0 VC (∞) = 0 ______________________________________________________________________________________ Problem 5.62 Solution: Known quantities: As described in Figure P5.62. Find: An expression for the inductor current for t ≥0. Assumptions: The switch has been closed for a long time. It is suddenly opened at Analysis: For 0 ≤ t ≤ 5 : Define clockwise mesh currents, equations are: t = 0 and then reclosed at t = 5 s t = 0 in the lower loop, and t = 0 in the upper loop. Then the mesh (5 + 5s ) I 1 − 3I 2 = 0 1 − 3I 1 + ( 3 + ) I 2 = 0 4s from which we determine that 1 )−9 = 0 4s s = 0.242 ± j 0.158 (5 + 5s )(3 + Therefore, the inductor current is of the form: i ( t ) = e − 0 .242 t [ A cos( 0 . 158 t ) + B sin( 0 . 159 t )] From the initial conditions: 6 =2= A 3 di di L |t =0 = 5 |t =0 = VC (0 + ) = −10 dt dt di |t =0 = −2 = −0.242 A + 0.158B dt i (0) = Solving the above equations: A=2 i (t ) = e B = −9.59 − 0.242 t [2 cos(0.158t ) − 9.59 sin(0.158t )] A The solution for capacitor voltage will have the same form. VC (t ) = e −0.242t [ A cos(0.158t ) + B sin(0.158t )]V From the initial conditions: VC (0) = 6 = A dVC 1 |t =0 = iC (0) = 0 − 0.242 A + 0.158 B = 0 dt C Solving the above equations: 5.51 for 0 ≤ t ≤ 5 s G. Rizzoni, Principles and Applications of Electrical Engineering A=6 Problem solutions, Chapter 5 B = 9.18 VC (t ) = e −0.242t [6 cos(0.158t ) + 9.18 sin(0.158t )]V for 0 ≤ t ≤ 5 s From the above results: VC (5) = 3.2807 V i (5) = −1.641 A These are the initial conditions for the solution after the switch recloses. For t ≥ 5 : The mesh equations are: 6 + 5i (5) s V (5) 1 − 3I 1 + ( 3 + ) I 2 = − C s 4s ( 3 + 5s ) I 1 − 3 I 2 = from which we determine that 60s 2 + 5s + 3 = 0 s = 0.041 ± j 0.220 Therefore, the inductor current is of the form: i (t ) = 2 + e −0.041t { A cos[0.220(t − 5)] + B sin[0.220(t − 5)] From the initial conditions: 2 + A = −1.641 A = −3.641 di V (5) 6 − 238 |t =5 = L = = 0.543 dt 5 5 − 0.41A + 0.220 B = 0.543 Solving the above equations: A = −3.641 B = 1.77 i (t ) = 2 + e −0.041t {−3.641 cos[0.220(t − 5) + 1.77 sin[0.220(t − 5)]} A for t ≥ 5 s This, together with the previous result, gives the complete solution to the problem. ______________________________________________________________________________________ Problem 5.63 Solution: Known quantities: As described in Figure P5.63. Find: Determine if the circuit is underdamped or overdamped. The capacitor value that results in critical dumping. Assumptions: The circuit initially stores no energy. The switch is closed at t = 0 . Analysis: a) For t ≥ 0 : The characteristic polynomial is: Ls 2 + Rs + 1 =0 C The damping ratio is: 5.52 G. Rizzoni, Principles and Applications of Electrical Engineering ξ= RC 2 Problem solutions, Chapter 5 1 400 ⋅10 −8 = 1010 = 0.2 < 1 LC 2 The system is underdamped, in fact we have the following complex conjugate roots: s1, 2 = −2 ×10 4 ± j (9.79 ×10 4 ) b) The capacitor value that results in critical damping is: 1 100 = 1 200C =1 LC C 100 1 200 2 C 2 =1 C = F = 0.25µF 4 × 10 6 C ξ= RC 2 ______________________________________________________________________________________ Problem 5.64 Solution: Known quantities: As described in Figure P5.63. Find: a) The capacitor voltage as t approaches infinity b) The capacitor voltage after 20 µs c) The maximum capacitor voltage. Assumptions: The circuit initially stores no energy. The switch is closed at t = 0. Analysis: For t ≥ 0 : The characteristic polynomial is: 0.01s 2 + 400s + 108 = 0 s = −2 ×10 4 ± j (9.79 ×10 4 ) Therefore, the solution is of the form: 4 VC (t ) = 10 + e −2×10 t [ A cos(9.79 × 10 4 t ) + B sin(9.79 × 10 4 t )] From the initial conditions: 10 + A = 0 − 2 × 10 4 A + 9.79 × 10 4 B = 0 Solving the above equations: A = −10 B = −2.04 4 VC (t ) = 10 + e −2×10 t [−10 cos(9.79 × 10 4 t ) − 2.04 sin(9.79 × 10 4 t )]V a) The capacitor voltage as t approaches infinity is: VC (∞) = 10V b) The capacitor voltage after 20 µs is: VC (20µs ) = 11.26V c) Graphically, the maximum capacitor voltage is: VCmax ≅ 15V 5.53 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 ______________________________________________________________________________________ Problem 5.65 Solution: Known quantities: As described in Figure P5.65. Find: An expression for the capacitor voltage for t ≥ 0. Assumptions: The circuit initially stores no energy, the switch S1 is open and the switch S 2 is closed. The switch S1 is closed at t = 0 and the switch is opened S 2 at t = 5 s . Analysis: This circuit has the same configuration during the interval 0 ≤ t ≤ 5 s as the one for Problem 5.39 did for t > 5 s . Therefore, the roots of the characteristic polynomial will be the same as those determined in that problem. They are: s = −0.041 ± j 0.220 Ad the general form of the capacitor voltage is VC (t ) = 6 + e −0.041t [ A cos(0.220t ) + B sin(0.220t )] For 0 ≤ t ≤ 5 s : The initial conditions are: VC (0) = 0 6 + A = 0 dVC 1 |t =0 = iC (0) = 0 dt C − 0.41A + 0.220 B = 0 Solving the above equations: 5.54 G. Rizzoni, Principles and Applications of Electrical Engineering A = −6 Problem solutions, Chapter 5 B = −0.904 VC (t ) = 6 + e −0.041t [−6 cos(0.220t ) − 0.904 sin(0.220t )]V Note that VC (5) = 3.127 V . for 0 ≤ t ≤ 5 s For t > 5 s : We have a simple RC decay: VC (t ) = 3.127e − t −5 12 V ______________________________________________________________________________________ Problem 5.66 Solution: Known quantities: Circuit shown in Figure P5.66, C = 1.6 nF ; After the switch is closed at t = 0 , the capacitor voltage reaches an initial peak value of 70V when t = 5π µs , a second peak value of 53.2 V when 3 t = 5π µs , and eventually approaches a steady-state of 50 V . Find: The values of R and L . Assumptions: The circuit is underdamped and the circuit initially stores no energy. Analysis: Using the given characteristics of the circuit step response and the assumption that the circuit is underdamped, the damping ratio and natural frequency can be determined as follows: ζ = 1 (π / ln(a / A))2 + 1 where a is the overshoot distance and A is the steady-state value ωn = 2π T 1−ζ 2 where T is the period of oscillation In our case, a = 70 − 50 = 20 and A = 50 ζ = 1 = 0.28 (π / ln(20 / 50) )2 + 1 The period of the waveform is 5π · 10π § −6 T = ¨ 5π − × 10 −6 ¸ × 10 = 3 ¹ 3 © 2π ωn = = 6.25 × 105 10π −6 2 × 10 1 − 0.28 3 Implying the characteristic polynomial for the circuit is s 2 + 2ζωn s + ωn 2 5.55 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Compare this with the standard form of the characteristic polynomial for a series RLC circuit: R 1 s+ L LC Matching terms yields L = 1.6 µH , R = 0.56 Ω . s2 + ______________________________________________________________________________________ Problem 5.67 Solution: Known quantities: Same as P5.66, but the first two peaks occur at 5π µs and 15π µs Find: Explain how to modify the circuit to meet the requirements. Assumptions: The capacitor value C cannot be changed. Analysis: Assuming we wish to retain the same peak amplitudes, we proceed as follows: The new period is T = 15π × 10 −6 − 5π × 10 −6 = 10π × 10 −6 Using the given characteristics of the circuit step response and the assumption that the circuit is underdamped, the damping ratio and natural frequency can be determined as follows: ζ = 1 (π / ln(a / A))2 + 1 where a is the overshoot distance and A is the steady-state value ωn = 2π T 1−ζ 2 where T is the period of oscillation In our case, a = 70 − 50 = 20 and A = 50 1 = 0.28 (π / ln(20 / 50) )2 + 1 2π ωn = = 2.17 × 105 −6 2 10π × 10 1 − 0.28 ζ = Implying the characteristic polynomial for the circuit is s 2 + 2ζωn s + ωn 2 Compare this with the standard form of the characteristic polynomial for a series RLC circuit: R 1 s+ L LC Matching terms yields L = 13.3 µH , R = 1.61Ω . s2 + 5.56 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Note that the frequency for this problem is one-third that of Problem 5.66, the inductance is 3 2 times that of Problem 5.66, and the resistance is 3 times that of Problem 5.66. ______________________________________________________________________________________ Problem 5.68 Solution: Known quantities: Circuit shown in Figure P5.68, Find: i (t ) for i (0) = 0 A,V (0) = 10V . t > 0. Assumptions: None. Analysis: The initial condition for the capacitor voltage is V (0 − ) = 10V . Applying KCL, i R + iC + i = 0 where iR = V , 1Ω i C = 0 .5 dV dt Therefore, i + V + 0 .5 dV =0 dt where V =2 di + 4i dt Thus, d 2i di + 4 + 5i = 0 2 dt dt Solving the differential equation: i (t ) = k1e ( −2+ j )t + k 2 e ( −2− j ) t t>0 i (0) = k1 + k 2 = 0 V (0) = 2 di (0) + 4i (0) = −(4 − j 2)k1 − (4 + j 2)k 2 = 10 dt Solving for k1 and k2 and substituting, we have 5 5 i (t ) = − j e ( −2+ j ) t + j e ( −2− j ) t A 2 2 for t > 0 ______________________________________________________________________________________ 5.57 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 Problem 5.69 Solution: Known quantities: As described in Figure P5.69. Find: The maximum value of V . Assumptions: The circuit is in steady state at t = 0− . Analysis: V ( 0 − ) = V (0 + ) = 0 Applying KVL: d 2V dV +4 + 4V = 48 2 dt dt Solving the differential equation: V = k1e −2t + k 2 te −2t + 12 From the initial condition: V (0) = 0 k1 = −12 k dV (0) 6 + 2 = 3 k 2 = −12 4 dt − 2t −2 t V (t ) = −12e − 12te + 12V for t > 0 The maximum value of V is: Vmax = V (∞) = 12V i L ( 0) = C _____________________________________________________________________________________ Problem 5.70 Solution: Known quantities: As described in Figure P5.70. Find: The value of t such that i = 2.5 A . Assumptions: The circuit is in steady state at t = 0− . Analysis: In steady state, the inductors behave as short circuits. Using mesh analysis, we can find the initial conditions. i ( 0 − ) = i (0 + ) = 5 A V (0 − ) = 0V After the switch is closed, the circuit is modified. Applying nodal analysis: V (0 + ) = 0V 5.58 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 d 2i di + 7 + 6i = 0 2 dt dt Solving the differential equation: i (t ) = k1e − t + k 2 e −6t for t > 0 i (0) = k1 + k 2 = 5 V (0) = − k1 − 6k 2 = 0 Solving for the unknown constants, using the initial conditions, we have: i (t ) = 6e −t − e −6t A for t > 0 Therefore, i (t ) = 6e −t − e −6t = 2.5 t1 = 873ms ______________________________________________________________________________________ Problem 5.71 Solution: Known quantities: As described in Figure P5.71. Find: The value of t such that i = 6 A . Assumptions: The circuit is in steady state at t = 0− . Analysis: In steady state, the inductors behave as short circuits. Using mesh analysis, we can find the initial conditions. i (0− ) = i (0+ ) = 12.5 A V (0 − ) = 0V After the switch is closed, the circuit is modified. Applying nodal analysis: V (0+ ) = −15V d 2i di + 7 + 6i = 0 2 dt dt Solving the differential equation: i (t ) = k1e − t + k 2e −6t for t > 0 i (0) = k1 + k2 = 12.5 V (0) = −k1 − 6k 2 = −15 Solving for the unknown constants, using the initial conditions, we have: i (t ) = 12e −t + 0.5e −6t A for t > 0 Therefore, i (t ) = 12e −t + 0.5e −6t = 6 t1 = 694ms ______________________________________________________________________________________ 5.59 G. Rizzoni, Principles and Applications of Electrical Engineering Problem 5.72 Solution: Known quantities: As described in Figure P5.72. Find: The value of t such that V = 7.5V . Assumptions: The circuit is in steady state at t = 0− . Analysis: − The circuit at t = 0 has the capacitors replaced by open circuits. By current division: 3 × 20 = 7.5 A 3+5 V (0 − ) = V (0 + ) = 2i2 = 15V i2Ω = i (0 − ) = 0 A After the switch opens, apply KCL: i (0 + ) = 0 A i + i1 + i2 = 0 dV 1 dV , = dt 6 dt 1 dV V i1 = − − 6 dt 2 i=C i2 = V 2 Applying KVL: 1 dV V − ) + V1 6 dt 2 1 dV V − ) V1 = V − 3(− 6 dt 2 1 dV1 1 d 2V 5 dV = + i1 = 6 dt 12 dt 2 12 dt V = 3i1 + V1 = 3(− Applying KCL: 1 d 2V 5 dV V 1 dV + + + =0 12 dt 2 12 dt 2 6 dt d 2V dV 2 +7 + 6V = 0 dt dt Solving the differential equation, V (t ) = k1e −t + k 2 e −6t for t > 0 V (0) = k1 + k 2 = 15 i (0) = 1 (− k1 − 6k 2 ) = 0 6 Solving for the unknown constants, using the initial conditions, we have: 5.60 Problem solutions, Chapter 5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 V (t ) = 18e −t − 3e −6t V for t > 0 Therefore, V (t ) = 18e −t − 3e −6t = 7.5 t1 = 873ms ______________________________________________________________________________________ Problem 5.73 Solution: Known quantities: As described in Figure P5.73. Find: The maximum value of V and the maximum voltage between the contacts of the switches. Assumptions: The circuit is in steady state at t = 0 − . L = 3 H. Analysis: At t = 0− : 10 = 2A 5 V ( 0 − ) = V (0 + ) = 0 V i ( 0 − ) = i (0 + ) = After the switch is closed: Applying KVL: t 1 1 dV V + =0 Vdt + ³ 12 dt 3 L −∞ d 2V dV 12 +4 + V =0 2 dt L dt The particular response is zero for t > 0 because the circuit is source-free. L = 3 H s 2 + 4s + 4 = 0 s1 = s 2 = −2 V (t ) = e −2t ( A + Bt ) for t > 0 From the initial condition: V (0 + ) = 0 = e 0 ( A + B(0)) A = 0 Substitute the solution into the original KCL equation and evaluate at 1 1 d − 2t ( 0) + [e ( A + Bt )] |t =0 +2 = 0 12 dt 3 1 0 + [ B e −2t − 2 Bte −2t )] |t =0 +2 = 0 B = −24 12 V (t ) = −24te −2t V for t > 0 5.61 t = 0+ : G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 The maximum absolute value of V is: Vmax = V (t = 0.5s ) = − 12 V ≅ 4.414V e The maximum voltage between the contacts of the switches is: MAX Vswitch = VS = 10V since the voltage between the contacts of the switches is a constant. ______________________________________________________________________________________ Problem 5.74 Solution: Known quantities: As described in Figure P5.74. Find: V at t > 0 . Assumptions: The circuit is in steady state at t = 0− . Analysis: At t = 0− : V (0 − ) = 12V iL ( 0 − ) = iL ( 0 + ) = 6 A iC (0 − ) = 0 A VC (0 − ) = VC (0 + ) = 4V VL (0 − ) = 0V For t > 0 : iC (0 + ) = −2 A VL (0 + ) = 4V V ( 0 + ) = 8V 5.62 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 5 −2 dV + dV i 0 = − C 0+ = − C = − =8 1 dt dt C 4 ( ) ( ) Using KVL and KCL, we can find the differential equation for the voltage in the resistor: diL =0 dt i = iL + iC iL = i − iC 12 − 2i − 0.8 dVC 1 d (12 − V ) = − 1 dV = dt 4 dt 4 dt d di d § 1 dV · 12 − 2i − 0.8 (i − ic ) = 0 12 − 2i − 0.8 + 0.8 ¨ − ¸=0 dt dt dt © 4 dt ¹ dV d 2V i = V / 2 12 − V − 0.4 − 0.2 2 = 0 dt dt 2 d V2 Ω dV 0.2 + 0.4 2Ω + V2Ω = 12 2 dt dt 2 d V2Ω dV + 2 2Ω + 5V2Ω = 60 2 dt dt iC = C Solving the differential equation: Homogeneous Solution: V2Ω ,h = K1e (−1+2 j )t + K 2e(−1−2 j )t t > 0 dV (0) = 8 dt K1 + K 2 = 8 V (0) = 8V , (− 1 + 2 j )K1 − (1 + 2 j )K 2 = 8 K1 = 4 − 4 j K2 = 4 + 4 j V2Ω ,h = (4 − 4 j )e(−1+2 j )t + (4 + 4 j )e(−1−2 j )t t>0 Particular Solution: V2Ω , p = (− 6 + 3 j )e (−1+ 2 j )t + (− 6 − 3 j )e (−1−2 j )t + 12 t >0 The Total Solution: V2Ω = V2Ω ,h + V2 Ω , p V2Ω = (4 − 4 j )e (−1+2 j )t + (4 + 4 j )e(−1−2 j )t (− 6 + 3 j )e(−1+2 j )t + (− 6 − 3 j )e(−1−2 j )t + 12 V2Ω = (− 2 − j )e(−1+2 j )t + (− 2 + j )e(−1−2 j )t + 12 t>0 ______________________________________________________________________________________ 5.63 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Chapter 6 Instructor Notes Chapter 6 has also seen substantive revisions with respect to the Third Edition. A section on Fourier analysis has been added (Section 6.2), and the material on Bode plots has been expanded (Section 6.4). The material on Laplace transforms has been moved to the Appendix B. These changes were prompted by comments and suggestions made by numerous users. Chapter 6 can be covered immediately following Chapter 4, or after completing Chapter 5. There is no direct dependence of Chapter 6 on Chapter 5. After the first section briefly introduces the notion of sinusoidal frequency response and motivates the use of sinusoidal signals, the Fourier Series method of representing signals is described in detail in Section 6.2. Further, the text and examples also illustrate the effect of a multi-components signal propagating through a linear system. Four examples accompany this presentation. Instructors who use this material will find some computing tools available on the website that will assist the students in developing an intuitive understanding of Fourier series. Section 6.3 introduces filters, and outlines the basic characteristics of low-, high- and band-pass filters. The concept of resonance is treated in greater depth than in the previous edition, and a connection is made with the natural response of second order circuits, which may be useful to those instructors who have already covered transient response of second-order circuits. Four detailed examples are included in this section, Further, the boxes Focus on Measurements: Wheatstone bridge filter (pp. 310-303), Focus on Measurements: AC line interference filter (pp. 303-305), and Focus on Measurements: Seismic displacement transducer (pp. 305-308) touch on additional application examples. The first and last of these boxes can be linked to related material in Chapters2, 3, and 4. The instructor who has already introduced the operational amplifier as a circuit element will find that section 8.3, on active filters, is an excellent vehicle to reinforce both the op-amp concept and the frequency response ideas. Another alternative (employed by this author) consists of introducing the opamp at this stage, covering sections 8.1 through 8.3. Finally, Section 6.4 expands the previous coverage of Bode plots, and illustrates how to create approximate Bode plots using the straight-line asymptotic approximation. The box Focus on Methodology: Bode Plots clearly outlines the method, which is further explained in two examples. The homework problems present several frequency response, Fourier Series, filter and Bode plot exercises of varying difficulty. The instructor who wishes to use one of the many available software aids (e.g., MATLAB® or Electronics Workbench® ) to analyze the frequency response of more complex circuits and to exploit more advanced graphics capabilities, will find that several advanced problems lend themselves nicely to such usage. More advanced problems could be used as a vehicle to introduce modern computer aids. The computer aided example solutions found in the Virtual Lab CD-ROM will guide the student in the solution of these more advanced problems. Learning Objectives 1. Understand the physical significance of frequency domain analysis, and compute the frequency response of circuits using AC circuit analysis tools. 2. Compute the Fourier spectrum of periodic signals using the Fourier series representation, and use this representation in connection with frequency response ideas to compute the response of circuits to periodic inputs. 3. Analyze simple first- and second-order electrical filters, and determine their frequency response and filtering properties. 4. Compute the frequency response of a circuit and its graphical representation in the form of a Bode plot. 6.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Section 6.1: Sinusoidal Frequency Response Problem 6.1 Solution: Known quantities: Resistance and inductance values, in the circuit of Figure P6.1, R = 200 kΩ and L = 0.5 H, respectively. Find: a) The frequency response for the circuit of Figure P6.1. b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. Analysis: a) Vout ( jω ) = Vin Vout = Vin R 1 1 = = R + jωL 1 + jωL / R 1 + j 2.5 × 10 −6 ω ( 1 1 + (ωL / R ) 2 = ϕ (ω ) = − arctan(2.5 × 10 −6 ω ) 1 1 + 6.25 ×10 −12 ω 2 The plots obtained using Matlab are shown below: b) c) 6.2 ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 d) ______________________________________________________________________________________ Problem 6.2 Solution: Known quantities: Resistance and capacitance values, in the circuit of Figure P6.2. Find: a) The frequency response for the circuit of Figure P6.2. b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. 6.3 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: RT = 500 500 = 250 Ω and v OC = a) v 500 v in = in 500 + 500 2 1 vout jω C 1 1 = = = vOC RT + 1 1 + jω RT C 1 + j (0.05ω ) jω C vout = vOC 1 1 + 0.0025ω 2 vout 1 vout = = vin 2 vOC 1 4 + 0.01ω 2 ϕ (ω ) = − arctan(0.05ω ) b) The plots obtained using Matlab are shown below: c) 6.4 Problem solutions, Chapter 6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 d) ______________________________________________________________________________________ Problem 6.3 Solution: Known quantities: Resistance and capacitance values, in the circuit of Figure P6.3. Find: a) The frequency response for the circuit of Figure P6.3. b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. 6.5 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: R T = 2000 2000 + 1000 = 2000 Ω and vOC = a) v 2000 vin = in 2000 + 2000 2 1 vout 1 1 jω C = = = 1 + jω RT C 1 + j (0.02ω ) vOC RT + 1 jω C vout = vOC 1 1 + 4 × 10 −4 ω 2 vout 1 vout = = vin 2 vOC 0 .5 1 + 4 × 10 −4 ω 2 ϕ (ω ) = − arctan(0.02 ω ) b) The plots obtained using Matlab are shown below: c) 6.6 Problem solutions, Chapter 6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 d) ______________________________________________________________________________________ Problem 6.4 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.4. Find: a) The frequency response for the circuit of Figure P6.4. b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. 6.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Analysis: a) Vout Vin R2 + jωL + 1 1 + jωCR2 + ( jω ) LC 1 − 0.0002ω 2 + j (0.1)ω jω C ( jω ) = = = 2 2 1 + jωC (R1 + R2 ) + ( jω ) LC 1 − 0.0002ω + j (0.15)ω R1 + R2 + jωL + 1 jω C 2 b) The plots obtained using Matlab are shown below: c) d) 6.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.5 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.5 Find: a) The frequency response for the circuit of Figure P6.5 b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. Assume: Assume that the output voltage is the voltage across the capacitor. Analysis: −1 a) Vout ( jω ) = Vin Vout Vin (Z L || Z C ) + R2 = (Z L || Z C ) + R2 + R1 § ¨¨ © · § 1 ¨¨ + jωC ¸¸ + R2 ¹ © jω L −1 · 1 + jωC ¸¸ + R2 + R1 jω L ¹ § jωL · ¨ ¸ + R2 R2 − CLR2ω 2 + jωL 1 − CLω 2 ¹ © ( jω ) = = R2 + R1 − CL(R2 + R1 )ω 2 + jωL § jωL · R R + + ¨ ¸ 2 1 2 © 1 − CLω ¹ Substituting the numerical values: Vout Vin ( jω ) = (1 − 4.5 ×10 (1 − 4.5 ×10 ω 2 ) + j (0.0015)ω −4 2 ω ) + j (0.0010)ω −4 b) The plots obtained using Matlab are shown below: 6.9 G. Rizzoni, Principles and Applications of Electrical Engineering c) d) 6.10 Problem solutions, Chapter 6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.6 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.6 Find: a) The frequency response for the circuit of Figure P6.6 b) Plot magnitude and phase of the circuit using a linear scale for frequency. c) Repeat part b., using semilog paper. d) Plot the magnitude response using semilog paper with magnitude in dB. Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: Z T = Z R 2 + (Z C1 || Z R1 ) = R2 + R1 1 jωC1 jωC1 + R1 and vOC = a) Z R1 R1 vin = Z R1 + Z C 1 R1 + 1 jωC1 vin = = R2 + R1 1 + jωC1 R1 jωC1 R1 vin 1 + jωC1 R1 1 vout ZC2 jω C 2 1 = = = vOC Z T + Z C 2 § · 1 § · R1 R1 ¨¨ R2 + ¸¸ + ¸ jω C 2 1 + ¨¨ R2 + jω C 2 1 + jωC1 R1 ¹ 1 + jωC1 R1 ¸¹ © © Therefore, 6.11 G. Rizzoni, Principles and Applications of Electrical Engineering v out jωC1 R1 = ⋅ vin 1 + jωC1 R1 = 1 § · R1 ¸ jω C 2 1 + ¨¨ R2 + 1 + jωC1 R1 ¸¹ © jωC1 R1 1 + jω [C1 R1 + C 2 (R1 + R2 )] + ( jω ) C1C 2 R1 R2 2 Substituting the numerical values: vout j (2 )ω = 2 vin 1 − ω + j (2.6)ω ( ) b) The plots obtained using Matlab are shown below: c) 6.12 Problem solutions, Chapter 6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 d) ______________________________________________________________________________________ Problem 6.7 Solution: Known quantities: Figure P6.7. Find: a) How the "driving point" impedance, [jω ] Z[jω ] = V i , behaves at extremely high or low I i [jω ] frequencies. b) An expression for the input (or driving point) impedance. c) Show that this expression can be manipulated into the form: Z[jω ] = Z o ( 1 ± j f[ ω ] ) 1 C = 0.5 µF R = 2 kΩ ωRC d) Determine the "cutoff" frequency ω = ωc at which f [ωc] = 1. e) Determine the magnitude and angle of Z [jω] at ω = 100 rad/s, 1000 rad/s, and 10,000 rad/s. f) Predict (without computing) the magnitude and angle of Z [jω] at ω = 10 rad/s and 100,000 rad/s. Where : Z o = R f[ ω ] = Analysis: a) As ω → ∞, Z C → 0 Short Z → R As ω → 0, Z C → ∞ Open Z → ∞ b) 6.13 G. Rizzoni, Principles and Applications of Electrical Engineering c) d) Problem solutions, Chapter 6 KVL : - V i + I i Z C + I i Z R = 0 1 V Z[jω ] = i = Z C + Z R = +R jω C Ii 1 1 Z[jω ] = R + j 2 = R [ 1- j ] ωRC j ωC 1 1 1 rad f[ ω c ] = = 1 ωc = = = 1000 −6 RC s [ 2000 ] [ 0.5 ⋅ 10 ] ω c RC e) 1 rad ] = R ( 1- j ) s [ 100 ] [ 2000 ] [ 0.5 ⋅10 −6 ] Z[ 100 = 2000 [ 1 - j 10 ] = 20.10 kΩ ∠ - 84.29 0 Z[ 1000 rad 1 ] = R ( 1- j ) s [ 1000 ] [ 2000 ] [ 0.5 ⋅ 10 −6 ] = 2000 [ 1 - j 1 ] = 2.828 kΩ ∠ − 45.00 0 Z[ 10 k rad 1 ] = R ( 1- j ) s [ 10000 ] [ 2000 ] [ 0.5 ⋅10 −6 ] = 2000 [ 1 - j 0.1 ] = 2.010 kΩ ∠ − 5.710 f) Z[ 10 rad ] ≈ 200 kΩ ∠ − 90 0 s Z[ 100 k rad ] ≈ 2 kΩ ∠0 0 s ______________________________________________________________________________________ Problem 6.8 Solution: Known quantities: Figure P6.8. Find: a) How the "driving point" impedance, [jω ] , behaves at extremely high or low Z[jω ] = V i I i [jω ] frequencies. 6.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 b) An expression for the driving point impedance. c) Show that this expression can be manipulated into the form: Z[jω ] = Z o ( 1 + j f[ ω ] ) ωL L = 2 mH R = 2 kΩ R Determine the "cutoff" frequency ω = ωc at which f [ωc] = 1. Determine the magnitude and angle of Z [jω] at ω = 100 krad/s, 1,000 krad/s, and 10,000 krad/s. Predict (without computing it) the magnitude and angle of Z [jω] at ω = 10 krad/s and 100,000 krad/s. Where : Z 0 = R d) e) f) f[ ω ] = Analysis: a) As ω → ∞, Z L → 0 Open Z → ∞ As ω → 0, Z L → 0 Short Z → R b) KVL : - V i + I i Z R + I i Z L = 0 Z[jω ] = c) d) e) Vi = Z L + Z R = jωL + R Ii In standard form: f[ ω c ] = Z[jω ] = R + j ωL = R [ 1 + j ωL ] R 2000 rad R ωc L = 1 = = 1000 k ωc = −3 s R L 2 ⋅ 10 The standard form can now be rewritten as: Z[jω ] = R [ 1 + j ω ω ] = 2000 [ 1+ j ] 1 ⋅10 6 ωc rad 100 ⋅103 Z[ 100 k ] = R ( 1+ j ) = 2000 [ 1 + j 0.1 ] = 2.01 kΩ ∠5.710 6 s 1 ⋅10 rad 1 ⋅10 6 Z[ 1 M ] = R ( 1+ j ) = 2000 [ 1 + j 1 ] = 2.82 kΩ ∠45.00 0 6 s 1 ⋅10 rad 10 ⋅10 6 Z[ 10M ] = R ( 1+ j ) = 2000 [ 1 + j 10 ] = 20.10 kΩ ∠84.29 0 6 s 1 ⋅10 Note, in particular, the behavior of the impedance one decade below and one decade above the cutoff frequency. f) 6.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 rad ] ≈ 200 kΩ ∠ 90o s rad Z[ 10 k ] ≈ 2 kΩ ∠ 0o s Z[ 100 M ______________________________________________________________________________________ Problem 6.9 Solution: Known quantities: With reference to Figure P6.9: L = 190 mH R1 = 2.3 kΩ C = 55 nF R2 = 1.1 kΩ Find: a) How the "driving point" impedance, [jω ] , behaves at extremely high or low Z[jω ] = V i I i [jω ] frequencies. b) An expression for the driving point impedance in the form: Z[jω ] = Z o [ 1+ j f 1 [ ω ] ] 1+ j f 2 [ ω ] Z o = R1 + L R2 C ω 2 R1 LC - R1 - R2 f 1[ω ] = ω [ R1 R 2 C + L ] ω 2 LC - 1 f 2 [ω ] = ωC R 2 c) Determine the four cutoff frequencies at which f1[ω] = +1 or -1 and f2[ω] = +1 or -1. d) Determine the resonant frequency of the circuit. e) Plot the magnitude of the impedance [in dB] as a function of the Log of the frequency, i.e., a Bode plot. Analysis: a) As ω → ∞, Z L → ∞ Open, Z C → 0 Short Z → R1 As ω → 0, Z C → ∞ Open, Z L → 0 Short Z → R1 + R2 b) 6.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 1 ] [ R 2 + jω L ] V[jω ] jω C jωC Z C [ Z R2 + Z L ] Z[jω ] = = Z R1 + = R1 + = 1 I[jω ] jω C Z C + [ Z R2 + Z L ] + [ R2 + jωL ] jω C [ = R1 + ( R1 [ 1 - ω 2 LC ] + R2 ) + j ( ω R1 R2 C + ωL ) (− j ) R 2 + j ωL = ⋅ (− j ) 1 - ω 2 LC + j ω R2 C [ 1 - ω 2 LC ] + j ω R2 C ω 2 R1 LC − R1 − R2 1+ j ω (R1 R2 C + L ) + j ω 2 R1 LC − R1 − R2 ω (R1 R2 C + L ) R1 R2 C + L Z [ jω ] = = ⋅ 2 R2 C ωR2 C + j ω LC − 1 ω 2 LC − 1 1+ j ωR2 C c) Both f1[ω] and f2[ω] can be positive or negative, and therefore equal to plus or minus one depending ( ( ) ) on the frequency; therefore, both cases must be considered. ω c [ R1 R2 C + L ] = ± 1 2 R1 [ 1 - ω c LC ] + R2 R2 + 1 ] R1 + R2 = 0 ω c2 ± [ ωc L R1 C R1 LC 1 rad R2 + 1 = 1100 + = 13.69 k -9 L R1 C 0.19 [ 2300 ] [ 55 ⋅10 ] s f 1[ωc ] = 3400 rad R1 + R2 = = 141.46 M 2 -9 [ 2300] [ 0.19 ] [ 55 ⋅10 ] s R1 LC 1 1 1/2 3 3 2 6 ω c = − [ ± 13.69 ⋅10 ] ± ( [ ± 13.69 ⋅10 ] - 4[1][ - 141.5 ⋅10 ] ) 2 2 = ± 6.845 ⋅103 ± 13.724 ⋅10 3 ω c1 = 6.879 k rad s ω c4 = 20.569 k rad s Where only the positive answers are physically valid, i.e., a negative frequency is physically impossible. ω c2 LC - 1 = ± 1 ω c R2 C R2 = 1100 = 5.79 k rad L 0.19 s f 2 [ω c ] = 1 R2 ] ωc = 0 L LC 1 1 rad = = 95.69 M 2 -9 LC [ 0.19 ] [ 55 ⋅10 ] s ω c2 ± [ 1 1 1/2 2 6 ω c = - [ ± 5790 ] ± ( [ ± 5790 ] + 4[1][ 95.69 ⋅10 ] ) = ± 2895 ± 10201 2 2 rad rad ω c2 = 7.31 k ω c3 = 13.09 k s s 6.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Again, the negative roots were rejected because they are physically impossible. d) Plotting the response in a Bode Plot: ______________________________________________________________________________________ Problem 6.10 Solution: Known quantities: In the circuit of Figure P6.10: R1 = 1.3 kΩ R2 = 1.9 kΩ C = 0.5182 µ F Find: a) How the voltage transfer function: V o [jω ] behaves at extremes of high and low H v [jω ] = V i [jω ] frequencies. b) An expression for the voltage transfer function, showing that it can be manipulated into the form: C f[ ω ] = ωR1 R2 R1 + R 2 c) The "cutoff" frequency at which f[ω] = 1 and the value of Ho in dB. H v [jω ] = Ho 1 + j f[ ω ] Where : H o = R2 R1 + R2 Analysis: a) As ω → ∞ : Z C → 0 ∠ − 90 0 Short VD : H v → 0 ∠ − 90 0 As ω → 0 : Z C → ∞ ∠ − 90 0 Open VD : H v → R2 ∠ 0 0 R1 + R2 6.18 G. Rizzoni, Principles and Applications of Electrical Engineering Z C Z R2 = Z eq = Z C + Z R2 b) [ Problem solutions, Chapter 6 1 ] [ R2 ] jωC jωC R2 = 1 1+ j ω R2 C jωC + R2 jωC R2 1 + jω R 2 C 1+ j ωR2 C = + j C ω 1 R 2 R 2 R1 + 1+ j ω R2 C 1 R2 = = R2 R1 + R2 + j ω R1 R2 C R1 + R2 1+ j ω R1 R2 C R1 + R2 Z eq V [jω ] = = VD : H v [jω ] = o V i [jω ] Z R1 + Z eq c) rad 1300 + 1900 ω c R1 R2 C = 1 = 2.5 k ωc = -6 s [ 1300 ] [ 1900] [ 0.5182 ⋅10 ] R1 + R2 1900 R2 = = 0.5938 = 20 ⋅ Log[0.5938] = - 4.527 dB Ho = 1300 + 1900 R1 + R2 f[ ω c ] = ______________________________________________________________________________________ Problem 6.11 Solution: Known quantities: Figure P6.11. Find: a) The behavior of the voltage transfer function or gain at extremely high and low frequencies. b) The output voltage Vo if the input voltage has a frequency where: V i = 7.07 V ∠ 45o c) The output voltage R1 = 2.2 kΩ R2 = 3.8 kΩ if the frequency of the X C = 5 kΩ X L = 1.25 kΩ input voltage X C = 2.5 kΩ X L = 2.5 kΩ d) The output voltage if the frequency of the input voltage again doubles so that: X C = 1.25 kΩ X L = 5 kΩ Analysis: a) As ω → 0 Z C → ∞ Open Z L → 0 Short Vo → 0 As ω → ∞ Z C → 0 Short Z L → ∞ Open [ 7.07] [ 3800] = 4.478 V ∠450 VD : V o = V i R2 = 2200 + 3800 + R1 R2 6.19 doubles so that: G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 b) Z R 2 Z L = [ R2 ] [ j X L ] Z eq 2 = Z R2 + Z L R2 + j X L j R2 X L V i Z eq 2 R2 + j X L R2 + j X L VD : V o = = Vi j R2 X L R2 + j X L Z eq1 + Z eq 2 R1 - j X C + R2 + j X L j R2 X L Vo = Vi [ R1 R2 + X C X L ] + j [ X L ( R1 + R2 ) - X C R2 ] Z eq1 = Z R1 + Z C = R1 - j X C 0 0 6 0 V i ⋅ [ j R2 X L ] = [ 7.07 V ∠45 ] [ ( 3.8 kΩ ) ( 1.25 kΩ ) ∠90 ] = 33.58 ⋅10 ∠135 6 R1 R2 + X C X L = [ 2200] [ 3800] + [ 5000] [ 1250] = 14.61 ⋅10 6 X L [ R1 + R2 ] - X C R2 = [ 1250] [ 6000] - [ 5000 ] [ 3800 ] = - 11.50 ⋅10 V0= 33.58 ∠1350 33.58 ⋅10 6 ∠1350 = 1.806V ∠173.2 0 = 6 6 0 18.59 ∠ − 38.2 14.61 ⋅10 - j 11.50 ⋅10 c) 0 0 6 0 V i ⋅ [ j R2 X L ] = [ 7.07 V ∠45 ] [ ( 3800) ( 2500) ∠90 ] = 67.17 ⋅10 ∠135 6 R1 R2 + X C X L = [ 2200] [ 3800] + [ 2500] [ 2500] = 14.61 ⋅10 6 X L [ R1 + R2 ] + X C R2 = [ 2500 ] [ 6000 ] - [ 2500 ] [ 3800] = 5.50 ⋅10 67.17 V ∠1350 67.17 ⋅10 6 ∠1350 = 4.303 V ∠114.4 0 = Vo = 6 6 0 15.61 ∠20.6 14.61 ⋅10 + j 5.50 ⋅10 d) 0 0 6 0 V i ⋅ [ j R2 X L ] = [ 7.07 V ∠45 ] [ ( 3800) ( 5000) ∠90 ] = 134.34 ⋅10 ∠135 6 R1 R2 + X C X L = [ 2200] [ 3800] + [ 1250] [ 5000] = 14.61 ⋅10 6 X L [ R1 + R2 ] + X C R2 = [ 5000 ] [ 6000 ] - [ 1250 ] [ 3800] = 25.25 ⋅10 Vo = 134.34 V ∠1350 134.34 ⋅10 6 ∠135 0 = 4.605 V ∠75.050 = 6 6 0 29.17 ∠59.94 14.61 ⋅10 + j 25.25 ⋅10 ______________________________________________________________________________________ Problem 6.12 Solution: Known quantities: Figure P6.12. Find: V o [jω ] = H vo . H v [jω ] = 1 ± j f[ ω ] V i [jω ] b) Plot the Bode diagram, i.e., a semilog plot where the magnitude [in dB] of the transfer function is a) The voltage transfer function in the form: plotted on a linear scale as a function of frequency on a log scale. 6.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Assume: The values of the resistors and of the capacitor in the circuit of Figure P6.12: R1 = 16 Ω R2 = 16 Ω C = 0.47 µF Analysis: a) R2 1 + R2 R1 + jωC 1 V o [jω ] R2 = H v [jω ] = 1 V i [jω ] R1 + R2 1 - j ωC [ R1 + R2 ] VD : V o = V i Z R2 = Vi Z R1 + Z C + Z R 2 b) ______________________________________________________________________________________ Problem 6.13 Solution: Known quantities: The values of the resistors and of the capacitor in the circuit of Figure P6.13: R1 = 100 Ω R L = 100 Ω R2 = 50 Ω Find: Compute and plot the frequency response function. Analysis: Using voltage division: 6.21 C = 80 nF G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 1 jω C jω C R2 = Z R2 Z C = = 1 jω C 1+ j ω R 2 C Z R2 + Z C R2 + jωC R2 Z eq VD : H v [jω ] = = V o [jω ] Z RL = = V i [jω ] Z R1 + Z eq + Z RL 1+ jωR2 C RL = + C 1 R j R ω 2 2 + RL R1 + 1 + jω R 2 C 1+ j ω R2 C R L [ 1 + j ωR2 C ] RL = R1 + R2 + R L + j [ R1 + R L ] ω R2 C R1 + R2 + R L 1+ j [ R1 + R L ] ω R2 C R1 + R 2 + R L Plotting the response in a Bode Plot: ______________________________________________________________________________________ 6.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Section 6.2: Fourier Analysis Problem 6.14 Solution: Find: Use trigonometric identities to show that the equalities in equations 6.16 and 6.17 hold. Analysis: Looking at figure 6.8, we can write the following equations: an = cn sin (θ n ) bn = cn cos(θ n ) and using the trigonometric identities sin 2 (θ n ) + cos 2 (θ n ) = 1 : an2 + bn2 = cn2 sin 2 (θ n ) + cn2 cos 2 (θ n ) = cn2 Finally, cn = an2 + bn2 bn cn cos(θ n ) = = cot (θ n ) = tan (ψ n ) an cn sin (θ n ) where, ψn = π − θn . 2 ______________________________________________________________________________________ Problem 6.15 Solution: Known quantities: The square wave of Figure 6.11(a) in the text. Find: A general expression for the Fourier series coefficients. Assume: None Analysis: The square wave is a function of time as follows: °° A x (t ) = ® °0 °¯ 1 1 (n − )T ≤ t ≤ (n + )T , n = ±0,±1,±2,... 4 4 1 3 (n + )T ≤ t ≤ (n + )T , n = ±0,±1,±2,... 4 4 We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22): a0 = A 1 T 1 T4 x t dt Adt = = ( ) ³ ³ T − T 0 T 4 2 6.23 G. Rizzoni, Principles and Applications of Electrical Engineering an = Problem solutions, Chapter 6 2 T 2 T4 § 2π · § 2π · x ( t ) cos n t dt A cos¨ n t ¸dt = = ¨ ¸ ³ ³ −T 0 T T 4 © T ¹ © T ¹ T 2 A ª § 2π · T º 4 A ª § nπ · § nπ sin ¨ n t¸ sin ¨ = = ¸ − sin ¨ − » « « T ¬ © T ¹ 2nπ ¼ −T nπ ¬ © 2 ¹ © 2 4 bn = ·º ¸» = 0 ¹¼ (∀n ) 2 T 2 T4 § 2π · § 2π · = ( ) sin x t n t dt A sin ¨ n t ¸dt = ¨ ¸ ³ ³ − T T 0 T 4 © T ¹ © T ¹ 2A ª § 2π = − cos¨ n « T ¬ © T A = nπ ª § nπ «− 2 cos¨ 2 © ¬ T A · T º 4 = t¸ » nπ ¹ 2nπ ¼ −T 4 2A º · °° nπ ¸» = ® ¹¼ ° °¯ 0 ª § nπ «− cos¨ 2 © ¬ (n even ) (n odd ) § nπ · ¸ + cos¨ − © 2 ¹ ·º ¸» = ¹¼ ______________________________________________________________________________________ Problem 6.16 Solution: Known quantities: The periodic function shown in Figure P6.16 and defined as: °A x(t ) = ® °0 ¯ 0≤t ≤ T 3 T ≤t ≤T 3 Find: A general expression for the Fourier series coefficients. Analysis: We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22): a0 = 1 T 1 T3 A ( ) Adt = x t dt = ³ ³ 0 0 T T 3 an = 2 T 2 T § 2π · § 2π · x(t ) cos¨ n t ¸ dt = ³ 3 A cos¨ n t ¸ dt = ³ T 0 T 0 © T ¹ © T ¹ 2 A ª § 2π sin ¨ n = T «¬ © T bn = T A · T º 3 t¸ = » nπ ¹ 2 nπ ¼ 0 ª §2 · º A §2 · «sin ¨ 3 nπ ¸ − 0» = nπ sin ¨ 3 nπ ¸ ¹ ¼ © ¹ ¬ © 2 T 2 T3 § 2π · § 2π · x t n t dt A sin ¨ n t ¸ dt = ( ) sin = ¨ ¸ ³ ³ 0 0 T T © T ¹ © T ¹ T 2A ª A ª § 2π · T º 3 § 2 ·º cos t¸ 1 − cos¨ nπ ¸» = − = ¨n « » « T ¬ nπ ¬ © T ¹ 2 nπ ¼ 0 © 3 ¹¼ Thus, the Fourier series expansion of the function is: 6.24 G. Rizzoni, Principles and Applications of Electrical Engineering x(t ) = A ∞ A § 2 · § 2π · ∞ A sin ¨ nπ ¸ cos¨ n +¦ t¸+ ¦ 3 n =1 nπ © 3 ¹ © T ¹ n=1 nπ Problem solutions, Chapter 6 ª § 2 · º § 2π · «1 − cos¨ 3 nπ ¸ » sin ¨ n T t ¸ © ¹¼ © ¹ ¬ ______________________________________________________________________________________ Problem 6.17 Solution: Known quantities: The periodic function shown in Figure P6.17 and defined as: § 2π · t °cos x(t ) = ® ¨© T ¸¹ °̄ 0 − T T ≤t≤ 4 4 else Find: A general expression for the Fourier series coefficients. Analysis: The function in Figure P6.17 is an even function. Thus, we only need to compute the an coefficients. We can compute the Fourier series coefficient using the integrals in equations (6.20) and (6.21): T 1 T 1 T 1 ª § 2π ·º 4 § 2π · sin ¨ a0 = ³ T2 x(t )dt = ³ T4 cos¨ t ¸dt = t¸ = 2π «¬ © T ¹»¼ −T T − 2 T − 4 © T ¹ 4 = 1 2π ª §π · § π ·º 1 «sin ¨ 2 ¸ − sin ¨ − 2 ¸ » = π © ¹¼ ¬ © ¹ 2 T2 2 T4 § 2π · § 2π · § 2π · x t n t dt t ¸ cos¨ n t ¸ dt = = ( ) cos cos¨ ¨ ¸ ³ ³ T T − − T 2 T 4 © T ¹ © T ¹ © T ¹ § nπ · cos¨ ¸ n −1 2 (n even ) 2 2 ¹ °(- 1) 2 © =− =® π n2 −1 2 π n −1 °̄ 0 (n odd ) an = ( ) ______________________________________________________________________________________ Problem 6.18 Solution: Known quantities: The periodic function shown in Figure P6.18 and defined as: T 2A ° T t 0≤t ≤ 2 x(t ) = ® T °A ≤t ≤T 2 ¯ Find: Compute the Fourier series expansion. Analysis: We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22): 6.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 § 2 T2 T T · 3A 1 T 1 § T 2 2A · A ¨ ªt º a0 = ³ x(t )dt = ¨ ³ t dt + ³T Adt ¸ = ¨ « » + T − ¸¸ = T 0 T© 0 T 2¸ 4 2 ¹ T ¨ ¬ T ¼0 © ¹ an = 2 T § 2π x(t ) cos¨ n ³ T 0 © T 2 § T 2A · § 2π t ¸ dt = ¨¨ ³ 2 t cos¨ n T© 0 T ¹ © T T · § 2π t ¸ dt + ³T A cos¨ n 2 ¹ © T · · t ¸ dt ¸¸ = ¹ ¹ T T · § 2 ª TA 2 ¨ ª TA § § 2π · 2π § 2π · · º § 2π · º ¸ = ¨« t¸+ n t sin ¨ n t ¸ ¸¸ » + « sin ¨ n t ¸» = ¨ cos¨ n T ¨ ¬ 2(nπ )2 ¨© © T ¹ T © T ¹ ¹¼ 0 © T ¹ ¼ T ¸¸ ¬ 2 nπ 2 ¹ © A [cos(nπ ) − 1 + 2nπ sin (nπ )cos(nπ )] = A 2 [cos(nπ ) − 1] = 2 (nπ ) (nπ ) bn = 2 T § 2π x(t ) sin ¨ n ³ T 0 © T 2 § T 2A · § 2π t ¸ dt = ¨¨ ³ 2 t sin ¨ n T© 0 T ¹ © T T · § 2π t ¸ dt + ³T A sin ¨ n 2 ¹ © T · · t ¸dt ¸¸ = ¹ ¹ T T · § 2 ª TA 2 ¨ ª TA § § 2π · 2π § 2π · ·º § 2π ·º ¸ = ¨« t¸−n t cos¨ n t ¸ ¸¸» + « − cos¨ n t ¸» = ¨ sin ¨ n T ¨ ¬ 2(nπ )2 ¨© © T ¹ T © T ¹ ¹¼ 0 © T ¹¼ T ¸¸ ¬ 2 nπ 2¹ © A A = sin (nπ ) − 2nπ cos 2 (nπ ) + nπ = 1 − 2 cos 2 (nπ ) 2 nπ (nπ ) [ ] [ ] Thus, the Fourier series expansion of the function is: x(t ) = 3A ∞ A § 2π · ∞ A § 2π · ( ) [ n ] t¸+ ¦ t¸ +¦ − cos π 1 cos 1 − 2 cos 2 (nπ ) sin ¨ n ¨n 2 4 n=1 (nπ ) © T ¹ n=1 nπ © T ¹ [ ] ______________________________________________________________________________________ Problem 6.19 Solution: Known quantities: The periodic function shown in Figure P6.19 and defined as: § 2π °°sin ¨ T x(t ) = ® © °0 °¯ · t¸ ¹ 0≤t≤ T 2 T ≤t ≤T 2 Find: Compute the Fourier series expansion. Analysis: The function in Figure P6.19 is an even function. Thus, we only need to compute the an coefficients. We can compute the Fourier series coefficient using the integrals in equations (6.20) and (6.21): 6.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 T 1 T 1 T 1 ª § 2π ·º 2 § 2π · cos a0 = ³ x(t )dt = ³ 2 sin ¨ t ¸dt = t ¸» = ¨ 2π «¬ T 0 T 0 ©T ¹ © T ¹¼ 0 1 [- cos(π ) + cos(0)] = 1 = π 2π 2 T § 2π · § 2π · t ¸ dt = ³ sin ¨ t ¸ cos¨ n T 0 ¹ © T ¹ © T 2 (n even ) cos(nπ ) + 1 °− =− = ® π n2 −1 2 π n −1 °̄ 0 (n odd ) an = 2 T § 2π x(t ) cos¨ n ³ 0 T © T ( ( ) · t ¸ dt = ¹ ) Thus, the Fourier series expansion of the function is: x(t ) = 1 ∞ cos(nπ ) + 1 § 2π cos¨ n −¦ π n =1 π n 2 − 1 © T ( ) · t¸ ¹ ______________________________________________________________________________________ Problem 6.20 Solution: Known quantities: The periodic function shown in Figure P6.20. Find: A complete expression for the function x(t ) and the Fourier coefficients. Analysis: The periodic function shown in Figure P6.20 can be defined as: V ° t ° t1 °V ° °V x(t ) = ® (T-t ) ° t1 °− V ° °V (t − 2T ) °t ¯1 0 ≤ t ≤ t1 t1 ≤ t ≤ T − t1 T − t1 ≤ t ≤ T + t1 T + t1 ≤ t ≤ 2T − t1 2T − t1 ≤ t ≤ 2T The function in Figure P6.19 is an odd function with period equal to the bn coefficients. 2T . Thus, we only need to compute We can compute the Fourier series coefficient using the integrals in equation (6.22): 6.27 G. Rizzoni, Principles and Applications of Electrical Engineering bn = 2 2T ³ 2T 0 Problem solutions, Chapter 6 T −t1 1 § t1 V § 2π · § 2π · § 2π · x(t ) sin ¨ n t ¸dt = ¨¨ ³ t sin ¨ n t ¸ dt + ³ V sin ¨ n t ¸ dt + t1 T © 0 t1 © T ¹ © T ¹ © T ¹ 2T −t1 2T · V V § 2π · § 2π · T − t )sin ¨ n t ¸ dt − ³ V sin ¨ n t ¸ dt + ³ ( (t − 2T )sin§¨ n 2π t ·¸ dt ¸¸ 2 T −t1 t T t T t + − 1 1 t © T ¹ © T ¹ © T ¹ ¹ 1 1 ª § V § π · § π · § π · § 2π · · = « 4T sin ¨ n T t1 ¸ cos¨ n T t1 ¸ − 2πnt1 cos¨ n T t1 ¸ +πnt1 ¨¨1 + cos¨ n T t1 ¸ ¸¸ 2 2(nπ ) t1 ¬ © ¹ © ¹ © ¹ © ¹¹ © +³ T + t1 § π · § π · § π · § π ·º − 12T cos 2 (nπ )sin ¨ n t1 ¸ cos¨ n t1 ¸ + 8T cos 4 (nπ )sin ¨ n t1 ¸ cos¨ n t1 ¸» © T ¹ © T ¹ © T ¹ © T ¹¼ ________________________________________________________________________ Problem 6.21 Solution: Known quantities: The periodic function shown in Figure P6.21. Find: A complete expression for the function x(t ) and the Fourier coefficients. Analysis: The periodic function shown in Figure P6.21 can be defined as: °° A x(t ) = ® °− A °¯ 0≤t ≤ T− T 4 T ≤t ≤T 4 The function in Figure P6.19 is an odd function. Thus, we only need to compute the bn coefficients. We can compute the Fourier series coefficient using the integrals in equation (6.22): bn = = 2 T2 § 2π x(t ) sin ¨ n ³ T − T 2 © T 2§ 0 · § 2π t ¸dt = ¨¨ − ³ T A sin ¨ n − T© 4 ¹ © T T § 2π · t ¸ dt + ³ 4 A sin ¨ n 0 © T ¹ · · t ¸ dt ¸¸ = ¹ ¹ 2A § § π ·· 2A ¨¨1 − cos¨ n ¸ ¸¸ = nπ © © 2 ¹ ¹ nπ ______________________________________________________________________________________ Problem 6.22 Solution: Known quantities: The periodic function defined as: x(t ) = 10 cos(10t + π / 6) 6.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Find: All Fourier series coefficients. Analysis: Using trigonometric identities we can expand the function x(t ) in the following way: x(t ) = 10 cos(10t + π / 6) = 10[cos(10t ) cos(π / 6) − sin(10t ) sin(π / 6)] = = 5 3 cos(10t ) − 5 sin(10t ) Now the function is already in Fourier series form, since it contains only sinusoidal terms! We recognize the following parameters: ω 0 = 10 a0 = 0 a1 = 5 3 b1 = 5 and all other coefficients are equal to zero. ______________________________________________________________________________________ 6.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Section 6.3: Filters Problem 6.23 Solution: Known quantities: The resistance of the RC high-pass filter. Find: Design an RC high-pass filter with a breakpoint at 200 kHz. Analysis: The frequency response of the RC high-pass filter is: Vo ( jω ) jωCR = Vi ( jω ) 1 + jωCR The cutoff frequency is: ω0 = 1 RC ω0 = 1 1 1 = 2π × 200000 C = = ≅ 53 pF RC Rω 0 2π ⋅ 15000 ⋅ 200000 Thus, ______________________________________________________________________________________ Problem 6.24 Solution: Known quantities: The resistance of the RC low-pass filter. Find: Design an RC low-pass filter that would attenuate a 120-Hz sinusoidal voltage by 20 the DC gain. Analysis: The frequency response of the RC low-pass filter is: H v ( jω ) = dB with respect to Vo ( j ω ) 1 = Vi ( jω ) 1 + jωCR u (t ) = A sin (ωˆ t + ϕ ) is: y∞ (t ) = H v ( jωˆ ) A sin (ωˆ t + ϕ + ∠H v ( jωˆ )) In order to attenuate the sinusoidal input by 20 dB (a factor of 10) with respect to the DC gain, 1 99 1 H v ( jωˆ ) = = 0.1 C = 10 2 − 1 = ≅ 26.4 µF 2 ˆ 500 2 120 ⋅ × R ω π ˆ 1 + (ωCR ) The response of the circuit to the periodic input ______________________________________________________________________________________ 6.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.25 Solution: Known quantities: The resistance and the inductance of the parallel LC resonant circuit. Find: Design a parallel LC resonant circuit to resonate at 500-kHz. Analysis: The frequency response of the parallel LC resonant circuit is: V ( jω ) H v ( jω ) = o = Vi ( jω ) 1 + ( jω ) LC L 2 1 + jω + ( jω ) LC R 2 The resonant frequency of the circuit is: ωn = 1 = 2π × 500 kHz LC Thus, C= 1 ≅ 1 pF L(π ×1012 ) 2 The damping ratio is, ξ= 1 / RC π 2 × 1011 = ≅ 0.3142 2ω n 500000 ⋅ 2(2π × 500000) The quality factor is, Q= 1 ≅ 1.5915 2ξ ______________________________________________________________________________________ Problem 6.26 Solution: Find: In an RLC circuit, show that Q= ωn ∆ω . Analysis: The frequency response of an RLC circuit is: § 2ξ · jω ¨ ω ¸ n¹ © 2 j ω ξ 2 § · § · 1 + jω ¨ ¸+¨ ¸ © ωn ¹ © ωn ¹ We can compute the half-power frequencies ω1 and ω 2 by equating the magnitude of the band-pass filter V ( jω ) = H v ( jω ) = o Vi ( jω ) 2 (this will result in a quadratic equation in ω , which can be solved for the ω two frequencies). Defining Ω = , we can write the following equation: ωn frequency response to 1 6.31 G. Rizzoni, Principles and Applications of Electrical Engineering H v ( jω ) = 2ξ ( jΩ ) = 2 1 + 2ξ ( jΩ ) + ( jΩ ) Ω 4 − 2(1 + 2ξ 2 )Ω 2 + 1 = 0 Finally, discarding the negative solutions: Ω1, 2 = ±ξ + 1 + ξ 2 Thus, ( 2ξΩ (1 − Ω ) + (2ξΩ) 2 2 2 = 1 2 Ω 2 = 1 + 2ξ 2 ± 2ξ 1 + ξ 2 ( )ω − (− ξ + ) 1 + ξ )ω ω1, 2 = ± ξ + 1 + ξ 2 ω n ∆ω = ω 2 − ω1 = ξ + 1 + ξ 2 and, Problem solutions, Chapter 6 n 2 n = 2ξω n ωn ω 1 = n = =Q ∆ω 2ξω n 2ξ ______________________________________________________________________________________ Problem 6.27 Solution: Known quantities: The resistance, inductance and capacitance of a series RLC resonant circuit. Find: a) Show that the impedance at the resonance frequency becomes a value of Q times the resistance at the resonance frequency. NOTE: The word inductive should not be in the problem statement. b) Determine the impedance at the resonance frequency, assuming L = 280 mH , C = 0.1 µF and R = 25 Ω . Assumptions: The circuit is as shown in the figure below with the output impedance across the inductor of the RLC circuit. Also, the output impedance is the impedance of interest. Analysis: a) The output impedance of the circuit is: § 1 · ¸ || jωL = Z out ( jω ) = ¨¨ R + jωC ¸¹ © 1 + jωCR (1 + jωCR ) jωL || jωL = = 2 jω C 1 + jωCR + ( jω ) LC and the quality factor is: 1 L ωn L 1 = = R C R ω n RC Thus, for ω → ω n : Q= 6.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ( 1 + jω n CR ) jω n L jω n L − ω n2 LRC = = Z out ( jω n ) = 2 2 1 + jω n CR + ( jω n ) LC 1 − ω n LC + jω n CR jω n L / R − (ω n L / R )(ω n RC ) jQ − Q / Q =R =R = 1 − (ω n L / R )(ω n RC ) + jω n CR 1− Q / Q + j / Q =R jQ − 1 = R Q 2 + jQ = RQ(Q + j ) j /Q ( ) For a high quality factor circuit, we have § L· Z out ( jω n ) = RQ 1 + Q 2 ≅ RQ 2 = R¨ ω n ¸Q = ω n LQ © R¹ Finally, the impedance at the resonance frequency becomes a value of Q times the inductive resistance at the resonance frequency. b) The quality factor is: Q= 1 L 1 = 2.8 × 10 6 ≅ 67 R C 25 The impedance at the resonance frequency is: Z out ( jω n ) = RQ 1 + Q 2 = 25 ⋅ 67 1 + 67 2 = 112.01 kΩ while, Q times the inductive resistance at the resonance frequency is: 1 ω n LQ = 0.28 ⋅ 67 = 112 kΩ 2.8 × 10 −8 ______________________________________________________________________________________ Problem 6.28 Solution: Known quantities: Frequency response H v ( jω ) of the circuit of Example 6.7. Find: The frequency at which the phase shift introduced by the circuit is equal to -10°. Analysis: The frequency response of the circuit is: H v ( jω ) = 1 1 + jωCR From Example 6.7: ω0 = 1 = 2,128rad / sec CR The phase shift introduced by the circuit is: Thus, §ω· ∠H v ( jω ) = − arctan −1 ¨¨ ¸¸ © ω0 ¹ § ω · ∠H v ( jω ) = − arctan −1 ¨ ¸ © 2128 ¹ 6.33 G. Rizzoni, Principles and Applications of Electrical Engineering § ω · ∠H v ( jω ) = − arctan¨ ¸ = −10 © 2128 ¹ Problem solutions, Chapter 6 ω = 2128 tan (10) = 375.2 rad/s ______________________________________________________________________________________ Problem 6.29 Solution: Known quantities: Frequency response H v ( jω ) of the circuit of Example 6.7. Find: The frequency at which the output of the circuit is attenuated by 10 percent. Analysis: The frequency response of the circuit is: 1 1 + jωCR H v ( jω ) = From Example 6.7: ω0 = 1 = 2,128rad / sec CR The attenuation introduced by the circuit is: H v ( jω ) = 1 1 + (ω / ω0 ) 2 Thus, H v ( jω ) = 2 1 1 + (ω / 2128) 2 § 1 · 2 = 0.9 ω = (2128) ¨ ¸ − 1 = 1031 rad/s 0 . 9 © ¹ ______________________________________________________________________________________ Problem 6.30 Solution: Known quantities: Frequency response H v ( jω ) of the circuit of Example 6.11. Find: The frequency at which the output of the circuit is attenuated by 10 percent. Analysis: The frequency response of the circuit is: H v ( jω ) = 2 1 + 0.2 jω The attenuation introduced by the circuit is: H v ( jω ) = 2 1 + (0.2ω ) 2 Thus, 6.34 G. Rizzoni, Principles and Applications of Electrical Engineering H v ( jω ) = Problem solutions, Chapter 6 2 2 1 + (0.2ω ) 2 1 § 2 · = 0.9 ω = ¨ ¸ − 1 = 9.9225 rad/s 0 .2 © 0 .9 ¹ ______________________________________________________________________________________ Problem 6.31 Solution: Known quantities: Frequency response H v ( jω ) of the circuit of Example 6.11. Find: The frequency at which the phase shift introduced by the circuit is equal to 20°. Analysis: The frequency response of the circuit is: H v ( jω ) = jωCR 2 1 + jωCR + ( jω ) LC The phase shift introduced by the circuit is: ∠H v ( jω ) = π § ωCR · − arctan¨ ¸ 2 2 © 1 − ω LC ¹ ∠H v ( jω ) = π ω CR § ω CR · − arctan¨ = tan 25° ¸ = 20° 2 2 1 − ω 2 LC © 1 − ω LC ¹ Thus, R = 1kΩ, C = 10 µF , L = 5mH : ω = 46.6rad / sec b) R = 10kΩ, C = 10 µF , L = 5mH : ω = 4.66rad / sec a) ______________________________________________________________________________________ Problem 6.32 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.1, the period T = 10 µs and the peak amplitude A = 1 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have x(t ) = 2 A § 2π · A § 4π · 2 1 sin ¨ t ¸ + sin ¨ t ¸ = sin 2 × 10 5 π t + sin 4 × 10 5 π t π π © T ¹ π © T ¹ π ( Thus, for this problem, 6.35 ) ( ) G. Rizzoni, Principles and Applications of Electrical Engineering c1 = 2 π ω1 = 2 × 10 5 π = 6.28 × 10 5 rad/s c2 = 1 π ω 2 = 4 ×105 π = 12.56 ×105 rad/s and, Problem solutions, Chapter 6 The frequency response of the system can be expressed in magnitude and phase form: ( ) 1 1 = ∠ − arctan 2.5 ×10 −6 ω −6 − 12 2 1 + j 2.5 ×10 ω 1 + 6.25 ×10 ω At this point, we could evaluate the frequency response of the system at the frequencies ω1 and ω 2 H v ( jω ) = ( ) analytically: H v ( jω1 ) = 1 1 + 6.25 × 10 ( −12 ω1 2 = 0.537 ) Φ( jω1 ) = − arctan 2.5 ×10 ω1 = −1.0039 rad = −57.52 , 1 = 0.3033 H v ( jω 2 ) = 2 −12 1 + 6.25 ×10 ω 2 ( −6 ) Φ( jω 2 ) = − arctan 2.5 ×10 −6 ω 2 = −1.2626 rad = −72.34 , Finally, we can compute the steady-state periodic output of the system: 2 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ( jω n )] = n =1 = 0.537 ( ) ( 2 1 sin 2 × 10 5 π t − 1.0039 + 0.3033 sin 4 × 10 5 π t − 1.2626 π π ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ 6.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.33 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.1, the period T = 10 µs and the peak amplitude A = 1 for the square wave of Figure 6.11(a). Find: Output of system y (t ) in response to input x(t ) . Analysis: The square wave can be defined as: 1 1 (n − )T ≤ t ≤ (n + )T , n = ±0,±1,±2,... 4 4 3 1 (n + )T ≤ t ≤ (n + )T , n = ±0,±1,±2,... 4 4 °° A x (t ) = ® °0 °¯ We can compute the Fourier series coefficient using the integrals in equations (6.20), (6.21) and (6.22): a0 = A 1 T 1 T4 ( ) = x t dt Adt = ³ ³ −T 0 T T 4 2 an = 2 T 2 T § 2π · § 2π · x(t ) cos¨ n t ¸dt = ³−T4 A cos¨ n t ¸dt = ³ T 0 T 4 © T ¹ © T ¹ 2 A ª § 2π = sin ¨ n T «¬ © T bn = T A · T º 4 t¸ = » ¹ 2nπ ¼ −T 4 nπ ª § nπ «sin ¨ 2 ¬ © · § nπ ¸ − sin ¨ − ¹ © 2 ·º ¸» = 0 ¹¼ (∀n ) 2 T 2 T4 § 2π · § 2π · = x t n t dt A sin ¨ n t ¸dt = ( ) sin ¨ ¸ ³ ³ −T 0 T T 4 © T ¹ © T ¹ T A ª 2A ª § 2π · T º 4 § nπ § nπ · t¸ = − = − cos¨ cos ¨n ¸ + cos¨ − « « » T ¬ nπ ¬ © T ¹ 2nπ ¼ −T © 2 © 2 ¹ 4 2A A ª § nπ ·º °° nπ = − 2 cos ¨ ¸» = ® nπ «¬ © 2 ¹¼ ° ¯° 0 (n even ) (n odd ) ·º ¸» = ¹¼ Using the first two terms of the Fourier series expansion of the square waveform, we have x (t ) = A A § 4π · 1 1 + sin ¨ t ¸ = + sin 4 × 105 π t T 2 π ¹ 2 π © ( Thus, for this problem, 1 2 c1 = 0 c0 = ω0 = 0 rad/s ω1 = 2 × 10 5 π = 6.28 × 10 5 rad/s and, c2 = 1 π ω 2 = 4 ×105 π = 12.56 ×105 rad/s 6.37 ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 The frequency response of the system can be expressed in magnitude and phase form: ( ) 1 1 = ∠ − arctan 2.5 ×10 −6 ω −6 2 −12 1 + j 2.5 ×10 ω 1 + 6.25 ×10 ω At this point, we could evaluate the frequency response of the system at the frequencies ω0 and ω2 H v ( jω ) = ( ) analytically: H v ( jω 0 ) = 1 H v ( jω 2 ) = 1 1 + 6.25 × 10−12 ω2 ( Φ( jω ) = − arctan(2.5 × 10 2 = 0.3033 ) ω ) = −1.2626 rad = −72.34 Φ( jω0 ) = − arctan 2.5 × 10 −6 ω0 = 0 rad = 0, 2 −6 2 , Finally, we can compute the steady-state periodic output of the system: 2 y (t ) = ¦ H v ( jωn ) cn sin[ωnt + θ n + Φ( jωn )] = n =1 = ( ) ( 1 0.3033 0.3033 sin (0) + sin 4 × 105 π t − 1.2626 = sin 4 × 105 π t − 1.2626 2 π π ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.34 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.1, the period T = 10 µs and the peak amplitude A = 1 for the pulse waveform. 6.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the pulse waveform of Example 6.4, we have ( ) ( x(t ) = 0.2 + 0.3027 cos 2 × 10 5 π t + 0.2199 sin 2 × 10 5 π t ( ) ( + 0.0935 cos 4 × 10 π t + 0.2879 sin 4 × 10 π t Thus, for this problem, 5 5 ) ) c0 = 0.2 c1 = 0.3742 , θ1 = 0.9425 rad = 54 , ω1 = 2 × 10 5 π = 6.28 × 10 5 rad/s c2 = 0.3027 , θ 2 = 0.3140 rad = 18, ω 2 = 4 ×105 π = 12.56 ×105 rad/s and, The frequency response of the system can be expressed in magnitude and phase form: ( ) 1 1 = ∠ − arctan 2.5 ×10 −6 ω −6 −12 2 1 + j 2.5 ×10 ω 1 + 6.25 ×10 ω At this point, we could evaluate the frequency response of the system at the frequencies ω1 and ω 2 H v ( jω ) = ( ) analytically: H v ( jω1 ) = 1 1 + 6.25 × 10 −12 ω1 ( 2 = 0.537 ) Φ( jω1 ) = − arctan 2.5 ×10 −6 ω1 = −1.0039 rad = −57.52 , 1 = 0.3033 H v ( jω 2 ) = 2 1 + 6.25 ×10 −12 ω 2 ( ) Φ( jω 2 ) = − arctan 2.5 × 10 −6 ω 2 = −1.2626 rad = −72.34 , Finally, we can compute the steady-state periodic output of the system: 2 y (t ) = c0 H v (0) + ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = n =1 ( ) ( = 0.2 + 0.2009 sin 2 × 10 5 π t − 0.0614 + 0.0918 sin 4 × 10 5 π t − 0.9484 The input and output steady state signals plot is shown below: 6.39 ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.35 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.2, the period T = 0.5 s and the peak amplitude A = 2 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have 2 A § 2π · A § 4π · 2 A § 6π t ¸ + sin ¨ t¸+ sin ¨ sin ¨ π © T ¹ π © T ¹ 3π ©T 4 2 4 = sin (4π t ) + sin (8π t ) + sin (12π t ) 3π π π x(t ) = · t¸ = ¹ Thus, for this problem, 4 π 2 c2 = π 4 c3 = 3π c1 = ω1 = 4π = 12.5664 rad/s ω 2 = 8π = 25.1327 rad/s ω3 = 12π = 37.6991 rad/s The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 1 = ∠ − arctan(0.05ω ) 2(1 + 0.05 jω ) 4 + 0.01ω 2 6.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 At this point, we could evaluate the frequency response of the system at the frequencies ω1 , ω 2 and ω3 analytically: H v ( jω1 ) = 1 4 + 0.01ω12 = 0.4234 Φ( jω1 ) = − arctan(0.05ω1 ) = −0.5610 rad = −32.14 , 1 H v ( jω 2 ) = = 0.3113 4 + 0.01ω 22 Φ( jω 2 ) = − arctan(0.05ω 2 ) = −0.8986 rad = −51.49 , 1 H v ( jω 3 ) = = 0.2343 4 + 0.01ω32 Φ( jω3 ) = − arctan(0.05ω3 ) = −1.0830 rad = −62.05 , Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = n =1 4 2 = 0.4234 sin (4π t − 0.561) + 0.3113 sin (8π t − 0.8986) π π 4 + 0.2343 sin (12π t − 1.083) 3π The input and output steady state signals plot is shown below: ______________________________________________________________________________________ 6.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.36 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.2, the period T = 0.5 s and the peak amplitude A = 2 for the square wave. Find: y (t ) in response to input x(t ) . Output of system Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the square waveform of P6.33, we have x(t ) = 4 A § 2π · 4 A § 6π · 8 8 sin ¨ t¸+ sin ¨ t ¸ = sin (4π t ) + sin (12π t ) π 3π © T ¹ 3π © T ¹ π Thus, for this problem, 8 π c2 = 0 8 c3 = 3π ω1 = 4π = 12.5664 rad/s c1 = ω 2 = 8π = 25.1327 rad/s ω3 = 12π = 37.6991 rad/s The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 1 = ∠ − arctan(0.05ω ) 2(1 + 0.05 jω ) 4 + 0.01ω 2 At this point, we could evaluate the frequency response of the system at the frequencies analytically: H v ( jω1 ) = 1 4 + 0.01ω12 = 0.4234 Φ( jω1 ) = − arctan(0.05ω1 ) = −0.5610 rad = −32.14 , 1 H v ( jω 3 ) = = 0.2343 4 + 0.01ω32 Φ( jω3 ) = − arctan(0.05ω3 ) = −1.0830 rad = −62.05 , Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = n =1 8 8 = 0.4234 sin (4π t − 0.561) + 0.2343 sin (12π t − 1.083) 3π π The input and output steady state signals plot is shown below: 6.42 ω1 and ω3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.37 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.2, the period T = 0.5 s and the peak amplitude A = 2 for the pulse waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the pulse waveform of Example 6.4, we have x(t ) = 0.4 + 0.6054 cos(4π t ) + 0.4398 sin (4π t ) + 0.1870 cos(8π t ) + 0.5758 sin (8π t ) − 0.1248 cos(12π t ) + 0.3838 sin (12π t ) Thus, for this problem, c0 = 0.4 c1 = 0.7484 , θ1 = 0.9425 rad = 54 , c2 = 0.6054 , θ 2 = 0.3140 rad = 18, c3 = 0.4036 , θ1 = −0.3144 rad = −18, ω1 = 4π = 12.5664 rad/s ω 2 = 8π = 25.1327 rad/s ω3 = 12π = 37.6991 rad/s The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 1 = ∠ − arctan(0.05ω ) 2(1 + 0.05 jω ) 4 + 0.01ω 2 At this point, we could evaluate the frequency response of the system at the frequencies analytically: H v ( jω1 ) = 1 4 + 0.01ω12 = 0.4234 6.43 ω1 , ω 2 and ω3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Φ( jω1 ) = − arctan(0.05ω1 ) = −0.5610 rad = −32.14 , 1 H v ( jω 2 ) = = 0.3113 4 + 0.01ω 22 Φ( jω 2 ) = − arctan(0.05ω 2 ) = −0.8986 rad = −51.49 , 1 H v ( jω 3 ) = = 0.2343 4 + 0.01ω32 Φ( jω3 ) = − arctan(0.05ω3 ) = −1.0830 rad = −62.05 , Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = c0 H v (0) + ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ( jω n )] = n =1 = 0.2 + 0.3168 sin (4π t + 0.3815) + 0.1885 sin (8π t − 0.5846 ) + 0.0946 sin (12π t − 1.3974 ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.38 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.3, the period T = 0.1 s and the peak amplitude A = 1 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first four terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have 6.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 2 A § 2π · A § 4π · 2 A § 6π · A § 8π · t ¸ + sin ¨ t¸+ t¸+ sin ¨ sin ¨ sin ¨ t ¸ π © T ¹ π © T ¹ 3π © T ¹ 2π ©T ¹ 2 1 2 1 = sin (20π t ) + sin (40π t ) + sin (60π t ) + sin (80π t ) 3π 2π π π x(t ) = Thus, for this problem, 2 π 1 c2 = π 2 c3 = 3π ω 2 = 40π = 125.6 rad/s 1 2π ω 4 = 80π = 251.2 rad/s ω1 = 20π = 62.8 rad/s c1 = and, c4 = ω3 = 60π = 188.4 rad/s The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 = 2(1 + 0.02 jω ) 1 4 + 0.0016ω 2 ∠ − arctan(0.02ω ) At this point, we could evaluate the frequency response of the system at the frequencies ω4 analytically: H v ( jω1 ) = 1 4 + 0.0016ω12 = 0.3113 Φ( jω1 ) = − arctan(0.02ω1 ) = −0.8986 rad = −51.49 , 1 H v ( jω 2 ) = = 0.1848 4 + 0.0016ω 22 Φ( jω 2 ) = − arctan(0.02ω 2 ) = −1.1921 rad = −68.3, 1 H v ( jω 3 ) = = 0.1282 4 + 0.0016ω32 Φ( jω3 ) = − arctan(0.02ω3 ) = −1.3115 rad = −75.14 , 1 H v ( jω 4 ) = = 0.0976 4 + 0.0016ω 42 Φ( jω 4 ) = − arctan(0.02ω 4 ) = −1.3744 rad = −78.75, Finally, we can compute the steady-state periodic output of the system: 4 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = n =1 = 0.1982 sin (20π t − 0.8986 ) + 0.0588 sin (40π t − 1.1921) + 0.0272 sin (60π t − 1.3115) + 0.0155 sin (80π t − 1.3744 ) The input and output steady state signals plot is shown below: 6.45 ω1 , ω 2 , ω3 and G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.39 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.3, the period T = 0.1 s and the peak amplitude A = 1 for the square waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first four terms of the Fourier series expansion of the square waveform of P6.33, we have x(t ) = 4 A § 2π sin ¨ π © T Thus, for this problem, · 4 A § 6π t¸+ sin ¨ ¹ 3π © T 4 · 4 t ¸ = sin (20π t ) + sin (60π t ) 3π ¹ π 4 ω1 = 20π = 62.8 rad/s π 4 c3 = ω3 = 60π = 188.4 rad/s 3π c2 = c4 = 0 c1 = The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 = 2(1 + 0.02 jω ) 1 4 + 0.0016ω 2 ∠ − arctan(0.02ω ) At this point, we could evaluate the frequency response of the system at the frequencies analytically: 6.46 ω1 and ω3 G. Rizzoni, Principles and Applications of Electrical Engineering H v ( jω1 ) = 1 4 + 0.0016ω12 Problem solutions, Chapter 6 = 0.3113 Φ( jω1 ) = − arctan(0.02ω1 ) = −0.8986 rad = −51.49 , 1 H v ( jω 3 ) = = 0.1282 4 + 0.0016ω32 Φ( jω3 ) = − arctan(0.02ω3 ) = −1.3115 rad = −75.14 , Finally, we can compute the steady-state periodic output of the system: 4 y (t ) = ¦ H v ( jω n ) c n sin[ω n t + θ n + Φ ( jω n )] = n =1 = 0.3964 sin (20π t − 0.8986 ) + 0.0544 sin (60π t − 1.3115) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.40 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.3, the period T = 0.1 s and the peak amplitude A = 1 for the pulse waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first four terms of the Fourier series expansion of the pulse waveform of Example 6.4, we have x(t ) = 0.2 + 0.3027 cos(20π t ) + 0.2199 sin (20π t ) + 0.0935 cos(40π t ) + 0.2879 sin (40π t ) − 0.0624 cos(60π t ) + 0.1919 sin (60π t ) − 0.0757 cos(80π t ) + 0.055 sin (80π t ) 6.47 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Thus, for this problem, c0 = 0.2 c1 = 0.3742 , c2 = 0.3027 , c3 = 0.2018 , θ1 = 0.9425 rad = 54 , θ 2 = 0.3140 rad = 18, θ 3 = −0.3144 rad = −18, ω1 = 20π = 62.8 rad/s ω 2 = 40π = 125.6 rad/s ω3 = 60π = 188.4 rad/s c4 = 0.0935 , θ1 = −0.9425 rad = −54, ω 4 = 80π = 251.2 rad/s and, The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = 1 = 2(1 + 0.02 jω ) 1 4 + 0.0016ω 2 ∠ − arctan(0.02ω ) At this point, we could evaluate the frequency response of the system at the frequencies ω4 analytically: H v ( jω1 ) = 1 4 + 0.0016ω12 = 0.3113 Φ( jω1 ) = − arctan(0.02ω1 ) = −0.8986 rad = −51.49 , 1 H v ( jω 2 ) = = 0.1848 4 + 0.0016ω 22 Φ( jω 2 ) = − arctan(0.02ω 2 ) = −1.1921 rad = −68.3, 1 H v ( jω 3 ) = = 0.1282 4 + 0.0016ω32 Φ( jω3 ) = − arctan(0.02ω3 ) = −1.3115 rad = −75.14 , 1 H v ( jω 4 ) = = 0.0976 4 + 0.0016ω 42 Φ( jω 4 ) = − arctan(0.02ω 4 ) = −1.3744 rad = −78.75, Finally, we can compute the steady-state periodic output of the system: 4 y (t ) = c0 H v (0) + ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ( jω n )] = n =1 = 0.1 + 0.1165 sin (20π t + 0.0439 ) + 0.0559 sin (40π t − 0.8781) + 0.0259 sin (60π t − 1.6259 ) + 0.00915 sin (80π t − 2.3169 ) The input and output steady state signals plot is shown below: 6.48 ω1 , ω 2 , ω3 and G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.41 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.4, the period T = 50 ms and the peak amplitude A = 2 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have x(t ) = 2 A § 2π sin ¨ π © T Thus, for this problem, · A § 4π t ¸ + sin ¨ ¹ π © T 2 · 4 t ¸ = sin (40π t ) + sin (80π t ) π ¹ π c1 = 4 π ω1 = 40π = 125.6 rad/s c2 = 2 π ω 2 = 80π = 251.2 rad/s H v ( jω ) = 1 − 0.0002ω 2 + 0.1 jω = 1 − 0.0002ω 2 + 0.15 jω and, The frequency response of the system can be expressed in magnitude and phase form: = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 ª 0.15ω 0.1ω § · § ·º ∠ «arctan¨ − arctan¨ 2 ¸ 2 ¸» © 1 − 0.0002ω ¹ © 1 − 0.0002ω ¹ ¼ ¬ 6.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 At this point, we could evaluate the frequency response of the system at the frequencies analytically: H v ( jω1 ) = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 1 2 2 1 2 1 2 1 ω1 and ω2 = 0.6720 · § § 0.15ω1 · 0.1ω1 ¸ − arctan¨¨ ¸ = 0.0561 rad = 3.21, Φ( jω1 ) = arctan¨¨ 2 ¸ 2 ¸ 1 0 . 0002 1 0 . 0002 − ω − ω 1 ¹ 1 ¹ © © H v ( jω 2 ) = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 2 2 2 2 = 0.7020 · § § 0.15ω 2 · 0.1ω 2 ¸ − arctan¨¨ ¸ = 0.1342 rad = 7.69, Φ( jω 2 ) = arctan¨¨ 2 ¸ 2 ¸ © 1 − 0.0002ω 2 ¹ © 1 − 0.0002ω 2 ¹ Finally, we can compute the steady-state periodic output of the system: 2 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ( jω n )] = n =1 4 2 = 0.6720 sin (40π t + 0.0561) + 0.7020 sin (80π t + 0.1342) π π The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.42 Solution: Known quantities: The frequency response peak amplitude H v ( jω ) of the circuit of P6.4, the periods T1 = 0.5 s and T2 = 5 ms and the A = 2 for the sawtooth waveform. 6.50 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Find: Output of system y (t ) in response to input x(t ) . Compare the plots with the one obtained in P6.41. Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have 2 A § 2π · A § 4π · t¸ sin ¨ t ¸ + sin ¨ x(t ) = π © T ¹ π © T ¹ Thus, for this problem, 4 2 sin (4π t ) + sin (8π t ) π π 4 2 x2 (t ) = sin (400π t ) + sin (800π t ) π π x1 (t ) = c1,1 = c1, 2 = 4 π ω1,1 = 4π = 12.56 rad/s ω1, 2 = 400π = 1256 rad/s c2,1 = c2, 2 = 2 π ω 2,1 = 8π = 25.12 rad/s ω 2, 2 = 800π = 2512 rad/s and, The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = = 1 − 0.0002ω 2 + 0.1 jω = 1 − 0.0002ω 2 + 0.15 jω [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 ª 0.15ω 0.1ω § · § ·º ∠ «arctan¨ − arctan¨ 2 ¸ 2 ¸» © 1 − 0.0002ω ¹ © 1 − 0.0002ω ¹ ¼ ¬ At this point, we could evaluate the frequency response of the system at the frequencies and ω 2, 2 ω1,1 , ω 2,1 , ω1, 2 analytically: H v ( jω1,1 ) = 0.7486 , Φ( jω1,1 ) = −0.1820 rad = −10.43, H v ( jω 2,1 ) = 0.6876 , Φ( jω 2,1 ) = −0.1068 rad = −6.12 , H v ( jω1, 2 ) = 0.9238 , Φ( jω1, 2 ) = 0.1597 rad = 9.15, H v ( jω 2, 2 ) = 0.9770 , Φ( jω 2, 2 ) = 0.0937 rad = 5.37 , and, Finally, we can compute the steady-state periodic outputs of the system: y1 (t ) = ¦ H v ( jω n ,1 ) cn,1 sin [ω n ,1t + θ n ,1 + Φ( jω n,1 )] = 2 n =1 and, 4 2 = 0.7486 sin (4π t − 0.182) + 0.6876 sin (8π t − 0.1068) π π y 2 (t ) = ¦ H v ( jω n , 2 ) c n, 2 sin [ω n , 2 t + θ n , 2 + Φ ( jω n, 2 )] = 2 n =1 4 2 = 0.9238 sin (400π t + 0.1597 ) + 0.977 sin (800π t + 0.0937 ) π π The input and output steady state signals plot for T1 = 0.5 s is shown below: 6.51 G. Rizzoni, Principles and Applications of Electrical Engineering The input and output steady state signals plot for Comparing the results with T2 = 5 ms is shown below: T = 50 ms , we have: 6.52 Problem solutions, Chapter 6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.43 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.4, the period T = 50 ms and the peak amplitude A = 2 for the square wave. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the square waveform of P6.33, we have x(t ) = 4 A § 2π sin ¨ π © T Thus, for this problem, c1 = and, 8 π · 8 t ¸ = sin (40π t ) ¹ π ω1 = 40π = 125.6 rad/s c2 = 0 The frequency response of the system can be expressed in magnitude and phase form: 1 − 0.0002ω 2 + 0.1 jω = H v ( jω ) = 1 − 0.0002ω 2 + 0.15 jω = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 ª 0.15ω 0.1ω § · § ·º ∠ «arctan¨ − arctan¨ 2 ¸ 2 ¸» © 1 − 0.0002ω ¹ © 1 − 0.0002ω ¹ ¼ ¬ At this point, we could evaluate the frequency response of the system at the frequency 6.53 ω1 analytically: G. Rizzoni, Principles and Applications of Electrical Engineering H v ( jω1 ) = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 1 2 2 1 2 1 2 1 Problem solutions, Chapter 6 = 0.6720 · § § 0.15ω1 · 0.1ω1 ¸ − arctan¨¨ ¸ = 0.0561 rad = 3.21, Φ( jω1 ) = arctan¨¨ 2 ¸ 2 ¸ 1 0 . 0002 1 0 . 0002 − ω − ω 1 ¹ 1 ¹ © © Finally, we can compute the steady-state periodic output of the system: 2 8 y (t ) = ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = 0.672 sin (40π t + 0.0561) π n =1 The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.44 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.4, the period T = 50 ms and the peak amplitude A = 2 for the pulse waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first two terms of the Fourier series expansion of the pulse waveform of Example 6.4, we have x(t ) = 0.4 + 0.6054 cos(40π t ) + 0.4398 sin (40π t ) + 0.1870 cos(80π t ) + 0.5758 sin (80π t ) Thus, for this problem, c0 = 0.4 c1 = 0.7484 , θ1 = 0.9425 rad = 54 , ω1 = 40π = 125.6 rad/s and, 6.54 G. Rizzoni, Principles and Applications of Electrical Engineering c2 = 0.6054 , θ 2 = 0.3140 rad = 18, Problem solutions, Chapter 6 ω 2 = 80π = 251.2 rad/s The frequency response of the system can be expressed in magnitude and phase form: H v ( jω ) = = 1 − 0.0002ω 2 + 0.1 jω = 1 − 0.0002ω 2 + 0.15 jω [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 ª 0.15ω 0.1ω § · § ·º ∠ «arctan¨ − arctan¨ 2 ¸ 2 ¸» © 1 − 0.0002ω ¹ © 1 − 0.0002ω ¹ ¼ ¬ At this point, we could evaluate the frequency response of the system at the frequencies analytically: H v ( jω1 ) = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 1 2 2 1 2 1 2 1 ω1 and ω2 = 0.6720 · § § 0.15ω1 · 0.1ω1 ¸ − arctan¨¨ ¸ = 0.0561 rad = 3.21, Φ( jω1 ) = arctan¨¨ 2 ¸ 2 ¸ 1 0 . 0002 1 0 . 0002 − ω − ω 1 ¹ 1 ¹ © © H v ( jω 2 ) = [1 − 0.0002ω ] + 0.01ω [1 − 0.0002ω ] + 0.0225ω 2 2 2 2 2 2 2 2 2 2 = 0.7020 · § § 0.15ω 2 · 0.1ω 2 ¸ − arctan¨¨ ¸ = 0.1342 rad = 7.69, Φ( jω 2 ) = arctan¨¨ 2 ¸ 2 ¸ © 1 − 0.0002ω 2 ¹ © 1 − 0.0002ω 2 ¹ Finally, we can compute the steady-state periodic output of the system: 2 y (t ) = c0 H v (0) + ¦ H v ( jω n ) cn sin[ω n t + θ n + Φ ( jω n )] = n =1 = 0.4 + 0.503 sin (40π t + 0.9986 ) + 0.425 sin (80π t + 0.4482 ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ 6.55 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.45 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.6, the period T = 5 s and the peak amplitude A = 1 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have 2 A § 2π · A § 4π · 2 A § 6π · sin ¨ t ¸ + sin ¨ t¸+ sin ¨ t ¸ π © T ¹ π © T ¹ 3π © T ¹ 2 1 2 = sin (0.4π t ) + sin (0.8π t ) + sin (1.2π t ) 3π π π x(t ) = Thus, for this problem, 2 π 1 c2 = π ω1 = 0.4π = 1.256 rad/s c1 = and, c3 = ω 2 = 0.8π = 2.512 rad/s 2 3π ω3 = 1.2π = 3.768 rad/s The frequency response of the system can be expressed in magnitude and phase form: ª § 2.6ω ·º ∠ «90 , − arctan¨ 2 ¸» 2 © 1 − ω ¹¼ 1 − ω 2 + 6.76ω 2 ¬ At this point, we could evaluate the frequency response of the system at the frequencies ω1 , ω 2 and ω3 H v ( jω ) = 2 jω = 1 − ω + 2.6 jω ( analytically: H v ( jω1 ) = 2 ) ( 2ω1 (1 − ω ) 2 2 1 + 6.76ω 2 1 2ω ) = 0.7575 § 2.6ω1 · ¸ = −0.1750 rad = −10 , Φ( jω1 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω1 ¹ 2ω 2 H v ( jω 2 ) = = 0.5969 (1 − ω22 )2 + 6.76ω22 § 2.6ω 2 · ¸ = −0.6826 rad = −39.11, Φ( jω 2 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω2 ¹ 2ω3 H v ( jω 3 ) = = 0.4585 2 2 2 (1 − ω3 ) + 6.76ω3 6.56 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 § 2.6ω3 · ¸ = −0.9322 rad = −53.41, Φ( jω3 ) = 90 , − arctan¨¨ 2 ¸ 1 ω − 3 ¹ © Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = ¦ H v ( jω n ) c n sin[ω n t + θ n + Φ( jω n )] = n =1 = 0.4822 sin (0.4π t − 0.175) + 0.19 sin (0.8π t − 0.6826 ) + 0.0973 sin (1.2π t − 0.9322 ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ Problem 6.46 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.6, the period T = 50 s and the peak amplitude A = 1 for the sawtooth waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the sawtooth waveform of Example 6.3, we have 2 A § 2π · A § 4π · 2 A § 6π · sin ¨ t ¸ + sin ¨ t¸+ sin ¨ t¸ π © T ¹ π © T ¹ 3π ©T ¹ 2 1 2 = sin (0.04π t ) + sin (0.08π t ) + sin (0.12π t ) 3π π π x(t ) = Thus, for this problem, 6.57 G. Rizzoni, Principles and Applications of Electrical Engineering 2 π 1 c2 = π ω1 = 0.04π = 0.1256 rad/s c1 = and, c3 = Problem solutions, Chapter 6 ω 2 = 0.08π = 0.2512 rad/s 2 3π ω3 = 0.12π = 0.3768 rad/s The frequency response of the system can be expressed in magnitude and phase form: ª § 2.6ω ·º ∠ «90 , − arctan¨ 2 ¸» © 1 − ω ¹¼ 1 − ω + 6.76ω 2 ¬ At this point, we could evaluate the frequency response of the system at the frequencies ω1 , ω 2 and ω3 H v ( jω ) = 2 jω = 1 − ω + 2.6 jω ( analytically: H v ( jω1 ) = 2 ) ( 2ω1 (1 − ω ) 2 2 1 + 6.76ω12 2ω ) 2 2 = 0.2422 § 2.6ω1 · ¸ = 1.2504 rad = 71.64 , Φ( jω1 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω1 ¹ 2ω 2 H v ( jω 2 ) = = 0.4399 2 2 2 (1 − ω2 ) + 6.76ω2 § 2.6ω 2 · ¸ = 0.9620 rad = 55.12 , Φ( jω 2 ) = 90 , − arctan¨¨ 2 ¸ 1 − ω 2 ¹ © 2ω3 H v ( jω 3 ) = = 0.5787 (1 − ω32 )2 + 6.76ω32 § 2.6ω3 · ¸ = 0.7193 rad = 41.21, Φ( jω3 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω3 ¹ Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = ¦ H v ( jω n ) c n sin[ω n t + θ n + Φ( jω n )] = n =1 = 0.1542 sin (0.4π t + 1.2504 ) + 0.14 sin (0.8π t + 0.9620 ) + 0.1228 sin (1.2π t + 0.7193) The input and output steady state signals plot is shown below: 6.58 G. Rizzoni, Principles and Applications of Electrical Engineering Comparing the results with Problem solutions, Chapter 6 T = 50 s , we have: ______________________________________________________________________________________ Problem 6.47 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.6, the period T = 5 s and the peak amplitude A = 1 for the square waveform. 6.59 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Find: y (t ) in response to input x(t ) . Output of system Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the square waveform of P6.33, we have x(t ) = 4 A § 2π · 4 A § 6π · 4 4 sin ¨ t¸+ sin ¨ t ¸ = sin (0.4π t ) + sin (1.2π t ) π 3π © T ¹ 3π © T ¹ π Thus, for this problem, 4 π 4 c3 = 3π c2 = 0 ω1 = 0.4π = 1.256 rad/s c1 = ω3 = 1.2π = 3.768 rad/s The frequency response of the system can be expressed in magnitude and phase form: ª § 2.6ω ·º ∠ «90 , − arctan¨ 2 ¸» © 1 − ω ¹¼ 1 − ω + 6.76ω 2 ¬ At this point, we could evaluate the frequency response of the system at the frequencies ω1 and ω3 H v ( jω ) = 2 jω = 1 − ω + 2.6 jω ( analytically: H v ( jω1 ) = 2 ) ( 2ω1 (1 − ω ) 2 2 1 + 6.76ω12 2ω ) 2 2 = 0.7575 § 2.6ω1 · ¸ = −0.1750 rad = −10 , Φ( jω1 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω1 ¹ 2ω3 H v ( jω 3 ) = = 0.4585 2 2 2 (1 − ω3 ) + 6.76ω3 § 2.6ω3 · ¸ = −0.9322 rad = −53.41, Φ( jω3 ) = 90 , − arctan¨¨ 2 ¸ 1 − ω 3 ¹ © Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = ¦ H v ( jω n ) c n sin[ω n t + θ n + Φ ( jω n )] = n =1 = 0.9644 sin (0.4π t − 0.175) + 0.1946 sin (1.2π t − 0.9322 ) The input and output steady state signals plot is shown below: 6.60 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.48 Solution: Known quantities: The frequency response H v ( jω ) of the circuit of P6.6, the period T = 5 s and the peak amplitude A = 1 for the pulse waveform. Find: Output of system y (t ) in response to input x(t ) . Analysis: According to the Fourier series definitions of the previous section, and using the first three terms of the Fourier series expansion of the pulse waveform of Example 6.4, we have x(t ) = 0.2 + 0.3027 cos(0.4π t ) + 0.2199 sin (0.4π t ) + 0.0935 cos(0.8π t ) + 0.2879 sin (0.8π t ) − 0.0624 cos(1.2π t ) + 0.1919 sin (1.2π t ) Thus, for this problem, c0 = 0.2 c1 = 0.3742 , c2 = 0.3027 , θ1 = 0.9425 rad = 54 , θ 2 = 0.3140 rad = 18, ω1 = 0.4π = 1.256 rad/s ω 2 = 0.8π = 2.512 rad/s c3 = 0.2018 , θ 3 = −0.3144 rad = −18, ω3 = 1.2π = 3.768 rad/s and, The frequency response of the system can be expressed in magnitude and phase form: ª § 2.6ω ·º ∠ «90 , − arctan¨ 2 ¸» © 1 − ω ¹¼ 1 − ω + 6.76ω 2 ¬ At this point, we could evaluate the frequency response of the system at the frequencies ω1 , ω 2 and ω3 H v ( jω ) = 2 jω = 1 − ω + 2.6 jω ( 2 ) ( 2ω ) 2 2 analytically: 6.61 G. Rizzoni, Principles and Applications of Electrical Engineering H v ( jω1 ) = 2ω1 (1 − ω ) 2 2 1 + 6.76ω 2 1 Problem solutions, Chapter 6 = 0.7575 § 2.6ω1 · ¸ = −0.1750 rad = −10 , Φ( jω1 ) = 90 , − arctan¨¨ 2 ¸ 1 − ω 1 ¹ © 2ω 2 H v ( jω 2 ) = = 0.5969 (1 − ω22 )2 + 6.76ω22 § 2.6ω 2 · ¸ = −0.6826 rad = −39.11, Φ( jω 2 ) = 90 , − arctan¨¨ 2 ¸ © 1 − ω2 ¹ 2ω3 H v ( jω 3 ) = = 0.4585 2 2 2 (1 − ω3 ) + 6.76ω3 § 2.6ω3 · ¸ = −0.9322 rad = −53.41, Φ( jω3 ) = 90 , − arctan¨¨ 2 ¸ 1 − ω 3 ¹ © Finally, we can compute the steady-state periodic output of the system: 3 y (t ) = c0 H v (0) + ¦ H v ( jω n ) cn sin [ω n t + θ n + Φ ( jω n )] = n =1 = 0.2834 sin (0.4π t + 0.7675) + 0.1807 sin (0.8π t − 0.3686 ) + 0.0925 sin (1.2π t − 1.2466 ) The input and output steady state signals plot is shown below: ______________________________________________________________________________________ 6.62 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.49 Solution: Known quantities: Resistance , capacitance, and inductance values, in the circuit of Figure P6.49. Find: The resonant frequency and the bandwidth fort the circuit. Analysis: Taking the output as the voltage across the parallel R-C subcircuit, Vo = VS § · ω n2 µ 1 / LC 3 / 64 ¨ ¸ = = 2 2 2 ¨ ¸ 1 1 2 + + + + ( j ω ) j ω 2 3 / 64 ( j ω ) j ω ( 2 ξω ) ω n n © ¹ ( jω ) + jω + RC LC The corresponding Bode diagrams are shown below: In this circuit, as frequency increases, the impedance of the capacitor decreases and the impedance of the inductor increases. Both effects cause the magnitude of the output voltage to decrease so this is a 2nd order low pass filter. The resonance frequency is, ωn = 1 64 = ≅ 4.6188 rad/s. LC 3 The damping ratio is, ξ= 1 / RC 3 = ≅ 0.2165 2ω n 8 The quality factor is, Q= 1 4 = ≅ 2.3094 2ξ 3 The bandwidth is, 6.63 G. Rizzoni, Principles and Applications of Electrical Engineering B= ωn 8 1 = =2 Q 3 4/ 3 Problem solutions, Chapter 6 rad/s. ______________________________________________________________________________________ Problem 6.50 Solution: Known quantities: Figure P6.50. Find: What kind of filters are the ones shown in Figure P6.50. Analysis: In a), as frequency increases, the impedance of the capacitor decreases and the impedance of the inductor increases. Both effects cause the magnitude of the output voltage to decrease so this is a 2nd order low pass filter. Note that L and C are connected neither in series nor parallel and do not form a resonant circuit. In b), L and C are connected in series and form a series resonant circuit with an impedance which is minimum at the resonant frequency and larger above and below the resonant frequency. This series resonant circuit is in series with the output giving, because of voltage division, a maximum output voltage at the resonant frequency and less at higher and lower frequencies. Therefore, b) is a band-pass filter. In c), L and C are connected in parallel and form a parallel resonant circuit with an impedance which is maximum at the resonant frequency and smaller above and below the resonant frequency. This parallel resonant circuit is in parallel with the output giving, because of voltage division, a maximum output at the resonant frequency and less at higher and lower frequencies. Therefore, c) is a band-pass filter. ______________________________________________________________________________________ Problem 6.51 Solution: Known quantities: Figure P6.51. Find: What kind of filters are the ones shown in Figure P6.51. Analysis: None of the inductors or capacitors is in series or parallel with any other. Therefore, there are no series or parallel resonant circuits and none of the circuits shown is band pass or band stop filters. Circuits a) and d): As frequency approaches infinity, the inductors can be modeled as open circuits and the capacitors as short circuits. Therefore, the voltage transfer function approaches zero. As frequency approaches zero, the inductors can be modeled as short circuits and the capacitors as open circuits. Then: VD : H v → RL Rs + R L Therefore, circuits a) and d) are low pass filters. Circuits b) and c) As frequency approaches infinity, the inductors can be modeled as open circuits and the capacitors as short circuits. Then, VD : H v → RL Rs + RL 6.64 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Therefore: As frequency approaches zero, inductors can be modeled as short circuits and the capacitors as open circuits. The voltage transfer function approaches zero. Therefore, circuits b) and c) are high pass filters. Note: Multiple capacitors and inductors give higher order low and high pass filter. Better performance is obtained outside the pass band where the response for these circuits decreases by 60 dB/decade. In first order filters, the response decreases by only 20 dB/decade. ______________________________________________________________________________________ Problem 6.52 Solution: Known quantities: Figure P6.52. Find: a) If this is a low-pass, high-pass, band-pass, or band-stop filter. b) Compute and plot the frequency response function if: L = 11 mH C = 0.47 nF R1 = 2.2 kΩ R2 = 3.8 kΩ Analysis: a) As ω → 0 : Z L → 0 Short Z C → ∞ Open VD : H v = V o → R2 R1 + R2 Vi As ω → ∞ : Z L → ∞ Open Z C → 0 Short Hv → 0 The filter is a low pass filter. b) First, we find the Thévenin equivalent circuit seen by the capacitor: Z T = (Z R1 + Z L ) || Z R 2 and vOC = § (R + jωL )R2 1 1 · = ¨¨ + ¸¸ = 1 R1 + jωL + R2 © R1 + jωL R2 ¹ −1 Z R2 R2 vin = vin Z R1 + Z L + Z R 2 R1 + jωL + R2 6.65 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 1 vout ZC R1 + jωL + R2 jω C = = = + R j L R ω ( ) vOC Z T + Z C R1 + jωL + R2 + (R1 + jωL ) jω CR2 1 2 + 1 j C ω R1 + jωL + R2 Therefore, vout R2 R1 + jωL + R2 = ⋅ vin R1 + jωL + R2 R1 + jωL + R2 + (R1 + jωL ) jω CR2 1 = 1+ § L · R1 2 + jω ¨¨ + CR1 ¸¸ + ( jω ) LC R2 © R2 ¹ Substituting the numerical values: vout 1 = −12 vin 1.579 − 5.17 × 10 ω 2 + j 3.929 × 10 −6 ω ( ) ( ) The corresponding Bode diagrams are shown below: ______________________________________________________________________________________ Problem 6.53 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.53. Find: Compute and plot the frequency response function. What type of filter is this? Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: 6.66 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 § · 1 1 ¸¸ Z T = Z C || [(Z Rs + Z L ) || (Z Rc + Z L )] = ¨¨ jωC + + R j ω L R j ω L + + S C © ¹ (RS + jωL )(RC + jωL ) = jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) −1 and, by node analysis, vOC = (RC + jωL ) Z C || [(Z Rs + Z L ) || (Z Rc + Z L )] vin = vin Z Rs + Z L jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) Z RL vout RL = = = vOC Z T + Z RL Z T + RL 1 = ZT 1+ RL jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) + (RS + jωL )(RC + jωL ) / RL Therefore, (RC + jωL ) vout = vin jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) + (RS + jωL )(RC + jωL ) / RL Substituting the numerical values: vout 8 ×1017 + j (2 × 1012 )ω = vin ( jω )3 + ( jω )2 9 × 10 5 + j (4.24 ×1012 )ω + 1.016 × 1018 The corresponding Bode diagrams are shown below: The magnitude of the voltage transfer function is highest at the resonant frequency and decreases at higher and lower frequencies. Therefore, this is a band-pass filter. 6.67 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 However, it is not a particularly good filter since the voltage gain [or actually insertion loss] is not very different at the resonant and lower frequencies. This is due to the large inductor losses modeled here as the equivalent resistance Rc. This causes a low "Q" circuit. ______________________________________________________________________________________ Problem 6.54 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.53. Find: Compute and plot the frequency response function. What type of filter is this? Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: · § 1 1 ¸ Z T = Z C || [(Z Rs + Z L ) || (Z Rc + Z L )] = ¨¨ jωC + + RS + jωL RC + jωL ¸¹ © (RS + jωL )(RC + jωL ) = jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) −1 and, by node analysis, vOC = (RC + jωL ) Z C || [(Z Rs + Z L ) || (Z Rc + Z L )] vin = vin Z Rs + Z L jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) Z RL vout RL = = = vOC Z T + Z RL Z T + RL 1 = ZT 1+ RL jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) + (RS + jωL )(RC + jωL ) / RL Therefore, (RC + jωL ) vout = vin jωC (RS + jωL )(RC + jωL ) + (RS + RC + j 2ωL ) + (RS + jωL )(RC + jωL ) / RL Substituting the numerical values: vout 8 ×1015 + j (2 × 1012 )ω = vin ( jω )3 + ( jω )2 5.04 ×10 5 + j (4.042 × 1012 )ω + 2.082 ×1017 The corresponding Bode diagrams are shown below: 6.68 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 The magnitude of the voltage transfer function is highest at the resonant frequency and decreases at higher and lower frequencies. Therefore, this is a band-pass filter. Note: The inductor or coil loss is much smaller [4 Ω] in this circuit, which gives much better band-pass filter performance or a higher "Q" circuit. The magnitude of the voltage transfer ratio or voltage gain [or insertion loss] is much higher at resonance than at higher or lower frequencies. ______________________________________________________________________________________ Problem 6.55 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.55: R s = 5 kΩ C = 56 nF R L = 100 kΩ L = 9 µH Find: a) An expression for the voltage transfer function: H v ( jω ) = b) The resonant frequency. c) The half-power frequencies. d) The bandwidth and Q. Analysis: a) 6.69 V0 ( jω ) Vi ( jω ) G. Rizzoni, Principles and Applications of Electrical Engineering Z eq = 1 ZC + 1 1 ZL + VD : H v ( jω ) = H v ( jω ) = 1 Problem solutions, Chapter 6 jωLR L 1 = 2 1 1 ( jω ) LCRL + jωL + RL jωC + + jωL RL = Z RL V0 ( jω ) Z eq = Vi ( jω ) Z RS + Z eq jωLR L ( jω ) LCRL + jωL + RL = jωLR L RS + 2 ( jω ) LCRL + jωL + RL 2 jωLR L 1 = ( jω ) LCRS RL + jωL(RL + RS ) + RS RL RS 2 jω L § ( jω )2 LC + jωL¨¨ RL + RS © RS RL · ¸¸ + 1 ¹ b) The resonance frequency is, 1 ≅ 1.4086 Mrad/s. LC ωn = c) H v ( jω hp ) = 1 RS jω hp L ( jω ) LC + jω 2 hp 1 RS hp § R + RS L¨¨ L © RS R L ω hp L 2 = · ¸¸ + 1 ¹ = 1 2 1 2 § § R + RS · · ¸¸ ¸ 1 − ω LC + ¨¨ ω hp L¨¨ L ¸ R R S L ¹¹ © © ω hp1 ≅ 1.4069 Mrad/s and ω hp 2 ≅ 1.41028 Mrad/s. ( 2 hp ) 2 d) The damping ratio is, ξ= ωn 2 § R + RS L¨¨ L © RS RL The quality factor is, Q= · ¸¸ ≅ 0.0013 ¹ 1 ≅ 375.62 2ξ The bandwidth is, B= ωn ≅ 3.75 Q Krad/s. Notes: 1. The absence of coil resistance caused the gain at the resonant frequency to be much higher than at high and low frequencies. 2. The bandwidth is small compared with the resonant frequency and the "Q" is quite large. These are dependent on the "loading" or power dissipation of the source and load resistors and the capacitance. 3. A circuit with a high Q is "selective" since it will pass a very narrow band of frequencies. "High" Q circuits have a Q = 10 or more. ______________________________________________________________________________________ 6.70 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.56 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.55: R s = 5 kΩ C = 0.5 nF R L = 100 kΩ L = 1 mH Find: a) H v ( jω ) = An expression for the voltage transfer function: V0 ( jω ) Vi ( jω ) b) The resonant frequency. c) The half-power frequencies. d) The bandwidth and Q. Analysis: a) Z eq = 1 ZC + 1 1 ZL + VD : H v ( jω ) = H v ( jω ) = 1 jωLR L 1 = 2 1 1 ( jω ) LCRL + jωL + RL jωC + + jωL RL = Z RL V0 ( jω ) Z eq = Vi ( jω ) Z RS + Z eq jωLR L ( jω ) LCRL + jωL + RL = jωLR L RS + 2 ( jω ) LCRL + jωL + RL 2 jωLR L 1 = ( jω ) LCRS RL + jωL(RL + RS ) + RS RL RS 2 jω L § ( jω )2 LC + jωL¨¨ RL + RS © RS RL b) The resonance frequency is, 1 ≅ 1.4142 Mrad/s. LC ωn = c) H v ( jω hp ) = 1 RS jω hp L ( jω ) LC + jω 2 hp 1 RS ω hp1 hp § R + RS L¨¨ L © RS R L ω hp L 2 = § § R + RS · · 2 ¸¸ ¸ LC + ¨¨ ω hp L¨¨ L ¸ R R © S L ¹¹ © ≅ 1.24 Mrad/s and ω hp 2 ≅ 1.62 Mrad/s. (1 − ω 2 hp ) d) The damping ratio is, ξ= ωn 2 § R + RS L¨¨ L © RS R L · ¸¸ ≅ 0.1485 ¹ 6.71 · ¸¸ + 1 ¹ = 1 2 1 2 · ¸¸ + 1 ¹ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 The quality factor is, Q= 1 ≅ 3.3672 2ξ The bandwidth is, B= ωn ≅ 420 Q Krad/s. ______________________________________________________________________________________ Problem 6.57 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.57. Find: Compute and plot the voltage frequency response function. What type of filter is this? Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: −1 § · 1 ¸¸ Z T = Z Rs + Z C || (Z Rc + Z L ) = RS + ¨¨ jωC + ω + R j L C © ¹ (RC + jωL ) = (RC + jωL ) + RS [1 + jωC (RC + jωL )] = RS + jωC (RC + jωL ) + 1 jωC (RC + jωL ) + 1 and vOC = vin Therefore, Z RL vout RL = = = vin Z T + Z RL Z T + RL = 1 = ZT 1+ RL jωC (RC + jωL ) + 1 = jωC (RC + jωL ) + 1 + {(RC + jωL ) + RS [1 + jωC (RC + jωL )]}/ RL 1 + jωCRC + ( jω ) LC = ª § RS · § RC + RS · § R · Lº 2 ¸¸ ¨¨1 + ¸¸ + jω «CRC ¨¨1 + S ¸¸ + » + ( jω ) LC ¨¨1 + RL ¹ © RL ¹ © © RL ¹ RL ¼ ¬ 2 Substituting the numerical values: ( ) vout 1 + j 2 ×10 −8 ω + ( jω ) 5 × 10 −15 = vin ( jω )2 5.5 × 10 −15 + j 2.22 ×10 −7 ω + 1.9 2 ( ) The corresponding Bode diagrams are shown below: 6.72 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 The magnitude of the voltage transfer function is lowest at the resonant frequency and increases at higher and lower frequencies. Therefore, this is a band stop or "notch" filter. At its resonant frequency, a parallel resonant circuit has a high equivalent resistance that is resistive. Connected here in series with the load, this high impedance reduces the magnitude of the voltage transfer function [or voltage gain or insertion loss] at the resonant frequency. The loading due to the inductor losses, modeled here as an equivalent "coil" resistance, is fairly small giving a substantially lower gain at the resonant frequency compared with the gain at higher or lower frequencies. Therefore this is a high "Q" circuit with good performance and selectivity. The inductor losses also affect only slightly the resonant frequency. The cutoff frequencies are difficult [but not impossible] to determine in circuits containing a parallel resonant circuit which includes inductor losses, so no attempt was made to do so. ______________________________________________________________________________________ Problem 6.58 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.58. Find: Compute and plot the frequency response function. Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: −1 Z T = Z Rs and [ § jωL + RS 1 + ( jω ) LC 1 · ¸¸ = + Z C || Z L = RS + ¨¨ jωC + 2 jω L ¹ 1 + ( jω ) LC © vOC = vin Therefore, 6.73 2 ] G. Rizzoni, Principles and Applications of Electrical Engineering Z RL vout RL = = = vin Z T + Z RL Z T + RL Problem solutions, Chapter 6 1 1 + ( jω ) LC = = 2 Z T 1 + ( jω ) LC + jωL + RS 1 + ( jω )2 LC / RL 1+ RL 2 { [ ]} 1 + ( jω ) LC § RS · § R · L 2 ¨¨1 + ¸¸ + jω + ( jω ) LC ¨¨1 + S ¸¸ RL © RL ¹ © RL ¹ 2 = Substituting the numerical values: vout 1 + ( jω ) 5 × 10 −15 = vin ( jω )2 5.5 ×10 −15 + j 2 × 10 −7 ω + 1.1 2 ( ) The corresponding Bode diagrams are shown below: ______________________________________________________________________________________ Problem 6.59 Solution: Known quantities: The filter circuit shown in Figure P6.58. Find: The equation for the voltage transfer function in standard form. Then, if: R s = 500 Ω R L = 5 kΩ ω n = 12.1278 M rad s determine the cutoff frequencies, the bandwidth, BW, and Q. Analysis First, we find the Thévenin equivalent circuit seen by the capacitor: 6.74 C = 68 nF L = 0.1 µH G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 [ § jωL + RS 1 + ( jω ) LC 1 · ¸¸ = + Z C || Z L = RS + ¨¨ jωC + 2 jω L ¹ 1 + ( jω ) LC © −1 Z T = Z Rs and 2 ] vOC = vin Therefore, 1 1 + ( jω ) LC = Z 1 + T §¨1 + RS ·¸ + jω L + ( jω )2 LC §¨1 + RS ·¸ ¨ R ¸ RL ¨© RL ¸¹ RL L ¹ © 2 RL RL 1 − ω LC 1 = = L L R S + R L 1 + jω RS + RL 1 − ω 2 LC + jω RS + R L (RS + RL ) 1 − ω 2 LC 2 Z RL vout RL = = = vin Z T + Z RL Z T + RL ( The corresponding Bode diagrams are shown below: Then, in order to calculate the half-power frequencies, we have to solve: H v ( jω hp ) RL = RS + RL 2 1 − ω hp LC § 2 L · ¸¸ LC + ¨¨ ω hp R + R S L ¹ © ω hp1 ≅ 12.061 Mrad/s and ωhp 2 ≅ 12.194 Mrad/s. (1 − ω 2 hp ) The damping ratio is, ξ= ωn L ≅ 1.1024 × 10 −4 2 RL + RS The quality factor is, Q= 1 ≅ 4535.4 2ξ The bandwidth is, 6.75 2 = 1 2 ) G. Rizzoni, Principles and Applications of Electrical Engineering B= ωn ≅ 2.6738 Q Problem solutions, Chapter 6 Krad/s. ______________________________________________________________________________________ Problem 6.60 Solution: Known quantities: The filter circuit shown in Figure P6.58. Find: The equation for the voltage transfer function in standard form. Then, if: rad s determine the cutoff frequencies, the bandwidth, BW, and Q. R s = 4 .4 k Ω R L = 6 0 k Ω ω r = 25 M Analysis First, we find the Thévenin equivalent circuit seen by the capacitor: and [ § jωL + RS 1 + ( jω ) LC 1 · ¸¸ = + Z C || Z L = RS + ¨¨ jωC + 2 jω L ¹ 1 + ( jω ) LC © −1 Z T = Z Rs C = 0.8 nF 2 L = 2 µH ] vOC = vin Therefore, 1 1 + ( jω ) LC = Z 1 + T §¨1 + RS ·¸ + jω L + ( jω )2 LC §¨1 + RS ·¸ ¨ R ¸ RL ¨© RL ¸¹ RL L ¹ © 2 RL RL 1 − ω LC 1 = = L L RS + RL 1 − ω 2 LC + jω R S + R L 1 + jω RS + R L (RS + RL ) 1 − ω 2 LC 2 Z RL vout RL = = = vin Z T + Z RL Z T + RL ( The corresponding Bode diagrams are shown below: 6.76 ) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Then, in order to calculate the half-power frequencies, we have to solve: H v ( jω hp ) = RL RS + RL 2 1 − ω hp LC § L · ¸ 1 − ω LC + ¨¨ ω hp RS + RL ¸¹ © ω hp1 ≅ 24.8609 Mrad/s and ω hp 2 ≅ 24,066 Mrad/s. ( ) 2 = 2 2 hp 1 2 The damping ratio is, ξ= ωn L ≅ 3.882 × 10 −4 2 RL + RS The quality factor is, Q= 1 = 1288 2ξ The bandwidth is, B= ωn ≅ 19.41 Krad/s. Q ______________________________________________________________________________________ Problem 6.61 Solution: Known quantities: The bandstop filter circuit shown in Figure P6.61, where: L = 0.4 mH Rc = 100 Ω C = 1 pF R s = R L = 3.8 kΩ Find: a) An expression for the voltage transfer function or gain in the form: 1+ j f 1 [ ω ] V o [jω ] = H v [jω ] = Ho 1+ j f 2 [ ω ] V i [jω ] 6.77 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 b) The magnitude of the function at high and low frequencies and at the resonant frequency. c) The resonant frequency. d) The half power frequencies. Analysis a) 1 ] ωC 1 1 ( Rc + j [ ωL ] ) RL ] Rc R L + j [ ωL N Z eq1 Z RL ω C C ω = = = Z eq = 1 1 D Z eq1 + Z RL ] ( R c + j [ ωL ] ) + RL R c + R L + j R L [ ωL ωC ωC N ω [j ] N Z eq V D D = (TRICKY ) VD : H v [jω ] = o = = N D D + N V i [jω ] R Z Rs + Z eq s Rs + D Z eq1 = Z Rc + Z L + Z C = Rc + j [ ωL - H v [jω ] = Rc R L + j R L [ ωL - 1 ] ωC 1 1 ] ) + ( Rc R L + j R L [ ωL ]) R s ( Rc + R L + j [ ωL ωC ωC 1 ] Rc R L + j R L [ ωL ωC = 1 ] R s [ R c + R L ] + R c R L + j [ R s + R L ] [ ωL ωC 1 1 1+ j [ ωL ] ωC R R c RL c H v [jω ] = 1 R s [ Rc + R L ] + Rc R L [ R s + R L ] [ ωL ] C ω 1+ j Rc [ R s + R L ] + R s R L b) As ω → ∞ : Z L → ∞ Open Z C → 0 Short VD : H v [jω ] → RL = 0.5 = - 6.021 dB Rs + R L As ω → 0 : Z C → ∞ Open Z L → 0 Short VD : H v [jω ] → RL = 0.5 = - 6.021 dB Rs + RL At resonance: 6.78 = G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Rc R L = R s [ Rc + R L ] + Rc R L [ 100 ] [ 3800] = = 0.025 = - 32.00 dB [ 3800 ] [ 100 + 3800 ] + [ 100 ] [ 3800 ] H[ jω r ] = H 0 = c) At the resonant frequency, the transfer function is real. This requires the two functions of frequency to be equal: [ Rs + RL ] 1 1 ] = [ ωr L ] ωr C ωr C Rc Rc [ R s + R L ] + R s R L 1 rad ωr L = 0 ω r = [ LC ] -1/2 = ([ 0.4 ⋅ 10 -3 ] [ 1 ⋅ 10 -12 ] )-1/2 = 50 M s ωr C f 1 [ω r ] = f 2 [ω r ] 1 [ ωr L - d) Then, in order to calculate the half-power frequencies, we have to solve: H v ( jω hp ) = Rc R L R s (R c + R L ) + R c R L ω hp1 ≅ 44.9922 Mrad/s and ω hp 2 § 1 § 1 ·· 1+ ¨¨ ¨ ωL ¸¸ ωC ¹ ¸¹ © Rc © 2 § 1 ·· § ¨ (Rc + R L )¨ ωL ¸¸ ωC ¹ ¸ © ¨ 1+ ¨ Rc (Rc + R L ) + R s R L ¸ ¸ ¨ ¹ © ≅ 55.565 Mrad/s. 2 = 1 2 ______________________________________________________________________________________ Problem 6.62 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.55: R s = 5 kΩ C = 5 nF R L = 50 kΩ L = 2 mH Find: a) An expression for the voltage transfer function: b) c) d) e) The resonant frequency. The half-power frequencies. The bandwidth and Q. Plot H v ( jω ) . H v ( jω ) = Analysis: a) 6.79 V0 ( jω ) Vi ( jω ) G. Rizzoni, Principles and Applications of Electrical Engineering Z eq = 1 1 1 + ZC 1 + ZL 1 = 1 1 jω C + + jω L R L = Z RL Problem solutions, Chapter 6 ( jω ) 2 jωLR L LCRL + jωL + RL jωLR L VD : H v ( jω ) = H v ( jω ) = ( jω ) 2 V0 ( jω ) ( jω ) LCRL + jωL + RL Z eq = = jωLR L Vi ( jω ) Z RS + Z eq RL + 2 ( jω ) LCRL + jωL + RL 2 jωLR L LCRS RL + jωL(RL + RS ) + RS RL 1 RS = jωL § ( jω )2 LC + jωL¨¨ RL + RS © RS R L b) The resonance frequency is, ωn = 1 ≅ 316.23 Krad/s. LC c) H v ( jω hp ) = 1 RS jω hp L ( jω ) LC + jω 2 hp 1 RS hp § R + RS L¨¨ L © RS R L ω hp L 2 = · ¸¸ + 1 ¹ 1 2 § § R + RS · · ¸¸ ¸ 1 − ω LC + ¨¨ ω hp L¨¨ L ¸ R R S L © ¹¹ © ω hp1 ≅ 1.4069 Mrad/s and ω hp 2 ≅ 1.41028 Mrad/s. ( ) 2 2 hp d) The damping ratio is, ξ= ωn 2 § R + RS L¨¨ L © RS R L The quality factor is, Q= · ¸¸ ≅ 0.0696 ¹ 1 ≅ 7.187 2ξ The bandwidth is, B= e) ωn ≅ 40 Q Krad/s. The Bode diagrams are shown below: 6.80 = 1 2 · ¸¸ + 1 ¹ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ 6.81 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Section 6.4: Bode Plots Focus on Methodology Bode plots This box illustrates the Bode plot asymptotic approximation construction procedure. The method assumes that there are no complex conjugate factors in the response, and that both the numerator and denominator can be factored into first-order terms with real roots. 1. Express the frequency response function in factored form, resulting in an expression similar to equation 6.57: j K 1 ... 1 H j j 3. 4. 5. 1 m 1 ... m 1 2. j j 1 n Select the appropriate frequency range for the semi-logarithmic plot, extending at least a decade below the lowest 3-dB frequency and a decade above the highest 3-dB frequency. Sketch the magnitude and phase response asymptotic approximations for each of the firstorder factors using the techniques illustrated in Figures 6.36 and 6.37. Add, graphically, the individual terms to obtain a composite response. If desired, apply the correction factors of Table 6.2. Problem 6.63 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.63: R1 = R2 = 1 kΩ C1 = 1 µF C 2 = 1 mF Find: a) The frequency response function H v ( jω ) = L =1H Vout ( jω ) for the circuit of Figure P6.63. Vin ( jω ) b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: −1 a) § 1 1 · Z T = Z R2 + Z C1 || Z L || Z R1 = R2 + ¨¨ jωC1 + + ¸¸ = jωL R1 ¹ © and jω VOC = L R1 ( jω ) LC1 + jω L + 1 R1 2 Vin 6.82 § · © R1 ¹ ( jω )2 LC1 + jω L +1 R1 ( jω )2 LC1 R2 + jωL¨¨1 + R2 ¸¸ + R2 G. Rizzoni, Principles and Applications of Electrical Engineering Vout VOC Thus, Problem solutions, Chapter 6 ( jω )2 LC1 + jω L +1 jω C 2 R1 = = ª § ·º § · ZT + 1 jω C 2 ( jω )3 LC1C 2 R2 + ( jω )2 L «C1 + C 2 ¨1 + R2 ¸ » + jω ¨ L + C 2 R2 ¸ + 1 ¨ ¸ ¨ ¸ R1 ¹ ¼ © © R1 ¹ ¬ Vout = Vin 1 jω L R1 ª § ·º ¬ © R1 ¹¼ § L · + C 2 R2 ¸¸ + 1 © R1 ¹ ( jω )3 LC1C2 R2 + ( jω )2 L «C1 + C2 ¨¨1 + R2 ¸¸» + jω ¨¨ b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = 10 −3 c) jω jω jω ·§ · + 1¸¨ + 1¸ © 968.361 ¹© 1031.638 ¹ ( jω + 1)§¨ The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.64 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.63: R1 = R2 = 1 kΩ C1 = 1 µF C 2 = 1 mF Find: a) The frequency response function H v ( jω ) = L =1H I out ( jω ) for the circuit of Figure P6.63. Vin ( jω ) 6.83 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. Analysis: a) Vout ( jω ) is (see P6.63 for details): Vin ( jω ) L jω R1 = ª § · § ·º ( jω )3 LC1C2 R2 + ( jω )2 L «C1 + C2 ¨¨1 + R2 ¸¸» + jω ¨¨ L + C2 R2 ¸¸ + 1 © R1 ¹ © R1 ¹¼ ¬ The frequency response function Vout Vin and, H v ( jω ) = I out = jωC 2Vout Thus, I out = Vin ( jω )2 LC2 R1 ª § ·º ¬ © R1 ¹¼ § L · + C 2 R2 ¸¸ + 1 © R1 ¹ ( jω )3 LC1C2 R2 + ( jω )2 L «C1 + C 2 ¨¨1 + R2 ¸¸» + jω ¨¨ b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H ( jω ) = 10 c) −6 ( jω )2 jω jω ·§ · + 1¸¨ + 1¸ © 968.361 ¹© 1031.638 ¹ ( jω + 1)§¨ The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ 6.84 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Problem 6.65 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.65: R1 = R2 = 1 kΩ C = 1 mF L =1H Find: a) The frequency response function H v ( jω ) = Vout ( jω ) for the circuit of Figure P6.65. I in ( jω ) b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: ( Z T = Z R1 + Z C1 a) and ) −1 § jωC § 1 · 1 · ¸¸ = ¸¸ || jωL = ¨¨ + || Z L = ¨¨ R1 + jω C ¹ © © 1 + jωCR1 jωL ¹ 2 ( jω ) LCR1 + jωL = ( jω )2 LC + jωCR1 + 1 VOC jωL = 2 I in ( jω ) LC + jωCR1 + 1 Vout R2 = = VOC Z T + R2 Thus, Vout = I in ( jω )2 LC + jωCR1 + 1 § · § © R2 ¹ © ( jω )2 LC ¨¨1 + R1 ¸¸ + jω ¨¨ CR1 + L · ¸ +1 R2 ¸¹ jωL § · § · ( jω )2 LC ¨¨1 + R1 ¸¸ + jω ¨¨ CR1 + L ¸¸ + 1 R2 ¹ © R2 ¹ © b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = c) jω ( jω + 1)§¨ jω + 1·¸ © 499.5 ¹ The sketch plots and the ones obtained using Matlab are shown below: 6.85 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.66 Solution: Known quantities: The values of the resistors, of the capacitance and of the inductance in the circuit of Figure P6.65: R1 = R2 = 1 kΩ C = 1 mF L =1H Find: a) The frequency response function H v ( jω ) = I out ( jω ) for the circuit of Figure P6.65. I in ( jω ) b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. Analysis: a) The frequency response function Vout = I in and, I out = H v ( jω ) = Vout ( jω ) is (see P6.65 for details): I in ( jω ) jωL § · § · ( jω )2 LC ¨¨1 + R1 ¸¸ + jω ¨¨ CR1 + L ¸¸ + 1 R2 ¹ © R2 ¹ © Vout R2 Thus, I out = I in jω L R2 § · § © R2 ¹ © ( jω )2 LC ¨¨1 + R1 ¸¸ + jω ¨¨ CR1 + L · ¸ +1 R2 ¸¹ 6.86 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = 10 −3 c) jω ( jω + 1)§¨ jω + 1·¸ © 499.5 ¹ The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.67 Solution: Known quantities: The values of the resistors and of the capacitances in the circuit of Figure P6.67: R1 = R2 = 1 kΩ C1 = 1 µF C 2 = 1 mF Find: a) The frequency response function H v ( jω ) = Vout ( jω ) for the circuit of Figure P6.65. I in ( jω ) b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. Analysis: First, we find the Thévenin equivalent circuit seen by the capacitor: −1 a) and Z T = Z R2 § R1 jωC1 R1 R2 + R1 + R2 1 · + Z C1 || Z R1 = R2 + ¨¨ jωC1 + ¸¸ = R2 + = R1 ¹ 1 + jωC1 R1 1 + jωC1 R1 © VOC R1 = I in 1 + jωC1 R1 6.87 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 1 Vout jωC 2 jω C 2 = = = VOC Z T + 1 ª jωC1 R1 R2 + R1 + R2 º jωC 2 1 + jωC 2 « » 1 + jωC1 R1 ¬ ¼ (1 + jωC1 R1 ) jωC2 = 2 ( jω ) C1C2 R1 R2 + jω [C1 R1 + C2 (R1 + R2 )] + 1 Thus, Vout jωC 2 R1 = 2 I in ( jω ) C1C 2 R1 R2 + jω [C1 R1 + C 2 (R1 + R2 )] + 1 b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = c) jω § jω ·§ jω · + 1¸¨ + 1¸ ¨ © 0.5 ¹© 2000 ¹ The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.68 Solution: Known quantities: The values of the resistors and of the capacitances in the circuit of Figure P6.67: R1 = R2 = 1 kΩ C1 = 1 µF C 2 = 1 mF Find: a) The frequency response function H v ( jω ) = Vout ( jω ) for the circuit of Figure P6.65. I in ( jω ) b) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. c) Use Matlab and the Bode command to generate the same plot. 6.88 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: a) The frequency response function H v ( jω ) = Problem solutions, Chapter 6 Vout ( jω ) is (see P6.67 for details): I in ( jω ) Vout jωC 2 R1 = 2 I in ( jω ) C1C 2 R1 R2 + jω [C1 R1 + C 2 (R1 + R2 )] + 1 and, I out = jωC 2Vout Thus, I out ( jωC 2 ) R1 = 2 I in ( jω ) C1C 2 R1 R2 + jω [C1 R1 + C 2 (R1 + R2 )] + 1 2 b) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = 10 c) −3 ( jω )2 § jω ·§ jω · + 1¸¨ + 1¸ ¨ © 0.5 ¹© 2000 ¹ The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.69 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.4. Find: a) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. b) Use Matlab and the Bode command to generate the same plot. 6.89 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: The frequency response function H v ( jω ) = Problem solutions, Chapter 6 Vout ( jω ) is (see P6.4 for details): Vin ( jω ) Vout 1 + jωCR2 + ( jω ) LC ( jω ) = 2 Vin 1 + jωC (R1 + R2 ) + ( jω ) LC 2 a) Substituting the numerical values and expressing the frequency response function in factored form, we have: § jω ·§ jω · + 1¸¨ + 1¸ ¨ 489.79 ¹© 10.21 ¹ H v ( jω ) = © § jω ·§ j ω · + 1 ¸¨ + 1¸ ¨ © 743.27 ¹© 6.72 ¹ b) The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.70 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.5. Find: a) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. b) Use Matlab and the Bode command to generate the same plot. Assume: Assume that the output voltage is the voltage across the resistor. 6.90 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: The frequency response function H V ( jω ) = a) H V ( jω ) = Problem solutions, Chapter 6 Vout ( jω ) is (see P6.5 for details): Vin ( jω ) Vout ( jω ) R2 − CLR2ω 2 + jωL = Vin ( jω ) R2 + R1 − CL(R2 + R1 )ω 2 + jωL Substituting the numerical values and expressing the frequency response function in factored form, we have: ω 2 ) + j (0.0015)ω −4 2 ω ) + j (0.0010)ω § jω ·§ jω · − 1.004¨ + 1¸¨ + 1¸ © 48.84 ¹© 45.54 ¹ H V ( jω ) = § jω ·§ j ω · + 1¸¨ + 1¸ ¨ © 48.26 ¹© 46.04 ¹ H V ( jω ) = (1 − 4.5 ×10 (1 − 4.5 ×10 −4 b) The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.71 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.5. Find: a) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. b) Use Matlab and the Bode command to generate the same plot. 6.91 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: The frequency response function Vout ( jω ) = ( jω ) = Vin Vin a) VC H v ( jω ) = Problem solutions, Chapter 6 Vout ( jω ) is (see P6.5 for details): Vin ( jω ) 1 − CLω 2 1 − CLω 2 + jωL(R2 + R1 ) Substituting the numerical values and expressing the frequency response function in factored form, we have: H V ( jω ) = (1 − 4.5 ×10 ω ) (1 − 4.5 ×10 ω ) + j (9000)ω −4 −4 2 2 § jω ·§ jω · + 1¸¨ + 1¸ ¨ 47.14 ¹© 47.14 ¹ H V ( jω ) = © § jω ·§ jω · + 1¸¨ + 1¸ ¨ © 2e7 ¹© 1.11e − 4 ¹ b) The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.72 Solution: Known quantities: Resistance, inductance and capacitance values, in the circuit of Figure P6.4. Find: a) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. b) Use Matlab and the Bode command to generate the same plot. Analysis: The frequency response function is: Vout 1 ( jω ) = 2 Vin 1 + jωC (R1 + R2 ) + ( jω ) LC 6.92 G. Rizzoni, Principles and Applications of Electrical Engineering a) Problem solutions, Chapter 6 Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = 1 § jω ·§ jω · + 1¸¨ + 1¸ ¨ © 743.27 ¹© 6.72 ¹ b) The sketch plots and the ones obtained using Matlab are shown below: ______________________________________________________________________________________ Problem 6.73 Solution: Known quantities: Resistance and capacitance values, in the circuit of Figure P6.6. Find: a) Manually sketch a magnitude and phase Bode plot of the system, using a five-cycle semilog paper. b) Use Matlab and the Bode command to generate the same plot. Analysis: The frequency response function H v ( jω ) = Vout ( jω ) is (see P6.6 for details): Vin ( jω ) jωC1 R1 Vout ( jω ) = 2 Vin 1 + jω [C1 R1 + C 2 (R1 + R2 )] + ( jω ) C1C 2 R1 R2 a) Substituting the numerical values and expressing the frequency response function in factored form, we have: H v ( jω ) = ( jω ) 2 jω 2 jω =2 § jω ·§ jω · + 2 .6 j ω + 1 + 1¸¨ + 1¸ ¨ © 2.13 ¹© 0.47 ¹ b) The sketch plots and the ones obtained using Matlab are shown below: 6.93 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 ______________________________________________________________________________________ Problem 6.74 Solution: Known quantities: Ratio of output amplitude to input amplitude, being proportional to 1 ω3 in a certain frequency range. Find: The slope of the Bode plot in this frequency range, expressed in dB per decade. Analysis: vout 1 1 ∝ 3 , it is seen that the amplitude is reduced by a factor of 1000, or multiplied by , vin ω 1000 1 is a – 60 dB gain, we speak of the every time the frequency increases by a factor of 10. Since 1000 dB slope. The term “decade” refers to a frequency factor of transfer function rolling of at a – 60 decade 10. If ______________________________________________________________________________________ Problem 6.75 Solution: The output amplitude of a given circuit as a function of frequency: V = Aω + B C + Dω 2 Find: a) The break frequency. b) The slope of the Bode plot (in dB per decade) above the break frequency. 6.94 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 c) The slope of the Bode plot below the break frequency. d) The high-frequency limit of V. Analysis: a) Given V = Aω + B C + Dω 2 , this is seen to rise from B A at zero frequency to at high C D frequencies. The corresponding complex phasor function is: A B A B + jω + jω jω A + B C = C = C = C V= ω j D C + j Dω 1+ 1+ j ω C C D A· B § ¨ 1 + jω ¸ B¹ C © jω 1+ C D which we recognize to have a break frequency (or cut-off frequency, or half-power frequency) of: ωCO = C D A D B c) At low frequencies the slope is zero and the magnitude is equal to C A d) At high frequencies, V → D b) At high frequencies the slope is zero and the magnitude is equal to ______________________________________________________________________________________ Problem 6.76 Solution: Known quantities: Figures P6.76a and P6.76b. Find: An expression for the equivalent impedance in Figure P6.76a in standard form. Choose the Bode plot, from Figure P6.76b, that best describes the behavior of the impedance as a function of frequency and describe how (a simple one line statement and no analysis is sufficient) you would obtain the resonant and cutoff frequencies and the magnitude of the impedance where it is constant over some frequency range. Label the Bode plot to indicate which feature you are discussing. Analysis: In standard form. 1 [ Rc + j ωL] jω C jω C Z C [ Z Rc + Z L ] = Z[jω ] = = 1 jω C Z C + Z Rc + Z L + R c + j ωL jω C ωL 1+ j f 1 [ ω ] Rc + j ωL Rc Rc = = = Z o 1 - ω 2 LC 1+ j ω Rc C 1+ j f 2 [ ω ] [ 1 - ω 2 LC ] + j ω Rc C 2 1 - ω LC 1+ j 6.95 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 6 Bode Plot [b]. The circuit is a parallel resonant circuit and should exhibit maxima of impedance and minima of impedance at low and high frequencies. 1. At the resonant frequency, the impedance is real, i.e., the 1. f 1 [ ω r ] = f 2 [ ω r ] ==> Solve for ω r reactive part is zero. 2. The magnitude of the impedance at the resonant frequency is Rc Zo evaluated at the resonant frequency. 2. Z o = 1 - ω 2r LC 3. There are three cutoff frequencies [the 3 dB] frequencies evaluated by making the functions of frequency equal to +1 or - 1. 3. f 1 [ ω c ] = 1 Gives ω c 3 . 4. The magnitude of the impedance when the frequency is low can be determined in two ways. First, the circuit can be f 2 [ ω c ] = ± 1 Gives ω c1 and ω c 2 . modeled at low frequencies by replacing the inductor with a short circuit and the capacitor with an open circuit. Under these conditions the impedance is equal to that of the resistor. Or the 4. Z o = lim Z[jω ]ω → 0 = Rc limit of the impedance as the frequency approaches zero can be determined. ______________________________________________________________________________________ 6.96 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Chapter 7 Instructor Notes Chapter 7 surveys all important aspects of electric power. Coverage of Chapter 7 can take place immediately following Chapter 4, or as part of a later course on energy systems or electric machines. The material in this chapter will be of particular importance to Aerospace, Civil, Industrial, and Mechanical engineers, who are concerned with the utilization of electric power. The chapter permits very flexible coverage, with sections 7.1 and 7.2 describing basic single-phase AC power ideas. A survey course might only use this introductory material. The next two sections discuss transformers and three-phase power. Two descriptive sections are also provided to introduce the ideas of residential wiring, grounding and safety, and the generation and distribution of AC power. These sections can be covered independent of the transformer and three-phase material. The section Focus on Measurements: The wattmeter provides a practical look at the measurement of power. The section Focus on Measurements: How Hall-effect current transducers work may be useful in the context of a power systems course, while Focus on Measurements: Power factor correction proposes a more applied look at the problem of improving the power factor of an industrial load. The homework problems present a few simple applications in addition to the usual exercises meant to reinforce the understanding of the fundamentals. Problems 7.19, 20, 22-25 present a variety of power factor correction problems. Two advanced problems (7.35, 7.36) discuss transformer test methods; these problems may be suitable in a second course in energy systems. Problem 7.48 illustrates the billing penalties incurred when electric loads have insufficient power factors (this problem is based on actual data supplied by Detroit Edison). Those instructors who plan to integrate the three-phase material into a course on power systems and electric machines, will find that problems 7.47 and 7.49-51 can be assigned in conjunction with the material covered in Chapter 17, as part of a more in-depth look at three-phase machines. 7.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.1 Solution: Known quantities: Resistance value, ~ R = 30 Ω , and the voltage across the soldering iron, V = 117 V . Find: The power dissipated in the soldering iron. Analysis: The power dissipated in the soldering iron is: ~ V 2 117 2 P= = = 456.3 W 30 R ______________________________________________________________________________________ Problem 7.2 Solution: Known quantities: Rated power, ~ P = 1000 W , and the voltage across the heating element, V = 240 V . Find: The resistance of the heating element. Analysis: The power dissipated in the electric heater is: R= ~ V 2 240 2 = = 57.6 Ω P 1000 ______________________________________________________________________________________ Problem 7.3 Solution: Known quantities: Resistance value, R = 50 Ω of the resistor. Find: The power dissipated in the resistor if the current source connected to the resistor is: a) i (t ) = 5 cos(50t ) A i (t ) = 5 cos(50t − 45°) A c) i (t ) = 5 cos(50t ) − 2 cos(50t − 50°) Α d) i (t ) = 5 cos(50t ) − 2 Α b) Analysis: The average power can be expressed as: 1 2 I R 2 52 ⋅ 50 Pav = = 625 W 2 Pav = a) 7.2 G. Rizzoni, Principles and Applications of Electrical Engineering b) c) Pav = Problem solutions, Chapter 7 5 2 ⋅ 50 = 625 W 2 By using phasor techniques: I = 5∠0° − 2∠ − 50° = 5 − 1.2856 + j1.5321 = 3.7144 + j1.5321 = 4.0180∠22.41° A Then, the instantaneous current can be expressed as: i (t ) = 4.0180 cos(100t + 22.41°) A Therefore, the average power is: Pav = 4.0180 2 ⋅ 50 = 403.6 W 2 d) The instantaneous voltage can be expressed as: v(t ) = Ri (t ) = 250 cos(50t ) − 100 V Then, the instantaneous power can be written as: p(t ) = v(t ) ⋅ i (t ) = [250 cos(50t ) − 100]⋅ [5 cos(50t ) − 2] = 1250 cos 2 (50t ) − 1000 cos(50t ) + 200 = 625 + 625 cos(100t ) − 1000 cos(50t ) + 200 W Therefore, the average power is: Pav = 625 + 200 = 825 W ______________________________________________________________________________________ Problem 7.4 Solution: Known quantities: The current values. Find: The rms value of each of the following currents. a) cos 450t + 2 cos 450t b) cos 5t + sin 5t c) cos 450t + 2 cos 5t + cos(5t + π 3) e) cos 200t + cos 400t d) Analysis: The rms current can be expressed as: I ~ if the current is periodic or if the current can be converted to a phasor quantity. I rms = I = 2 a) Otherwise, the rms current must be calculated using integration techniques. Summing the common cosine terms leads to I = 3 cos 450t 3 ~ I = = 2.1213 A 2 b) Using phasor analysis: I = cos 5t + cos(5t − 90°) = 1∠0° + 1∠ − 90° = 1 − j = 2∠ − 45° A 2 ~ I = = 1A 2 7.3 G. Rizzoni, Principles and Applications of Electrical Engineering c) Problem solutions, Chapter 7 Since the second term is not periodic, integration techniques must be used: ~ I = 2π 2π 450 (cos 450t + 2) 2 dt = ³ 0 450 2π 2π 450 (cos 2 450t + 4 cos 450t + 4)dt = ³ 0 450 ~ I = 2π 2π 450 (cos 2 450t + 4 cos 450t + 4)dt = 450 ³0 ~ I = 2π 450 ª 9 1 4 3 º − sin 900t − sin 450t » = = 2.1213 A « 450 2π ¬ 4 1800 450 2 ¼ 2π 2π 450 §1 1 · + cos 900t + 4 cos 450t + 4 ¸dt = ¨ ³ 450 0 © 2 2 ¹ d) Using phasor analysis: I = 1∠0° + 1∠60° = 1 + 0.5 + j 0.866 = 1.732∠30° A ~ 1.732 I = = 1.225 A 2 e) Can’t use phasor analysis because phasor analysis does not work for different frequencies. Must integrate as in part c: 2 ~ I = = 1.414 A 2 ______________________________________________________________________________________ Problem 7.5 Solution: Known quantities: The current rms value, 4 A, the voltage source rms value, 110 V, the lag between the current and the voltage, 60°. Find: The power dissipated by the circuit and the power factor Analysis: The average power drawn by the circuit is: P= VI 110 2 ⋅ 4 2 cos(θ ) = cos(60°) = 220 W 2 2 The power factor is: pf = cos(60°) = 0.5 ______________________________________________________________________________________ Problem 7.6 Solution: Known quantities: The voltage source rms value, 120 V, the source frequency, 60 Hz, the power consumption, 1.2 kW, and the power factor, 0.8. Find: a) The rms current. b) The phase angle. c) The impedance. d) The resistance. 7.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Analysis: a) The power is expressed as: ~~ P = V I cos(θ ) Thus, the rms current is: P 1200 ~ I = ~ = = 12.5 A V cos(θ ) 120 ⋅ 0.8 b) The power factor is: pf = cos(θ ) Thus, the phase angle θ is: θ = cos −1 (0.8) = 36.87° c) The impedance Z is: ~ V 120 Z=~= = 9.6 Ω I 12.5 d) The resistance R is: R = Z cos(θ ) = 7.68 Ω ______________________________________________________________________________________ Problem 7.7 Solution: Known quantities: The rms values of the supply voltage and current, 110 V and 14 A, the power requirement, 1 kW, the machine efficiency, 90%, and the power factor, 0.8. Find: The AC machine efficiency. Analysis: The efficiency is: ηmotor = Mechanical Power 1 kW 0.9 1111 W = ~~ = = 0 .9 Electrical Power V I cos(θ ) 1232 W ______________________________________________________________________________________ Problem 7.8 Solution: Known quantities: The waveform of a voltage source shown in Figure P7.8. Find: a) The steady DC voltage that would cause the same heating effect across a resistance. b) The average current supplied to a 10-Ω resistor connected across the voltage source. c) The average power supplied to a 1-Ω resistor connected across the voltage source. Analysis: a) ~ VDC = V = b) I av (1⋅1) + (9 ⋅1) = 10 = 2.24 V 2 2 (1 ⋅1) + (− 3 ⋅1) = − 2 = −0.1 A = 2 ⋅10 20 7.5 G. Rizzoni, Principles and Applications of Electrical Engineering c) Problem solutions, Chapter 7 ~ V2 2 Pav = = = 5W R 1 ______________________________________________________________________________________ Problem 7.9 Solution: Known quantities: The current and the voltage values. Find: The average power, the reactive power and the complex power. Analysis: 450 50 ~~ P = V I cos(θ ) = ⋅ ⋅ cos(20°) = 21140 W 2 2 450 50 ~~ Q = V I sin(θ ) = ⋅ ⋅ sin(20°) = 7696 VAR 2 2 ~~ S = V I * = 11250∠20° VA ~~ b) P = V I cos(θ ) = 140 ⋅ 5.85 ⋅ cos (− 30°) = 709.3 W ~~ Q = V I sin(θ ) = 140 ⋅ 5.85 ⋅ sin(− 30°) = −409.5 VAR ~~ S = V I * = 819∠ − 30° VA ~~ c) P = V I cos (θ ) = 50 ⋅19.2 ⋅ cos (− 45.8°) = 668.8 W ~~ Q = V I sin(θ ) = 50 ⋅19.2 ⋅ sin(− 45.8°) = −688.7 VAR ~~ S = V I * = 960∠ − 45.8° VA ~~ d) P = V I cos (θ ) = 740 ⋅10.8 ⋅ cos(− 85.9° + 45°) = 6040.8 W ~~ Q = V I sin(θ) = 740 ⋅10.8 ⋅ sin(− 85.9° + 45°) = −5232.7 VAR ~~ S = V I * = 7992∠ − 40.9° VA a) ______________________________________________________________________________________ Problem 7.10 Solution: Known quantities: The current and the voltage values or the impedance. Find: The power factor and state if it is leading or lagging. Analysis: a) pf = cos(θi − θv ) = cos (21.2°) = 0.932 Leading pf = cos(θi − θv ) = cos(− 40.6°) = 0.759 Lagging c) iL (t ) = 48.7 sin(ωt + 2.74 ) = 48.7 sin(ωt + 2.74 − 90°) pf = cos(θi − θv ) = cos(67°) = 0.391 Leading b) 7.6 G. Rizzoni, Principles and Applications of Electrical Engineering d) Problem solutions, Chapter 7 §8· θ = tan −1 ¨ ¸ = 33.7° pf = cos(θ i − θ v ) = cos(− 33.7°) = 0.832 Lagging © 12 ¹ ______________________________________________________________________________________ Problem 7.11 Solution: Known quantities: The power factor or the values of the current and the voltage. Find: The kind of the load (capacitive or inductive). Analysis: a) Capacitive. b) Capacitive. c) Since iL (t ) = 1.8 cos (ωt − 90°) , Inductive. d) Since the phase difference is zero, Resistive. ______________________________________________________________________________________ Problem 7.12 Solution: Known quantities: Circuit shown in Figure P7.12, the values of the resistance, inductance, R = 4 Ω , the capacitance, C = 1 18 µF , the L = 2 H , and the voltage source. Find: The real and reactive power supplied by the following sources. a) vS (t ) = 10 cos(3t ) V b) vS (t ) = 10 cos(9t ) V Analysis: 10 ~ ω = 3, ZT = j 6 − j 6 + 4 = 4 + j 0 Ω, I = = 1.77 A 2 ⋅4 ~ ~ P = I 2 R = 12.5 W , Q = I 2 X = 0 VAR 10 ~ = 0.42 A b) ω = 9 , ZT = j18 − j 2 + 4 = 4 + j16 Ω, I = 2 ⋅16.5 ~ ~ P = I 2 R = 0.7 W, Q = I 2 X = 2.82 VAR a) ______________________________________________________________________________________ Problem 7.13 Solution: Known quantities: Circuit shown in Figure P7.13, the values of the resistances, R1 = 8 Ω , R2 = 6 Ω , the reactances, ~ ~ X C = −12 Ω , X L = 6 Ω , and the voltage sources, VS1 = 36∠ − π 3 V , VS 2 = 24∠0.644 V . Find: a) The active and reactive current for each source 7.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 b) The total real power. Analysis: a) From Figure P7.13: VS 1 = R1I1 + jX L (I1 − I 2 ) = (8 + j 6)I1 − j 6I 2 − VS 2 = − jX L (I1 − I 2 ) + R2I 2 + X C I 2 = − j 6I1 + (6 − j 6)I 2 Substituting the values for the voltages sources gives: 18 − j 31.2 = (8 + j 6)I1 − j 6I 2 ® ¯− 19.2 − j14.4 = − j 6I1 + (6 − j 6)I 2 Solving for I1 and I2 yields: I1 = 0.398 − j 3.38 A ® ¯I 2 = 1.091 − j 0.911 A Therefore, the active and reactive currents for each source are: I A1 = 0.398 A I = 3.38 A and ® R1 ® ¯I A2 = 1.091 A ¯I R 2 = 0.91 A 2 2 2 2 b) P = R2Ι 2 + R1I1 = 6 ⋅1.421 + 8 ⋅ 3.403 = 105 W ______________________________________________________________________________________ Problem 7.14 Solution: Known quantities: Circuit shown in Figure P7.14, the values of the resistors, RL = 25 Ω , R = 1Ω , the capacitor, ~ C = 0.1 µF , the voltage source, VS = 230 V , and the frequency, f = 60 Hz . Find: a) The source power factor. b) The current IS. c) The apparent power delivered to the load. d) The apparent power supplied by the source. e) The power factor of the load. Analysis: a) pf source = R = Z Rline + Rload (Rline + Rload )2 + X C2 = 26 676 + 7.036e8 = 0.00098 Leading ~ 230 ~ V = 8.67m A IS = S = Z 26525∠ − 90° Therefore, I S = 8.67∠ − 90° mA 2 ~2 2 2 2 c) Sload = Pload + Qload = I S Rload + X C = 1.994 VA ~~ d) S source = I SVS = 1.994 VA 25 R e) pf load = load = = 0.00094 Z load 26525 b) ______________________________________________________________________________________ 7.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.15 Solution: Known quantities: Circuit shown in Figure P7.14, the e values of the resistors, RL = 25 Ω , R = 1Ω , the inductor, ~ L = 0.1 H , the voltage source, VS = 230 V , and the frequency, f = 60 Hz . Find: a) The apparent power supplied by the source. b) The apparent power delivered to the load. c) The power factor of the load. Analysis: a) ~ V2 S= = Z ~ V2 (Rline + Rload )2 + X L2 = 230 2 26 2 + 37.7 2 = 1.155 kVA 2 ~ V2 ~2 § 230 · 2 2 2 2 b) S load = I S Z load = 2 Rload + X L = ¨ ¸ 25 + 37.7 = 1.141 kVA Z © 45.8 ¹ 25 R = 0.55 c) pf load = load = Z load 45.2 ______________________________________________________________________________________ Problem 7.16 Solution: Known quantities: Circuit shown in Figure P7.14, the values of the resistors, RL = 25 Ω , R = 1Ω , the capacitor, ~ C = 0.1 mF , the inductor, L = 70.35 mH , the voltage source, VS = 230 V , and the frequency, f = 60 Hz . Find: a) The apparent power delivered to the load. b) The real power supplied by the source. c) The power factor of the load. Analysis: a) ~ V2 S= = Z b) XC = X L c) XC = X L ~ V2 (Rline + Rload )2 + ( X L − X C )2 = ~ V 2 230 2 P= = = 2.03 kW R 25 + 1 pf = 1 230 2 26 2 + (26.5 − 26.5) 2 = 2.03 kVA ______________________________________________________________________________________ 7.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.17 Solution: Known quantities: Circuit shown in Figure P7.17, the values of the resistor, ~ voltage source, VS = 50 V . R = 20 Ω , the capacitor, C = 0.1 mF , the Find: The apparent power, the real power, and the reactive power; draw the power triangle. Analysis: S= ~ V2 = ( ωC ) R2 + 1 2 P = S ⋅ cos(θ ) = S ⋅ 50 2 20 2 + 26.52 = 2500 = 75.3 VA 33.2 R 20 = 75.3 = 45.36 W Z 33.2 Q = S 2 − P 2 = −60 VAR −1 § R · For the power triangle, θ = cos ¨ ¸ = 53° ©Z ¹ Therefore, the power triangle can be drawn as shown below: P = 45.36 W 53° S = 75.3 VA Q = -60 VAR ______________________________________________________________________________________ Problem 7.18 Solution: Known quantities: Circuit shown in Figure P7.17, the values of the resistor, ~ voltage source, VS = 50 V . R = 20 Ω , the capacitor, C = 0.1 mF , the Find: The apparent power, the real power, and the reactive power, in the cases of f = 50 and 0 Hz. Analysis: For the frequency of 0 Hz, S0 Hz = 0, P0 Hz = 0, Q0 Hz = 0 For the frequency of 50 Hz, 7.10 G. Rizzoni, Principles and Applications of Electrical Engineering S50 Hz = ~ V2 = ( ωC ) 2 R2 + 1 P50 Hz = S ⋅ cos(θ ) = S ⋅ Problem solutions, Chapter 7 2500 = 67.15 VA 37.2 20 R = 67.15 = 36 W 37.2 Z Q50 Hz = S 2 − P 2 = 56.7 VA ______________________________________________________________________________________ Problem 7.19 Solution: Known quantities: A single-phased motor connected across a 220-V source at 50 Hz as shown in Figure P7.19, power factor pf = 1.0, I = 20 A, and I1 = 25 A. Find: The capacitance required to give a unity power factor when connected in parallel with the load. Analysis: ~ The magnitude of the current I2 is: ~ ~ ~ I 2 = I12 − I 2 = 625 − 400 = 15 A The voltage source can be expressed as: ~ ~ V = I2 ⋅ XC Therefore, the required capacitor is: ~ I 15 C= ~2 = = 217 µF V ⋅ ω 220 ⋅ 314 ______________________________________________________________________________________ Problem 7.20 Solution: Known quantities: The currents and voltages required by an air-conditioner, a freezer, a refrigerator, and their power factors. Find: The power to be supplied by an emergency generator to run all the appliances. Analysis: In this problem we will use the following equations: ~~ ~~ P = I V cos(θ ) , Q = I V sin(θ ) , pf = cos(θ ) The real and reactive power used by the air conditioner are: P1 = 9.6 ⋅120 ⋅ 0.9 = 1036.8 W , Q1 = 9.6 ⋅120 ⋅ sin(cos −1 (0.9)) = 502.15 VAR The real and reactive power used by the freezer are: P2 = 4.2 ⋅120 ⋅ 0.87 = 438.48 W , Q2 = 4.2 ⋅120 ⋅ sin(cos −1 (0.87 )) = 248.5 VAR The real and reactive power used by the refrigerator are: P3 = 3.5 ⋅120 ⋅ 0.8 = 336 W , Q3 = 3.5 ⋅ 120 ⋅ sin(cos −1 (0.8)) = 252 VAR The total real and reactive power P are: P = P1 + P2 + P3 = 1811.28 W , Q = Q1 + Q2 + Q3 = 1002.65 VAR Therefore, the following power must be supplied: 7.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 S = P + jQ = 1811.28 + j1002.65 VA = 2070.3∠28.97 o VA ______________________________________________________________________________________ Problem 7.21 Solution: Known quantities: The schematics of the power supply module consisting of two 25-kV single-phase power stations shown in Figure P7.21, the power consumption by the train, the DC power supply at a low speed operation, the average power factor in AC operation, the over-head line equivalent specific resistance, and negligible rail resistance. Find: a) The equivalent circuit. b) The locomotive current in the condition of a 10% voltage drop. c) The reactive power. d) The supplied real power, over-head line losses, and the maximum distance between two power station supplied in the condition of a 10% voltage drop when the train is located at the half distance between the stations. e) Over-head line losses in the condition of a 10% voltage drop when the train is located at the half distance between the stations, assuming pf = 1 (The French TGV is designed with the state of art power compensation system). f) The maximum distance between the two power station supplied in the condition of a 10% percent voltage drop when the train is located at the half distance between the stations, assuming the DC (1.5 kV) operation at a quarter power. Analysis: a) The equivalent circuit is: I1 Line Line I2 ILOC VS1 VS2 Train b) The locomotive current for the 10% voltage drop is: PLOC 11 MW ~ ~ ~ ~ ~ I LOC = I1 + I 2 = 2 I1 = 2 I 2 = ~ = = 611 A VS − 10% cos(θ ) 22.5 kV ⋅ 0.8 c) ( The reactive power is: 2 2 Q = S LOC − PLOC = = ) 2 2 (V~S − 10%)2 I LOC − PLOC (13.75 MVA )2 − (11 MW )2 = 8.25 MVAR d) The supplied real power is: P = S 2 − Q2 = (V~S ⋅ ~I )2 − Q 2 = 12.85 MW The over-head line power loss is: 7.12 G. Rizzoni, Principles and Applications of Electrical Engineering PLine = − PLOC + Problem solutions, Chapter 7 (V~S ⋅ ~I )2 − Q 2 = −11 MW + (15.27 MVA )2 − (8.25 MVAR )2 = 1.85 MW The maximum distance between the two power stations is: P 2 RLine RLine || RLine = ~Line = 5 Ω Distance max = = 100 km 2 0.2 Ω/km I LOC e) f) The over-head line power loss is: ~ PLOC = 10% ⋅VS ⋅ I LOC ⋅ cos(θ ) = 2500 V ⋅ 489 A = 1.22 MW 0.25PLOC 2.75 MW I LOC = = = 2037 A 90% ⋅VS ,DC 1350 V PLine = 10% ⋅ VS ,DC ⋅ I LOC = 305 kW PLine RLine || RLine = 2 I LOC = 0.0735 Ω The maximum distance between the two power stations is: Distance max = 2 RLine = 1.5 km 0.2 Ω/km ______________________________________________________________________________________ Problem 7.22 Solution: Known quantities: One hundred 40-W lamps supplied by a 120-V and 60-Hz source, the power factor of 0.65, the penalty at billing, and the average prices of the power supply and the capacitors. Find: Number of days of operation for which the penalty billing covers the price of the power factor correction capacitor. Analysis: The capacitor value for pf = 0.85 is: 2 ( ~ ~ ~ I X ,0.65 − I X ,0.85 IC C= ~ = ω ⋅VC VC ⋅ ω ) 2 2 § P · § P· § P · § P· ¨~ ¸ −¨ ¸ − ¨ ~ ¸ −¨ ¸ © V ⋅ 0.65 ¹ © V ¹ © V ⋅ 0.85 ¹ © V ¹ = ω ⋅VC 2 51.32 − 33.32 − 39.22 − 33.32 19 = = = 420 µF 337 ⋅120 377 ⋅120 Therefore, the number of days of operation for which the penalty billing covers the price of the power factor correction capacitor is: $ mF Number of Days = = 88 Days 4 kW $ hr ⋅1 hr ⋅ 0.01 ⋅ 24 4 kW day 420 µF ⋅ 50 ______________________________________________________________________________________ 7.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.23 Solution: Known quantities: Reference to the problem 7.22, and the network current decreasing with the power factor correction. Find: a) The capacitor value for the unity power factor. b) The maximum number of lamps that can be installed supplementary without changing the cable network if a local compensation capacitor is used. Analysis: a) ~ ~ IC = IL 2 2 § P · §P· ~ ¨~ ¸ −¨ ~ ¸ IL 51.32 − 33.32 © V ⋅ 0.65 ¹ © V ¹ = 862 µF C= ~ = = ~ 377 ⋅120 Vω Vω b) Initial cable network is: 4000 P ~ I = ~ = = 51.3 A V cos(θ ) 120 ⋅ 0.65 One lamp current for pf = 1 is: PLamp 40 = 0.333 A ~ = 120 V ~ I = 154 The total number of lamps = ~ I Lamp I Lamp = Therefore, the number of supplementary lamps = 154 − 100 = 54 ______________________________________________________________________________________ Problem 7.24 Solution: Known quantities: The voltage and the current supplied by a source, ~ ~ VS = 7∠50° V , IS = 13∠ − 20° A . Find: a) The power supplied by the source which is dissipated as heat or work in the load b) The power stored in reactive components in the load. c) Determine if the circuit is an inductive or a capacitive load. Analysis: ~ ~ * S = VS IS* = (7∠50° V )(13∠ − 20° A ) = 91∠70° VA = 31.12 + j85.51VA Pav = 31.12 W b) Q = 85.51VAR c) θ = (θ I − θV ) = −70° pf = cos (θ ) = 0.342 Lagging a) The load is inductive. ______________________________________________________________________________________ 7.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.25 Solution: Known quantities: Circuit shown in Figure P7.25, the voltage supplied by a power plant, vS (t ) = 450 cos(ωt ) V , ω = 377 rad s , and the impedances of the plant, power plant, Z G Z = 7 + j Ω and of the = 3 + j 0.11Ω . Find: Determine C so that the plant power factor is corrected to 1. Analysis: Note: ZG influences only the phase difference between VS and V0 and not the one between V0 and Z. For this reason, the result does not depend from ZG. Z eq = (R + jX )(− jX C ) = XX C − jRX C R − j ( X − X C ) = ZZ C = Z + Z C (R + jX ) + (− jX C ) R + j ( X − X C ) R − j ( X − X C ) ( XX C R − RX C ( X − X C )) − j (R 2 X C − XX C ( X − X C )) = = Req − jX eq R 2 + ( X − X C )2 IS = V0 Z eq If IS and V0 are in phase, they have the same phase angle. For this reason, it must be: Z eq = Req − jX eq = Req − j 0 X eq = 0 R 2 X C − XX C ( X − X C ) = 0 XC = R2 + X 2 1 1 1 = 50 Ω = C= = = 51.3 µF X ωC ωX C (377 rad s )(50 Ω ) ______________________________________________________________________________________ Problem 7.26 Solution: Known quantities: Circuit shown in Figure P7.25, the voltage supplied by a power plant, vS (t ) = 450 cos(ωt ) V , ω = 377 rad s , and the impedances of the plant, Z = 7∠10° Ω . Find: Determine C so that the plant power factor is corrected to 1. Analysis: Note: ZG influences only the phase difference between VS and V0 and not the one between V0 and Z. For this reason, the result does not depend from ZG. Z = 7∠10° Ω = 6.89 + j1.21 Ω (R + jX )(− jX C ) = XX C − jRX C R − j ( X − X C ) = ZZ C Z eq = = Z + Z C (R + jX ) + (− jX C ) R + j ( X − X C ) R − j ( X − X C ) = ( XX C R − RX C ( X − X C )) − j (R 2 X C − XX C ( X − X C )) = R − jX eq eq R 2 + ( X − X C )2 7.15 G. Rizzoni, Principles and Applications of Electrical Engineering IS = Problem solutions, Chapter 7 V0 Z eq If IS and V0 are in phase, they have the same phase angle. For this reason, it must be: Z eq = Req − jX eq = Req − j 0 X eq = 0 R 2 X C − XX C ( X − X C ) = 0 XC = R2 + X 2 1 1 1 = 40.31 Ω = C= = = 65.8 µF X ωC ωX C (377 rad s )(40.31 Ω ) ______________________________________________________________________________________ Problem 7.27 Solution: Known quantities: ~ V0 = 450∠0° V , f = 60 Hz , and the current ~ through it without the capacitance in parallel with the plant, IS = 17∠ − 10° A , the value of the capacitance in parallel with the plant, C = 17.40 µF . Circuit shown in Figure P7.25, the voltage across a plant, Find: The reduction of current which resulted from connecting the capacitor into the circuit. Assumptions: The impedance of the power plant is very small, Z G ≈ 0 . Analysis: Without capacitor: ~ ~ ~ IS = Ig = 17∠ − 10° A , V0 = 450∠0° V . Z G ≈ 0 , the voltage across the plant does not change, as well as the current: ~ ~ V0 = 450∠0° V Ig = 17∠ − 10° V ~ ~ ~ − IS + Ig + IC = 0 ~ 450∠0° V0 ~ ~ ~ ~ = 17∠ − 10° + IS = Ig + IC = Ig + 1 jωC ZC With capacitor, being KCL: ( ) = 16.74 − j 2.952 + j 377 ⋅ 17.4 ⋅ 10− 6 = 16.74 + j 0 A = 16.74∠0° A ______________________________________________________________________________________ Problem 7.28 Solution: Known quantities: Circuit shown in Figure P7.25, the voltage across a plant, v0 (t ) = 170 cos (ωt ) V , f = 60 Hz , the current through it without the capacitance in parallel with the plant, iS (t ) = 130 cos (ωt − 11°) A , and the value of the capacitance in parallel with the plant, C = 387 µF . Find: The reduction of current which resulted from connecting the capacitor into the circuit. Assumptions: The impedance of the power plant is very small, Z G ≈ 0 . 7.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Analysis: Without capacitor: ~ ~ IS = Ig = 130∠ − 11° A , V̂0 = 170∠0° V . Z G ≈ 0 , the voltage across the plant does not change, as well as the current: ~ ~ V0 = 170∠0° V Ig = 130∠ − 11° V ~ ~ ~ − IS + Ig + IC = 0 ~ V0 170∠0° ~ ~ ~ ~ = 130∠ − 11° + IS = I g + IC = I g + ZC 1 jωC With capacitor, being KCL: ( ) = 127.61 − j 24.81 + j 377 ⋅ 17.4 ⋅ 10 −6 = 127.61 + j 0 A = 127.61∠0° A ______________________________________________________________________________________ Problem 7.29 Solution: Known quantities: Circuit shown in Figure P7.29, the values of the voltages and all the impedances. Find: The total average power, the real power dissipated and the reactive power stored in each of the impedances. Analysis: S1 = ~ VS1 Z1* ~ VS 2 (170 = )2 (170 = )2 2V = 20.643∠30° kVA = 17.88 + j10.32 kVA = Pav1 + jQ1 0.7∠ − 30° Ω 2V S2 = * = 9.633∠7° kVA = 9.56 + j1.17 kVA = Pav 2 + jQ2 1.5∠ − 7° Ω Z2 ~ ~ 2 (170∠0° + 170∠90° V )2 V +V S3 = S1 * S 2 = 0.3 − j 0.4 Ω Z3 ( = ) (240.42∠45° V )2 0.5∠ − 53.13° Ω = 57.8∠53.13° kVA = 34.68 − j 46.24 kVA = Pav 3 + jQ3 ______________________________________________________________________________________ Problem 7.30 Solution: Known quantities: The voltage and the current supplied by a source, ~ ~ VS = 170∠ − 9° V , IS = 13∠16° A . Find: a) The power supplied by the source which is dissipated as heat or work in the load b) The power stored in reactive components in the load. c) Determine if the circuit is an inductive or a capacitive load. Analysis: a) ~ ~ * S = VS IS* = (170∠ − 9° V )(13∠16° A ) = 2210∠ − 25° VA = 2003 − j 934.0 VA Pav = 2003 W 7.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Q = −934 VAR c) θ = (θ I − θV ) = 25° pf = cos (θ ) = 0.906 Leading b) The load is capacitive. ______________________________________________________________________________________ Problem 7.31 Solution: Known quantities: Circuit shown in Figure P7.31, 3ach secondary connected to 5-kW resistive load, the primary connected to 120-V rms. Find: a) Primary power. b) Primary current. Analysis: a) Pprim = Psec 1 + Psec 2 = 10 kW b) Pprim 10000 ~ I prim = ~ = = 83.3 A 120 V ~ I prim = 83.3∠0 A ______________________________________________________________________________________ Problem 7.32 Solution: Known quantities: ~ Vsec Circuit shown in Figure P7.31, the ratio between the secondary and the primary, ~ = n and V prim 1~ ~ ~ Vsec1 = Vsec 2 = Vsec . 2 Find: a) Vsec and Vsec1 if Vprim = 220 V rms and n = 11. b) n if Vprim = 110 V rms and Vsec2 = 5 V rms. Analysis: a) Vsec = V prim = 220 = 20 V 11 n Vsec Vsec1 = = 10 V 2 V prim 110 = = 11 b) n = 2 ⋅ Vsec 2 2 ⋅ 5 ______________________________________________________________________________________ 7.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.33 Solution: Known quantities: The circuit shown in Figure P7.33 and vg = 120 V rms. Find: a) The total resistance seen by the voltage source. b) The primary current. c) The primary power. Analysis: From the circuit shown on the right hand side: v g = i1 R1 + v1 i2 = v2 / R2 i1 = ni2 v2 = nv1 v g R1i1 + v1 R = = R1 + 22 = 2 Ω a) Rtot = i1 i1 n v g 120 ~ = = 60 A b) I1 = 2 Rtot ~ ~ c) P1 = I1 ⋅ V1 = 7.2 kW ______________________________________________________________________________________ Problem 7.34 Solution: Known quantities: The circuit shown in Figure P7.33 and vg = 120 V rms. Find: a) The secondary current. b) The installation efficiency Pload Psource . c) The value of the load resistance which can absorb the maximum power from the given source. Analysis: From the circuit shown on the right hand side: I1 60 = = 15 A 4 4 Pload I 22 ⋅ R2 225 ⋅16 3.6 kW = = = = 0.5 b) η = Psource v g ⋅ I1 120 ⋅ 60 7.2 kW a) I2 = c) For the maximum power transfer: R prim = R' sec R1 = 1 Rload Rload = R1 ⋅ n 2 = 1Ω ⋅ 16 = 16 Ω 2 n ______________________________________________________________________________________ 7.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.35 Solution: Known quantities: Circuit shown in Figure P7.35, the voltage and the power that a transformer is rated to deliver to a ~ customer, V1 = 380 V , Pin = 460 kW . Find: a) The current that the transformer supply to the customer. b) The maximum power that the customer can receive if the load is purely resistive. c) The maximum power that the customer can receive if the power factor is 0.8, lagging. d) The maximum power that the customer can receive if the power factor is 0.7, lagging. e) The minimum power factor to operate if the customer requires 300 kW. Analysis: a) ~~ ~ ~ Sin = V1I1 = V2 I 2 = Sout : 460 ⋅1000 S I1 = ~in = = 1.21 kA 380 V1 From b) For an ideal transformer: For c) For d) For e) For ~~ Pout = Pin cos(θ ) = V1 I1 cos(θ ) cos(θ ) = 1 : ~~ Pout = V1I1 = 460 kW cos(θ ) = 0.8 , the maximum power is: ~~ Pout = V1I1 cos(θ) = 368 kW cos(θ ) = 0.7 , the maximum power is: ~~ Pout = V1I1 cos(θ) = 322 kW Pout = 300 kW , the minimum power factor is: P 300 kW cos(θ) = out = = 0.65 Pin 460 kW ______________________________________________________________________________________ Problem 7.36 Solution: Known quantities: Circuit shown in Figure P7.36, the voltage, vS (t ) = 294 cos(377t ) V , the resistances in a circuit containing a transformer and the ratio n= v0 (t ) 1 = . v S (t ) 2.5 Find: a) Primary current. b) v0 (t ) . c) Secondary power. d) The installation efficiency Pload Psource . 7.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Analysis: a) The primary circuit is described in the figure left-hand side. The primary current is: ( ) ~ ~ ~ VS = I1 (R1 + R' 2 ) = I1 R1 + n 2 R2 ~ V1 294 1 ~ I1 = = ⋅ = 0.82 A 2 2 100 + 6.25 ⋅ 25 R1 + n R2 i (t ) = 1.144 cos(377t ) A b) The output voltage is: ~ ~ ~ ~ V1 VS − R1I1 208 − 100 ⋅ 0.82 V0 = = = = 50.4 V n 2.5 n v0 (t ) = 71cos(377t ) V c) For the secondary power, if pf = 1 : ~~ ~ ~ P2 = I 2V0 = I1 ⋅V0 ⋅ n = 0.82 ⋅ 50.4 ⋅ 2.5 = 103.3 W d) The installation efficiency is: η= Pload P 103.3 W = ~ 2~ = = 0 .6 Psource VS ⋅ I1 208 V ⋅ 0.82 A ______________________________________________________________________________________ Problem 7.37 Solution: Known quantities: Circuit shown in Figure P7.37, the resistances, RS = 1800 Ω , RL = 8 Ω . Find: The turn’s ratio that will provide the maximum power transfer to the load. Analysis: From Equation (7.41) for the reflected source impedance circuit, we have: RS eq = N 2 RS Therefore, the power is maximized if: RS eq = RL N = 1 n = RL RS = 0.067 ∴ n = 15 ______________________________________________________________________________________ Problem 7.38 Solution: Known quantities: The voltage source and the resistances in the circuit shown in Figure P7.38. Find: a) Maximum power dissipated by the load. b) Maximum power absorbing from the source. c) The installation efficiency. Analysis: All the impedances are resistances, and therefore it is possible to consider the modules of voltages and currents. 7.21 G. Rizzoni, Principles and Applications of Electrical Engineering a) Problem solutions, Chapter 7 To maximize the power delivered to the 8-Ω resistance, n must be selected to maximize the load current I 2 > 0 . Note that 1 V1 = V2 and I1 = nI 2 . n KVL Mesh 1: 1 1 v g = 3I1 + V2 + 4(I1 − I 2 ) = 3nI 2 + V2 + 4(nI 2 − I 2 ) n n KVL Mesh 2: V2 = 8I 2 + 4(I 2 − I1 ) = 8I 2 + 4(I 2 − nI 2 ) Rearranging the two mesh equations: 1 1 °(7 n − 4 )I 2 + V2 = v g (7 n − 4 )I 2 + (12 − 4n )I 2 = v g n ® n °̄(12 − 4n )I 2 = V2 I2 = n vg 7n − 8n + 12 d 7n 2 − 8n + 12 − n(14n − 8) − 7n 2 + 12 I2 = Vg = v 2 2 g dn 7n 2 − 8n + 12 7n 2 − 8n + 12 dI 2 For the maximum value of the load current I2, = 0: dn 12 − 7n 2 + 12 = 0 n = = 1.31 7 2 ( ) ( ) ( ) The maximum load current is: I2 = n 2 7n − 8n + 12 v g = 0.122v g = 13.47 A The maximum power dissipated by the load is: Pload = Rload I 22 = 2.45 kW b) The maximum power absorbing from the source is: Psource = Pload + 3 Ω ⋅ I12 + 4 Ω ⋅ (I1 − I 2 )2 Psource = Pload + 3 Ω ⋅ nI 22 + 4 Ω ⋅ (nI 2 − I 2 )2 Psource = 2540 W + 934 W + 70 W = 3.54 kW c) The installation efficiency is: η= Pload 2.45 kW = = 0 .7 Psource 3.54 kW ______________________________________________________________________________________ Problem 7.39 Solution: Known quantities: The current and the voltage delivered by the transformer, and the circuit of the transformer shown in Figure P7.39. 7.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Find: The efficiency of the installation. Analysis: ~ ~ ~ VW = nVsec = 2Vsec = 220 V rms 1~ 1~ ~ IW = I sec = I sec = 25 A rms 2 n ~ V 220 ~ IL = W = ∠ − 90° = 11∠ − 90°A rms j 20 20 Since the currents are exactly 90° out of phase, the current is the square root of the sum of the phasor magnitudes squared: ~ ~ ~ I S = I L2 + IW2 = 27.3 A rms ~ PS = 1Ω ⋅ I S2 + Psec = 745 W + 5.5 kW = 6.245 kW Therefore, the efficiency of the installation is: η= Pload P 5.5 kW = sec = = 0.88 Psource Psource 6.245 kW ______________________________________________________________________________________ Problem 7.40 Solution: Known quantities: The model for the circuit of a transformer shown in Figure P7.40 and the results of two tests performed at ω = 377 rad s : ~ ~ Voc = 241 V , I oc = 0.95 A , Poc = 32 W . ~ ~ 2. Short-circuit test: Vsc = 5 V , I sc = 5.25 A , Psc = 26 W . 1. Open-circuit test: Find: The value of the impedances in the equivalent circuit. Analysis: The power factor during the open circuit test is: P pf oc = cos(θoc ) = ~ oc ~ = 0.1398 Lagging Voc I oc The excitation admittance is given by: ~ I oc 0.95 Yc = ~ ∠ cos −1 ( pf oc ) = ∠ − 81.96° S = 0.0005511 − j 0.003903 S 241 Voc Rc = 1.8 kΩ Rc = 1.8 kΩ ° ® Xc ® ¯ X c = 256.2 Ω °̄ Lc = ω = 0.68 H The power factor during the short circuit test is: P pf sc = cos(θsc ) = ~ sc~ = 0.9905 Leading Vsc I sc The series impedance is given by: 7.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 ~ Vsc 5 Z w = ~ ∠ cos −1 ( pf sc ) = ∠7.914° Ω = 0.9476 + j 0.1311Ω 5.25 I sc Rw = 0.9476 Ω Rw = 0.9476 Ω ° ® Xw ® ¯ X w = 0.1311 Ω °̄ Lw = ω = 0.348 mH ______________________________________________________________________________________ Problem 7.41 Solution: Known quantities: The model for the circuit shown in Figure P7.40 of a 460 kVA transformer and the results of two tests f = 60 Hz : ~ ~ 1. Open-circuit test: Voc = 4600 V , I oc = 0.7 A , Poc = 200 W . ~ 2. Short-circuit test: Vsc = 5.2 V , Psc = 50 W . performed at Find: The value of the impedances in the equivalent circuit. Analysis: The power factor during the open circuit test is: P pf oc = cos(θoc ) = ~ oc ~ = 0.062 Lagging Voc I oc The excitation admittance is given by: ~ I oc 0.7 Yc = ~ ∠ cos −1 ( pf oc ) = ∠ − 86.45° S = 9.4 ⋅10 −6 − j 0.152 ⋅10−3 S 4600 Voc Rc = 106.38 kΩ Rc = 106.38 kΩ ° ® Xc ® ¯ X c = 6.58 kΩ °̄ Lc = ω = 17.46 H The power factor during the short circuit test: pf sc = cos(θsc ) = Psc ≈1 S sc since it is a high power transformer. The series impedance has therefore imaginary part ≈ 0: ~ Vsc2 Zw = ∠ cos −1 ( pf sc ) = 0.54 Ω Psc Rw = 0.54 Ω Rw = 0.54 Ω ° ® Xw ® ¯X w = 0Ω °̄ Lw = ω = 0 Therefore, the equivalent circuit is shown besides. ______________________________________________________________________________________ 7.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.42 Solution: Known quantities: Circuit shown in Figure P7.42 of the single-phase transformer with the high voltage regulation from five different slots in the primary winding, the secondary voltage regulation in the range of 10%, and the number of turns in the secondary coil. Find: The number of turns for each slot. Analysis: The secondary voltages are: ~ ~ V21 = V23 − 0.12 V = 1.08 V rms ~ ~ V22 = V23 − 0.06 V = 1.14 V rms ~ V23 = 1.20 V rms ~ ~ V24 = V23 + 0.06 V = 1.26 V rms ~ ~ V25 = V23 + 0.12 V = 1.32 V rms Therefore, the number of turns for each slot is: n prim ~ V prim = ~ ⋅ nsec Vsec n11 = 203.7 × 2 ≅ 408 turns °n = 192.9 × 2 ≅ 386 turns °° 12 ®n13 = 183.3 × 2 ≅ 367 turns °n = 174.6 × 2 ≅ 349 turns ° 14 ¯°n15 = 166.6 × 2 ≅ 333 turns ______________________________________________________________________________________ Problem 7.43 Solution: Known quantities: The pipe’s resistance = 0.0002 Ω, the secondary resistance = 0.00005 Ω, the primary current = 28.8 A, and pf = 0.91. Find: a) The slot number. NOTE: Typo in problem statement. b) The secondary reactance. c) The installation efficiency. Analysis: a) The secondary current is: ~ ~ ~2 Pprim = Psec V prim ⋅ I prim ⋅ cos(θ ) = Rsec ⋅ I sec ~ ~ V prim ⋅ I prim ⋅ cos(θ ) 220 ⋅ 28.8 ⋅ 0.91 I sec = = = 4800 A Rsec 0.00025 Therefore, the slot number is: N= I sec 4800 = = 166.6 Slot Number 5 I prim 28.8 b) The secondary reactance is: 7.25 G. Rizzoni, Principles and Applications of Electrical Engineering Rsec pf prim = pf sec = X sec = Rsec c) 2 2 + X sec Rsec 1 pf Problem solutions, Chapter 7 2 − 1 = 0.00025 1 0.912 − 1 = 114 µΩ The installation efficiency is: η= 2 Pload Rload ⋅ I sec Rload 200 µΩ = = = = 0.8 2 Pprim Rsec 250 µΩ Rsec ⋅ I sec ______________________________________________________________________________________ Problem 7.44 Solution: Known quantities: A single-phase transformer converting 6 kV to 230 V with 0.95 efficiency, the pf of 0.8, and the primary apparent power of 30 KVA. Find: a) The secondary current. b) The transformer’s ratio. Analysis: a) The secondary current is: Psec = Pprim ⋅ η = S prim ⋅ cos(θ ) ⋅ η = 30 ⋅ 0.8 ⋅ 0.95 = 22.8 kW P ~ I sec = ~ sec = 124 A Vsec ⋅ cos(θ ) b) The primary current is: S prim 30000 ~ = = 5A I prim = ~ 6000 V prim Therefore, the transformer’s ratio is: ~ 1 I sec = ~ = 24.8 N = 0.04 N I prim ______________________________________________________________________________________ Problem 7.45 Solution: Known quantities: The magnitude of the phase voltage of a three-phase wye system, 220 V rms. Find: The expression of each phase in both polar and rectangular coordinates. Analysis: The phase voltages in polar form are: ~ ~ ~ Van = 220∠0o V , Vbn = 220∠ − 120° V , Vcn = 220∠120° V The rectangular forms are: ~ ~ ~ Van = 220 V , Vbn = −110 − j190.52 V , Vcn = −110 + j190.5 V The line voltages in polar form are: 7.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 ~ ~ Vab = 3Van ∠30° = 380∠30° V ~ Vbc = 380∠ − 90° V ~ Vca = 380∠150° V The line voltages in rectangular form are: ~ ~ ~ Vab = 329 + j190 V , Vbc = − j 380 V , Vca = −329 + j190 V ______________________________________________________________________________________ Problem 7.46 Solution: Known quantities: The phase currents, ~ ~ ~ Ian = 10∠0 , Ibn = 12∠150° , Icn = 8∠165° . Find: The current in the neutral wire. Analysis: The neutral current is: ~ ~ ~ ~ In = Ian + Ibn + Icn = 5∠0° + 12∠150° + 8∠165° = −13.11 + j8.07 = 15.39∠148.4° A ______________________________________________________________________________________ Problem 7.47 Solution: Known quantities: Circuit shown in Figure P7.47, the voltage sources, ~ VB = 120∠ 240° V . ~ ~ VR = 120∠0° V , VW = 120∠120° V , Find: ~ ~ ~ VRW , VWB , VBR . ~ ~ ~ ~ ~ b) The voltages, VRW , VWB , VBR , using Vxy = Vx 3∠ − 30° . a) The voltages, c) Compare the results obtained in a and b. Analysis: a) ~ ~ ~ VRW = VR − VW = 120∠0° − 120∠120° = 120 + 60 − j103.92 = 207.8∠ − 30° V ~ ~ ~ VWB = VW − VB = 120∠120° − 120∠240° = −60 + j103.92 + 60 − j103.92 = 207.8∠90° V ~ ~ ~ VBR = VB − VR = 120∠240° − 120∠0° = −60 − j103.92 − 120 = 207.8∠ − 150° V ~ ~ b) VRW = VR 3∠ − 30° = 120 3∠ − 30° = 207.8∠ − 30° V ~ ~ VWB = VW 3∠ − 30° = 120∠120° 3∠ − 30° = 207.8∠90° V ~ ~ VBR = VB 3∠ − 30° = 120∠240° 3∠ − 30° = 207.8∠ 210° V = 207.8∠ − 150° V c) The two calculations are identical. ______________________________________________________________________________________ 7.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.48 Solution: Known quantities: Circuit shown in Figure P7.48, the voltage sources, ~ ~ VR = 110∠0° V , VW = 110∠120° V , ~ VB = 110∠ 240° V , and the three loads, Z R = 50 Ω , ZW = − j 20 Ω , Z B = j 45 Ω . Find: a) The current in the neutral wire. b) The real power. Analysis: ~ VR 110∠0° ~ a) IR = = = 2.2∠0° A ZR 50 ~ VB 110∠240° ~ IB = = = 2.44∠150° A ZB j 45 ~ VW 110∠120° ~ IW = = = 5.5∠210° A ZW − j 20 ~ ~ ~ ~ I N = I R + IW + I B = 2.2 + 5.5∠ 210° + 2.44∠150° = 4.92∠ − 161.9° A ~2 2 b) P = R ⋅ IB = 50 ⋅ 2.2 = 242 W ______________________________________________________________________________________ Problem 7.49 Solution: Known quantities: Circuit shown in Figure P7.49, the voltage sources, ~ ~ VR = 220∠0° V , VW = 220∠120° V , ~ VB = 220∠240° V , and the impedances, RW = RB = RR = 10 Ω . Find: a) The current in the neutral wire. b) The real power. Analysis: a) ~ V 220∠0° ~ IR = R = = 22∠0° A RR 10 ~ V 220∠120° ~ IW = W = = 22∠120° A RW 10 ~ V 220∠240° ~ IB = B = = 22∠240° A RB 10 Therefore, the current in the neutral wire is: ~ ~ ~ ~ I N = I R + IW + I B = 0 A b) The real power is: 7.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 ~ VR2 220 2 ~2 ~2 ~2 ~2 P = I R ⋅ R + IW ⋅ R + IB ⋅ R = 3R ⋅ IR = 3 =3 = 14.52 kW R 10 ______________________________________________________________________________________ Problem 7.50 Solution: Known quantities: A three-phase electric oven with a phase resistance of 10 Ω, connected at 3 × 380 V AC. Find: a) The current flowing through the resistors in Y and ∆ connections. b) The power of the oven in Y and ∆ connections. Analysis: a) In Y-connection: ~ V phase 380 3 220 ~ I RY = = = = 22 A R 10 10 In ∆-connection: ~ Vline 380 ~ = = 12.7 A I RD = R 30 b) In Y-connection: ~ ~ ~ ~ P = 3 ⋅ Vline ⋅ I line = 3 ⋅ Vline ⋅ I RY = 3 ⋅ 380 ⋅ 22 = 14.5 kW In ∆-connection: ~ ~ ~ ~ P = 3 ⋅ Vline ⋅ I line = 3 ⋅Vline ⋅ I RD = 3 ⋅ 380 ⋅12.7 = 8.36 kW ______________________________________________________________________________________ Problem 7.51 Solution: Known quantities: Apparent power of 50 kVA and supplied voltage of 380 V for a synchronous generator. Find: The phase currents, the active powers, and the reactive powers if: a) The power factor is 0.85. b) The power factor is 1. Analysis: a) For the power factor of 0.85: 50000 S ~~ ~ S = 3V I I = = 76 A ~= 3V 3 ⋅ 380 P = S ⋅ cos(θ ) = 50000 ⋅ 0.85 = 42.5 kW Q = S 2 − P 2 = 50 2 − 42.52 = 26.3 kVAR b) For the power factor of 1.00: ~ S = P I = 76 A P = S ⋅ cos(θ ) = 50000 ⋅ 1.00 = 50.0 kW Q = S 2 − P2 = 0 ______________________________________________________________________________________ 7.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.52 Solution: Known quantities: Circuit shown in Figure P7.52, the voltage sources, v s1 (t ) = 170 cos(ωt ) V , v s 2 (t ) = 170 cos(ωt + 120°) V , v s 3 (t ) = 170 cos(ωt − 120° )V , and the impedances, Z1 = 0.5∠20° Ω , Z 2 = 0.35∠0° Ω , Z 3 = 1.7∠ − 90° Ω , the frequency, f = 60 Hz . Find: The current through Z1, using: a) Loop/mesh analysis. b) Node analysis. c) Superposition. Analysis: a) Applying KVL in the upper mesh: ( ) ~ ~ ~ ~ ~ ~ ~ ~ ~ Vs 2 − Vs1 + I1 Z1 + I1 − I2 Z 2 = 0 I1 (Z1 + Z 2 ) + I2 (− Z 2 ) = Vs1 − Vs 2 Applying KVL in the lower mesh: ( ) ~ ~ ~ ~ ~ ~ ~ ~ ~ Vs 3 − Vs 2 + I2 − I1 Z 2 + I2 Z 3 = 0 I1 (− Z 2 ) + I2 (Z 2 + Z 3 ) = Vs 2 − Vs 3 For each mesh equation: ~ ~ Vs1 − Vs 2 = 170∠0° − 170∠120° = 170 − (− 85 + j147 ) = 294∠ − 30° V ~ ~ Vs 2 − Vs 3 = 170∠120° − 170∠ − 120° = (− 85 + j147 ) − (− 85 − j147 ) = 294∠90° V Z1 + Z 2 = 0.47 + j 0.171 + 0.35 = 0.838∠11.8° Ω Z 2 + Z 3 = 0.35 − j1.7 = 1.74∠ − 78.4° Ω Therefore, the current through Z1 is: ~ ~ Vs1 − Vs 2 − Z 2 294∠ − 30° − 0.35∠0° ~ ~ 294∠90° 1.74∠ − 78.4° ~ Vs 2 − Vs 3 Z 2 + Z 3 I1 = = Z1 + Z 2 − Z 2 0.838∠11.8° − 0.35∠0° − Z2 Z2 + Z3 − 0.35∠0° 1.74∠ − 78.4° = 512∠ − 108.4° + 103∠90° 415.9∠ − 112.9° = = 293∠ − 41.8° A 1.46∠ − 66.6° − 0.123∠0° 1.416∠ − 71.2° b) Choose the ground at the center of the three voltage source, and let a be the center of the three loads. The voltage between the node a and the ground is unknown. Applying KCL at the node a: ~ ~ ~ ~ ~ ~ Va − Vs1 Va − Vs 2 Va − Vs 3 + + =0 Z1 Z2 Z3 Rearranging the equation: ~ ~ ~ Vs1 Vs 2 Vs 3 170∠0° 170∠120° 170∠ − 120° + + + + Z1 Z2 Z3 ~ 0 5 20 0 35 10 1.7∠ − 90° . ∠ ° . ∠ ° Va = = 1 1 1 1 1 1 + + + + 0.5∠20° 0.35∠10° 1.7∠ − 90° Z1 Z 2 Z 3 = 340∠ − 20° + 486∠120° + 100∠330° 303∠57.3° = = 63.9∠58.5° V 2∠ − 20° + 2.86∠0° + 0.59∠90° 4.74∠ − 1.2° Applying KVL, the current through Z1 is: 7.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 ~ ~ ~ ~ ~ ~ Vs1 − Va 170∠0° − 63.9∠58.5° Va − Vs1 − I1Z1 = 0 I1 = = = 293∠ − 41.8° A Z1 0.5∠20° c) Superposition is not the method of choice for its great complexity. ______________________________________________________________________________________ Problem 7.53 Solution: Known quantities: Circuit shown in Figure P7.13, the voltage sources, v1 (t ) = 170 cos(ωt ) V , v2 (t ) = 170 cos(ωt + 120°) V , v3 (t ) = 170 cos(ωt − 120° )V , and the impedances, R = 100 Ω , C = 0.47 µF , L = 100 mH , the frequency, f = 400 Hz . Find: The current through R. Analysis: For each impedance: Z1 = 100 Ω = 100∠0° Ω 1 1 Z2 = − j =−j = − j846.6 Ω = 846.6∠ − 90° Ω ωC 2πf ⋅ C Z 3 = jωω = − j 2πf ⋅ L = j 251.3 Ω = 251.3∠90° Ω Applying KVL in the upper mesh: ( ) ~ ~ ~ ~ ~ ~ ~ ~ ~ V2 − V1 + I1 Z1 + I1 − I2 Z 2 = 0 I1 (Z1 + Z 2 ) + I2 (− Z 2 ) = V1 − V2 Applying KVL in the lower mesh: ( ) ~ ~ ~ ~ ~ ~ ~ ~ ~ V3 − V2 + I2 − I1 Z 2 + I2 Z 3 = 0 I1 (− Z 2 ) + I2 (Z 2 + Z 3 ) = V2 − V3 For each mesh equation: ~ ~ V1 − V2 = 170∠0° − 170∠120° = 170 − (− 85 + j147 ) = 294∠ − 30° V ~ ~ V2 − V3 = 170∠120° − 170∠ − 120° = (− 85 + j147 ) − (− 85 − j147 ) = 294∠90° V Z1 + Z 2 = 100 − j846.6 = 852.5∠ − 83.3° Ω Z 2 + Z 3 = − j846.6 + j 251.3 = 595.3∠ − 90° Ω Therefore, the current through R is: ~ ~ V1 − V2 − Z 2 294∠ − 30° − 846.6∠ − 90° ~ ~ 294∠90° 595.3∠ − 90° ~ V2 − V3 Z 2 + Z 3 I1 = = Z1 + Z 2 − Z 2 852.5∠ − 83.3o − 846.6∠ − 90 o − Z2 Z2 + Z3 − 846.6∠ − 90 o 595.3∠ − 90 o 175.0 ⋅ 10 3 ∠ − 120° + 248.9 ⋅ 10 3 ∠0° 221.4∠ − 43.2° = 3 3 507.5 ⋅ 10 ∠ − 173.3° − 716.7 ⋅ 10 ∠ − 180° 220.8∠ − 15.6° = 1.003∠ − 27.6° A = ______________________________________________________________________________________ 7.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Problem 7.54 Solution: Known quantities: Circuit shown in Figure P7.54, the voltage sources, vs1 (t ) = 170 cos(ωt ) V , vs 2 (t ) = 170 cos(ωt + 120°) V , vs 3 (t ) = 170 cos(ωt − 120°) V , and the impedances, Z1 = 3∠0° Ω , Z 2 = 7∠90° Ω , Z 3 = 0 − j11Ω , the frequency, f = 60 Hz . Find: ~ ~ ~ The currents, I1 , I2 , I3 . Analysis: Applying KVL in the upper mesh: ~ ~ ~ Vs 2 − Vs1 + I1Z1 = 0 ~ ~ ~ Vs1 − Vs 2 170∠0° − 170∠120° 294∠ − 30° I1 = = = = 98.1∠ − 30° A Z1 3∠0° 3∠0° Applying KVL in the right-side mesh: ~ ~ ~ Vs3 − Vs1 + I2 Z 2 = 0 ~ ~ ~ Vs1 − Vs 3 170∠0° − 170∠ − 120° 294∠30° I2 = = = = 42.1∠ − 60° A Z2 7∠90° 7∠90° Applying KVL in the lower mesh: ~ ~ ~ Vs3 − Vs 2 + I3 Z 3 = 0 ~ ~ ~ Vs 2 − Vs3 170∠120° − 170∠ − 120° 294∠90° I3 = = = = 26.8∠ − 180° A Z3 11∠ − 90° 11∠ − 90° ______________________________________________________________________________________ Problem 7.55 Solution: Known quantities: Circuit shown in Figure P7.55, the voltage sources, ~ ~ VRW = 416∠ − 30° V , VWB = 416∠210° V , ~ VBR = 416∠90° V , and the impedances, R1 = R2 = R3 = 40 Ω , L1 = L2 = L3 = 5 mH . The frequency of each of the sources, f = 60 Hz . Find: ~ The currents, IW ~ ~ ~ , IB , IR , IN . Analysis: The line voltages are: ~ ~ ~ VRW = 416∠ − 30° V , VWB = 416∠210° V , VBR = 416∠90 V The phase voltages are: ~ ~ ~ VR = 240∠0° V , VW = 240∠120° V , VB = 240∠ − 120° V The currents are: 7.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 ~ ~ ~ VR VR VR 240∠0° ~ = = = = 6 ∠ − 2 .7 ° A IR = Z1 R1 + jωω1 R1 + j 2πL1 40 + j 377 ⋅ 5 ⋅10 −3 ~ VW 240∠120° ~ = = 6∠117.3° A IW = Z 2 40 + j 377 ⋅ 5 ⋅10 −3 ~ V 240∠ − 120° ~ = 6∠ − 122.7° A IB = B = Z 3 40 + j 377 ⋅ 5 ⋅10 −3 ~ ~ ~ ~ IN = IR + IW + IB = 0 ______________________________________________________________________________________ Problem 7.56 Solution: Known quantities: Circuit shown in Figure P7.54, the voltage sources, ~ ~ VRW = 416∠ − 30° V , VWB = 416∠210° V , ~ VBR = 416∠90° V , and the impedances, R1 = R2 = R3 = 40 Ω , L1 = L2 = L3 = 5 mH . The frequency of each of the sources, f = 60 Hz . Find: a) The power delivered to the motor. b) The motor's power factor. c) The reason for which it is common in industrial practice not to connect the ground lead to motors of this type. Analysis: a) The power delivered to the motor is: ~ ~ P = 3VR I R cos(θ ) = 3 ⋅ 240 ⋅ 6 ⋅ cos(− 2.7°) = 4315 W b) The motor's power factor is: pf = cos(− 2.7°) = 0.9988 Lagging c) The circuit is balanced and no neutral current flows; thus the connection is unnecessary. ______________________________________________________________________________________ Problem 7.57 Solution: Known quantities: A three-phase induction motor designed not only for Y connection operation in general but also for ∆ connection at the nominal Y voltage for a short time operation. Find: The ratio between the powers. Analysis: The power for Y connection operation is: ~ Vline ~ ~ ~ ~ ~ PY = 3 ⋅ V phaseMOTOR ⋅ IlineY = 3 ⋅ ⋅ IlineY = 3 ⋅ Vline ⋅ IlineY 3 The power for ∆ connection operation is: 7.33 G. Rizzoni, Principles and Applications of Electrical Engineering ~ ~ ~ P∆ = 3 ⋅ Vline ⋅ I phaseMOTOR = 3 ⋅ Vline ⋅ Problem solutions, Chapter 7 ~ Vline Z phaseMOTOR ~ V phaseMOTOR ~ ~ ~ = 3 ⋅ Vline ⋅ 3 ⋅ = 3 ⋅ Vline ⋅ 3 ⋅ IlineY = 3 ⋅ PY Z phaseMOTOR Therefore, the ratio between the powers is: Ratio = P∆ =3 PY ______________________________________________________________________________________ Problem 7.58 Solution: Known quantities: Circuit shown in Figure P7.58, the voltage sources, ~ ~ VR = 120∠0° V , VW = 120∠120° V , ~ VB = 120∠240° V , and the impedances, R1 = R2 = R3 = 5 Ω , X L1 = X L2 = X L3 = 6 Ω of the motor. Find: a) The total power supplied to the motor. b) The power converted to mechanical energy if the motor is 80% efficient. c) The power factor. d) The risk for the company to face a power factor penalty if all the motors in the factory are similar to this one. Analysis: a) By virtue of the symmetry of the circuit, we can solve the problem by considering just one phase. ~ The current IR is: ~ IR = ~ VR 120∠0° = = 15.36∠ − 50.19° A R1 + jX 1 5 + j6 The total power supplied to the motor is: ~ ~ P = 3VR I R cos(θ ) = 3 ⋅120 ⋅15.36 ⋅ cos(50.19°) = 3541.3 W b) The mechanical power is: Pm = 0.8P = 2832.23 W c) The power factor is: pf = cos(θ ) = 0.64 d) The company will face a 25% penalty. ______________________________________________________________________________________ Problem 7.59 Solution: Known quantities: The voltage source at 220 V rms of a residential four-wire system supplying power to the single-phase appliances; ten 75-W bulbs on the 1st phase, one 750-W vacuum cleaner with pf = 0.87 on the 2nd phase, ten 40-W lamps with pf = 0.64 on the 3rd phase. Find: a) The current in the neutral wire. b) The real, reactive, and apparent power for each phase. 7.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 7 Analysis: a) The current in the neutral wire is: P 3⋅P 3 ⋅10 ⋅ 75 ~ IA = ~ A = ~ A = = 3.4∠0 A 380 V phase Vline 3 ⋅ PB 3 ⋅ 750 P ~ IB = ~ B = ~ = = 3.92∠ − 150° A = −3.4 − j1.95 A V phase Vline ⋅ cos(θ ) 380 ⋅ 0.87 3 ⋅ PC 3 ⋅10 ⋅ 40 P ~ IC = ~ C = ~ = = 2.8∠ − 290° A = 0.95 + j 2.63 A V phase Vline ⋅ cos(θ ) 380 ⋅ 0.64 ~ ~ ~ ~ ~ IN = I A + IB + IC = 0.95 − j 0.68 A I N = 1.16 A b) The real, reactive, and apparent powers for the 1st phase are: S A = PA = 750 W , QA = 0 The real, reactive, and apparent powers for the 2nd phase are: PB = 750 W , S B = PB 750 = = 862 VA , QB = S B ⋅ sin(θ B ) = 431 VAR cos(θ B ) 0.87 The real, reactive, and apparent powers for the 3rd phase are: PC = 400 W , SC = PC 400 = = 625 VA , QC = SC ⋅ sin(θC ) = 478 VAR cos(θC ) 0.64 ______________________________________________________________________________________ 7.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Chapter 8 Instructor Notes Chapter 8 introduces the notion of integrated circuit electronics through the most common building block of electronic instrumentation, the operational amplifier. This is, in practice, the area of modern electronics that is most likely to be encountered by a practicing non-electrical engineer. Thus, the aim of the chapter is to present a fairly complete functional description of the operational amplifier, including a discussion of the principal limitations of the op-amp and of the effects of these limitations on the performance of op-amp circuits employed in measuring instruments and in signal conditioning circuits. The material presented in this chapter lends itself particularly well to a series of laboratory experiments (see for example1), which can be tied to the lecture material quite readily. After a brief introduction, in which ideal amplifier characteristics are discussed, open- and closed- loop models of the op-amp are presented in section 8.2; the use of these models is illustrated by application of the basic circuit analysis methods of Chapters 2 and 3. Thus, the Instructor who deems it appropriate can cover the first two sections in conjunction with the circuit analysis material. A brief, intuitive discussion of feedback is also presented to explain some of the properties of the op-amp in a closed-loop configuration. The closed-loop models include a fairly detailed introduction to the inverting, non-inverting and differential amplifier circuits; however, the ultimate aim of this section is to ensure that the student is capable of recognizing each of these three configurations, so as to be able to quickly determine the closed loop gain of practical amplifier circuits, summarized in Table 8.1 (p. 402). The section is sprinkled with various practical examples, introducing practical op-amp circuits that are actually used in practical instruments, including the summing amplifier (p. 938), the voltage follower (p. 400)), a differential amplifier (Focus on Measurements: Electrocardiogram (EKG) Amplifier, pp. 402-404), the instrumentation amplifier (pp. 404405), the level shifter (p.406), and a transducer calibration circuit (Focus on Measurements: Sensor calibration circuit, pp. 407-409). Two features, new in the third edition, will assist the instructor in introducing practical design considerations: the box Practical Op-Amp Design Considerations (p. 410) illustrates some standard design procedures, providing an introduction to a later section on op-amp limitations; the box Focus on Methodology: Using Op-amp Data Sheets (pp. 410-412) illustrates the use of device data sheets for two common op-amps. The use of Device Data Sheets is introduced in this chapter for the first time. In a survey course, the first two sections might be sufficient to introduce the device. Section 8.3 presents the idea of active filters; this material can also be covered quite effectively together with the frequency response material of Chapter 6 to reinforce these concepts. Section 8.4 discusses integrator and differentiator circuits, and presents a practical application of the op-amp integrator in the charge amplifier (Focus on Measurements: Charge Amplifiers, pp. 420-421). The latter example is of particular relevance to the non-electrical engineer, since charge amplifiers are used to amplify the output of piezo-electric transducers in the measurement of strain, force, torque and pressure (for additional material on piezo-electric transducers, see, for example2). A brief section (8.5) is also provided on analog computers, since these devices are still used in control system design and evaluation. Coverage of sections 8.4 and 8.5 is not required to complete section 8.6. The last section of the chapter, 8.6, is devoted to a discussion of the principal performance limits of the operational amplifier. Since the student will not be prepared to fully comprehend the reason for the saturation, limited bandwidth, limited slew rate, and other shortcomings of practical op-amps, the section focuses on describing the effects of these limitations, and on identifying the relevant parameters on the data sheets of typical op-amps. Thus, the student is trained to recognize these limits, and to include them in the design of practical amplifier circuits. Since some of these limitations are critical even in low frequency applications, it is easy (and extremely useful) to supplement this material with laboratory exercises. The box Focus on Methodology: Using Op-amp Data Sheets – Comparison of LM 741 and LMC 6061, pp. 441446) further reinforces the value of data sheets in realizing viable designs. The homework problems present a variety of interesting problems at varying levels of difficulty; many of these problems extend the ideas presented in the text, and present practical extensions of the circuits discussed in the examples. In a one-semester course, Chapter 8 can serve as a very effective capstone to a first course in circuit analysis and electronics by stimulating curiosity towards integrated circuit electronics, 1 2 Rizzoni, G., A Practical Introduction to Electronic Instrumentation, 3rd Ed. Kendall-Hunt, 1998 Doebelin E. O., Measurement Systems, McGraw-Hill, Fourth Edition, 1987. 8.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 and by motivating the student to pursue further study in electronics or instrumentation. In many respects this chapter is the centerpiece of the book. Learning Objectives 1. Understand the properties of ideal amplifiers, and the concepts of gain, input impedance, and output impedance. Section 1. 2. Understand the difference between open-loop and closed-loop op-amp configuration, and compute the gain (or complete the design of) simple inverting, non-inverting, summing and differential amplifiers using ideal op-amp analysis. Analyze more advanced op-amp circuits, using ideal op-amp analysis, and identify important performance parameters in op-amp data sheets. Section 2. 3. Analyze and design simple active filters. Analyze and design ideal integrator and differentiator circuits. Sections 3 and 4. 4. Understand the structure and behavior of analog computers, and design analog computer circuits to solve simple differential equations. Section 5. 5. Understand the principal physical limitations of an op-amp. Section 6. Section 8.1: Ideal Amplifiers Problem 8.1 Solution: Known quantities: For the circuit shown in Figure P8.1: G = Po R s = 0.6 kΩ R L = 0.6 kΩ PS Ro1 = 2 kΩ Ro 2 = 2 kΩ Ri1 = 3 kΩ Ri 2 = 3 kΩ Avo1 = 100 Find: The power gain G -1 G M 2 = 350 mΩ Vi 2 = Vl 2 = Po/PS, in dB. Analysis: Starting from the last stage and going backward, we get 8.2 Vi1 = Vl1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 V02 Ro 2 R L = 161.5 V P = V I = = 43.49 Vl 22 V O = GM 2 V l 2 l2 0 0 0 + R Ro 2 R L L Ri 2 Vl 2 = Av 01Vl1 = 60 Vl1 P0 = 1.566 ⋅ 10 5 Vl12 Ri 2 + R01 Vl1 = VS Ri1 = 0.833 VS P0 = 1.0875 ⋅ 10 5 VS2 Ri1 + RS I S = VS 1 = 2.778 ⋅10 −4 VS PS = VS I S = 2.778 ⋅ 10 −4 VS2 Ri1 + RS P0 1.0875 ⋅ 10 5 G= = = 3.915 ⋅ 10 8 = 20 dB Log10 [ 3.915 ⋅ 10 8 ] = 171.85 dB −4 PS 2.778 ⋅10 ______________________________________________________________________________________ Problem 8.2 Solution: Known quantities: The temperature sensor shown in Figure P8.2 produces a no load (i.e., sensor current = 0) voltage: vs = V so cos ωt R s = 400 Ω V so = 500 mV ω = 6.28 k rad s The temperature is monitored on a display (the load) with a vertical line of light emitting diodes. Normal conditions are indicated when a string of the bottommost diodes 2 cm in length are on. This requires that a voltage be supplied to the display input terminals where: = 12 kΩ = cos ω t = 6 V . vo V o Vo RL The signal from the sensor must therefore be amplified. Therefore, a voltage amplifier is connected between the sensor and CRT with: Ri = 2 kΩ Ro = 3 kΩ . Find: The required no load gain of the amplifier. Analysis: The overall loaded voltage gain, using the amplitudes of the sensor voltage and the specified CRT voltage must be: V o = 6 V = 12 AV = 500 mV V so An expression for the overall voltage gain can also be obtained using two voltage divider relationships: Ri ] R L RL = Avo [ V so + + Ri R s R o + R L Ro R L [ 12 ] [ 0.4 + 2 ] [ 3 + 12 ] [ + ] [ Ro + R L ] = 18 = = Av R s Ri [ 2 ] [ 12 ] Ri R L V o = Avo V io Vo Avo = V so The loss in gain due to the two voltage divisions or "loading" is characteristic of all practical amplifiers. Ideally, there is no reduction in gain due to loading. This would require an ideal signal source with a source resistance equal to zero and a load resistance equal to infinity. ______________________________________________________________________________________ 8.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.3 Solution: Known quantities: For the circuit shown in Figure P8.3: Po Pi Ri1 = 1.1 kΩ G= 2 vI1 Pi = Ri1 Ri 2 = 19 kΩ R s = 0.7 kΩ Ro1 = 2.9 kΩ Avo1 = 65 Find: The power gain G R L = 16 Ω Ro 2 = 22 Ω G m 2 = 130 mS = Po/Pi, in dB. Analysis: Ro 2 R L 1.204 ⋅ vI 2 Ro 2 + R L [ 65 ] [ 19 ] vO Ri 2 [ 1.204 ] = 67.91 = ] [1.204 ] VD : vO = [ Avo1 v I 1 2.9 + 19 vI1 Ro1 + Ri 2 OL : vO = G m 2 v I 2 2 vO 1100 v = 317.1 ⋅10 3 = 20 dB Log10 [ 317.1 ⋅10 3 ] = 110.02 dB G = R L2 = [ o ] 2 Ri1 = [ 67.91 ]2 16 vI1 RL vI1 Ri1 ______________________________________________________________________________________ Problem 8.4 Solution: Known quantities: For the circuit shown in Figure P8.4: R s = 0.3 kΩ R L = 2 kΩ Ri1 = Ri 2 = 7.7 kΩ Ro1 = Ro 2 = 1.3 kΩ vO = 149.9 vI1 Avo1 = Avo 2 = 17 Find: a) The power gain in, dB. b) The overall voltage gain vO/vS. Analysis: a) vO2 7.7 = 86509 = 20 Log10 [ 86509 ] = 98.74 dB G = PO = R L2 = [ vO ] 2 Ri1 = [ 149.9 ]2 2 vI1 R L PI vI1 Ri1 8.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 b) The input voltage of the first stage can be determined in terms of the source voltage using voltage division: Av = vO vO1 v I 1 vO = = vS v I 1 v S v I 1 vS Ri1 Ri1 + R s vS = 149.9 7.7 = 144.3= 20 Log10 [ 144.3 ] = 43.18 dB 0.3 + 7.7 ______________________________________________________________________________________ Problem 8.5 Solution: Known quantities: Figure P8.5. Find: What approximations are usually made about the voltages and currents shown for the ideal operational amplifier [op-amp] model. Analysis: iP ≈ 0 iN ≈ 0 vD ≈ 0 . ______________________________________________________________________________________ Problem 8.6 Solution: Known quantities: Figure P8.6. Find: What approximations are usually made about the circuit components and parameters shown for the ideal operational amplifier [op-amp] model. Analysis: µ ≈ ∞ ri ≈ ∞ ro ≈ 0 . ______________________________________________________________________________________ 8.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Section 8.2: The Operational Amplifier Problem 8.7 Solution: Known quantities: The circuit shown in Figure P8.7. Find: The resistor Rs that will accomplish the nominal gain requirement, and state what the maximum and minimum values of Rs can be. Will a standard 5 percent tolerance resistor be adequate to satisfy this requirement? Analysis: For a non-inverting amplifier the voltage gain is given by: Av = vout R f + Rs R f = = +1 ; vin Rs Rs Avnom = 16 = To find the maximum and minimum RS we note that the minimum Av: Rs max = Rf Rs Rs ∝ +1 Rs = Rf 16 − 1 = 1kΩ 1 , so to find the maximum RS we consider Av 15 kΩ = 1.02 kΩ . 16(1 − 0.02) − 1 15 kΩ = 980Ω . 16(1 + 0.02) − 1 Since a standard 5% tolerance 1-kΩ resistor has resistance 950 < R < 1050, a standard resistor will not Conversely, to find the minimum RS we consider the maximum Av: Rs min = suffice in this application. ______________________________________________________________________________________ Problem 8.8 Solution: Known quantities: The values of the two 10 percent tolerance resistors used in an inverting amplifier: RF = 33 kΩ ; RS = 1.2 kΩ . Find: a. The nominal gain of the amplifier. b. The maximum value of Av . c. The minimum value of Av . Analysis: Rf − 33 = −27.5 . Rs 1.2 b. First we note that the gain of the amplifier is proportional to Rf and inversely proportional to RS. This tells us that to find the maximum gain of the amplifier we consider the maximum Rf and the minimum RS. R f max 33 + 0.1(33) Av max = = 33.6 . = RS min 1.2 − 0.1(1.2) a. The gain of the inverting amplifier is: Av = − 8.6 = G. Rizzoni, Principles and Applications of Electrical Engineering c. To find Av we consider the opposite case: min Av min = R f min RS max Problem solutions, Chapter 8 = 33 − 0.1(33) = 22.5 . 1.2 + 0.1(1.2) ______________________________________________________________________________________ Problem 8.9 Solution: Known quantities: For the circuit shown in Figure P8.9, let v1 (t ) = 10 + 10 −3 sin (ωt ) V , RF = 10 kΩ , Vbatt = 20 V . Find: a. The value of RS such that no DC voltage appears at the output. b. The corresponding value of vout(t). Analysis: a. The circuit may be modeled as shown: Applying the principle of superposition: For the 20-V source: vo 20 = − 10kΩ (20) RS For the 10-V source: vo 10 § 10kΩ · = ¨¨ + 1¸¸(10) R ¹ © S The total DC output is: Solving for RS b. vo DC = vo 20 + vo 10 10,000 (10 - 20) = -10 RS =− § 10,000 · 10,000 (20) + ¨¨ + 1¸¸(10) = 0 RS ¹ © RS RS = 10 kΩ . Since we have already determined RS such that the DC component of the output will be zero, we can simply treat the amplifier as if the AC source were the only source present. Therefore, § Rf · + 1¸¸ = 0.001sin (ωt ) ⋅ (1 + 1) = 2 ⋅10 −3 sin (ωt ) V . vo (t ) = 0.001sin (ωt ) ⋅ ¨¨ © RS ¹ ______________________________________________________________________________________ Problem 8.10 Solution: Known quantities: For the circuit shown in Figure P8.10, let: R = 2 MΩ RS = 1kΩ AV (OL ) = 200,000 R0 = 50Ω R1 = 1kΩ R2 = 100kΩ RLOAD = 10kΩ 8.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Find: AV = The gain v0 vi Analysis: The op-amp has a very large input resistance, a very large "open loop gain" [ΑV(OL)], and a very small output resistance. Therefore, it can be modeled with small error as an ideal op-amp. The amplifier shown in Figure P8.10 is a noninverting amplifier, so we have AV = 1 + R2 100 ⋅103 = 1+ = 101 R1 1⋅103 ______________________________________________________________________________________ Problem 8.11 Solution: Known quantities: vout (t ) = −(2 sin ω1t + 4 sin ω 2t + 8 sin ω 3t + 16 sin ω 4t ) V, RF = 5 kΩ . Find: Design an inverting summing amplifier to obtain vout (t ) and determine the require source resistors. Analysis: The inverting summing amplifier is shown in the following figure. By superposition and by selecting 4 vout = −¦ i =1 RSi = RF , vSi = sin ω i t , i = 1 4 , the output voltage is 2i 4 4 RF vSi = −¦ 2i vSi = −¦ 2i sin ω i t RSi i =1 i =1 that coincides with the desired output voltage. So, the required source resistors are RS1 = RF R R R = 2.5kΩ, RS 2 = F = 1.25kΩ, RS 3 = F = 625Ω, RS 4 = F = 312.5Ω . 2 4 8 16 ______________________________________________________________________________________ Problem 8.12 Solution: Known quantities: For the circuit shown in Figure P8.12: 8.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 ri = 2 MΩ µ = 200,000 r0 = 25Ω RS = 2.2kΩ R1 = 1kΩ RF = 8.7kΩ RL = 20Ω Find: a. An expression for the input resistance vi/ii including the effects of the op-amp. b. The value of the input resistance in including the effects of the op-amp. c. The value of the input resistance with ideal op-amp. Analysis: a. The circuit in Figure P8.12 can be modeled as in the following figure where vi = −vd + R1 ii , vd = − ri (ii − iF ) , iF = − vd + v0 , RF v0 = µ vd + r0 (iF − v0 ). RL Substituting the expression for vd in all the other equations, we obtain vi = ( R1 + ri ) ii − ri iF , iF = ri ii − v0 , RF + ri v0 (1 + r0 ) = − µ ri ii + ( µ ri + r0 )iF . RL By solving the second equation above for v0 and substituting in the third equation vi = ( R1 + ri ) ii − ri iF , § · r ¨¨1 + 0 + µ ¸¸ri © RL ¹ ii iF = § r0 · ¨¨1 + ¸¸( RF + ri ) + µ ri + r0 © RL ¹ and finally, 8.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 § · r ¨¨1 + 0 + µ ¸¸ ri 2 vi © RL ¹ = ( R1 + ri ) − ii § r · ¨¨1 + 0 ¸¸( RF + ri ) + µ ri + r0 © RL ¹ b. The value of the input resistance is § · r § 25 · ¨¨1 + 0 + µ ¸¸ ri 2 + 2 ⋅105 ¸ 4 ⋅1012 ¨1 + vi © RL ¹ © 20 ¹ = ( R1 + ri ) − = 2.001 ⋅106 − = ii § 25 · § r0 · 6 11 ¨1 + ¸ 2.0087 ⋅10 + 4 ⋅10 + 25 ¨¨1 + ¸¸( RF + ri ) + µ ri + r0 © 20 ¹ © RL ¹ = 1.0001kΩ c. In the ideal case: vi = R1= 1kΩ ii ______________________________________________________________________________________ Problem 8.13 Solution: Known quantities: For the circuit shown in Figure P8.13: vS (t ) = 0.02 + 10 −3 cos(ωt ) V , RF = 220 kΩ , R1 = 47 kΩ , R2 = 1.8kΩ . Find: a) expression for the output voltage. b) The corresponding value of vo(t). Analysis: a) the circuit is a noninverting amplifier, then § R · v0 = ¨¨1 + F ¸¸vS R2 ¹ © § 220 · b) v0 = ¨1 + ¸vS = 2.464 + 0.1232 cos(ωt ) V © 1 .8 ¹ ______________________________________________________________________________________ Problem 8.14 Solution: Known quantities: For the circuit shown in Figure P8.13: vS (t ) = 0.05 + 30 ⋅10 −3 cos(ωt ) V , RS = 50Ω , RL = 200Ω . Find: The output voltage vo. 8.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: It is a particular case of a noninverting amplifier where v0=vS. ______________________________________________________________________________________ Problem 8.15 Solution: Known quantities: For the circuit shown in Figure P8.15: vS 1 (t ) = 2.9 ⋅10−3 cos(ωt ) V , R1 = 1kΩ , R2 = 3.3kΩ vS 2 (t ) = 3.1 ⋅10−3 cos(ωt ) V , R3 = 10kΩ , R4 = 18kΩ . Find: The output voltage vo and a numerical value. Analysis: By using superposition, R3 R4 § R3 · § 10 · 18 ¨¨1 + ¸¸vS 2 = ¨1 + ¸ vS 1 + 3.1 ⋅10 −3 cos(ωt ) − 2.9 ⋅10 −2 cos(ωt ) = R1 R2 + R4 © R1 ¹ 1 ¹ 18 + 3.3 © = −1.83 ⋅10 −3 cos(ωt ) V v0 = − ______________________________________________________________________________________ Problem 8.16 Solution: Known quantities: For the circuit shown in Figure P8.15: vS1 = 5 mV , R1 = 1kΩ , R2 = 15kΩ vS 2 = 7 mV , R3 = 72kΩ , R4 = 47 kΩ Find: The output voltage vo analytically and numerically. Analysis: From the solution of Problem 8.15, v0 = − R3 R4 § R3 · 47 ¨¨1 + ¸¸vS 2 = (1 + 72) vS 1 + 7 ⋅10 −3 − 3.60 ⋅10 −1 = 27.37mV R1 R2 + R4 © R1 ¹ 15 + 47 ______________________________________________________________________________________ Problem 8.17 Solution: Known quantities: If, in the circuit shown in Figure P8.17: vS1 = v S 2 = 7 mV R1 = 850 Ω R2 = 1.5 kΩ R F = 2.2 kΩ Op Amp : Motorola MC1741C r i = 2 MΩ µ = 200,000 r o = 25 Ω 8.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Find: a) The output voltage. b) The voltage gain for the two input signals. Analysis: a) The op amp has a very large input resistance, a very large "open loop gain" [µ], and a very small output resistance. Therefore, it can be modeled with small error as an ideal op amp with: KVL : v D + v N = 0 vN ≈ 0 vD ≈ 0 KCL : v N - v S 2 + v N - v S 1 + v N - vO = 0 R2 R1 RF ≈ 0 iN ≈ 0 iN + vN 2.2 2.2 ] [ 7 mV ] + [ ] [ 7 mV ] vO = - R F v S 1 - R F v S 2 = [ 0.85 1.5 R1 R2 = [ - 2.588 ] [ 7 mV ] + [ - 1.467 ] [ 7 mV ] = - 28.38 mV b) Using the results above: AV1 = -2.588 AV2 = -1.467. Note: The output voltage and gain are not dependent on either the op amp parameters or the load resistance. This result is extremely important in the majority of applications where amplification of a signal is required. ______________________________________________________________________________________ Problem 8.18 Solution: Known quantities: For the circuit shown in Figure P8.15: vS1 = kT1 , R1 = 11kΩ , R2 = 27 kΩ vS 2 = kT2 , R3 = 33kΩ , R4 = 68kΩ T1 = 35 C , T2 = 100 , C , k = 50mV / , C , Find: a) The output voltage. b) The conditions required for the output voltage to depend only on the difference between the two temperatures. Analysis: a) From the solution of Problem 8.15, 33 R4 § R3 · 33 68 R3 ¨¨1 + ¸¸kT2 = §¨1 + ·¸ 5 − 1.75 = kT1 + 11 R2 + R4 © R1 ¹ R1 © 11 ¹ 68 + 27 = 9.065 V v0 = − b) For R3=R4, R1=R2= k R3 , the output voltage is 8.12 G. Rizzoni, Principles and Applications of Electrical Engineering v0 = − Problem solutions, Chapter 8 R3 R4 § R3 · 1 R3 § R1 + R3 · ¨¨1 + ¸¸ kT2 = − kT1 + ¨ ¸ kT2 = T2 − T1 kT1 + R1 R2 + R4 © R1 ¹ k R1 + R3 ¨© R1 ¸¹ ______________________________________________________________________________________ Problem 8.19 Solution: Find: In a differential amplifier, if: Av1 = - 20 Av 2 = + 22 , derive expressions for and then determine the value of the common and differential mode gains. Analysis: There are several ways to do this. Using superposition: vO-C v S -C v = O- D vS -D If : v S - D = 0 : vO = vO-C + vO- D = vO-C Av-C = If : v S -C = 0 : vO = vO-C + vO- D = vO- D Av- D Assume that signal source #2 is connected to the non-inverting input of the op-amp. The common-mode output voltage can be obtained using the gains given above but assuming the signal voltages have only a common-mode component: vO = v S1 Av1 + v S 2 Av 2 v S1 = v S -C - 1 vS -D 2 v S 2 = v S -C + 1 vS - D 2 Let : v S - D = 0 vO-C = v S -C Av1 + v S -C Av 2 = v S -C [ Av1 + Av 2 ] = v S -C Av-C Av-C = Av1 + Av 2 = [ - 20 ] + [ + 22 ] = 2 The difference-mode output voltage can be obtained using the gains given but assuming the signal voltage have only a difference-mode component: Let : v S -C = 0 1 1 1 vO- D = [ - v S - D ] Av1 + [ v S - D ] Av 2 = v S - D [ - Av1 + Av 2 ] = v S - D Av- D 2 2 2 1 1 Av- D = [ Av 2 - Av1 ] = ( [ + 22 ] - [ - 20 ] ) = + 21 2 2 Note: If signal source #1 were connected to the non-inverting input of the op amp, then the difference mode gain would be the negative of that obtained above. ______________________________________________________________________________________ Problem 8.20 Solution: Known quantities: For the circuit shown in Figure P8.15: vS1 = 1.3V , R1 = R2 = 4.7kΩ vS 2 = 1.9V , R3 = R4 = 10kΩ Find: a) The output voltage. b) The common-mode component of the output voltage. 8.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 c) The differential-mode component of the output voltage. Analysis: a) From the solution of Problem 8.15, v0 = − R4 § R3 · 10000 R3 10000 · 10000 ¨¨1 + ¸¸vS 2 = §¨1 + vS 1 + 1.9 − 1.3 = 1.28V ¸ R1 R2 + R4 © R1 ¹ 4700 ¹ 4700 + 10000 4700 © b) The common-mode component is zero. c) The differential-mode component is v0. ______________________________________________________________________________________ Problem 8.21 Solution: Known quantities: For the circuit shown in Figure P8.15: vS1, 2 = A + BP1,2 , A = 0.3 , B = 0.7 V/psi R1 = R2 = 4.7 kΩ , R3 = R4 = 10kΩ , RL = 1.8kΩ P1 = 6kPa , P2 = 5kPa Find: a) The common-mode input voltage. b) The differential-mode input voltage. Analysis: a) The common-mode input voltage is vin+ 3 vS 2 + vS 1 P1 + P2 -4 11 ⋅ 10 = = A+ B = 0.3 + 0.7 ⋅1.4504 ⋅10 = 0.858V 2 2 2 b) The differential-mode input voltage is vin− = vS 2 − vS 1 = B( P2 − P1 ) = −0.7 ⋅1.4504 ⋅10 -4 ⋅103 = −0.1015V ______________________________________________________________________________________ Problem 8.22 Solution: Known quantities: A linear potentiometer [variable resistor] Rp is used to sense and give a signal voltage vY proportional to the current y position of an x-y plotter. A reference signal vR is supplied by the software controlling the plotter. The difference between these voltages must be amplified and supplied to a motor. The motor turns and changes the position of the pen and the position of the "pot" until the signal voltage is equal to the reference voltage [indicating the pen is in the desired position] and the motor voltage = 0. For proper operation the motor voltage must be 10 times the difference between the signal and reference voltage. For rotation in the proper direction, the motor voltage must be negative with respect to the signal voltage for the polarities shown. An additional requirement is that iP = 0 to avoid "loading" the pot and causing an erroneous signal voltage. Find: a) Design an op amp circuit which will achieve the specifications given. Redraw the circuit shown in Figure P8.22 replacing the box [drawn with dotted lines] with your circuit. Be sure to show how the signal voltage and output voltage are connected in your circuit. b) Determine the value of each component in your circuit. The op amp is an MC1741. 8.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: a) The output voltage to the motor must be dependent on the difference between two input voltages. A difference amp is required. The signal voltage must be connected to the inverting input of the amplifier. However, the feedback path is also connected to the inverting input and this will caused loading of the input circuit, ie, cause an input current. This can be corrected by adding an isolation stage [or voltage follower] between the input circuit and the inverting input of the 2nd stage. The isolation stage must have a gain = 1 and an input current = 0. The difference amplifier, the second stage, must give an output voltage: § 10 kΩ · + 1¸¸(10 ) v0 10 = ¨¨ R ¹ © S vM = 10 [ v R - vY ]= [ 10 ] v R + [ - 10 ] vY Avr = 10 Avy = - 10 The circuit configuration shown will satisfy these specifications. b) In this first approximation analysis, assume the op amps can be modeled as ideal op amps: vD ≈ 0 iP = i N ≈ 0 Consider the first or isolation or voltage follower stage: Ideal : i P ≈ 0 KVL : - vY + v D + vO1 = 0 v D ≈ 0 KVL : - v N - v D + v P = 0 vN = vP vD ≈ 0 KCL : v N - vO1 R1 + v N - vM R2 + iN = 0 -0 KCL : v P v R + v P + i P = 0 R3 R4 iN ≈ 0 iP ≈ 0 vO1 = vY vO1 + v M R1 R2 R1 R2 vN = 1 1 R1 R2 + R1 R2 vR R3 R4 R3 vP = 1 1 R3 R4 + R3 R4 The second stage: vO1 R2 + v M R1 = v R R4 R2 + R1 R2 + R1 R4 + R3 R2 ] + R4 [ R2 + R1 ] = vM = vY [ vR vY Avy + v R Avr R1 R1 [ R4 + R3 ] R2 = - 10 Choose : R 2 = 100 kΩ R1 = 10 kΩ Avy = R1 R4 [ R2 + R1 ] = 10 Choose : R3 = 10 kΩ R4 = 100 kΩ Avr = R1 [ R4 + R3 ] vO1 = vY vN = vP The resistances chosen are standard values and there is some commonality in the choices. Moderately large values were chosen to reduce currents and the resistor power ratings. Cost of the resistors will be determined primarily by power rating and tolerance. ______________________________________________________________________________________ 8.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.23 Solution: Known quantities: For the circuit shown in Figure P8.15: vS1 = 13mV , R1 = 1kΩ , R2 = 13kΩ vS 2 = 19 mV , R3 = 81kΩ , R4 = 56kΩ Find: The output voltage vo. Analysis: From the solution of Problem 8.15, v0 = R 56 R4 § R3 · ¨¨1 + ¸¸vS 2 − 3 vS1 = (1 + 81) 19 ⋅10 −3 − 81⋅13 ⋅10 −3 = 0.211 V R2 + R4 © R1 ¹ R1 56 + 13 ______________________________________________________________________________________ Problem 8.24 Solution: Known quantities: The circuit shown in Figure P8.24. Find: Show that the current Iout through the light-emitting diode is proportional to the source voltage VS as long as : VS > 0. Analysis: Assume the op amp is ideal: V − ≈ V + = VS ½° ¾ if °¿ I− ≈0 VS > 0 I out = V − VS = . R2 R2 ______________________________________________________________________________________ Problem 8.25 Solution: Known quantities: The circuit shown in Figure P8.25. Find: Show that the voltage Vout is proportional to the current generated by the CdS solar cell. Show that the transimpedance of the circuit Vout/Is is –R. Analysis: v + ≅ v − = 0 Vout = − RI S V The transimpedance is given by Rtrans = out = − R IS Assuming an ideal op-amp : ______________________________________________________________________________________ 8.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.26 Solution: Known quantities: The op-amp voltmeter circuit shown in Figure P8.26 is required to measure a maximum input of E = 20 mV. The op-amp input current is IB = 0.2 µA, and the meter circuit has Im = 100 µA full-scale deflection and rm = 10 kΩ. Find: Determine suitable values for R3 and R4. Analysis: Vout = I m rm = (100µA )(10kΩ) = 1V Vout R4 + R3 R E E − Vout From KCL at the inverting input, + = 0 or = = 1+ 4 . E R3 R3 R3 R4 R4 Vout 1 Then, = −1 = − 1 = 49 . Now, choose R3 and R4 such R3 E 20 × 10 −3 E IB = ≤ 0.2µ. R3 (R4 + rm ) At the limit, ( 20 × 10 −3 R3 49 R3 + 10 ×10 3 ) that = 0.2 × 10 −6 . Solving for R3, we have R3 ≈ 102kΩ . Therefore, R4 ≈ 5MΩ . ______________________________________________________________________________________ Problem 8.27 Solution: Known quantities: Circuit in Figure P8.27. Find: The output voltage vo. Analysis: The circuit is a cascade of a noninverting op-amp with an inverting op-amp. Assuming ideal op-amps, the input-output voltage gain is equal to the product of the single gains, therefore 8.17 G. Rizzoni, Principles and Applications of Electrical Engineering v0 = − Problem solutions, Chapter 8 RF 2 § RF 1 · ¸ vS ¨1 + RS 2 ¨© RS1 ¸¹ ______________________________________________________________________________________ Problem 8.28 Solution: Known quantities: Circuit in Figure P8.27. Find: Select appropriate components using standard 5% resistor values to obtain a gain of magnitude approximately equal to 1,000. How closely can you approximate the gain? Compute the error in the gain assuming that the resistors have the nominal value. Analysis: From the solution of Problem 8.27, the gain is given by AV = − RF 2 § RF 1 · ¸ ¨1 + RS 2 ¨© RS1 ¸¹ From Table 2.2, if we select RS1 = 1.8kΩ , RS 2 = 1kΩ RF 1 = 8.2kΩ , RF 2 = 180kΩ we obtain AV = − RF 2 § RF 1 · 180 § 1.8 + 8.2 · ¸¸ = − ¨¨1 + ¨ ¸ = −1000 RS 2 © RS 1 ¹ 1 © 1.8 ¹ So, we can obtain a nominal error equal to zero. ______________________________________________________________________________________ Problem 8.29 Solution: Known quantities: Circuit in Figure P8.27. Find: Same as in Problem 8.28, but use the ±5% tolerance range to compute the possible range of gains for this amplifier. Analysis: From the solution of Problem 8.27, the gain is given by AV = − RF 2 § RF 1 · ¸ ¨1 + RS 2 ¨© RS1 ¸¹ From Table 2.2, if we select RS1n = 1.8kΩ , RS 2 n = 1kΩ RF 1n = 8.2kΩ , RF 2 n = 180kΩ we obtain 8.18 G. Rizzoni, Principles and Applications of Electrical Engineering AVn = − Problem solutions, Chapter 8 RF 2 n § RF 1n · 180 § 1.8 + 8.2 · ¸¸ = − ¨¨1 + ¸ = −1000 ¨ RS 2 n © RS1n ¹ 1 © 1.8 ¹ So, we can obtain a nominal error equal to zero. The maximum gain is given by AV + = − RF 2 § RF 1 · R (1 + 0.05) § RF 1n (1 + 0.05) · ¸¸ = − F 2 n ¸ = −1200 ¨¨1 + ¨1 + RS 2 © RS1 ¹ RS 2 n (1 − 0.05) ¨© RS1n (1 − 0.05) ¸¹ while the minimum gain is AV − = − RF 2 § RF 1 · R (1 − 0.05) § RF 1n (1 − 0.05) · ¸¸ = − F 2 n ¸ = −834 ¨¨1 + ¨1 + RS 2 © RS1 ¹ RS 2 n (1 + 0.05) ¨© RS 1n (1 + 0.05) ¸¹ ______________________________________________________________________________________ Problem 8.30 Solution: Known quantities: For the circuit in Figure P8.30, RL = 20kΩ , v0 = 0 ÷ 10 V iinmax = 1mA Find: The resistance R such that iinmax = 1mA . Analysis: iin = v0 v0 10 R = max = −3 = 10 kΩ R iinmax 10 ______________________________________________________________________________________ Problem 8.31 Solution: Known quantities: Circuit in Figure P8.13. Find: Select appropriate components using standard 5% resistor values to obtain a gain of magnitude approximately equal to 200. How closely can you approximate the gain? Compute the error in the gain assuming that the resistors have the nominal value. Analysis: From the solution of Problem 8.13, the gain is given by § R · AV = ¨¨1 + F ¸¸ R2 ¹ © From Table 2.2, if we select R2 = 33Ω RF = 6.8kΩ we obtain 8.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 § R · 6800 AVn = ¨¨1 + F ¸¸ = 1 + = 207 R2 ¹ 33 © The error in the gain is εA = V 207 − 200 = 3 .5 % 200 ______________________________________________________________________________________ Problem 8.32 Solution: Known quantities: Circuit in Figure P8.13. Find: Same as in Problem 8.31, but use the ±5% tolerance range to compute the possible range of gains for this amplifier. Analysis: From the solution of Problem 8.31, the gain is given by § R · AV = ¨¨1 + F ¸¸ R2 ¹ © From Table 2.2, if we select R2 n = 33Ω RFn = 6.8kΩ we obtain § R · 6800 AVn = ¨¨1 + Fn ¸¸ = 1 + = 207 33 © R2 n ¹ The maximum gain is given by § R (1 + 0.05) · 6800 ⋅1.05 ¸¸ = 1 + AV + = ¨¨1 + Fn = 228.75 33 * 0.95 © R2 n (1 − 0.05) ¹ while the minimum gain is § R (1 − 0.05) · 6800 ⋅ 0.95 ¸¸ = 1 + AV − = ¨¨1 + Fn = 187.44 33 *1.05 © R2 n (1 + 0.05) ¹ ______________________________________________________________________________________ Problem 8.33 Solution: Known quantities: For the circuit in Figure P8.15, R1 = R2 , R3 = R4 Find: Select appropriate components using standard 1% resistor values to obtain a differential amplifier gain of magnitude approximately equal to 100. How closely can you approximate the gain? Compute the error in the gain assuming that the resistors the have nominal value. Analysis: From the solution of Problem 8.15, the gain is given by 8.20 G. Rizzoni, Principles and Applications of Electrical Engineering AV = Problem solutions, Chapter 8 R3 R1 From Table 2.2, if we select R1 = 1kΩ , R3 = 100kΩ we obtain AV = R3 = 100 R1 and the error for the gain with nominal values is zero. ______________________________________________________________________________________ Problem 8.34 Solution: Known quantities: For the circuit in Figure P8.15, R1 = R2 , R3 = R4 Find: Same as in Problem 8.33, but use the ±1% tolerance range to compute the possible range of gains for this amplifier. Analysis: From the solution of Problem 8.33 AV = R3 100(1 + 0.01) AV max = = 102 R1 1(1 − 0.01) AV = R3 100(1 − 0.01) AV min = = 98 R1 1(1 + 0.01) ______________________________________________________________________________________ 8.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Section 8.3: Active Filters Problem 8.35 Solution: Known quantities: For the circuit shown in Figure P8.35: C = 1 µF R = 10 kΩ RL = 1kΩ . Find: a) The gain [in dB] in the pass band. b) The cutoff frequency. c) If this is a low or high pass filter. Analysis: a) Assume the op-amp is ideal. Determine the transfer function in the form: 1 V o [j ] = Ho 1 + j f[ ] V s [j ] - Vn - Vd = 0 0 Vd Vn - Vs + Vn - Vo = 0 In + 1 R j C H v [j ] = KVL : KCL : Vn 0 V o [j ] = j RC V s [j ] This is not in the standard form desired but is the best that can be done. There are no cutoff frequencies and no clearly defined pass band. The gain [i.e., the magnitude of the transfer function] and output voltage increases continuously with frequency, at least until the output voltage tries to exceed the DC supply voltages and clipping occurs. In a normal high pass filter, the gain will increase with frequency until the cutoff frequency is reached above which the gain remains constant. b) There is no cutoff frequency. c) This filter is best called a high pass filter; however, since the output will be clipped and severely distorted above some frequency, it is not a particularly good high pass filter. It could even be called a terrible filter with few redeeming graces. ______________________________________________________________________________________ In 0 Vn 0 H v [j ] = Problem 8.36 Solution: Known quantities: For the circuit shown in Figure P8.36: C = 1 µF 8.22 R1 = 1.8 kΩ R2 = 8.2kΩ RL = 333Ω . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Find: a) Whether the circuit is a low- or high-pass filter. b) The gain V0 /VS in decibel in the passband. c) The cutoff frequency. Analysis: a) From Figure 8.24, it results that the amplifier in Figure P8.36 is a high-pass filter. In fact, the output voltage is V0 ( jω ) = − b) c) lim ω →∞ jω CR2 VS ( jω ) 1 + jω CR1 V0 ( jω ) R 8.2 = 20 Log 2 = 20 Log = 13.17 dB VS ( jω ) dB R1 1.8 ω0 = 1 1 = = 5555 rad/s −6 CR1 1 ⋅10 ⋅1.8 ⋅103 ______________________________________________________________________________________ Problem 8.37 Solution: Known quantities: For the circuit shown in Figure P8.36: C = 200 pF R1 = 10 kΩ R2 = 220kΩ RL = 1kΩ . Find: a) Whether the circuit is a low- or high-pass filter. b) The gain V0 /VS in decibel in the passband. c) The cutoff frequency. Analysis: a) From Figure 8.24, it results that the amplifier in Figure P8.36 is a high-pass filter. In fact, the output voltage is V0 ( jω ) = − jω CR2 VS ( jω ) 1 + jω CR1 b) V0 ( jω ) R 220 = 20 Log 2 = 20 Log = 26.84 dB ω →∞ V ( jω ) R 10 1 S dB c) ω0 = lim 1 1 = = 5 ⋅105 rad/s 3 −12 CR1 200 ⋅10 ⋅10 ⋅10 ______________________________________________________________________________________ Problem 8.38 Solution: Known quantities: For the circuit shown in Figure P8.38: C = 100 pF R1 = 4.7 kΩ R2 = 68kΩ RL = 220kΩ . Find: Determine the cutoff frequencies and the magnitude of the voltage frequency response function at very low and at very high frequencies. Analysis: The output voltage in the frequency domain is 8.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 § · ¨ ¸ R2 ¨ ¸ V ( jω ) = 1 + jω C ( R1 + R2 ) V ( jω ) V0 ( jω ) = 1 + i ¨ 1 ¸ i 1 + jω CR1 + R 1 ¨ ¸ jω C © ¹ The circuit is a high-pass filter. The cutoff frequencies are ω1 = 1 1 = = 2.127 ⋅10 6 rad/s 3 −12 CR1 100 ⋅10 ⋅ 4.7 ⋅10 ω2 = 1 1 = = 1.375 ⋅105 rad/s −12 C ( R1 + R2 ) 100 ⋅10 ⋅ 72.7 ⋅103 For high frequencies we have 68 V0 ( jω ) R = 1+ 2 = 1+ = 15.46 ω →∞ V ( jω ) 4.7 R1 i A∞ = lim At low frequencies A0 = lim ω →0 V0 ( jω ) =1 Vi ( jω ) ______________________________________________________________________________________ Problem 8.39 Solution: Known quantities: For the circuit shown in Figure P8.39: Find: a) An expression for H υ ( jω ) = C = 20 nF R1 = 1kΩ R2 = 4.7kΩ R3 = 80kΩ . V0 ( jω ) . Vi ( jω ) b) The cutoff frequencies. c) The passband gain. d) The Bode plot. Analysis: a) The frequency response is H υ ( jω ) = V0 ( jω ) = 1+ Vi ( jω ) b) The cutoff frequencies are ω1 = RR 1 1 + jω C 2 3 R2 + R3 jωC R2 + R3 + jωCR2 R3 § R3 · = = ¨¨1 + ¸¸ R2 R2 + jωCR2 R3 © R2 ¹ 1 + jωCR3 R3 || 1 1 = = 625 rad/s −9 CR3 20 ⋅10 ⋅ 80 ⋅103 ( R2 + R3 ) 84.7 ⋅103 ω2 = = 11263 rad/s = C R2 R3 20 ⋅10 −9 ⋅ 80 ⋅103 ⋅ 4.7 ⋅103 c) The passband gain is obtained by evaluating the frequency response at low frequencies, A0 = lim Hυ ( jω ) = 1 + ω →0 R3 80 = 1+ = 18 R2 4 .7 d) The magnitude Bode plot for the given amplifier is as shown. 8.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 ______________________________________________________________________________________ Problem 8.40 Solution: Known quantities: For the circuit shown in Figure P8.40: C = 0.47 µF R1 = 9.1kΩ R2 = 22kΩ RL = 2.2kΩ . Find: a) Whether the circuit is a low- or high-pass filter. b) An expression in standard form for the voltage transfer function. c) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff frequency. Analysis: a) From Figure 8.21, it results that the amplifier in Figure P8.40 is a low-pass filter. In fact, the output voltage is R2 1 VS ( j ω ) R1 1 + jω CR 2 V ( jω ) R 1 H υ ( jω ) = 0 =− 2 Vi ( jω ) R1 1 + jω CR 2 V0 ( jω ) = − b) c) The gain in decibel is obtained by evaluating Hυ ( j 0) dB = 20 Log Hυ ( jω ) at ω=0, i.e. R2 22 = 20 Log = 7.66 dB . R1 9 .1 The cutoff frequency is ω0 = 1 1 = = 96.71 rad/s −6 CR2 0.47 ⋅10 ⋅ 22 ⋅103 ______________________________________________________________________________________ 8.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.41 Solution: Known quantities: For the circuit shown in Figure P8.40: C = 0.47 nF R1 = 2.2 kΩ R2 = 68kΩ RL = 1kΩ . Find: a) An expression in standard form for the voltage frequency response function. b) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff frequency. Analysis: a) From Figure 8.21, it results that the amplifier in Figure P8.40 is a low-pass filter. The voltage frequency response function is H υ ( jω ) = V0 ( jω ) R 1 =− 2 Vi ( jω ) R1 1 + jω CR 2 b) The gain in decibel is obtained by evaluating Hυ ( j 0) dB = 20 Log Hυ ( jω ) at ω=0, i.e. R2 68 = 20 Log = 29.8 dB . R1 2 .2 The cutoff frequency is ω0 = 1 1 = = 31289 rad/s −9 CR2 0.47 ⋅ 10 ⋅ 68 ⋅ 103 ______________________________________________________________________________________ Problem 8.42 Solution: Known quantities: For the circuit shown in Figure P8.42: C1 = C2 = 0.1 µF R1 = R2 = 10 kΩ . Find: a) The passband gain. b) The resonant frequency. c) The cutoff frequencies. d) The circuit Q. e) The Bode plot. Analysis: a) From Figure 8.26, we have ABP ( jω ) = − Z2 jωC1 R2 jωC1 R1 jωC1 R1 =− =− =− 2 2 (1 + jω C1 R1 )(1 + jω C2 R2 ) (1 + jω C1 R1 ) Z1 ( jω ) (C1 R1 ) 2 + 2 jω C1 R1 + 1 The magnitude of the frequency response is ABP ( jω ) = ωC1 R1 2 1 + ω 2 (C1 R1 ) The passband gain is the maximum over ω of the magnitude of the frequency response, i.e. ABP ( j 1 1 )= C1 R1 2 b) The resonant frequency is c) ωn = 1 = 1000 rad/s R1C1 The cutoff frequencies are obtained by solving with respect to ω the equation 8.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 ωC1 R1 1 = ω 2 − 2 2ω nω + ω n2 = 0 ω1, 2 = ω n ( 2 ± 1) 2 2 1 + ω (C1 R1 ) 2 2 ω1 = 2414 rad/s ω 2 = 414 rad/s ABP ( jω ) = d) In this case ζ=1, which implies e) Q= 1 1 = 2ζ 2 The Bode plot of the frequency response is as shown. ______________________________________________________________________________________ Problem 8.43 Solution: Known quantities: For the circuit shown in Figure P8.43: C = 0.47 nF R1 = 220Ω R2 = 68kΩ RL = 1kΩ . Find: a) An expression in standard form for the voltage frequency response function. b) The gain in decibels in the passband, that is, at the frequencies being passed by the filter, and the cutoff frequency. Analysis: a) The voltage frequency response function is H υ ( jω ) = VO ( jω ) R 1 =− 2 R1 1 + jω CR2 Vi ( jω ) b) The gain in decibel is obtained by evaluating Hυ ( j 0) dB = 20 Log Hυ ( jω ) at ω=0, i.e. R2 68000 = 20 Log = 49.8 dB . R1 220 The cutoff frequency is ω0 = 1 1 = = 31289 rad/s −9 CR2 0.47 ⋅ 10 ⋅ 68 ⋅ 103 ______________________________________________________________________________________ 8.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.44 Solution: Known quantities: For the circuit shown in Figure P8.44: C1 = 2.2 µF C2 = 1 nF R1 = 2.2kΩ R2 = 100kΩ . Find: Determine the passband gain. Analysis: The voltage frequency response is ABP ( jω ) = − Z2 ωC1 R2 jωC1 R2 =− ABP ( jω ) = 2 2 2 Z1 (1 + jω C1 R1 )(1 + jω C2 R2 ) 1 + ω (C1 R1 ) 1 + ω 2 (C2 R2 ) The cutoff frequencies are ω1 = 1 1 = = 206.6 rad/s −6 C1R1 2.2 ⋅10 ⋅ 2.2 ⋅103 ω2 = 1 1 = = 10000 rad/s −9 C2 R2 1 ⋅10 ⋅100 ⋅103 The passband gain can be calculated approximately by evaluating the magnitude of the frequency response at frequencies greater than ω1 and smaller than ω2, i.e. ω1 << ω << ω 2 1 + ω 2 (C1 R1 ) ≈ ω 2 (C1 R1 ) ωC1 R2 R 100 = 2 = = 45.45 ABP ≅ 2 R1 2.2 ω 2 (C1 R1 ) 2 2 , 1 + ω 2 (C2 R2 ) ≈ 1 2 ______________________________________________________________________________________ Problem 8.45 Solution: Known quantities: For the circuit shown in Figure P8.45, let C = C1 = C2 = 220 µF Find: Determine the frequency response. Analysis: With reference to the Figure shown below, we have 8.28 R = R1 = R2 = RS = 2kΩ . G. Rizzoni, Principles and Applications of Electrical Engineering I R 2 ( jω ) = Problem solutions, Chapter 8 1 1 1 VO ( jω ) VC 2 ( jω ) = I R 2 ( jω ) = VO ( jω ), R2 jω C 2 jωC 2 R2 I R 1 ( jω ) = − 1 1 VC 2 ( jω ) = − VO ( jω ), R1 jωC 2 R2 R1 § 1 · 1 ¨¨ R2 + ¸¸ VO ( jω ) ω j C § · 1 2 ¹ R2 ¸ jωC 1 VO ( jω ), = ¨¨1 + I C1 ( jω ) = © 1 jωC 2 R2 ¸¹ © jω C 1 § · 1 1 C ¸ VO ( jω ) = I in ( jω ) = I C1 ( jω ) + I R 2 ( jω ) − I R1 ( jω ) = ¨¨ jωC 1+ 1 + + C 2 R2 R2 jωC 2 R2 R1 ¸¹ © §2 1 · ¸ VO ( jω ), = ¨¨ + jωC + jωCR 2 ¸¹ ©R Vin ( jω ) = − VC 2 ( jω ) − RS I in ( jω ) = − § 2 · ¸VO ( jω ) = −¨¨ 2 + jωCR + jωCR ¸¹ © V ( jω ) 1 =− H υ ( jω ) = O Vin ( jω ) 2 + jωCR + § 1 1 · ¸VO ( jω ) = VO ( jω ) − ¨¨ 2 + jωCR + jωCR jωCR ¸¹ © 2 jωCR =− jωCR ( jω ) (CR )2 + 2 jωCR + 2 2 ______________________________________________________________________________________ Problem 8.46 Solution: Known quantities: The inverting amplifier shown in Figure P8.46. Find: a) The frequency response of the circuit. b) If R1 = R2 = 100 kΩ and C = 0.1 µF, compute the attenuation in dB at ω = 1,000 c) Compute gain and phase at ω = 2,500 rad/s. d) Find range of frequencies over which the attenuation is less than 1 dB. Analysis: a) Applying KCL at the inverting terminal: R2 + 1 jω C vOUT 1 + jωR2 C =− =− v IN R1 jωR1C b) Gain = 0.043 dB o c) Gain = 0.007 dB ; Phase = 177.71 d) To find the desired frequency range we need to solve the equation: 8.29 rad/s. G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 1 + jωR2 C < 0.8913 since 20 log10 (0.8913) = −1dB . jωR1C This yields a quadratic equation in ω, which can be solved to find ω > 196.5 rad/s . ______________________________________________________________________________________ Problem 8.47 Solution: Known quantities: For the circuit shown in Figure P8.47, let C = C1 = C2 = 100 µF R1 = 3kΩ R2 = 2kΩ . Find: Determine an expression for the gain. Analysis: With reference to the Figure shown above, we have VC1 ( jω ) = VO ( jω ) I C1 ( jω ) = jωC1 VO ( jω ), VA ( jω ) = R2I C1 ( jω ) + VO ( jω ) = (1 + jωC1R2 )VO ( jω ), I C 2 ( jω ) = jωC2 (VO ( jω ) − VA ( jω ) ) = −( jω ) C1C2 R2 VO ( jω ), 2 ( ) Vin ( jω ) = VA ( jω ) − R1 (I C 2 ( jω ) − I C1 ( jω ) ) = 1 + jωC1 ( R1 + R2 ) + ( jω ) C1C2 R1 R2 VO ( jω ) And finally, the expression for the gain is Av ( jω ) = 2 VO ( jω ) 1 1 = = 2 2 Vin ( jω ) 1 + jωC ( R1 + R2 ) + ( jω ) C R1 R2 (1 + jωCR1 )(1 + jωCR2 ) ______________________________________________________________________________________ Problem 8.48 Solution: Known quantities: The circuit shown in Figure P8.48. Find: Sketch the amplitude response of V2 ideal. / V1, indicating the half-power frequencies. Assume the op-amp is 8.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: From KCL at the inverting input: Vx − V1 Vx + =0 ZC R § ¨ 1 Vx = V1 ¨ ¨ 1 ¨ 1 + jωRC © § 1 · ¸ = V1 Vx ¨¨1 + jωRC ¸¹ © 1· § Vx ¨ jωC + ¸ = jωCV1 R¹ © Similarly, from KCL at the output of the op-amp: V2 − Vx V2 + =0 R ZC §1 · V V2 ¨ + jωC ¸ = x Vx = V2 (1 + jωRC ) ©R ¹ R V2 jωRC = Combining the above results, we find V1 (1 + jωRC )2 V2 ωRC = V1 1 + (ωRC )2 or This function has the form of a band-pass filter, with maximum value determined as follows: G= V2 ωRC = V1 1 + (ωRC )2 [ ] [ dG 1 + (ωRC ) 2 RC − ωRC 2ω ( RC ) 2 = 2 dω 1 + (ωRC )2 [ ] ] Setting the derivative equal to zero and solving for the center frequency, RC + ω 2 R 3C 3 − 2ω 2 R 3C 3 = 0 Then ω= 1 1 = , and the half-power frequencies are given by: 1+1 2 1 1 1 = Gmax = ω 2 R 2 C 2 + 1 = 2 2ωRC 2 22 1 RC Gmax = ωRC 1 + (ωRC )2 ω= 1 − ω 2 R 2C 2 = 0 2 2 RC ± 8R 2 C 2 − 4 R 2 C 2 2 ±1 = 2 2 RC 2R C The curve is sketched below. 8.31 R 2 C 2ω 2 − 2 2 RCω + 1 = 0 · ¸ ¸ ¸ ¸ ¹ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 ______________________________________________________________________________________ Problem 8.49 Solution: Known quantities: The circuit shown in Figure P8.49. Find: Determine an analytical expression for the output voltage for the circuit shown in Figure P8.49. What kind of filter does this circuit implement? Analysis: The resistance RF1 does not influence the output voltage, so 1 1 jωC F VO ( jω ) = − VS1 ( jω ) = − VS 1 ( jω ) RS 1 jωC F RS1 The circuit is an integrator (low-pass filter). ______________________________________________________________________________________ Problem 8.50 Solution: Known quantities: The circuit shown in Figure P8.50. Find: Determine an analytical expression for the output voltage for the circuit shown in Figure P8.50. What kind of filter does this circuit implement? Analysis: Figure P8.50 shows a noninverting amplifier, so the output voltage is given by 1 § ¨ RF + jωC F VO ( jω ) = ¨1 + ¨ 1 ¨ jωCS © · ¸ ¸V ( jω ) = §¨1 + CS + jωC R ·¸ V ( jω ) S F ¸ S ¨ C ¸ S F © ¹ ¸ ¹ The filter is clearly a high-pass filter. ______________________________________________________________________________________ 8.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Sections 8.4, 8.5: Integrator and Differentiator Circuits, Analog Computers Problem 8.51 Solution: Known quantities: Figure P8.50. Find: If: C = 1 µ F R = 10 kΩ function of time. Analysis: RL = 1kΩ , determine an expression for and plot the output voltage as a KVL : - v N - v D = 0 vD ≈ 0 vN ≈ 0 dvC = C d[ v N - v S ] iC = C dt dt d[ v N - v S ] KCL : C + i N + v N vO = 0 dt R dv S i N ≈ 0 v N ≈ 0 vO = - RC dt The derivative is the slope of the curve for the source voltage, which is zero for: t < 2.5 ms, 5 ms < t < 7.5 ms, and t > 15 ms 3 -0 3 −6 For 2.5 ms < t < 5 ms: vO = [ 10 ⋅ 10 ] [ 1 ⋅10 ] = - 12 V −3 5 ⋅10 - 2.5 ⋅ 10 −3 [ - 1.5 ] - [ + 3 ] −3 =+6V For 7.5 ms < t < 15 ms: vO = - 10 ⋅10 15 ⋅10 −3 - 7.5 ⋅ 10 −3 ______________________________________________________________________________________ Problem 8.52 Solution: Known quantities: Figure P8.52(a) and Figure P8.52(b). Find: If: C = 1 µ F R = 10 kΩ RL = 1kΩ a) An expression for the output voltage. 8.33 G. Rizzoni, Principles and Applications of Electrical Engineering b) The value of the output voltage at t = 5, 7.5, 12.5, 15, and 20 a function of time. Analysis: a) As usual, assume the op amp is ideal so: Problem solutions, Chapter 8 ms and a plot of the output voltage as KVL : - v N - v D = 0 vN ≈ 0 vD ≈ 0 d[ v N - vO ] iC = C dt d[ v N - vO ] KCL : v N v S + i N + C =0 R dt 1 d vO = vN ≈ 0 iN ≈ 0 v S dt RC t 1 f Integrating: vO [ t f ] = vO [ t i ] ³ v S dt RC ti b) Integrating gives the area under a curve. Recall area of triangle = 1/2[base x height] and area of rectangle = base x height. Integrating the source voltage when it is constant gives an output voltage which is a linear function of time. Integrating the source voltage when it is a linear function of time gives an output voltage which is a quadratic function of time. 1 1 rad = 100 = 3 −6 s RC [ 10 ⋅ 10 ] [ 1 ⋅10 ] 1 -3 vO [5ms] = 0 V - [ 100 ] [ 2.5 ⋅10 ] [ 3 V ] = - 375 mV 2 -3 -3 vO [7.5ms] = - 375 ⋅10 - [ 100 ] [ 2.5 ⋅10 ] [ 3 ] = - 1125 mV 1 -3 -3 vO [12.5ms] = - 1124 ⋅10 - [ 100 ] [ 5 ⋅10 ] [ 3 ] = - 1875 mV 2 1 -3 -3 vO [15ms] = - 1875 ⋅10 - [ 100 ] [ 2.5 ⋅10 ] [ - 1.5 ] = - 1687.5 mV 2 -3 -3 vO [20ms] = - 1687.5 ⋅10 - [ 100 ] [ 5 ⋅10 ] [ - 1.5 ] = - 937.5 mV ______________________________________________________________________________________ Problem 8.53 Solution: Known quantities: In the circuit shown in Figure P8.53, the capacitor is initially uncharged, and the source voltage is: vin (t ) = 10 ⋅10 −3 + sin (2,000πt ) V . Find: a) At t = 0, the switch S1 is closed. How long does it take before clipping occurs at the output if RS = 10 kΩ and C F = 0.008 µ F ? b) At what times does the integration of the DC input cause the op-amp to saturate fully? 8.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: vout = − a) =− 1 1 τ vin (t )dt =− ³ ³ [0.01 + sin (2000πt )]dt RS C F RS C F 0 1 τ 1 τ 0.01dt − ³ ³ sin (2000πt )dt RS C F 0 RS C F 0 The peak amplitude of the AC portion of the output is: The output will begin to clip when: vp = 1 § 1 ¨ RS C F © 2000π · ¸ = 1.989 V ¹ v0 (DC ) − v p = −15 V so we need to find at what time the 1 τ ³ 0.01dt = −13 V is satisfied. The answer is found below: RS C F 0 13RS C F 1 − 0.01τ = −13 τ = = 104 ms 0.01 RS C F 15 RS C F τ= = 120 ms b) Using the results obtained in part a.: 0.01 condition: − ______________________________________________________________________________________ Problem 8.54 Solution: Known quantities: The circuit shown in Figure 8.21. Find: a) If RS = 10 kΩ , RF = 2 MΩ , C F = 0.008 µ F , and vS (t ) = 10 + sin (2,000πt ) V , find vout using phasor analysis. b) Repeat part a if RF = 200 kΩ , and if R F = 20 kΩ . c) Compare the time constants with the period of the waveform for part a and b. What can you say about the time constant and the ability of the circuit to integrate? Analysis: a) Replacing the circuit elements with the corresponding impedances: Zf vin Zf =− Rf 1 + jω R f C f ZS + Z S = RS For the signal component at - ω = 2,000π : 8.35 vout G. Rizzoni, Principles and Applications of Electrical Engineering vout Rf § 1 ¨ =− RS ¨© 1 + jωR f C f = vin 200 ( ( § · · ¸ 1 ¸vin = −200¨¨ ¸vin ¸ ¨ 1 + jω ¸ ¹ 62.5 ¹ © ( ∠ 180 o − arctan ω ) 2 1+ ω 62.5 For the signal component at Problem solutions, Chapter 8 ω =0 ( 62.5 )) = 1.9839∠90.57 o V vout = −200vin = −2,000 V . Thus, (DC): ) vout (t ) = −2000 + 1.9839 sin 2,000πt + 90.57 o V ≈ −2000 + 2 cos(2,000πt ) V . b) RF = 200 kΩ ω = 2,000π : · Rf § 1 ¨ ¸vin = 1.9797∠95.68o V vout = − RS ¨© 1 + jωR f C f ¸¹ For the signal component at ω = 0 (DC): vout = −20vin = −200 V . For the signal component at ( ) Thus, vout (t ) = −200 + 1.9797 sin 2,000πt + 95.68 V ≈ −200 + 2 cos(2,000πt ) V RF = 20 kΩ o ω = 2,000π : · Rf § 1 ¨ ¸vin = 1.41∠134.8 o V vout = − RS ¨© 1 + jωR f C f ¸¹ For the signal component at ω = 0 (DC): vout = −2vin = −20 V . For the signal component at ( ) ( Thus, ) vout (t ) = −20 + 1.41sin 2,000πt + 135 V ≈ −20 + 1.41cos 2,000πt + 45o V . o c) Rf τ T 2 MΩ 16 ms 1 ms 200 kΩ 1.6 ms 1 ms 20 kΩ 0.16 ms 1 ms In order to have an ideal integrator, it is desirable to have τ >> T. ______________________________________________________________________________________ Problem 8.55 Solution: Known quantities: v S (t ) = 10 ⋅10 −3 sin (2,000πt ) V , C S = 100 µ F , C F = 0.008 µ F , RF = 2 MΩ , and RS = 10 kΩ . For the circuit of Figure 8.26, assume an ideal op-amp with Find: a) The frequency response, v0 ( jω ) . vS b) Use superposition to find the actual output voltage (remember that DC = 0 8.36 Hz). G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: a) v (t) S Zf : ZS Z S = RS + + vo (t) Rf 1 Zf = jω C S 1 + R f C f jω Rf Zf 1 + jω R f C f jω R f C S vo =− =− =− 1 ( jωRS C S + 1) jωR f C f + 1 vS ZS RS + jω C S jω vo 5 ⋅10 −3 ( jω ) = − vS ( jω + 1)§¨ jω + 1·¸ © 62.5 ¹ v j 0v S = 0V b) vo = v S o ( jω ) . By superposition, vo 10 mV = vS vS 1+1 ( vo ω =2000π = ) j1.257 ⋅10 6 v S = 20 ⋅10 −3 ∠ − 89.43o V (1 + j 6283)(1 + j100) ( ) vo (t ) = 20 ⋅10 −3 sin 2000πt − 89.43o V We can say that the practical differentiator is a good approximation of the ideal differentiator. ______________________________________________________________________________________ Problem 8.56 Solution: Find: Derive the differential equation corresponding to the analog computer simulation circuit of Figure P8.56. Analysis: x(t ) = −200³ z dt or z = − Therefore, Therefore 1 dx .Also, z = −20 y . 200 dt dy 1 dx . Also, y = ³ (4 f (t ) + x (t )) dt or = 4 f (t ) + x(t ) . dt 4000 dt 1 d 2x d 2x = 4 f (t ) + x(t ) or − 4000 x(t ) − 16000 f (t ) = 0 . 4000 dt 2 dt 2 y= ______________________________________________________________________________________ 8.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.57 Solution: Find: Construct the analog computer simulation corresponding to the following differential equation: dx d 2x + 100 + 10 x = −5 f (t ) . 2 dt dt Analysis: d 2x dt 2 -5 f(t) -100 - dx dt x(t) dx dt -100 -10 ______________________________________________________________________________________ 8.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Section 8.6: Physical Limitations of Operational Amplifiers Problem 8.58 Solution: Known quantities: For the circuit shown in Figure 8.8: RS = RF = 2.2kΩ . Find: Find the error introduced in the output voltage if the op-amp has an input offset voltage of 2 mV. Assume zero bias currents and that the offset voltage appears as in Figure 8.48. Analysis: By superposition, the error is given by § R · ∆vO = ¨¨1 + F ¸¸VOffset = 4mV © RS ¹ ______________________________________________________________________________________ Problem 8.59 Solution: Known quantities: For the circuit shown in Figure 8.8: RS = RF = 2.2kΩ . Find: Repeat Problem 8.58 assuming that in addition to the input offset voltage, the op-amp has an input bias current of 1 µA. Assume that the bias currents appear as in Figure 8.49. Analysis: By superposition, the effect of the bias currents on the output is v+ = − RI B+ = v− § § 1 1 · ·¸ ¸¸ I B+ ∆vO ,IB = RF ¨¨ I B− − R¨¨ + ¸ © RF RS ¹ ¹ © −6 If R = RF || RS then ∆vO ,IB = RF (I B− − I B+ ) = − RF I OS = −2200 ⋅1 ⋅10 = −2.2mV ∆vO ,IB − v− v− − = I B− RF RS The effect of the bias voltage is § R · ∆ v O ,VB = ¨¨ 1 + F ¸¸V Offset = 4 mV RS ¹ © So, the total error on the output voltage is ∆vO = ∆vO ,VB + ∆vO ,IB = 1.8mV ______________________________________________________________________________________ Problem 8.60 Solution: Known quantities: The circuit shown in Figure P8.60. The input bias currents are equal and the input bias voltage is zero. 8.39 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Find: The value of Rx that eliminates the effect of the bias currents in the output voltage. Analysis: We can write v+ = − Rx I B+ = v− , I B+ = I B− ∆vO ,IB − v− v− − = I B− RF R1 By selecting § § 1 1· · ∆vO ,IB = RF ¨¨ I B− − Rx ¨¨ + ¸¸ I B+ ¸¸ © RF R1 ¹ ¹ © Rx = RF || R1 , a zero output voltage error is obtained. ______________________________________________________________________________________ Problem 8.61 Solution: Known quantities: For the circuit shown in Figure P8.60: RF = 3.3kΩ R1 = 1kΩ vS = 1.5 sin(ωt ) V Find: The highest-frequency input that can be used without exceeding the slew rate limit of 1V/µs. Analysis: The maximum slope of a sinusoidal signal at the output of the amplifier is S0 = A ⋅ ω = V 1 6 RF ω = 10 6 ω max = 10 = 3.03 ⋅105 rad/s s 3 .3 R1 ______________________________________________________________________________________ Problem 8.62 Solution: Known quantities: The Bode plot shown in Figure 8.45: A0 = 10 6 ω 0 = 10π rad/s Find: The approximate bandwidth of a circuit that uses the op-amp with a closed loop gain of A1 =75 and A2 =350. Analysis: The product of gain and bandwidth in any given op-amp is constant, so ω1 = A0ω 0 107 π = = 4.186 ⋅105 rad/s A1 75 ω2 = A0ω 0 107 π = = 8.971⋅10 4 rad/s A2 350 ______________________________________________________________________________________ 8.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.63 Solution: Known quantities: For the practical charge amplifier circuit shown in Figure P8.63, the user is provided with a choice of three time constants - τ long = RL C F , τ medium = RM C F , τ short = RS C F , which can be selected by means of a switch. Assume that RL = 10 MΩ , RM = 1MΩ , RS = 0.1MΩ , and C F = 0.1 µF . Find: Analyze the frequency response of the practical charge amplifier for each case, and determine the lowest input frequency that can be amplified without excessive distortion for each case. Can this circuit amplify a DC signal? Analysis: Applying KCL at the inverting terminal: i+ or, Vout =0 1 R+ jω C Vout jωC =− i 1 + jωRC This response is clearly that of a high-pass filter, therefore the charge amplifier will never be able to amplify a DC signal. The low end of the (magnitude) frequency response is plotted below for the three time constants. The figure illustrates how as the time constant decreases the cut-off frequency moves to the right (solid line: R = 10 MΩ; dashed line: R =1 MΩ; dotted line: R = 0.1 MΩ). -100 response of practical charge amplifier -110 dB -120 -130 -140 -150 -160 10-1 100 101 102 frequency, rad/s From the frequency response plot one can approximate the minimum useful frequency for distortionless response to be (nominally) 1 Hz for the 10 MΩ case, 10 Hz for the 1 MΩ case, and 100 Hz for the 0.1 MΩ case. ______________________________________________________________________________________ 8.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.64 Solution: Find: Consider a differential amplifier. We would desire the common-mode output to be less than 1% of the differential-mode output. Find the minimum dB common-mode rejection ratio (CMRR) that fulfills this requirement if the differential mode gain Adm = 1,000. Let v1 = sin (2,000πt ) + 0.1sin (120π ) V ( ) v2 = sin 2,000πt + 180 o + 0.1sin (120π ) V §v +v · v0 = Adm (v1 − v2 ) + Acm ¨ 1 2 ¸ © 2 ¹ Analysis: We first determine which is the common mode and which is the differential mode signal: v1 − v2 = 2 sin (2,000πt ) v1 + v2 = 0.1sin (120πt ) 2 Therefore, vout = Adm 2 sin (2000πt ) + Acm 2 sin (120πt ) Since we desire the common mode output to be less than 1% of the differential mode output, we require: Acm (0.1) ≤ 0.01(2) or Acm ≤ 0.2 . CMRR = Adif Acm So CMRRmin = 1000 = 5000 = 74 dB . 0 .2 ______________________________________________________________________________________ Problem 8.65 Solution: Known quantities: As indicated in Figure P8.65, the rise time, tr, of the output waveform is defined as the time it takes for the waveform to increase from 10% to 90% of its final value, i.e., t r ≡ t b − t a = −τ (ln 0.1 − ln 0.9) = 2.2τ , where τ is the circuit time constant. Find: Estimate the slew rate for the op-amp. Analysis: dv out Vm (0.9 − 0.1) 15 × 0.8 V V . Therefore, the slew rate is approximately = = ≈ 2.73 −6 4.4 µs µs dt max (14.5 − 10.1) × 10 2.73 V . µs ______________________________________________________________________________________ Problem 8.66 Solution: Find: 5 Consider an inverting amplifier with open-loop gain 10 . With reference to Equation 8.18, 8.42 G. Rizzoni, Principles and Applications of Electrical Engineering a) If Problem solutions, Chapter 8 RS = 10 kΩ and RF = 1 MΩ , find the voltage gain AV (CL ) . RS = 10 kΩ and RF = 10 MΩ . c) Repeat part a if RS = 10 kΩ and RF = 100 MΩ . d) Using the resistors values of part c, find AV (CL ) if AV (OL ) → ∞ . b) Repeat part a if Analysis: RF RS ACL = − 1 1+ R F + RS RS AOL ACL = - 99.899 b) ACL = - 990 c) ACL = -9091 d) As AOL → ∞ , ACL = −10,000 . a) ______________________________________________________________________________________ Problem 8.67 Solution: Known quantities: Figure P8.67. Find: a) If the op-amp shown in Figure P8.67 has an open-loop gain of 45 X 105, find the closed-loop gain for RS = RF = 7.5 kΩ . b) Repeat part a if RF = 5 RS = 37.5 kΩ . Analysis: a) Rf RS vv+ + - vin vin = v + and ( v0 = AOL vin − v − - Writing KCL at v : ) + + A (v+-v - ) OL v out - · § v v − = −¨¨ 0 − vin ¸¸ . ¹ © AOL v − − 0 v − − v0 + =0. RS RF Substituting, − v0 − v0 + vin + vin AOL AOL v + = 0 RS RF RF § § 1 1 1 1 · 1 · ¸¸ = −vin ¨¨ ¸¸ v0 ¨¨ − − − + © AOL RS AOL RF RF ¹ © RS RF ¹ 8.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 § 1 1 · ¸ − ¨¨ + RS RF ¸¹ § 1 v0 1 · © ¸¸ = = AOL RS RF ¨¨ + 1 1 1 vin K K 2 ¹ © 1 − − − AOL RS AOL RF RF 2 where: K1 = RF RS + RS + RS AOL RF K 2 = RF RS + RF 2 + RS AOL RF For the conditions of part a we obtain: ACL = b) 1 2 ACL = 5 45 5 ×10 + 1 + 1 6 45 5 × 10 + 1 1 2 + 45 ×105 + 1 = 1.999 1 6 45 § · § · ¨ ¸ ¨ ¸ v0 RF ¨ 1 1 ¸ ¨ ¸ Æ = + ¨ ¸ ¨ ¸ + + R R R R vin RS S F F +1¸ ¨ S +1¸ ¨ © AOL RS ¹ © AOL RS ¹ × 10 5 + 1 = 5.999 . ______________________________________________________________________________________ Problem 8.68 Solution: Find: Given the unity-gain bandwidth for an ideal op-amp equal to 5.0 MHz, find the voltage gain at frequency of f = 500 kHz. Analysis: A0ω 0 = K = A1ω1 K = 1× 2π × 5.0MHz = 10π × 10 6 A1 = K 10π × 10 6 = = 10,000 ω1 2π × 500 ______________________________________________________________________________________ Problem 8.69 Solution: Find: Determine the relationship between a finite and frequency-dependent open-loop gain closed-loop gain AV (OL ) (ω ) and the AV (CL ) (ω ) of an inverting amplifier as a function of frequency. Plot AV (CL ) versus ω. Analysis: As shown in Equation 8.84, if we consider a real op-amp: 8.44 G. Rizzoni, Principles and Applications of Electrical Engineering A0 AV (OL ) (ω ) AV (CL ) (ω ) = AC 1 + jω 1 + jω ω0 , where AC = - RF / RS, and A0 is the low-frequency open-loop gain. ω1 If we choose A0 ω1 = Problem solutions, Chapter 8 = 106, and ω0 = 10 π, and since A0ω 0 = ACω1 : 10 6 ⋅10π . RF RS Therefore: 20Log10 AVCL 20Log10 AC ω1 Log10 ω ______________________________________________________________________________________ Problem 8.70 Solution: Find: A sinusoidal sound (pressure) wave impinges upon a condenser microphone of sensitivity S (mV / Pa). The voltage output of the microphone vS is amplified by two cascaded inverting amplifiers to produce an amplified signal v0. Determine the peak amplitude of the sound wave (in dB) if v0 = 5 VRMS. Estimate the maximum peak magnitude of the sound wave in order that v0 not contain any saturation effects of the op-amps. Analysis: p (t ) = P0 sin (ωt ) Pa ; v S (t ) = S ⋅ p (t ) mV ½ ° A1 A2 A1 A2 ° v0 (t ) = v S (t ) V = S ⋅ P0 sin (ωt ) V ¾ A1 A2 ⋅ S ⋅ P0 = 5000 mV 1000 1000 ° v0 = 5 VRMS °¿ 5000 dB P0 = 20 log10 A1 A2 ⋅ S A1 A2 ⋅ S ⋅ P0 ≤ 12 V P0 ≤ 12,000 ⋅ A1 A2 ⋅ S Pa . 1000 ______________________________________________________________________________________ 8.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Problem 8.71 Solution: Known quantities: If, in the circuit shown in Figure P8.71 vS1 = 2.8 + 0.01cos(ωt ) V ; vS 2 = 3.5 − 0.007 cos(ωt ) V Av1 = −13 ; Av 2 = 10 ; ω = 4 krad s Find: a) The common and difference mode input signals. b) The common and differential mode gains. c) The common and difference mode components of the output voltage. d) The total output voltage. e) The common mode rejection ratio. Analysis: a) 1 1 [ v S1 + v S 2 ] = ( [ 2.8 V + 10 mV cos ωt ] + [ 3.5 V - 7 mV cos ωt ] )= 2 2 = 3.15 V + 1.5 mV cos ωt v S -C = vS - D = vS1 - v S 2 = [ 2.8 V + 10 mV cos ωt ] - [ 3.5 V - 7 mV cos ωt ]= = - 0.7 V + 17 mV cos ωt Note that the expression for the differential input voltage (and the differential gain below) depends on which of the two sources (in this case, vS1) is connected to the non-inverting input. 1 [ Av1 - Av 2 ] = - 11.5 2 vO-C = v S -C Avc = [ 3.15 V + 1.5 mV cos ωt ] [ - 3 ] = - 9.450 V - 4.5 mV cos ωt c) vO- D = v S - D Avd = [ - 0.7 V + 17 mV cos ωt ] [ - 11.5 ] = 8.050 V - 195.5 mV cos ωt d) vO = vO-C + vO - D = - 1.4 V - 200.0 mV cos ωt | Avc | 3 ] = 20 dB Log10 [ ] = - 11.67 dB . e) CMRR = 20 dB Log10 [ | Avd | 11.5 b) Avc = Av1 + Av 2 = - 3 Avd = ______________________________________________________________________________________ Problem 8.72 Solution: Known quantities: If, in the circuit shown in Figure P8.71: vS1 = 3.5 + 0.01 cos ωt V Avc = 10 dB Avd = 20 dB v S 2 = 3.5 - 0.01 cos ωt krad ω = 4 s Find: a) The common and differential mode input voltages. b) The voltage gains for vS1 and vS2. c) The common mode component and differential mode components of the output voltage. d) The common mode rejection ratio [CMRR] in dB. 8.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 8 Analysis: a) 1 1 [ v S1 + v S 2 ] = ( [ 3.5 V + 10 mV cos ωt ] + [ 3.5 V - 10 mV cos ωt ] )= 3.5 V 2 2 = v S1 - v S 2 = [ 3.5 V + 10 mV cos ωt ] - [ 3.5 V - 10 mV cos ωt ] = 20 mV cos ωt vS -C = vS - D Note the expression for the difference mode voltage depends on which signal [in this case, vS1] is connected to the non-inverting input. This is also true for the expression for the differential gain below. b) Avc = 10 dB 10 20 dB 20 dB 10 20 dB = 3.162 = 10 Avd = 1 1 Avc = Av1 + Av 2 Av1 = Avc + Avd = [ 3.162 ] + 10 = 11.581 2 2 1 1 1 Avd = [ Av1 - Av 2 ] Av 2 = Avc - Avd = [ 3.162 ] - 10 = - 8.419 2 2 2 vO-C = vS-C Avc = [ 3.5 V ] [ 3.162 ] = 11.07 V c) vO- D = v S - D Avd = [ 20 mV cosωt ] [ 10 ] = 200 mV cosωt 3.162 Avc | ] = 20 dB d) CMRR = 20 dB Log10 [ | ] = - 10 dB . Log10 [ 10 Avd ______________________________________________________________________________________ Problem 8.73 Solution: Known quantities: If, in the circuit shown in Figure P8.73, the two voltage sources are temperature sensors with T Temperature [Kelvin] and: vS1 = k T 1 vS 2 = k T 2 V Where : k = 120 µ K R1 = R3 = R4 = 5 kΩ R 2 = 3 kΩ R L = 600 Ω Find: a) The voltage gains for the two input voltages. b) The common mode and differential mode input voltage. c) The common mode and difference mode gains. d) The common mode component and the differential mode component of the output voltage. e) The common mode rejection ratio [CMRR] in dB. Analysis: a) 8.47 = G. Rizzoni, Principles and Applications of Electrical Engineering Assume the op amp is ideal.: v N - v S1 KVL : - v N - v D + v P = 0 vN = vP vD ≈ 0 v N - vO = 0 iN ≈ 0 : KCL : i P + v P vS 2 + v P vO = 0 R4 R2 iP ≈ 0 : KCL : i N + R1 + Problem solutions, Chapter 8 R3 v S 1 vO + R R3 1 vN = 1 1 + R1 R3 v S 2 vO + R4 R 2 vP = 1 1 + R 2 R4 R1 R3 R1 R3 R 2 R4 R 2 R4 v S1 R3 + vO R1 = v S 2 R4 + vO R2 R3 + R1 R4 + R 2 R4 + [ - R3 ] vS 2 v S1 R R3 + R1 [ R4 + R2 ] [ R3 + R1 ] 4 + R2 vO = [ R4 + R2 ] [ R3 + R1 ] R1 - R2 R3 + R1 R4 + R2 R4 [ R3 + R1 ] R3 [ R4 + R 2 ] = vS 2 + v S1 [ ] R1 [ R4 + R2 ] - R2 [ R3 + R1 ] R1 [ R4 + R2 ] - R2 [ R3 + R1 ] [ 5 ] [ 10 ] R4 [ R3 + R1 ] = 5 = Av 2 = [ 5 ] [ 8 ] - [ 3 ] [ 10 ] R1 [ R4 + R2 ] - R2 [ R3 + R1 ] - R3 [ R4 + R 2 ] -[ 5 ][ 8 ] = =-4 Av1 = [ 5 ] [ 8 ] - [ 3 ] [ 10 ] R1 [ R4 + R2 ] - R2 [ R3 + R1 ] Equating: b) V ] [ 310 K ] = 37.20 mV K V vS 2 = k T 2 = [ 120 µ ] [ 335 K ] = 40.20 mV K 1 1 vS -C = [ v S1 + v S 2 ] = [ 37.20 mV + 40.20 mV ] = 38.70 mV 2 2 vS - D = vS 2 - v S1 = 40.20 mV - 37.2 mV = + 3 mV vS1 = k T 1 = [ 120 µ Note that the expression for the differential mode voltage (and the differential mode gain below) depends on Source #2 being connected to the non-inverting input. c) vO = v S 2 Av 2 + v S1 AS1 = [ v S -C + 1 1 v S - D ] Av 2 + [ v S -C - v S - D ] Av1 2 2 1 [ Av 2 - Av1 ] = v S -C Av-c + v S - D Avd 2 Avc = Av 2 + Av1 = 5 + [- 4] = 1 ° ® 1 1 °̄ Avd = 2 [ Av 2 - Av1 ] = 2 [ 5 - (- 4) ] = 4.5 = v S -C [ Av 2 + Av1 ] + v S - D 8.48 G. Rizzoni, Principles and Applications of Electrical Engineering d) e) Problem solutions, Chapter 8 vO-C = v S -C Avc = [ 38.70 mV ] [ 1 ] = 38.70 mV vO- D = v S - D Avd = [ 3 mV ] [ 4.5 ] = 13.5 mV An ideal difference amplifier would eliminate all common mode output. This did not happen here. A figure of merit for a differential amplifier is the Common Mode Rejection Ratio [CMRR]: 1 CMRR = 20 dB Log10 [ Avc ] = 20 dB Log10 [ ] = - 13.06 dB . 4.5 Avd ______________________________________________________________________________________ Problem 8.74 Solution: Known quantities: If, for the differential amplifier shown in Figure P8.73: vS1 = 13 mV v S 2 = 9 mV v0C = 33 mV v0 D = 18 V v 0 = v0 C + v 0 D Find: a) The common mode gain. b) The differential mode gain. c) The common mode rejection ratio in dB. Analysis: a) 1 1 [ v S1 + vS 2 ] = [ 13 mV + 9 mV ] = 11 mV 2 2 33 mV AVC = vOC = = 3 11 mV v SC vSC = v SD = v S 2 - vS1 = 9 mV - 13 mV = - 4 mV b) 18 V v AVD = OD = = - 4500 - 4 mV v SD | AVC | c) CMRR = 20 dB Log10 [ ] = - 63.52 dB . | AVD | ______________________________________________________________________________________ 8.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Chapter 9 Instructor Notes After a brief introduction to semiconductors materials in Section 9.1, Section 9.2 introduces the pn junction and the semiconductor diode equation and i-v curves. These sections could be bypassed in favor of a more intuitive presentation, provided in Section 9.3 with an explanation of the basic circuit behavior of the diode, beginning with large-signal models. This explanation is supplemented by the sidebar Make the Connection: Hydraulic Check Valves (pp. 466-467), which continues the electric-hydraulic analogy introduced in earlier chapters, and provides an intuitive explanation of the operation of the semiconductor diodes1. The box Focus on Methodology: Determining the Conduction State of a Diode (p. 467) helps the student understand how diode state of conduction can be determined. The large-signal circuit model material is probably sufficient for the purposes of most introductory courses. 5nstructors who are interested in introducing the subject of small-signal models, in preparation for a the study of transistor small-signal amplifiers, may find the rest of the section, covering small-signal diode models (pp. 474-483), of interest. The concept of operating point is introduced, with a review of the load-line equation and the definition of the quiescent operating point of a device. The box Focus on Methodology: Determining the Operating Point of a Diode (p. 475) summarizes this material. Section 3.8 can be recalled (or introduced for the first time) at this point. The solution methods used in Examples 9.5, 9.6, 9.7 emphasize the use of simple circuit models for the diode, together with the use of Thèvenin equivalent circuits; this method is quite general, and will reinforce the importance (and understanding) of the concept of equivalent circuits in analyzing more advanced circuits. The use of Device Data Sheets is continued in this chapter with a summary data sheet on a general purpose diode. The box Focus on Methodology: Using device data sheets (pp. 476-477) is designed to familiarize the student with the basic contents and function of device data sheets. The instructor may wish to expand on this introductory presentation by asking students to locate data sheets in the accompanying CD-ROM or on the web, and to identify specific devices and their parameters for use in homework problems. Section 9.4 discusses various diode rectifier circuits; Section 9.5 introduces DC power supplies; Zener diode circuits and voltage regulation. Section 9.6 analyzes various signal processing circuits, and introduces two application examples in the boxes Focus on Measurements: Peak Detector Circuit for Capacitive Displacement Transducer (pp. 499-501), which is tied to two earlier boxeson the capacitive displacement transducer (pp. 147-148 and pp. 175-177), and Focus on Measurements: Diode Thermometer (pp. 502-503). The latter example can be tied to a laboratory experiment2. Finally, Section 9.7 introduces photodiodes and solar cells and includes the box Focus on Measurements: Opto-Isolators (p. 506) The homework problems present a graded variety of problems, mostly related to the 17 examples and application examples presented in the text. Learning Objectives 1. Understand the basic principles underlying the physics of semiconductor devices in general, and of the pn junction in particular. Become familiar with the diode equation and i-v characteristic. Sections 1, 2. 2. Use various circuit models of the semiconductor diode in simple circuits. These are divided into two classes: large signal model, useful to study rectifier circuits, and small signal models, useful in signal processing applications. Section 2. 3. Study practical full-wave rectifier circuits and learn to analyze and determine the practical specifications of a rectifier using large-signal diode models. Section 3. 4. Understand the basic operation of Zener diodes as voltage references, and use simple circuit models to analyze elementary voltage regulators. Section 4. 5. Use the diode models presented in Section 2 to analyze the operation of various practical diode circuits in signal processing applications. Section 5. 6. Understand the basic principle of operation of photodiodes, including solar cells, photosensors and light-emitting diodes. Section 6. 1 2 With many thanks to Bill Ribbens, who first suggested this idea to me some 20 years ago. G. Rizzoni, A Practical Introduction to Electronic Instrumentation, 3rd Edition, Kendall-Hunt, 1998. 9.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Section 9.1: Electrical Conduction in Semiconductor Devices Section 9.2: The pn Junction and the Semiconductor Diode Problem 9.1 Solution: Known quantities: The Ionized Acceptor Density for a doped Silicon: N a = N a− = 1017 1 , Nd = 0 m3 Find: a) If this material is an N or P type extrinsic semiconductor. b) Which are the majority and which the minority charge carriers. c) The density of majority and minority carriers. Analysis: a) Each acceptor dopant atom introduces an additional positive charge carrier and a negative atomic ion. The ion is NOT a charge carrier. The density of positive carriers [holes] increases because of the doping so the material is extrinsic P type Silicon. b) The majority carriers are the positive carriers or valence band holes; the minority carriers are the negative carriers or conduction band free electrons. c) CNE : p po + 0 − n po − N a− = 0 , p po − ( nio2 − N a− = 0 , p po Using the quadratic equation: p po = − ( ) CPE : n po = nio2 p po ) ( ) p 2po + p po − N a− + − nio2 = 0 (( 1 1 − N a− ± − N a− 2 2 ) 2 ( − 4 − nio2 = 51016 ± 5.221016 = 1.0221017 )) 12 =− ( ) (( 1 1 − 1017 ± − 1017 2 2 ) 2 ( − 4 − 2.2510 32 )) 12 1 m3 where the negative answer is physically impossible. Now use the CPE to obtain the minority carrier density: n po = nio2 2.2510 32 1 = = 2.2021015 3 17 p po 1.02210 m Note that because of the doping, the hole density is now about 100 times the electron density. The thermally produced carriers present in the intrinsic Silicon before doping has a small effect on the carrier densities in the extrinsic Silicon. At higher doping levels, the effect becomes negligible. As temperature increases, the densities of the thermally produced carriers increase and their effect on the final carrier densities increase. At very high temperatures [about 175 C for Silicon] the thermally produced carriers primarily determine the final carrier densities and the doping has a negligible effect [ie, the semiconductor behaves as an intrinsic material]. This is why semiconductors cannot operate in high temperature environments. ______________________________________________________________________________________ 9.2 = G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.2 Solution: Known quantities: The intrinsic Silicon is doped: N a = N a− = 1017 1 1 18 + , N d = N d = 510 3 m m3 Find: a) If the silicon is an N or P type extrinsic semiconductor. b) Which are the majority and which the minority charge carriers. c) The density of majority and minority carriers. Analysis: a) N Type b) Majority = Conduction band free electrons = Negative carriers. Minority = Valence band holes = Positive carriers. c) n no ≈ 4.91018 1 , m3 p no = 4.591013 1 m3 ______________________________________________________________________________________ Problem 9.3 Solution: Find: Describe the microscopic structure of semiconductor materials. What are the three most commonly used semiconductor materials? Analysis: Semiconductor materials are crystalline with the atoms arranged in a repeated three dimensional array. The distance between atoms in the array is the "lattice constant". Each atom of a semiconductor has four valence electrons. These electrons participate in covalent bonds with the valance electrons of other atoms. For certain materials with the properties above, quantum/wave mechanics predicts that the valance electrons may have a total energy [kinetic plus Coulombic potential energy] within certain "allowed" bands. The two most important bands are the valence band containing the valence electrons in covalent bonds and the conduction band containing conduction or free electrons which have obtained enough energy to escape from its covalent bond. Separating these two allowed bands is the "energy gap" extending over those energies which the electrons are "forbidden" to have. For semiconductor materials, the energy gap is on the order of 1 electron-Volt [eV]. Silicon is the most common semiconductor material and is used in a variety of applications and devices. Germanium is used in some optical devices and other special purpose devices. Gallium Arsenide is a compound III-V semiconductor material. One atom has three and the other has five valence electrons giving an average of four per atom. It is used in microwave, optical, and very high speed digital devices. ______________________________________________________________________________________ Problem 9.4 Solution: Find: Describe the thermal production of charge carriers in a semiconductor and how this process limits the operation of a semiconductor device. 9.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Analysis: At a temperature of absolute zero, ALL valence electrons in a semiconductor are contained in a covalent bond and there are NO charge carriers. The internal or thermal energy of a solid material is caused by the vibration of the atoms and electrons about their equilibrium position. As the temperature of the material increases, its thermal vibrational energy increases. Some electrons will gain sufficient energy to escape the covalent bond in the valence band, and "jump" past the energy gap into the conduction band. As a consequence, TWO charge carriers are generated. The conduction or free electron in the conduction band is a negative charge carrier. The vacancy in the valence band covalent bond or "hole" is a positive charge carrier. A conduction band electron may also give up energy and recombine with a valence band hole. The generation and recombination rates both increase with temperature. At any particular temperature, they are equal and produce equal equilibrium densities of electrons and holes. The equilibrium carrier densities increase with temperature. For Silicon at T = 300 K [approximately room temperature]: nio = pio = 1.51010 carriers carriers = 1.51016 3 cm m3 [A number of carriers is a dimensionless quantity and may be omitted from the units.] Almost all semiconductors devices are "doped" to achieve DIFFERENT densities of positive and negative carriers. A "P-type" semiconductor has a higher density of positive carriers and an "N-type" semiconductor has a higher density of negative carriers. However, at high temperatures the density of thermally produced carriers becomes very large and significantly reduces or nullifies the effects of the doping, ie, the positive and negative carrier densities become nearly equal. For this reason, semiconductor devices cannot be used in high temperature applications. The limit in temperature depends on the semiconductor material. ______________________________________________________________________________________ Problem 9.5 Solution: Find: Describe the properties of donor and acceptor dopant atoms and how they affect the densities of charge carriers in a semiconductor material. Analysis: An "intrinsic" semiconductor material is undoped. When dopant atoms are added the material becomes an "extrinsic" semiconductor. Doping results in the replacement of an intrinsic atom with a dopant atom. As few as one out of every million intrinsic atoms may be replaced. A "donor" dopant atom has 5 valence electrons. Only 4 are required to complete the bonding structure in the semiconductor material. The 5th electron requires very little energy to escape to the conduction band and become a negative charge carrier. This leaves behind a donor atom with one missing electron or a negative atomic ion. The ion is immobile and cannot move through the material; therefore, IT IS NOT A CHARGE CARRIER. Each donor contributes an additional negative carrier to the material. The increased density of negative carriers results in an increased recombination rate which reduces the density of positive carriers. [The PRODUCT of the two densities remains constant.] Materials doped with donor atoms are N type extrinsic semiconductors. The majority carriers are conduction band electrons, the minority carriers are valence band holes. An "acceptor" dopant atom has 3 valence electrons; however, 4 are required to complete the bonding structure in the semiconductor material. The "missing" 4th electron causes a vacancy or hole in the bonding structure. Another valence electron may move to and occupy this hole thus eliminating it and generating another hole and a negative atomic ion. The ion is immobile and cannot move through the material; therefore, IT IS NOT A CHARGE CARRIER. Each acceptor contributes an additional positive carrier to the material. The increased density of positive carriers results in an increased recombination rate which reduces the density of negative carriers. [The PRODUCT of the two densities remains constant.] Materials doped with acceptor atoms are P type 9.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 extrinsic semiconductors. The majority carriers are valance band holes, the minority carriers are conduction band electrons. ______________________________________________________________________________________ Problem 9.6 Solution: Find: Physically describe the behavior of the charge carriers and ionized dopant atoms in the vicinity of a semiconductor PN junction that causes the potential [energy] barrier that tends to prevent charge carriers from crossing the junction. Analysis: Semiconductor atoms are not shown in the two figures. The circles represent ionized dopant atoms and the uncircled plus and minus signs represent charge carriers. The dotted line in the first figure represents how far the depletion/space charge region extends into the P and N regions. Near the junction, the negative carriers in the N material recombine with the positive carriers in the P material. This forms a small region on either side of the junction which is depleted of charge carriers; however, the ionized dopant atoms are immobile and remain. Therefore, in the N material the region has a net positive charge and in the P material the region has a net negative charge. This region is called either a "depletion" region [depleted of carriers] or a "space charge" region [due to the dopant ions]. If a conduction band free electron [a majority carrier] in the N material tries to cross into the P material, it encounters and is repelled by the net negative charge [due to the ionized acceptor atoms] in the depleted part of the P material. If a valence band hole [a majority carrier] in the P material tries to cross into the the N material, it encounters and is repelled by the net positive charge [due to the ionized donor atoms] in the depleted part of the N material. The repulsion of the carriers is characterized as a Coulombic "potential [energy] barrier". With no voltage applied across the junction, Ohm's law requires the current to be zero. Actually, very, very small equal but opposite currents do flow across the junction but the net curent is zero. The barrier can be decreased by applying a "forward bias" voltage across the junction. This allows more carriers to cross the junction and when this voltage is greater than a certain value [0.7 V for Silicon] a significant current [milliamps] flows. The barrier can be increased by applying a "reverse bias" voltage across the junction. This increases the barrier and fewer majority carriers have sufficient energy to cross the junction, ie, the current essentially ceases. However, there is a VERY, VERY small "reverse saturation current" [in the femtoamps range] due to the minority carriers. Since the minority carriers are thermally produced, this current is dependent on temperature. ______________________________________________________________________________________ 9.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Section 9.3: Circuit Models for the Semiconductor Diode Focus on Methodology: Determining the conduction state of an ideal diode 1. 2. 3. 4. Assume a diode conduction state (on or off). Substitute the ideal diode circuit model into the circuit (short circuit if “on”, open circuit if “off”). Solve for diode current and voltage using linear circuit analysis techniques. If the solution is consistent with the assumption, then the initial assumption was correct; if not, the diode conduction state is opposite to that initially assumed. For example, if the diode has been assumed “off”, but the diode voltage computed after replacing the diode with an open circuit is a forward bias, then it must be true that the actual state of the diode is “on”. Focus on Methodology: Determining the operating point of a diode 1. 2. 3. Reduce the circuit to a Thévenin or Norton equivalent circuit, with the diode as the load. Write the load line equation, 9.15. Solve numerically two simultaneous equations in two unknowns (the load-line equation and the diode equation) for the diode current and voltage. 4. Solve graphically by finding the intersection of the diode curve (e.g., from a a data sheet) with the loadline. The intersection of the two curves is the diode operating point. or Problem 9.7 Solution: Known quantities: The circuit of Figure P9.7. Find: A plot of vL versus vS. Analysis: For v S < 0 , the diode is reverse biased, and v L = 0 . For vS ≥ 0 , the diode is forward biased, and § RL R1 · ¸. v L = vS ¨¨ ¸ + R R R L 1 ¹ © S vL α 0 Where § RL || R1 · ¸¸ α = tan −1 ¨¨ © RS + RL || R1 ¹ vS ______________________________________________________________________________________ 9.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.8 Solution: Known quantities: The circuit of Figure P9.7 using the offset diode model. Find: A plot of vL versus vS. Analysis: The circuit can be represented as shown in the following figure. For v1 < Vγ , the diode is reverse biased, and v L = 0 . In term of vS , we have § R · v L = 0 ⇔ vS < Vγ ¨¨ 1 + S ¸¸ R1 ¹ © § R · For v1 ≥ Vγ , i.e. v S ≥ Vγ ¨¨1 + S ¸¸ the diode is forward biased, and R1 ¹ © vS Vγ + RS RL vL v1 = 1 1 1 + + RS R1 RL § 1 vS 1· − Vγ ¨¨ + ¸¸ R © RS R1 ¹ v L = v1 − Vγ = S 1 1 1 + + RS R1 RL . α 0 1 § ¨ § R · RS −1 Where VS = Vγ ¨¨ 1 + S ¸¸, α = tan ¨ 1 1 1 ¨ R1 ¹ © ¨R +R +R L 1 © S VS · ¸ ¸. ¸ ¸ ¹ vS ______________________________________________________________________________________ Problem 9.9 Solution: Known quantities: For the circuit of Figure P9.7: v S = 6 V , and the resistances R1 = RS = RL = 1kΩ . 9.7 G. Rizzoni, Principles and Applications of Electrical Engineering Find: Determine Problem solutions, Chapter 9 i D and v D graphically. Assumptions: Use the diode characteristic of the 1N461A. Analysis: Replace the diode with an open circuit, and v DOC = 3 V . Replace the diode with a short circuit, and iDSC = 1 6 = 2 mA . 2 1500 These are the end points of the load line. The load line is superimposed on the diode characteristic in the figure. Load line From the intersection of the load line and the diode characteristic, we see that iD ≈ 1.5 mA and v D ≈ 0.75 V. ______________________________________________________________________________________ Problem 9.10 Solution: Known quantities: The current I = 1 mA , that make the diode to be above the knee of its i − v characteristic. Find: a) The value of R to establish a 5 mA current in the circuit. b) With the value of R established in the preceding part, what is the minimum value to which the voltage E could be reduced and still maintain diode current above the knee. Assumptions: Vγ = 0.7 V . Analysis: a) R= 5 − 0 .7 = 860 Ω 5 10 −3 b) E min − 0.7 = 1 10 −3 860 E min = 0.86 + 0.7 = 1.56 V I= ______________________________________________________________________________________ Problem 9.11 Solution: Known quantities: The circuit of Figure P9.11 driven by a sinusoidal source of Find: a) The maximum forward current. 9.8 50 V rms , Vγ = 0.7 V , R = 220 Ω . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 b) The peak inverse voltage across the diode. Analysis: 50 2 − 0.7 = 318 mA 220 = 50 2 = 70.7 V a) I Fmax = b) Vrevmax ______________________________________________________________________________________ Problem 9.12 Solution: Known quantities: The configurations shown if Figure P9.12. Find: Which diode are forward biased, and which are reverse biased. Analysis: a) reverse-biased b) forward-biased c) reverse-biased d) forward-biased e) forward-biased _____________________________________________________________________________________ Problem 9.13 Solution: Known quantities: The configuration shown if Figure P9.13. Find: The range of Vin for which D1 is forward-biased. Analysis: The diode D1 is clearly forward-biased for any Vin > 0. ______________________________________________________________________________________ Problem 9.14 Solution: Known quantities: The configurations of Figure P9.14. Find: Determine which diodes are forward-biased and which are reverse-biased. Determine the output voltage. Assumptions: The drop across each forward biased diode is 0.7 V. Analysis: a) D2 and D4 are forward biased; D1 and D3 are reverse biased. vout = −5 + 0.7 = −4.3 V b) D1 and D2 are reverse biased; D3 is forward biased. v out = −10 + 0.7 = −9.3 V c) D1 is reverse biased; D2 is reverse biased. ______________________________________________________________________________________ 9.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.15 vo Solution: (V) Known quantities: The circuit of Figure P9.15; v S (t ) = 10 sin(2,000πt ) . Find: The output waveform and the voltage transfer characteristic. Assumptions: The diode is ideal. Analysis: 5 4 0 0.5 1 1.5 2 For vS <8 V, vo=4 V. For vS ≥ 8 V, vo= vS /2. The voltage transfer characteristic is vo 8 0 8 16 vS __________________________________________________________________________________ Problem 9.16 Solution: Known quantities: The circuit of Figure P9.15: vS (t ) = 10 sin(2,000πt ). Find: The output waveform and the voltage transfer characteristic. Assumptions: The diode has an offset Vγ = 0.6 V. Analysis: For vS <6.8 V, vo=4-Vγ = 3.4 V. For vS ≥ 6.8 V, vo= vS /2. The voltage transfer characteristic is vo 6.8 0 6.8 13.6 vS ______________________________________________________________________________________ 9.10 t (ms) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.17 Solution: Known quantities: Same as Problem 9.15 but with vS (t ) = 1.5 sin (2,000πt ) , the battery equal to 1 V and the resistors of 1kΩ. Find: The output waveform and the voltage transfer characteristic. Assumptions: The diode has an offset Vγ = 0.6 V , and rD = 200 Ω. Analysis: 1 − Vγ vS + v +2 rD 1000 = S For vS < 2 (1- Vγ) = 0.8 V, vo = 1 1 1 7 + + 1000 rD 1000 For vS ≥ 0.8 V, vo= vS /2. The voltage transfer characteristic is vo 0.8 0.4 0 0.8 1.6 vS ______________________________________________________________________________________ Problem 9.18 Solution: Known quantities: § VVD · The circuit of Figure P9.18; the diode is fabricated from Silicon and I D = I 0 ¨ e T − 1¸ . At ¨ ¸ © ¹ kT T = 300 K , I 0 = 250 10 −12 A , VT = ≈ 26 mV , v S = 4.2 V + 110 cos(ωt ) mV , q rad ω = 377 , R = 7 kΩ . s Find: Determine, using superposition, the DC or Q-point current through the diode: a) Using the DC offset model for the diode. b) By iteratively solving the circuit characteristic and device characteristic. 9.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Analysis: a) Suppress the AC component of the source voltage. Construct the DC equivalent circuit using the threshold (or offset voltage model). The Q point is at the intersection of the device (diode) characteristic and the circuit characteristic given by the KVL below. Here, the device characteristic is approximated by the threshold voltage model giving the approximate Q point at the upper right. Assume the diode is on. Then: VD = VD −on = 0.7 V , KVL − VS + I DQ R + V D = 0 , I DQ = VS − V D −on = 0.5 mA R b) In the forward biased region with significant conduction: Device: ID = I D = I 0e VD VT , VD = VT ln ID , I0 Circuit, KVL: − VS + I D R + V D = 0 , VS − V D R A simultaneous solution (the lower left Q point) of the device and circuit characteristics is required. To do this iteratively, initially assume a value for the diode voltage, say 0.7 V for a Silicon device. Then: 1. Using the initial or new diode voltage and the circuit characteristic, determine a new diode current. 2. Using this new diode current and the device characteristic, determine a new diode voltage. ITERATE or REPEAT until convergence is obtained. Voltage --------700 mV 377.2 mV 379.5 mV New Current ---------0.5 ma 0.5461 ma 0.5458 ma New Voltage ----------377.2 mV 379.5 mV 379.5 mV <<< Convergence 9.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Convergence is obtained after only 3 iterations and: I DQ = 546 µA , V DQ = 379.5 mV This is a much more accurate solution than that of Part a). The two solutions differ significantly because the reverse saturation current given is atypically large. ______________________________________________________________________________________ Problem 9.19 Solution: Known quantities: § VVD · The circuit of Figure P9.18; the diode is fabricated from Silicon and I D = I 0 ¨ e T − 1¸ . At ¨ ¸ © ¹ kT T = 300 K , I 0 = 2.03 10 −15 A , VT = ≈ 26 mV , v S = 5.3 V + 7 cos(ωt ) mV , q rad ω = 377 , R = 4 .6 k Ω . s Find: Determine, using superposition and the offset voltage model for the diode, the DC or Q-point current through the diode. Analysis: Suppress the AC component of the source voltage. Construct the DC equivalent circuit using the threshold (or offset voltage model). The Q point is at the intersection of the device (diode) characteristic and the circuit characteristic given by the KVL below. Here, the device characteristic is approximated by the threshold voltage model giving the approximate Q point at the upper right. The DC source voltage will tend to make the diode conduct. Assume the diode is on. Then: VD = VD − on = 0.7 V , KVL − VS + I DQ R + V D = 0 , I DQ = VS − V D −on = 1.0 mA R The current is positive so the assumption above that the diode is on is valid. ______________________________________________________________________________________ Problem 9.20 Solution: Known quantities: § VVD · The circuit of Figure P9.18; the diode is fabricated from Silicon and I D = I 0 ¨ e T − 1¸ . At ¨ ¸ © ¹ kT T = 300 K , I 0 = 250 10 −12 A , VT = ≈ 26 mV , v S = 4.2 V + 110 cos(ωt ) mV , q rad ω = 377 , R = 7 kΩ . The DC operating point is: I DQ = 0.546 mA , V DQ = 379.5 mV . s Find: The equivalent small-signal AC resistance of the diode at room temperature at the Q point given. 9.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Analysis: rD = dV D dI D = Q 1 dI D dV D = Q 1 1 I 0e VT ≈ VD VT VT ID = Q VT = 47.64 Ω I DQ Q ______________________________________________________________________________________ Problem 9.21 Solution: Known quantities: § VVD · The circuit of Figure P9.18; the diode is fabricated from Silicon and I D = I 0 ¨ e T − 1¸ . At ¨ ¸ © ¹ kT T = 300 K , I 0 = 2.03 10 −15 A , VT = ≈ 26 mV , v S = 5.3 V + 70 cos(ωt ) mV , q rad ω = 377 , R = 4.6 kΩ . The DC operating point is: I DQ = 1.0 mA , V DQ = 0.7 V . s Find: The equivalent small-signal AC resistance of the diode at room temperature at the Q point given. Analysis: rD = dV D dI D = Q 1 dI D dV D = Q 1 1 I 0e VT ≈ VD VT VT ID = Q VT = 26 Ω I DQ Q 9.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 ______________________________________________________________________________________ Problem 9.22 Solution: Known quantities: A diode with the i-v characteristic in Figure 9.8, connected to a 5 V source and a load resistance of 200 Ω. Find: a) The load current and voltage b) The power dissipated by the diode. c) The load current and voltage if the load is changed to 100 Ω and 500 Ω. Analysis: a) The operating point can be determined by using the load-line analysis The load line is iD = Diode i−v curve 0.05 v 5 − vD 5 = − D 200 200 RL 0.045 0.04 The load voltage is The load current is obtained by the figure as intersection of the two characteristics and is equal to 0.021 A. Diode current, A vL = 5 − v D ≅ 5 − 0.74 = 4.26 V 0.035 0.03 0.025 0.02 0.015 0.01 0.005 b) The power dissipated by the diode is 0 0 PD = v D iD = 0.74 ⋅ 0.021 = 15 mV 0.2 0.4 0.6 i−v Vcurve DiodeDiode voltage, 0.05 0.8 1 0.045 c) For RL=100 Ω, we have 0.04 vD ≅ 0.757 V 0.03 0.025 0.02 0.015 Similarly, for RL = 500 Ω, we have vD ≅ 0.717 V L 0.035 Diode current, A 5 − 0.757 = 0.0424 A 100 vL = 5 − 0.757 = 4.24 V iD = iL = R =100 Ω RL=500 Ω 0.01 0.005 0 0 5 − 0.717 iD = iL = = 0.04283 A 100 vL = 5 − 0.717 = 4.283 V 0.2 0.4 0.6 Diode voltage, V 0.8 ______________________________________________________________________________________ 9.15 1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.23 Solution: Known quantities: A diode with the i-v characteristic in Figure 9.32, connected in series to a 2 V source and a load resistance of 200 Ω. Find: a) The load current and voltage b) The power dissipated by the diode. c) The load current and voltage if the load is changed to 100 Ω and 300 Ω. Analysis: a) The operating point can be determined by using the load-line analysis The load line is iD = 2 − vD 2 v = − D = 0.01 − 0.005 vD 200 200 RL By drawing this line on the top of Figure 9.32, the following operating point is obtained iD = iL ≈ 6 mA; v D ≈ 0.73 V v L = 2 − vD = 2 − 0.73 = 1.27 V b) PD = vDiD = 0.73 ⋅ 0.006 = 4.38 mV c) For RL=100 Ω, we have v D ≈ 0.825 V 2 − 0.825 = 11.75 mA 100 v L ≈ 2 − 0.825 = 1.175 V iD = i L ≅ For RL=300 Ω, we have v D ≈ 0 .7 V 2 − 0 .7 = 4.3 mA 100 v L ≈ 2 − 0 .7 = 1 .3 V iD = i L ≅ ______________________________________________________________________________________ Problem 9.24 Solution: Known quantities: § VVD · The circuit of Figure P9.18; the diode is fabricated from Silicon and I D = I 0 ¨ e T − 1¸ . At ¨ ¸ © ¹ kT T = 300 K , I 0 = 250 10 −12 A , VT = ≈ 26 mV , v S = 4.2 V + 110 cos(ωt ) mV , q 9.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 rad , R = 7 kΩ . The DC operating point is: I DQ = 0.548 mA , V DQ = 0.365 V , s rd = 47.45 Ω . ω = 377 Find: Determine, using superposition the AC voltage across the diode and the AC current through it. Analysis: Suppress the DC component of the source voltage. Replace the diode with its AC equivalent resistance; then: Vd = V S rd R + rd 47.45 = 740.6∠0° µ V , v d (t ) = 740.6 cos(ωt ) µ V 7 ,000 + 47.45 V 740.6∠0° Id = d = = 15.61∠0° µ A , id = 15.61cos(ωt ) µ A . rd 47.45 Vd = 110∠0° The total solution is then: id = 0.548 10 −3 + 15.61 cos (ωt ) 10 −6 A v d (t ) = 0.365 + 740.6 cos(ωt ) 10 −6 V ______________________________________________________________________________________ Problem 9.25 Solution: Known quantities: The circuit of Figure P9.25. The diode is fabricated from Silicon and Find: The minimum value of R = 2 .2 k Ω , V S 2 = 3 V . VS 1 at and above which the diode will conduct with a significant current. Analysis: A diode fabricated from silicon will conduct with a significant current if it has a forward bias equal to or larger than about 0.7 V. KVL − VS 1 + I D R + V D + VS 2 = 0 , VS 1 = I D R + V D + VS 2 . At point of conduction: I D = 0 , VD = VD − on = 0.7 V , VS1 = 0 + 0.7 + 3 = 3.7 V ______________________________________________________________________________________ 9.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Section 9.4: Rectifier Circuits Problem 9.26 Solution: Known quantities: The circuit of Figure P9.26. The input voltage is sinusoidal with an amplitude of 5 V. Find: The average value of the output voltage. Assumptions: Vγ = 0.7 V . Analysis: The capacitor will charge to 5 V − 0.7 V = 4.3 V and, therefore, the input sine wave will be shifted up 4.3 V to produce the output. As a result, after the cycle (the capacitor builds up its stored charge during the third quarter cycle), the average value of the output will be 4.3 V. ______________________________________________________________________________________ Problem 9.27 Solution: Known quantities: The rectifier circuit of Figure P9.27; v(t ) = A sin(2π 100t ) V . The conduction must begin during each positive half-cycle at an angle no greater than 5° . Find: The minimum peak value A that the AC source must produce. Assumptions: Vγ = 0.7 V . Analysis: ( ) Amin sin 5 ° = 0.7 Amin = 0 .7 0 .7 = = 8.03V ° 0.0872 sin 5 ( ) ______________________________________________________________________________________ Problem 9.28 Solution: Known quantities: A half-wave rectifier is to provide an average voltage of 50 Find: a) Draw a schematic diagram of the circuit. b) Sketch the output voltage waveshape. c) Determine the peak value of the input voltage. d) Sketch the input voltage waveshape. e) The rms voltage at the input. Analysis: a) 9.18 V at its output. G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 b) c) v ave = 0.318 v peak = 50 v peak = 157.2 V d) e) Vinrms = 157.2 2 = 111.2 V ______________________________________________________________________________________ Problem 9.29 Solution: Known quantities: A rectifier circuit similar to that of Figure 9.25. Load Resistance 100 Ω, AC source voltage 30 V (rms). Find: The peak and average current in the load. Assumptions: Ideal diode. Analysis: The peak voltage is VR peak = 30 2 V I R peak = VR peak R = 30 2 = 0.424 A 100 The average current in the load is 9.19 G. Rizzoni, Principles and Applications of Electrical Engineering I RAV = I R peak π Problem solutions, Chapter 9 = 0.135 A ______________________________________________________________________________________ Problem 9.30 Solution: Known quantities: A rectifier circuit similar to that of Figure 9.25. Load Resistance 220 Ω, AC source voltage 25 V (rms). Find: The peak and average current in the load. Assumptions: Ideal diode. Analysis: The peak voltage is VR peak = 25 2 V I R peak = VR peak R = 25 2 = 0.161 A 220 The average current in the load is I RAV = I R peak π = 51.8 mA ______________________________________________________________________________________ Problem 9.31 Solution: Known quantities: The full-wave power supply of Figure P9.31. The diodes are 1N461 with a rated peak reverse voltage equal to 25 V, and are fabricated from Silicon. n = 0.05883 , C = 80 µF , Vline = 170 cos(377t ) V . Find: a) The actual peak reverse voltage across each diode. b) The reasons for which these diodes are or are not suitable for the specification given. Analysis: a) At ωt = 0, D1 is on. At ωt = π, D1 is off and the reverse voltage across it is maximum. Vso = Vio n = 170 0.05883 = 10 V KVL : − v s1 (t ) + v D1 + v L (t ) = 0 At ωt = 0, − Vso + V D −on + Vm = 0 Vm = Vso − V D −on = 10 − 0.7 = 9.3 V At ωt = π, − (− Vso ) + V D1 + Vm = 0 V D1 = −Vso − Vm = −10 − 9.3 = −19.3 V b) The actual peak reverse voltage (19.3 V) is less than the rated peak reverse voltage (25 V) by a barely adequate margin of safety. Therefore, the diodes are suitable for the specifications given. ______________________________________________________________________________________ Problem 9.32 Solution: Known quantities: The full-wave power supply of Figure P9.31; n = 0.05883 , C = 80 µF , 9.20 Vline = 170 cos(377t ) V . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 The diodes are 1N4727 switching diodes fabricated from Silicon and with the following rated performances: 4 Pmax = 500 mW at T = 25 °C ; derated 3 mW T = 125 ÷ 175 °C ; V pk − rev = 30 V . °C mW T = 25 ÷ 125 °C and °C Find: a) The actual peak reverse voltage across each diode. b) The reasons for which these diodes are or are not suitable for the specification given. Analysis: a) At ωt = 0, D1 is on. At ωt = π, D1 is off and the reverse voltage across it is maximum. Vso = Vio n = 170 0.1 = 17 V KVL : − v s1 (t ) + v D1 + v L (t ) = 0 At ωt = 0, − Vso + V D −on + Vm = 0 Vm = Vso − V D −on = 17 − 0.7 = 16.3 V = π, − (− Vso ) + VD1 + Vm = 0 VD1 = −Vso − Vm = −17 − 16.3 = −33.3 V b) The actual peak reverse voltage (33.3 V) is greater than the rated peak reverse voltage (30 V). At ωt Therefore, the diodes are not suitable for the specifications given. ______________________________________________________________________________________ Problem 9.33 Solution: Known quantities: The full-wave DC power supply of Figure P9.31; the load voltage of Figure P9.33; VL = 5 V , Vr = 5 % , vline = 170 cos(ωt ) V , ω = 377 I L = 60 mA , rad . s Find: a) The turns ratio. b) The capacitor C. Analysis: 1 1 Vr = 5 + 0.125 = 5.125 V , V L − min = V L − Vr = 5 − 0.125 = 4.875 V , 2 2 = Vs 0 cos(ωt ) . a) Vm = V L + Vs1 = Vs 2 KVL : − v s1 (t ) + v D1 + v L (t ) = 0 At ωt = 0, − Vso + V D − on + V L = 0 V s 0 = V D −on + Vm = 0.7 + 5.125 = 5.825 V n= Vso = 0.0343 Vio 9.21 G. Rizzoni, Principles and Applications of Electrical Engineering b) KVL : At Problem solutions, Chapter 9 v s 2 (t ) + v D 2 + v L (t ) = 0 . t = t 2 , Vso cos(ωt 2 ) = −VD 2−on − VL − min . t 2 = § V + VL − min 1 cos −1 ¨¨ − D 2− on Vso ω © The exponential discharge of the capacitor can be expressed: v L (t ) = v L (∞ ) + (v L (0) − v L (∞ ))e v L (t 2 ) = VL − min = Vm e − I Lt 2 VL C , − C=− t TC = 0 + (Vm − 0)e I Lt2 §V VL ln¨¨ L − min © Vm · ¸¸ ¹ − t RL C = Vm e − · ¸¸ = 7.533 ms . ¹ I Lt VL C = 1812 µF Note: An approximate but conservative value of C can be obtained by using the approximation: ωt 2 ≈ π . Then C≈− I L (ωt 2 ) §V ωVL ln¨¨ L −min © Vm · ¸¸ ¹ = 2000 µ F . This value is conservative because it gives a smaller ripple voltage than that specified. ______________________________________________________________________________________ Problem 9.34 Solution: Known quantities: The full-wave DC power supply of Figure P9.31; vline = 170 cos(ωt ) V , ω = 377 I L = 600 mA , VL = 50 V , Vr = 8 % , rad . s Find: a) The turns ratio. b) The capacitor C. Analysis: 1 1 Vr = 50 + 2 = 52 V , V L − min = V L − Vr = 50 − 2 = 48 V , 2 2 = Vs 0 cos(ωt ) . a) Vm = V L + Vs1 = Vs 2 KVL : − v s1 (t ) + v D1 + v L (t ) = 0 At ωt = 0, − Vso + V D −on + V L = 0 Vs 0 = V D − on + Vm = 0.7 + 52 = 52.7 V 9.22 G. Rizzoni, Principles and Applications of Electrical Engineering n= Problem solutions, Chapter 9 Vso = 0.31 Vio v s 2 (t ) + v D 2 + v L (t ) = 0 . b) KVL : At t = t 2 , Vso cos(ωt 2 ) = −VD 2−on − VL − min . t 2 = § V + VL − min 1 cos −1 ¨¨ − D 2− on Vso ω © The exponential discharge of the capacitor can be expressed: v L (t ) = v L (∞ ) + (v L (0) − v L (∞ ))e v L (t 2 ) = VL − min = Vm e I t − L2 VL C , − C=− t TC = 0 + (Vm − 0)e I Lt2 §V V L ln¨¨ L − min © Vm · ¸¸ ¹ − t RL C = Vm e − · ¸¸ = 7.29 ms . ¹ I Lt VL C = 1093 µ F ______________________________________________________________________________________ Problem 9.35 Solution: Known quantities: The full-wave DC power supply of Figure P9.31; vline = 170 cos(ωt ) V , ω = 377 I L = 5 mA , VL = 10 V , Vr = 20 % , rad . s Find: a) The turns ratio. b) The capacitor C. Analysis: 1 1 Vr = 10 + 1 = 11 V , V L − min = V L − Vr = 10 − 1 = 9 V , 2 2 = Vs 0 cos(ωt ) . a) Vm = V L + Vs1 = Vs 2 KVL : − v s1 (t ) + v D1 + v L (t ) = 0 At ωt n= = 0, − Vso + VD −on + VL = 0 Vs 0 = VD −on + Vm = 0.7 + 1 = 6.7 V Vso = 0.0688 Vio 9.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 v s 2 (t ) + v D 2 + v L (t ) = 0 . b) KVL : At t = t 2 , Vso cos(ωt 2 ) = −VD 2−on − VL − min . t 2 = § V + VL − min 1 cos −1 ¨¨ − D 2− on Vso ω © The exponential discharge of the capacitor can be expressed: v L (t ) = v L (∞ ) + (v L (0) − v L (∞ ))e v L (t 2 ) = VL − min = Vm e − I Lt 2 VL C , − C=− t TC = 0 + (Vm − 0)e I Lt2 §V VL ln¨¨ L − min © Vm · ¸¸ ¹ − t RL C = Vm e − · ¸¸ = 6.76 ms . ¹ I Lt VL C = 16.84 µ F Note: An approximate but conservative value of C can be obtained by using the approximation: ωt 2 ≈ π . Then C≈− I L (ωt 2 ) §V ωVL ln¨¨ L − min © Vm · ¸¸ ¹ = 20.76 µ F . This value is conservative because it gives a smaller ripple voltage than that specified. ______________________________________________________________________________________ Problem 9.36 Solution: Known quantities: The full-wave rectifier of Figure P9.36, with a 12 V rms supply. Find: a) Sketch the input source voltage v S (t ) , and the output voltage v L (t ) , and state which diodes are on and which are off if the diodes have an offset voltage of 0.6 V and the frequency of the source is 60 Hz. b) Sketch the outpu voltage if R L = 1,000 Ω and a capacitor, placed across R L to provide some filtering, has a value of 8 µF. c) As part b, with the capacitance equal to 100 µF. Analysis: a) The input source voltage is shown below, together with the rectified load voltage. (12 V peak) 9.24 V rms = 16.97 G. Rizzoni, Principles and Applications of Electrical Engineering v (V) Problem solutions, Chapter 9 vL vS 16.97 15.77 D1D4 D2D3 D1D4 on on on . . b) The time constant, τ = CR, is: CR = 1000 8 10 -6 t = 8 ms. 1 1 = = 16.7 ms . f 60 Since the capacitor initial voltage is: vC (0 ) = 16.97 − 1.2 = 15.77 V , and the final value is vC (∞ ) = 0 V The period of the input sinusoid is: T= vC (t ) is given by: vC (t ) = 15.77 e − t τ . Therefore, at t = T, we have vC (T ) = 15.77 e −T τ = 1.96 V . The output waveform is shown below: v (V) vS vL 16.97 15.77 1.96 0 D1 D4 D 2D3 D1 D4 on on on . . t -6 (c) The time constant is CR = 1000 100 10 = 100 ms. Note that CR >> T; vC (0 ) = 16.97 − 1.2 = 15.77 V , and the final value is given by: vC (t ) = 15.77 e −t τ and therefore vC (T ) = 15.77 e The output waveform is shown below. 9.25 −T τ vC (∞ ) = 0 V ; vC (t ) is = 13.34 V . G. Rizzoni, Principles and Applications of Electrical Engineering v (V) vS Problem solutions, Chapter 9 vL 16.97 15.77 13.34 D1 D4 D2D3 D1 D4 on on on t ______________________________________________________________________________________ Problem 9.37 Solution: Known quantities: The full-wave bridge power supply of Figure P9.37; the diodes are 1N659 with a rated peak reverse voltage 50 V. n = 0.2941 , R L = 2.5 kΩ , C = 700 µ F , vline = 170 cos(ωt ) V , ω = 377 rad . s Find: a) The actual peak reverse voltage across the diodes. b) Explain why these diodes are or are not suitable for the specifications given. Analysis: vline = nvline = 50 cos(ωt ) V , Vso = 50 V At ωt = 0, the source voltage has the polarity shown; therefore, D1 and D3 are conducting, and D2 and D4 are off. At ωt = π, D1 and D3 are off and the voltage across them is the peak reverse voltage. KVL : − v s (t ) + v D1 + v L (t ) + v D 3 = 0 At ωt = 0: v L (0 ) = Vm = V D1 + V D 3 + Vso = 48.6 V a) = π: v D1 (π ) + v D 3 (π ) = Vso cos(π ) − v L (π ) = −Vso − Vm 1 = VD 3 , V D1,3 = (− 50 − 48.6 ) = −49.3 V 2 At ωt VD1 b) The diodes are not suitable because the rated and actual peak reverse voltages are about the same. ______________________________________________________________________________________ Problem 9.38 Solution: Known quantities: The full-wave bridge power supply of Figure P9.37; the diodes are T151 with a rated peak reverse voltage 10 V and are fabricated from Silicon. n = 0.0423 , VL = 5.1 V , Vr = 0.2 V , I L = 2.5 mA , vline = 156 cos(ωt ) V , ω = 377 rad . s Find: a) The actual peak reverse voltage across the diodes. b) Explain why these diodes are or are not suitable for the specifications given. 9.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Analysis: 1 Vso = nVio = 6.6 V , v s (t ) = 6.6 cos(ωt ) V . Vm = V L + Vr = 5.2 V 2 At ωt = 0, the source voltage has the polarity shown; therefore, D1 and D3 are conducting, and D2 and D4 are off. KVL : − v s (t ) − v D 4 − v L (t ) − v D 2 = 0 At ωt = 0: v D 2 (0 ) + v D 4 (0 ) = −v L (0 ) − V so cos (0 ) = −Vm − V so 1 VD 2 = VD 4 , V D 2 ,4 = (− 6.6 − 5.2 ) = −5.9 V 2 b) The diodes are suitable because the actual PRV (5.9 V) is significantly less than the rated PRV (10 V). a) ______________________________________________________________________________________ Problem 9.39 Solution: Known quantities: The full-wave bridge power supply of Figure P9.37; the diodes are fabricated from Silicon. φ = 23.66 deg , VL = 10 V , Vr = 1 V , I L = 650 mA , vline = 170 cos(ωt ) V , ω = 377 rad . s Find: The value of the average and the peak current through each diode. Analysis: Diodes D1 and D3 will conduct half of the load current and Diodes D2 and D4 will conduct the other half. Therefore: I D − ave = 1 I L = 325 mA 2 The waveforms of the diode currents are complex but can be roughly approximated as triangular (recall area of triangle = bh/2): I L = (I D1,3 + I D 2 ,4 )ave = I D − pk = 1 2π π ³ (I (ωt ) + I 2 0 D1,3 D 2 ,4 (ωt ))d (ωt ) = 2πI L 2πI L = = 1.28 A 1 1 φ φ+ φ 2 2 1 2π § φI D − pk φI D − pk ¨¨ + 2 © 2 · ¸¸ ¹ ______________________________________________________________________________________ Problem 9.40 Solution: Known quantities: The full-wave bridge power supply of Figure P9.37; the diodes are fabricated from Silicon. Vr = 0.6 V , I L = 85 mA , vline = 156 cos(ωt ) V , ω = 377 Find: a) The turns ratio n. 9.27 rad . s VL = 5.3 V , G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 b) The capacitor C. Analysis: a) First determine the maximum and minimum voltage across the load resistance and capacitor: 1 1 Vm = V L + Vr = 5.3 + 0.3 = 5.6 V , V L − min = V L − Vr = 5.3 − 0.3 = 5 V . 2 2 The amplitude of the supply voltage can now be determined. KVL : − Vso cos(ωt ) + v D1 + v D 3 + v L (t ) = 0 At t = 0, − Vso + V D −on + V D −on + V L = 0 Vs 0 = 2 V D −on + Vm = 0.7 + 0.7 + 5.6 = 7 V Vso = 0.04487 Vio b) KVL : v s (t ) + v D 2 + v D 4 + v L (t ) = 0 . At t = t 2 , Vso cos (ωt 2 ) = −V D −on − V D −on − V L − min . n= § 2V + VL −min · ¸¸ = 2.725 rad . ωt 2 = cos −1 ¨¨ − D 2−on Vso ¹ © The exponential discharge of the capacitor can be expressed: v L (t ) = v L (∞ ) + (v L (0) − v L (∞ ))e v L (t 2 ) = VL −min = Vm e − I Lωt 2 ωVL C , − t TC C=− = 0 + (Vm − 0)e I Lωt 2 §V ωVL ln¨¨ L − min © Vm · ¸¸ ¹ − t RL C = Vm e − I Lωt ωVL C = 1023 µ F ______________________________________________________________________________________ Problem 9.41 Solution: Known quantities: The full-wave bridge power supply of Figure P9.37; the diodes are fabricated from Silicon. Vr = 2.4 V , I L = 250 mA , vline = 156 cos(ωt ) V , ω = 377 VL = 10 V , rad . s Find: a) The turns ratio n. b) The capacitor C. Analysis: a) First determine the maximum and minimum voltage across the load resistance and capacitor: 1 1 Vm = V L + Vr = 10 + 1.2 = 11.2 V , V L − min = V L − Vr = 10 − 1.2 = 8.8 V . 2 2 The amplitude of the supply voltage can now be determined. KVL : − Vso cos(ωt ) + v D1 + v D 3 + v L (t ) = 0 At t = 0, − Vso + V D −on + V D −on + V L = 0 Vs 0 = 2 V D −on + Vm = 0.7 + 0.7 + 11.2 = 12.6 V 9.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Vso = 0.08077 Vio b) KVL : v s (t ) + v D 2 + v D 4 + v L (t ) = 0 . At t = t 2 , Vso cos (ωt 2 ) = −V D −on − V D −on − V L − min . n= § 2V + VL −min · ¸¸ = 1.505 rad . ωt 2 = cos −1 ¨¨ − D 2−on Vso ¹ © The exponential discharge of the capacitor can be expressed: v L (t ) = v L (∞ ) + (v L (0) − v L (∞ ))e v L (t 2 ) = VL −min = Vm e − I Lωt 2 ωVL C , − t TC C=− = 0 + (Vm − 0)e I L ωt 2 §V ωVL ln¨¨ L − min © Vm · ¸¸ ¹ − t RL C = Vm e − I Lωt ωVL C = 691.3 µ F ______________________________________________________________________________________ 9.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Section 9.5: DC Power Supplies, Zener Diodes and Voltage Regulations Problem 9.42 Solution: Known quantities: The piecewise characteristic that passes through the points (-10 V, -5 (1 V, 50 mA). µA), (0, 0), (0.5 V, 5 mA) and Find: Determine the piecewise linear model, and, using that model, solve for i and v. Analysis: Assume that the diode is forward-biased, and operating in the region between (0.5V, 5mA) and (1V, 50mA). If this is true, then the diode can be modeled by the resistance ∆v D 1 − 0.5 0.5 = = = 11.11 Ω in series with a battery having value −3 ∆i D (50 − 5)10 4510 −3 Vbat = 0.5 − 510 −3 (11.11) = 0.444 V . 2 − 0.444 Then, i = = 14 mA and v = 0.444 + 0.014(11.11) = 0.6 V . 111.11 rD = This solution is within the range initially assumed, justifying the assumption. ______________________________________________________________________________________ Problem 9.43 Solution: Known quantities: The output voltage at 5.6 V. Find: Determine the minimum value of RL for which the output voltage remains at just 5.6 V. Analysis: RLmin RLmin + 1800 (18) = 5.6 12.4RL min = 10080 RLmin = 812.9 Ω ______________________________________________________________________________________ Problem 9.44 Solution: Known quantities: The output voltage 25 V. The input voltage that varies from 35 to 40 mA. The maximum current, 250 mA, for the Zener diode used. V. The maximum load current is 75 Find: Determine the minimum and the maximum value for the series resistance. 9.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Analysis: i S max = 35 − 25 40 − 25 , i S min = , i Z min = 0 , i Z max = −250 mA , i L = 75 mA RS max RS min 15 = 46.2 Ω, 325×10−3 10 = = 133.3 Ω 75×10−3 iSmax = 250+ 75= 325 mA RSmin = iSmin = 0 + 75 = 75 mA RSmax ______________________________________________________________________________________ Problem 9.45 Solution: Known quantities: The i-v characteristic of a semiconductor; the minimum current at the "knee" of the curve, 5 maximum current, 90 mA. mA, and the Find: Determine the Zener resistance and Zener voltage of the diode. Analysis: The Zener voltage is evaluated at the middle of the rated region of operation, ie, midway between the knee of the curve and maximum rated current. VZ = 5 V The Zener resistance is determined from the slope of the i-v characteristic over the rated region of operation. rZ = ∆v v − vD2 1 1 2 = = 25 Ω = = D = D1 ∆ iD Slope ∆i D i D1 − i D 2 810 −3 ∆v D Note: The maximum Zener current is directly related to the maximum power the Zener diode can dissipate without breaking out in smoke and flames. In the specification sheet, either or both may be specified. ______________________________________________________________________________________ 9.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.46 Solution: Known quantities: The Zener diode used, the 1N5231B; the ripple component of the source voltage, obtained from a DC power supply, v S = V S + Vr , where VS = 20 V , Vr = 250 mV , R = 220 Ω , I L = 65 mA , VL = 5.1 V , RZ = 17 Ω , PRated = 0.5 W , I Z −min = 10 mA . Find: Determine the maximum rated current the diode can handle without exceeding its power limitation. Analysis: Only the rated power is required from the information given. Use the DC model of the Zener. Note that power is dissipated in the equivalent Zener resistance and in the equivalent source. Note the direction of the current and the polarity of the source. P = I Z2 RZ + I Z VZ Let P = PRated Then: I Z = I Z − max PRated = I Z2 − max R Z + I Z −maxVZ I Z −max 1§ V = ¨¨ − Z 2 © RZ · 1 §¨ § VZ ¸¸ ± ¨¨ ¹ 2 ¨© © RZ 2 · P ¸¸ − 4 Rated RZ ¹ · ¸ ¸ ¹ 12 = −150 ± 227.8 = 77.8 mA where the negative answer is rejected because the Zener current by definition flows in the direction shown. ______________________________________________________________________________________ Problem 9.47 Solution: Known quantities: The Zener diode used, the 1N963; iZK = 0.25 mA , rZK curve, VZ = 12 V , RZ = 11.5 Ω , PRated = 0.4 W , and at the knee of the = 700 Ω . Find: Determine the maximum rated current the diode can handle without exceeding its power limitation. Analysis: Only the rated power is required from the information given. Use the DC model of the Zener. Note that power is dissipated in the equivalent Zener resistance and in the equivalent source. Note the direction of the current and the polarity of the source. P = I Z2 RZ + I Z VZ . Let P = PRated Then: I Z = I Z −max , PRated = I Z2 − max R Z + I Z −maxVZ I Z −max 1§ V = ¨¨ − Z 2 © RZ · 1 §¨ § VZ ¸¸ ± ¨¨ ¹ 2 ¨© © RZ 2 · P ¸¸ − 4 Rated RZ ¹ · ¸ ¸ ¹ 12 = −521.5 ± 554.1 = 32.6 mA where the negative answer is rejected because the Zener current by definition flows in the direction shown. ______________________________________________________________________________________ 9.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Problem 9.48 Solution: Known quantities: VZ = 5 V ± 10% , RZ = 15 Ω , I Z −min = 3.5 mA , I Z −max = 80 mA , VS = 12 ± 3 V , I L = 70 ± 20 mA . Find: Determine the maximum and minimum value of R to maintain the Zener diode current within its specified limits. Analysis: Construct DC equivalent circuit: V L − VS + IZ + IL = 0 R KVL − VZ − I Z RZ + VL = 0 KCL Then: V L = VZ + I Z RZ V − V L V S − VZ − I Z RZ R= S = IZ + IL IZ + IL A maximum Zener current is caused by: Minimum value of R Maximum source voltage Minimum load current Minimum Zener voltage. Rmin = VS −max − VZ −min − I Z −max RZ = 71.54 Ω I Z −max + I L−min A minimum Zener current is caused by: Maximum value of R Maximum load current Rmax = VS −min − VZ −max − I Z −min RZ = 36.87 Ω I Z −min + I L−max Minimum source voltage Maximum Zener voltage. Note that the minimum value of R EXCEEDS the maximum value of R. This means that there is no value of R for which all the specifications will be met under all conditions. A value of R can be chosen but conditions may occur where the Zener current exceeds its maximum value or falls below its minimum value. This problem can be solved by: 1. Choosing another Zener diode with different minimum and maximum currents. 2. Relaxing the specifications on source voltage and load current. Note that the relationships between minimum and maximum values ALWAYS STARTS WITH THE QUESTION OF WHAT WILL CAUSE A MINIMUM OR MAXIMUM ZENER CURRENT !!! If the Zener current exceeds its maximum rated value, it will burn up; if it falls below its minimum value, the diode will leave the Zener region and cease to regulate the voltage. These represent WORST CASE conditions, a procedure frequently used in design. Note that a Zener diode acts like an electrical "surge tank" for current; however, the analogy is not exact. A liquid surge tank regulates, ie, maintains constant, pressure and fluid flow rates. It does this by temporarily storing excess fluid when flow rates increase or supplying extra fluid when flow rates decrease. The Zener "surge tank" primarily regulates load voltage as the load resistance [and therefore the load current] or source voltage [and therefore the current supplied by the source] changes. It temporarily "stores" 9.33 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 excess current when load current decreases or source voltage increases and "supplies" extra current when load current increases or source voltage decreases. ______________________________________________________________________________________ Problem 9.49 Solution: Known quantities: Circuit shown in Figure P9.46, VZ = 12 V ± 10% , RZ = 9 Ω , I Z −min = 3.25 mA , I Z −max = 80 mA , VS = 25 ± 1.5 V , I L = 31 ± 21.5 mA . Find: Determine the maximum and minimum value of R to maintain the Zener diode current within its specified limits. Analysis: Construct DC equivalent circuit: KCL V L − VS + I Z + I L = 0 , KVL − VZ − I Z RZ + VL = 0 R Then: V L = VZ + I Z RZ , R = VS − V L V S − VZ − I Z RZ = IZ + IL IZ + IL A maximum Zener current is caused by: Minimum value of R Minimum load current Rmin = VS −max − VZ −min − I Z −max RZ = 14.98 Ω I Z −max + I L −min A minimum Zener current is caused by: Maximum value of R Maximum load current Rmax = VS −min − VZ −max − I Z −min RZ = 184.2 Ω I Z −min + I L−max Maximum source voltage Minimum Zener voltage. Minimum source voltage Maximum Zener voltage. ______________________________________________________________________________________ Problem 9.50 Solution: Known quantities: The diode used, 1N4740A, in the circuit shown in Figure P9.46. I Z −min = 10 mA , I Z −max VZ = 10 V ± 5% , RZ = 7 Ω , = 91 mA , VS = 14 ± 2 V , R = 19.8 Ω , PRated = 1 W . Find: Determine the maximum and minimum value of the load current to maintain the Zener diode current within its specified limits. Analysis: Construct DC equivalent circuit: KCL V L − VS + I Z + I L = 0 , KVL − VZ − I Z RZ + VL = 0 R Then: 9.34 G. Rizzoni, Principles and Applications of Electrical Engineering V L = VZ + I Z RZ , I L = Problem solutions, Chapter 9 VS − V L V − V Z − I Z RZ − IZ = S − IZ R R A minimum and maximum Zener current is caused by [respectively]: Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage Substituting with these extreme values and solving for the load current: V S −min − VZ −max − I Z − min R Z − I Z −min = 62.22 mA R V − VZ −min − I Z − max R Z = S −max − I Z −max = 205.1 mA R I L −max = I L −min The minimum load current EXCEEDS the maximum load current. ______________________________________________________________________________________ Problem 9.51 Solution: Known quantities: The diode used, 1N963, in circuit shown in Figure P9.46. I Z −min = 2.5 mA , I Z −max VZ = 12 V ± 10% , RZ = 11.5 Ω , = 32.6 mA , VS = 25 ± 2 V , R = 470 Ω , PRated = 0.4 W . Find: Determine the maximum and minimum value of the load current to maintain the Zener diode current within its specified limits. Analysis: Construct DC equivalent circuit: KCL V L − VS + I Z + I L = 0 , KVL − VZ − I Z RZ + VL = 0 R Then: V L = VZ + I Z RZ , I L = VS − V L V − V Z − I Z RZ − IZ = S − IZ R R A minimum and maximum Zener current is caused by [respectively]: Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage Substituting with these extreme values and solving for the load current: V S −min − V Z −max − I Z −min RZ − I Z −min = 18.29 mA R V − VZ −min − I Z − max R Z = S −max − I Z −max = 1.07 mA R I L −max = I L −min ______________________________________________________________________________________ Problem 9.52 Solution: Known quantities: The diode used, 1N4740A, in circuit shown in Figure P9.46. I Z −min = 10 mA , I Z −max VZ = 10 V ± 5% , RZ = 7 Ω , = 91 mA , I L = 35 ± 10 mA , R = 80 Ω ± 5% , PRated = 1 W . 9.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Find: Determine the maximum and minimum value of the source voltage to maintain the Zener diode current within its specified limits. Analysis: Construct DC equivalent circuit: KCL V L − VS + I Z + I L = 0 , KVL − VZ − I Z RZ + VL = 0 R Then: V L = V Z + I Z R Z , VS = V L + R(I Z + I L ) = VZ + I Z R Z + R(I Z + I L ) Minimum and Maximum Zener currents are caused by [respectively]: Minimum source voltage. Maximum source voltage. Maximum load current. Minimum load current. Maximum Zener voltage. Minimum Zener voltage. Maximum R. Minimum R. Substituting with these extreme values and solving for the source voltage: VS −min = VZ −max + I Z −min RZ + Rmax (I Z −min + I L −max ) = 15.19 V VS −max = VZ −min + I Z −max RZ + Rmin (I Z −max + I L−min ) = 18.95 V ______________________________________________________________________________________ Problem 9.53 Solution: Known quantities: The diode used, 1N4740A, in circuit shown in Figure P9.46. The source voltage is obtained from a DC power supply that has a DC and a ripple component v S = V S + Vr , where VS = 16 V , Vr = 2 V , I Z −min = 10 mA , I Z −max = 91 mA , I L = 35 mA , RZ = 7 Ω , R = 80 Ω , VL = 10 V , VZ = 10 V . Find: Determine the ripple voltage across the load. Analysis: Construct the AC equivalent circuit. Suppress the DC component of all voltages and currents in the circuit. Suppress also the DC source in the Zener diode model: VL RZ R L = 285.7 Ω , Req = = 6.83 Ω IL RZ + R L Note: v S = Vr . Req Vl = V L = 157.4 mV R + Req RL = ______________________________________________________________________________________ Problem 9.54 Solution: Known quantities: The diode used, 1N5231B, in circuit shown in Figure P9.46. The source voltage is obtained from a DC power supply that has a DC and a ripple component v S = V S + Vr , where VS = 20 V , 9.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Vr = 250 mV , I Z −min = 10 mA , PRated = 0.5 W , I L = 65 mA , RZ = 17 Ω , R = 220 Ω , VL = 5.1 V , VZ = 5.1 V . Find: Determine the ripple voltage across the load. Analysis: Construct the AC equivalent circuit. Note that in this case the load resistance is much smaller than the Zener resistance and will cause a small reduction in the ripple voltage across the load. VL RZ R L = 78.46 Ω , Req = = 13.97 Ω IL RZ + R L Note: v S = Vr . Req Vl = V L = 14.93 mV R + Req RL = ______________________________________________________________________________________ Problem 9.55 Solution: Known quantities: The diode used, 1N970, in circuit shown in Figure P9.46. The source voltage is obtained from a DC power supply that has a DC and a ripple component v S = V S + Vr , where VS = 30 V , Vr = 3 V , I Z −min = 1.5 A , I Z −max = 15 A , I L = 8 A , RZ = 33 Ω , R = 1 Ω , VL = 24 V , VZ = 24 V . Find: Determine the ripple voltage across the load. Analysis: Construct the AC equivalent circuit. Note that in this case the load resistance is much smaller than the Zener resistance and will cause a small reduction in the ripple voltage across the load. VL RZ R L = 3 Ω , Req = = 2.750 Ω IL RZ + R L Note: v S = Vr . Req Vl = V L = 2. 2 V R + Req RL = ______________________________________________________________________________________ 9.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Section 9.6: Signal Processing Applications Section 9.7: Photodiodes Problem 9.56 Solution: Known quantities: The range for the input voltage 0 ≤ v ≤ 10 V for the circuit of Figure P9.56. Find: Determine the i-v characteristic of the circuit. Analysis: With the variables defined in the circuit below, we can compute the following currents: v v v , I2 = for v ≥ 4 V , I 3 = for v ≥ 6 V 100 100 100 For 0 ≤ v ≤ 4 V , I = 0.01v For 4 ≤ v ≤ 6 V , I = 0.02v − 0.04 For 6 ≤ v ≤ 10 V , I = 0.03v − 0.1 I1 = The resulting i-v characteristic is shown below: i (mA) 100 80 60 40 20 6 8 10 v (V) 0 2 4 ______________________________________________________________________________________ Problem 9.57 Solution: Known quantities: The input voltage waveform and the circuit of Figure P9.57. Find: Determine the output voltage. Analysis: For vin < 50.7 V , vout = vin . When vin ≥ 50.7 V , v out = 50.7 + ( ) 0.6 v in = 50.7 + 6.085 × 10 −3 vin 98 + 0.6 The input and output voltage waveforms are sketched below: 9.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 ______________________________________________________________________________________ Problem 9.58 Solution: Known quantities: The circuit of Figure P9.58 where the switch is close at time Find: The currents a) t = t1 . I S , I B , I SW for the following conditions: − 1 t = t , b) t = t1+ , c) what will happen to the battery after the switch closes. Repeat all considering the circuit of Figure P9.58(b) if the diode has an offset voltage of 0.6 V. Analysis: For Figure P9.58 (a): a) At t = t1− , before the switch S1 closes, we have I SW = 0 , I S = I B = b) At VS − VBattery RS + R B = 0.31 A t = t1+ , we have I S = 13 A , I B = −0.96 A , I SW = I S − I B = 13.96 A c) The battery voltage will drop quickly because of the small resistance in the circuit. For Figure P9.58 (b): a) At t = t1− , we have 9.39 G. Rizzoni, Principles and Applications of Electrical Engineering I SW = 0 , I S = I B = b) At VS − VBattery − Vγ RS + R B Problem solutions, Chapter 9 = 0.25 A t = t1+ , we have I S = I SW = 13 A , IB =0 c) The battery will not be drained, because of the large reverse resistance of the diode. ______________________________________________________________________________________ Problem 9.59 Solution: Known quantities: The circuit in Figure P9.59. The diode has Vγ = 0.7 V . The input voltage is sinusoidal with an amplitude of 6, 1.5, 0.4 V and zero average value. Find: Determine the average value of the output voltage. Analysis: The capacitor will charge to VS − 0.7 V . 6 V: Therefore, the input sine wave will be shifted up 5.3 V to produce the output. As a result, after the cycle (the capacitor builds up its stored charge during the third quarter cycle), the average value of the output will be 5.3 V. 1.5 V: Therefore, the input sine wave will be shifted up 0.8 V to produce the output. As a result, after the cycle (the capacitor builds up its stored charge during the third quarter cycle), the average value of the output will be 0.8 V. 0.4 V: Since the input is less than 0.7 V, the capacitor does not charge and the output voltage is 0 V. ______________________________________________________________________________________ Problem 9.60 Solution: Known quantities: The LED circuit in Figure 9.14. Diode operating point: VLED =1.7 V; ILED=20 mA; VS=5 V. Find: Determine the LED power consumption and the power required by the voltage source. Analysis: The power consumption is PLED = VLED I LED = 34 mW The power required at the source is PS = VS I LED = 100 mW ______________________________________________________________________________________ Problem 9.61 Solution: Known quantities: The LED circuit in Figure 9.14. Diode operating point: VLED =1.5 V; ILED=30 mA; VS=5 V. 9.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 9 Find: Determine the LED power consumption and the power required by the voltage source. Analysis: The power consumption is PLED = VLED I LED = 45 mW The power required at the source is PS = VS I LED = 150 mW 9.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Chapter 10 Instructor Notes Chapter 10 introduces bipolar junction transistors. The material on transistors has been reorganized in this 4th Edition, and is now divided into two independent chapters, one on bipolar devices, and one on field-effect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect devices. Section 10.2 introduces the fundamental ideas behind the operation of bipolar transistors, and illustrates the calculation of the state and operating point of basic transistor circuits. The discussion of the properties of the BJT in Section 10.2 is centered around a description of the base and collector characteristics, and purposely avoids a detailed description of the physics of the device, with the intent of providing an intuitive understanding of the transistor as an amplifier and electronic switch. Example 10.5 introduces the use of Electronics Workbench (EWB) as a tool for analyzing electronics circuits; the CDROM contains an introduction to the software package and a number of solved problems. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 535-537). Example 10.7 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 539-541) provide two application examples that include EWB solutions. Section 8.4 introduces the analysis of BJT switches and presents TTL gates. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. Learning Objectives 1. Understand the basic principles of amplification and switching. Section 1. 2. Understand the physical operation of bipolar transistors; determine and select the operating point of a bipolar transistor circuit; understand the principle of small signal amplifiers. Section 2. 3. Understand the large-signal model of the bipolar transistor, and apply it to simple amplifier circuits. Section 3. 4. Understand the operation of bipolar transistor as a switch and analyze basic analog and digital gate circuits. Section 4. 10.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Section 10.2: Operation of the Bipolar Junction Transistor Problem 10.1 Solution: Known quantities: Transistor diagrams, as shown in Figure P10.1: (a) pnp, VEB = 0.6 V and VEC = 4.0 V (b) npn, VCB = 0.7 V and VCE = 0.2 V (c) npn, VBE = 0.7 V and VCE = 0.3 V (d) pnp, VBC = 0.6 V and VEC = 5.4 V Find: For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward or reverse biased, and determine the operating region. Analysis: (a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. (b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased. VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the cutoff region. (c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the saturation region. (d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased. VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in the active region. ______________________________________________________________________________________ Problem 10.2 Solution: Known quantities: Transistor type and operating characteristics: a) npn, VBE = 0.8 V and VCE = 0.4 V b) npn, VCB = 1.4 V and VCE = 2.1 V c) pnp, VCB = 0.9 V and VCE = 0.4 V d) npn, VBE = - 1.2 V and VCB = 0.6 V Find: The region of operation for each transistor. Analysis: a) Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus, the CB junction is forward-biased. Therefore, the transistor is in the saturation region. 10.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased. VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. c) VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased. VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in the b) saturation region. = - 1.2 V, the BE junction is reverse-biased. VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff region. d) With VBE ______________________________________________________________________________________ Problem 10.3 Solution: Known quantities: The circuit of Figure P10.3. β= IC = 100 . IB Find: The operating point and the state of the transistor. Analysis: VBE = 0.6 V and the BE junction is forward biased. V − VBE 12 − 0.6 = = 13.9µA I B = CC R1 820 I C = β ⋅ I B = 1.39 mA Writing KVL around the right-hand side of the circuit: − VCC + I C RC + VCE + I E RE = 0 VCE = VCC − I C RC − (I B + I C )RE VBC = VBE VCE > VBE = 12 − (1.39)(2.2) − (1.39 + 0.0139)(0.910) = 7.664 V + VCE = 0.6 + 7.664 = 8.264 V The transistor is in the active region. ______________________________________________________________________________________ 10.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.4 Solution: Known quantities: The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitter-base and collector-base junctions: IE = 6 mA, IB = 0.1 mA and VEB = 0.65 V, VCB = 7.3 V. Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as P = VCE I C + VBE I B . Analysis: a) VCE = VCB - VEB = 7.3 - 0.65 = 6.65 V. b) IC = IE - IB = 6 - 0.1 = 5.9 mA. c) The total power dissipated in the transistor can be found to be: P ≈ VCEIC = 6.65×5.9×10-3 = 39 mW. ______________________________________________________________________________________ Problem 10.5 Solution: Known quantities: The circuit of Figure P10.5, assuming the BJT has Vγ = 0.6 V . Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit, Then, on the left-hand side, assuming β >> 1 : (0.6 + 15) ) = −520µA § V + 15 · I E = −¨ BE ¸=− 30000 © 30000 ¹ VCB = 10 − I C ⋅ RC − 10 + I C RC + VCB = 0 = 10 − (−520) ⋅ (15) ⋅ 10 −3 = 17.8 V ______________________________________________________________________________________ Problem 10.6 Solution: Known quantities: The circuit of Figure P10.6, assuming the BJT has VBE = 0.6 V and β =150. Find: The operating point and the region in which the transistor operates. 10.4 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: Define RC Problem solutions, Chapter 10 = 3.3kΩ, RE = 1.2kΩ, R1 = 62kΩ, R2 = 15kΩ, VCC = 18 V By applying Thevenin’s theorem from base and mass, we have RB = R1 || R2 = 12.078 kΩ VBB = IB = R2 VCC ≅ 3.5 V R1 + R2 VBB − VBE ≅ 15 µA RB + RE (1 + β ) I C = β I B = 2.25 mA VCE = VCC − RC I C − RE I E = 18 − 3300 ⋅ 2.25 ⋅ 10 −3 − 1200 ⋅ 151 ⋅ 15 ⋅ 10 −6 = 7.857 V From the value of VCE it is clear that the BJT is in the active region. ______________________________________________________________________________________ Problem 10.7 Solution: Known quantities: The circuit of Figure P10.7, assuming the BJT has Vγ = 0.6 V . Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit, − VCC + I E RE + VEB = 0 VCC − VEB 20 − 0.6 = = 497.4µA . Since β >> 1 , I C ≈ I E = 497.4µA RE 39 ⋅10 3 Applying KVL to the left-hand side: VCB + I C RC − VDD = 0 IE = VCB = VDD − I C RC = 20 − 497.4 ⋅ 20 ⋅10 −3 = 10.05 V ______________________________________________________________________________________ Problem 10.8 Solution: Known quantities: The circuit of Figure P10.7, assuming the emitter resistor is changed to 22 kΩ and the BJT has Vγ = 0.6 V . Find: The operating point of the transistor. Analysis: IE = VCC − VEB 20 − 0.6 = = 881.8µ A , I C ≈ I E = 881.8µA RE 22 ⋅10 3 10.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 VCB = VDD − I C RC = 20 − 881.8 ⋅ 20 ⋅10 − 3 = 2.364 V ______________________________________________________________________________________ Problem 10.9 Solution: Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9. Find: a) The ratio IC / IB for VCE = 10 V and IB = 100 µA, 200 µA, and 600 µA b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for IB = 500 µA. Analysis: = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. -3 b) For IB = 500 µA, and if we consider an average β from a., we have IC = 159·500 10 = 79.5 mA. The power dissipated by the transistor is P = VCE I C + VBE I B ≈ VCE I C , therefore: 0 .5 P = = 6.29 V . VCE ≈ I C 79.5 ⋅ 10 −3 a) For IB ______________________________________________________________________________________ Problem 10.10 Solution: Known quantities: Figure P10.10, assuming both transistors are silicon-based with Find: a) IC1, b) IC2, β = 100 . VC1, VCE1. VC2, VCE2. Analysis: a) From KVL: − 30 + I B1 RB1 + VBE1 = 0 I B1 = I C1 = β ⋅ I B1 = 3.907 mA VC1 = 30 − RC1 I C1 = 30 − 3.907 ⋅ 6.2 = 5.779 V VCE1 = VC1 = 5.779 V . b) Again, from KVL: and I C 2 − 5.779 + VBE 2 + I E 2 RE 2 = 0 § β · § 100 · ¸¸ = 1.081 ⋅ ¨ = I E 2 ¨¨ ¸ = 1.07 mA . © 101 ¹ © β +1¹ 10.6 30 − 0.7 = 39.07µA 750 ⋅ 10 3 I E2 = 5.779 − 0.7 = 1.081 mA 4.7 ⋅ 10 3 G. Rizzoni, Principles and Applications of Electrical Engineering Also, Problem solutions, Chapter 10 − 30 + I C 2 ( RC 2 + RE 2 ) + VCE 2 = 0 VCE 2 = 30 − (1.07) ⋅ (20 + 4.7) = 3.574 V . Finally, I C 2 = 30 − VC 2 RC 2 VC 2 = 30 − (1.07) ⋅ (20) = 8.603 V . ______________________________________________________________________________________ Problem 10.11 Solution: Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 536. Find: The operating point of the transistor in Figure P10.11, and the value of β at this point. Analysis: Construct a load line. Writing KVL, we have: − 50 + 5000 ⋅ I C + VCE = 0 . Then, if I C = 0 , VCE = 50 V ; and if VCE = 0 , I C = 10 mA . The load line is shown superimposed on the collector characteristic below: load line The operating point is at the intersection of the load line and the Therefore, I B = 20µA line of the characteristic. I CQ ≈ 5 mA and VCEQ ≈ 20 V . Under these conditions, an 5 µA increase in I B yields an increase in I C of approximately 6 − 5 = 1 mA . Therefore, ∆I C 1 ⋅ 10 −3 β≈ = = 200 ∆I B 5 ⋅ 10 −6 The same result can be obtained by checking the hFE gain from the data-sheets corresponding to 5 mA. 10.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 _____________________________________________________________________________________ Problem 10.12 Solution: Known quantities: The circuit shown in Figure P10.12. With reference to Figure 10.20, assume Vγ = 0.6 V , Vsat = 0.2 V . Find: The operating point of the transistor, by computing the ratio of collector current to base current. Analysis: 10 − 0.2 = 9.8 mA RC 5 .7 − 0 .6 = Vγ = 0.6 V , therefore I B = = 102µA RB VCE = Vsat = 0.2 V , therefore I C = VBE I C 9.8 ⋅10 −3 = = 96.08 << β I B 102 ⋅ 10 −6 ______________________________________________________________________________________ Problem 10.13 Solution: Known quantities: For the circuit shown in Figure P10.13, Find: a) VB b) c) IB IE Analysis: a) VEB = VE VE = 1V and Vγ = 0.6 V . d) IC e) β f) α . − VB = Vγ = 0.6 V VB = VE − VEB = 1 − 0.6 = 0.4 V 10.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 VB 0 .4 = = 20µA RB 20 ⋅ 10 3 5 − VE 5 − 1 c) I E = = = 800µA RE 5000 d) I C = I E − I B = 800 − 20 = 780µA b) e) f) IB = I C 780 = = 39 IB 20 I 780 α= C = = 0.975 . I E 800 β= ______________________________________________________________________________________ Problem 10.14 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 20 V β = 130 R1 = 1.8 MΩ RL = 1kΩ RS = 0.6 kΩ vS = cos 6.28 ×103 ⋅ t mV . [ R2 = 300 kΩ ] Find: The Thèvenin equivalent of the part of the circuit containing terminals of RC = 3 kΩ RE = 1kΩ R1 , R2 , and VCC with respect to the R2 . Redraw the schematic using the Thèvenin equivalent. Analysis: Extracting the part of the circuit specified, the Thèvenin equivalent voltage is the open circuit voltage. The equivalent resistance is obtained by suppressing the ideal independent voltage source: Note that VCC must remain in the circuit because it supplies current to other parts of the circuit: ______________________________________________________________________________________ 10.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.15 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 15 V RL = 1.5 kΩ β = 100 R1 = 68 kΩ RS = 0.9 kΩ [ R2 = 11.7 kΩ RC = 200Ω 3 ] RE = 200Ω vS = cos 6.28 ×10 ⋅ t mV . Find: VCEQ and the region of operation. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network in the base circuit: 15 ⋅11.7 V CC R 2 = = 2.202 V 68 + 11.7 R1 + R2 R1 R 2 = 68 ⋅11.7 = 9.982 kΩ R B = Req = R1 + R 2 68 + 11.7 VD : V BB = V TH = Suppress V CC : Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ I EQ = (β + 1) I BQ Then: 10.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 KVL: - V BB + I BQ RB +V BEQ+ I EQ RE = 0 - V BB + I BQ RB +V BEQ+ [ β +1 ] I BQ RE = 0 2.202- 0.7 V BB - V BEQ = 49.76 µA = I BQ = RB + [ β +1 ] RE 9982+ (100 + 1)(200) −6 I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA −6 I EQ = (β + 1)⋅ I BQ = (100 + 1) ⋅ 49.76 ⋅ 10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V CC - I CQ RC - I EQ R E = 15 − 4.976 ⋅ 0.2 − 5.026 ⋅ 0.2 = 13.00 V The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. If the collector-emitter voltage were less than its saturation value, the transistor would be operating in its "saturation" or "ohmic" region. In this case, the collector-emitter voltage is equal to its saturation value (0.3 V for Silicon). The solution for the base current remains valid. However, the parameter β and the solution for the collector and emitter curents become invalid. In the saturation region, the base-emitter junction is still forward biased and its voltage remains the same. A solution for the collector and emitter currents is possible using the collector-emitter saturation voltage (0.3 V for Silicon) in a KVL. In the saturation region, the base current has no control over the collector or emitter currents (since β is invalid). Therefore, amplification is impossible. ______________________________________________________________________________________ 2. Problem 10.16 Solution: Known quantities: For the circuit shown in Figure P10.14: VCC = 15 V β = 100 R1 = 68 kΩ RL = 1.5 kΩ RS = 0.9 kΩ vS = cos 6.28 ×103 ⋅ t mV . [ R2 = 11.7 kΩ RC = 4 kΩ ] Find: VCEQ and the region of operation. Analysis: 10.11 RE = 200Ω G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network in the base circuit: Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ I EQ = (β + 1) I BQ KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 - V BB + I BQ R B + V BEQ + [ β + 1 ] I BQ R E = 0 I BQ = 2.202 - 0.7 V BB - V BEQ = 49.76 µA = 9982 + (100 + 1)(200) RB + [ β +1 ] RE −6 I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA −6 I EQ = (β + 1)⋅ I BQ = (100 + 1) ⋅ 49.76 ⋅10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V CC - I CQ RC - I EQ R E = 15 − 4.976 ⋅ 4 − 5.026 ⋅ 0.2 = − 5.91 V The collector-emitter voltage is less (more negative) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was incorrect and the solution is not valid. The device is operating in the saturation region with a saturation collector-emitter voltage equal to 0.3 V. The solutions for the collector and emitter currents are invalid. THE VALID SOLUTION: KVL : - I EQ R E - vCE -SAT - I CQ RC + V CC = 0 I EQ = I BQ + I CQ - [ I BQ + I CQ ] R E - vCE -SAT - I CQ RC + V CC = 0 I CQ = 15 − 49.76 ⋅ 0.2 ⋅10 −3 − 0.3 V CC - I BQ R E - vCE - SAT = = 3.498 mA 200 + 4000 R E + RC The solution for the base current is valid. The value of beta given is not valid. ______________________________________________________________________________________ 10.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.17 Solution: Known quantities: For the circuit shown in Figure P10.17: VCC = 12 V β = 130 R1 = 82 kΩ R2 = 22 kΩ RE = 0.5 kΩ RL = 16Ω . Find: VCEQ at the DC operating point. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit: 12 ⋅ 22 V CC R2 = = 2.538 V R1 + R2 82 + 22 R1 R2 = 82 ⋅ 22 = 17.35 kΩ R B = Req = 82 + 22 R1 + R2 VD : V BB = V TH = V OC = Suppress V CC : Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ ≈ 700 mV [Si] I EQ = [ β +1 ] I BQ KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 - V BB + I BQ R B + V BEQ + [ β + 1 ] I BQ R E = 0 I BQ = 2.538 − 0.7 V BB - V BEQ = = 22.18µA 17350 + (130 + 1) ⋅ 500 R B + (β + 1) ⋅ R E −6 I EQ = (β + 1 ) I BQ = (130 + 1 ) ⋅ 22.18 ⋅10 = 2.906 mA KVL : - I EQ R E - V CEQ + V CC = 0 V CEQ = V CC - I EQ R E = 12 − 2.906 ⋅ 0.5 = 10.55 V The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. ______________________________________________________________________________________ Problem 10.18 Solution: Known quantities: For the circuit shown in Figure P10.18: 10.13 G. Rizzoni, Principles and Applications of Electrical Engineering VCC = 12 V RL = 6 kΩ VEE = 4 V Rs = 0.6 kΩ β = 100 [ RB = 100 kΩ 3 ] Problem solutions, Chapter 10 RC = 3 kΩ RE = 3 kΩ vS = cos 6.28 ×10 ⋅ t mV . Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the base-emitter junction is forward biased and: V BEQ ≈ 700 mV [Si] I CQ = β ⋅ I BQ I EQ = (β + 1) I BQ KVL : - V EE + I BQ R B + V BEQ + I EQ R E = 0 4 - 0.7 V EE - V BEQ = 8.189 µ A = I BQ = 100000 + (100 + 1)(3000) RB + [ β +1 ] RE I CQ = β ⋅ I BQ = (100 ) ⋅ 8.189 ⋅10 −6 = 818.9 mA I EQ = (β + 1) ⋅ I BQ = (100 + 1) ⋅ 8.189 ⋅ 10 −6 = 827.0 mA KVL : + V EE - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V EE + V CC - I CQ RC - I EQ R E = 4 + 12 − 818.9 ⋅ 3 ⋅ 10 −3 − 827.0 ⋅ 3 ⋅10 −3 = 11.06 V The collector-emitter voltage is greater (more positive) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. DC power may be supplied to an npn BJT circuit by connecting the positive terminal of a DC source to the collector circuit, or, by connecting the negative terminal of a DC source to the emitter circuit, or, as was done here, both. 2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and positive to emitter. ______________________________________________________________________________________ Problem 10.19 Solution: Known quantities: For the circuit shown in Figure P10.19: VCC = 12 V β = 130 RB = 325 kΩ RC = 1.9 kΩ 10.14 RE = 2.3 kΩ G. Rizzoni, Principles and Applications of Electrical Engineering RL = 10 kΩ Rs = 0.5 kΩ [ Problem solutions, Chapter 10 ] vS = cos 6.28 ×103 ⋅ t mV . Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the base-emitter junction is forward biased. The base and collector currents both flow through the collector resistor in this circuit. V BEQ ≈ 700 mV [Si] I CQ = β I BQ I EQ = [ β + 1 ] I BQ KCL : I BQ + I CQ - I RC = 0 I RC = I CQ + I BQ = β I BQ + I BQ = [ β + 1 ] I BQ KVL : - I EQ R E - V BEQ - I BQ R B - I RC RC + V CC = 0 - [ β + 1 ] I BQ [ R E + RC ] - V BEQ - I BQ R B + V CC = 0 12 − 0.7 V CC - V BEQ = = 12.91µA [325 + (130 + 1) ⋅ (2.3 + 1.9) ]⋅103 R B + [ β + 1 ] [ R E + RC ] = I EQ = [ β + 1 ] I BQ = (130 + 1) ⋅12.91 ⋅10 −6 = 1.691mA I BQ = I RC KVL : - I EQ R E - V CEQ - I RC RC + V CC = 0 V CEQ = V CC - I RC RC - I EQ R E = 12 − 1.691 ⋅1.9 − 1.691 ⋅ 2.3 = 4.896 V The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. ______________________________________________________________________________________ Problem 10.20 Solution: Known quantities: For the circuit shown in Figure P10.19: VCC = 15 V C = 0.5µF β = 170 RL = 1.7 kΩ Rs = 70Ω vS = cos 6.28 ×103 ⋅ t mV . [ RB = 22 kΩ ] 10.15 RC = 3.3 kΩ RE = 3.3 kΩ G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Find: VCEQ and the region of operation. Analysis: When β is very high (say, greater than 50) the "high beta approximation" can be used, i.e.: IC ≈ I E ≈ β I B Using this approximation in this problem: I BQ ≈ 12.5µA I CQ ≈ I EQ ≈ 2.125 mA V CEQ ≈ 0.975 V The collector-emitter voltage is greater than the saturation value (but not by much) so the original assumption (operation in the active region) and the solution are valid. However, when a signal is introduced into the circuit, the collector-emitter voltage will vary about its Q point value. Since the Q point is close to saturation, saturation and severe distortion due to clipping is likely. ______________________________________________________________________________________ Problem 10.21 Solution: Known quantities: • For the circuit shown in Figure P10.14: VCC = 15 V RE = 710Ω RL = 3 kΩ • • C = 0.47µ F R1 = 220 kΩ R2 = 55 kΩ Rs = 0.6 kΩ vS = Vi 0 ⋅ sin (ωt ) RC = 3 kΩ Vi 0 = 10 mV . DC operating point: I BQ = 19.9µ A V CEQ = 7.61 V Transfer characteristic and β of the npn silicon transistor: v BE V BEQ + vbe IC ≈ I s eVT = I s e VT β = 100 • Device i-v characteristic plotted in Figure P9.21. Find: a) The no-load large signal gain [v0/vi]. b) Sketch the waveform of the output voltage as a function of time. c) How the output voltage is distorted compared with the input waveform. Analysis: a) Simplify the circuit by determining the Thèvenin equivalent of the biasing network in the base circuit. VD : V BB = V TH = 15 ⋅ 55 V CC R2 = = 3.00 V R1 + R2 220 + 55 Suppress V CC : 3 R1 R2 = 220 ⋅ 55 ⋅10 = 44.0 kΩ R B = Req = 220 + 55 R1 + R2 First, determine the collector and emitter Q point: 10.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = − 2.010 ⋅ 0.710 − 1.990 ⋅ 3 + 15 = 7.603 V −6 I CQ = β I BQ = 100 ⋅19.9 ⋅10 = 1.99 mA −6 I EQ = [ β + 1 ] I BQ = 101 ⋅19.9 ⋅10 = 2.01mA Now, determine the DC load line (circuit characteristic). KVL : - i E R E - vCE - iC RC + V CC = 0 iC = iE = β +1 iC β V CC - vCE V CC - vCE = β +1 3717 R E + RC β If : vCE = V CC = 15 V [Intercept 1] = 0 ° =® 15 °̄ = 3717 = 4.035 mA If : vCE = 0 [Intercept 2] Plot the DC load line on the same plot as the transistor i-v characteristic. Note that the collector current is a linear function of the collector-emitter voltage. Only two points (the two intercepts shown as Points 1 and 2 on the plot) are required to plot a linear function. The DC load line should pass through the Q point. The AC load line can now be plotted. It will pass through the Q point as does the DC load line; however, its intercepts cannot be directly determined. Instead, its slope will be determined. The slope and the Q point can then be used to plot the AC load line. Slope AC = - 1 = - RC 1 mA 2mA ∆ iC = - 0.3333 = = 3000 V 6V ∆ vCE For AC the capacitors act as short circuits. The bypass capacitor in parallel with the emitter resistor shorts out any AC voltage across it. The collector coupling capacitor acts as a short and connects the load resistor to the circuit. However, no-load conditions are specified, i.e., the load current is zero or the load resistor is replaced by an open circuit. With no signal, the circuit operates at its Q point. When a signal is introduced, the operating point changes along the AC load line as a function of the signal voltage. Unfortunately, the device i-v characteristic is plotted as a function of input (base) current instead of voltage. But, there is the static characteristic that relates collector current to base-emitter voltage: v BE iC = I s e V T = I s e V BEQ + v be VT = [ Is e V BEQ VT v be v be ] eV T = I CQ eV T v be iC = I CQ vbe = iB = eV T I BQ eV T β β The base-emitter signal voltage is related to the input signal voltage. At sufficiently high frequencies (the "mid"-frequency range) the capacitors can be modeled as short circuits for AC. This means that there is no AC voltage across either the AC coupling (DC blocking) capacitor or the capacitor bypassing the emitter resistor. Using superposition to sum only AC voltages around the loop: 10.17 G. Rizzoni, Principles and Applications of Electrical Engineering KVL : - vi + vc + vbe + ve = 0 iB = vbe I BQ eV T vbe = vi vc = ve = 0 = [ 19.9 µA ] Problem solutions, Chapter 10 vi e 26 mV This transfer function can now be used to give the base current as a function of the input voltage. The intersection of the base current curves with the AC load line give the corresponding values of the collector current and collector-emitter voltage (points 3, Q and 4 on the plot): ωt [rad] 0,π,2π π/2 3π/2 vi [mV] 0 +10 -10 iB [µ µA] 19.9 29.23 13.54 iC [mA] 1.99 2.92 1.35 vCE [V] 7.61 4.55 9.15 vo [V] 0 -3.06 +1.54 The capacitor connecting the load to the circuit is a DC blocking [and AC coupling] capacitor. Remember that for no load conditions, the load resistor is assumed to be open. Again using superposition to sum only AC voltages around the loop and modeling the capacitors as shorts: - ve + vce + vc + vo = 0 ve = vc = 0 vo = vce = vCE - V CEQ The no-load voltage gain can be determined using the maximum excursion in the input and output voltages: Avo = ∆ vo (− 1.54) − (+ 3.06) = −230 = ∆ vi 10 ⋅10 −3 − − 10 ⋅10 −3 ( ) ( ) When the input voltage becomes more positive, the output voltage becomes more negative and vice versa. This is called "phase inversion" and is why the gain is negative. In normal amplification this is not important. If the loading effect of the load resistance is included, the AC load line will be steeper and the gain will be reduced. b) A plot is a large number of calculated points plotted and connected by a smooth curve. In a sketch, only the most significant points [maxima, minima, intercepts, etc] are calculated, plotted, and connected by a carefully drawn smooth curve. c) The output waveform is very distorted, i.e., it is not a true linearly amplified copy of the input waveform. This is most obvious when the positive and negative peaks of the two curves are compared: Input voltage +10 mV -10 mV Output voltage -3.06 V +1.94 V The two peaks of the input voltage are equal; however, the positive peak is amplified less than the negative peak. The nonlinear (exponential) behavior of the transistor causes the amplification to be nonlinear and the output waveform to be very distorted. This is characteristic of large signal amplifiers. Even more serious distortion could occur if the input voltage (also called the "excitation" or "drive") were increased so that saturation or cutoff occurs. This causes "clipping" and a severely distorted output. ______________________________________________________________________________________ 10.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Section 10.3: BJT Large-Signal Model Problem 10.22 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B = 5 mA, RB = 1 kΩ, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, β = 95, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW Find: Range of RC. Analysis: RC = VCC − Vγ LED − VCEsat I LED ≤ 5 − 1.4 − 0.2 = 340 Ω 0.01 From the maximum power I LED max = RC > Pmax 0.1 = = 71 mA Vγ LED 1.4 VCC − Vγ LED − VCEsat I LED max = 47 Ω Therefore, RC ∈[47, 340] Ω ______________________________________________________________________________________ Problem 10.23 Solution: Known quantities: For the circuit shown in Figure 10.26: VD = 1.1 V, RB = 33 kΩ, VCC = 12 V, VBE = 0.75 V, VCEQ = 6 V, β = 188.5, RS = 500 Ω Find: The resistance RC. Analysis: The current through the resistance RB is given by IB = VD − VBEQ RB = 1.1 − 0.75 = 10.6 µA 33000 The current through RS is IS = VCEQ − VD RS = 6 − 1.1 = 9.8 mA 500 It follows that the current through the resistance RC is I CQ = β I B + I S = 11.8 mA Finally, RC = VCC − VCEQ I CQ = 12 − 6 = 508.5 Ω 0.0118 ______________________________________________________________________________________ 10.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Problem 10.24 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 5 mA, RC = 340 Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, β = 95, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW Find: Range of RB. Analysis: If the BJT is in saturation IC = VCC − Vγ LED − VCEsat RC = 10 mA In order to guarantee that the BJT is in saturation RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 5 − 0.7 = 40.85 kΩ 0.01 95 = 860 Ω ______________________________________________________________________________________ Problem 10.25 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 5 mA, RB = 10 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: IB = β min Von − Vγ RB I = LED min IB 4.3 = 0.43 mA 10000 0.01 = = 23.25 0.43 ⋅ 10 −3 = ______________________________________________________________________________________ Problem 10.26 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 3.3 V, I B max = 5 mA, RB = 10 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, Vγ LED = 1.4 V, I LED ≥ 10mA, Pmax = 100mW 10.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Find: Minimum value of β that will ensure the correct operation of the LED. Analysis: IB = β min Von − Vγ RB I = LED min IB 3.3 − 0.7 = 0.26 mA 10000 0.01 = = 38.5 0.26 ⋅ 10 −3 = ______________________________________________________________________________________ Problem 10.27 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 1 mA, RB = 1 kΩ, R = 12 Ω, VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: Minimum value of β that will ensure the correct operation of the fuel injector. Analysis: IC = β min VCC − VCEsat 13 − 1 = = 1A 12 R I 1 = C = = 1000 I B max 1 ⋅ 10 −3 ______________________________________________________________________________ Problem 10.28 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 5 V, I B max = 1 mA, β = 2000, R = 12 Ω, VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: The range of RB that will ensure the correct operation of the fuel injector. 10.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Analysis: If the BJT is in saturation IC = VCC − VCEsat = 1A R Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 5 − 0.7 = 8.6 kΩ 1 2000 = 4.3 kΩ ______________________________________________________________________________________ Problem 10.29 Solution: Known quantities: For the circuit shown in Figure 10.24: Voff = 0 V, Von = 3.3 V, I B max = 1 mA, β = 2000, R = 12 Ω, VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V, IC ≥ 1A Find: The range of RB that will ensure the correct operation of the fuel injector. Analysis: If the BJT is in saturation IC = VCC − VCEsat = 1A R Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation RB ≤ RB ≥ Von − Vγ IC / β Von − Vγ I B max = 3.3 − 0.7 = 5.2 kΩ 1 2000 = 2.6 kΩ ______________________________________________________________________________________ 10.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Section 10.4: BJT Switches and Gates Problem 10.30 Solution: Known quantities: The circuit given in Figure P10.30. Find: Show that the given circuit functions as an OR gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . v1 v2 Q1 Q2 Q3 vo1 vo 2 0 0 off off on 0 5V 0 5V off on off 5V 0 5V 0 on off off 5V 0 5V 5V on on off 5V 0 ______________________________________________________________________________________ Problem 10.31 Solution: Known quantities: The circuit given in Figure P10.30. Find: Show that the given circuit functions as a NOR gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.30. This table clearly describes a NOR gate when the output is taken at vo 2 . ______________________________________________________________________________________ Problem 10.32 Solution: Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as an AND gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . 10.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 v1 v2 Q1 Q2 Q3 vo1 vo 2 0 0 off off on 0 5V 0 5V off on on 0 5V 5V 0 on off on 0 5V 5V 5V on on off 5V 0 ______________________________________________________________________________________ Problem 10.33 Solution: Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as a NAND gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.32. This table clearly describes a NAND gate when the output is taken at vo 2 . ______________________________________________________________________________________ Problem 10.34 Solution: Known quantities: In the circuit given in Figure P10.34 the minimum value of vin for a high input is 2.0 transistor Q1 has a β of at least 10. V. Assume that the Find: The range for resistor RB that can guarantee that the transistor is on. Analysis: 5 − 0 .2 = 2.4 mA , therefore, iB = iC/β = 0.24 mA. 2000 (vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL: -vin +RB iB + 0.6 = 0 v − 0 .6 or . Substituting for (vin)min and (vin)max , we find the following range for RB: RB = in iB ic = 5.833 kΩ ≤ RB ≤ 18.333k Ω ______________________________________________________________________________________ Problem 10.35 Solution: Known quantities: For the circuit given in Figure P10.35: R1C = R2C = 10 kΩ , R1B = R2 B = 27 kΩ . Find: a) vB, vout, and the state of the transistor Q1 when vin is low. b) vB, vout, and the state of the transistor Q1 when vin is high. 10.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 Analysis: vin is low Q1 is cutoff vB = 5 V Q2 is in saturation vout = low = 0.2 V. b) vin is high Q1 is in saturation vB = 0.2 V Q2 is cutoff vout = high = 5 V. a) ______________________________________________________________________________________ Problem 10.36 Solution: Known quantities: For the inverter given in Figure P10.36: RC1 = RC 2 = 2 kΩ , RB = 5 kΩ . Find: The minimum values of β1 and β2 to ensure that Q1 and Q2 saturate when vin is high. Analysis: 5 − 0 .2 2.5 = 2.4 mA , therefore, ic = mA . Applying KVL: 2000 β − 5 + RB i B1 + 0.6 + 0.6 + 0.6 = 0 + i or 0.64 ⋅ β1 = 1.2 + 2.5 Therefore, iB1= 0.64 mA. i E1 = β1 ⋅ i B1 = 600 500 B 2 β2 ic = Choose β2 = 10 β1 = 2.27. ______________________________________________________________________________________ Problem 10.37 Solution: Known quantities: For the inverter given in Figure P10.36: RC1 = 2.5 kΩ , Find: Show that Q1 saturates when vin is high. Find a condition for RC 2 = 2 kΩ , β1 = β 2 = 4 . RC 2 to ensure that Q2 also saturates. Analysis: 3 .2 = 0.8 mA iC1 = 3.2 mA 4000 600 Applying KCL: + i B 2 = 3.2 i B 2 = 2 mA ; iC 2 = β ⋅ i B 2 = 8 mA 500 Applying KVL: 5 − 0.2 = 0.008 ⋅ RC 2 RC 2 = 600Ω i B1 = ______________________________________________________________________________________ Problem 10.38 Solution: Known quantities: The basic circuit of a TTL gate, shown in Figure P10.38. Find: The logic function performed by this circuit. Analysis: The circuit performs the function of a 2-input NAND gate. The analysis is similar to Example 10.8. 10.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 10 ______________________________________________________________________________________ Problem 10.39 Solution: Known quantities: The circuit diagram of a three-input TTL NAND gate, given in Figure P10.39. Find: vB1, vB2, vB3, vC2, and vout, assuming that all the input voltages are high. Analysis: Q2 and Q3 conduct, while Q4 is cutoff. vB1 = 1.8 V, vB2 = 1.2 V, vB3 = 0.6 V, and vC2 = vout = 0.2 V. ______________________________________________________________________________________ Problem 10.40 Solution: Known quantities: Figure P10.40. Find: Show that when two or more emitter-follower outputs are connected to a common load, as shown in Figure P10.54, the OR operation results; that is, v0 = v1 OR v2. Analysis: v2 v1 Q1 Q2 v0 L L L L L L H H L H H L L H H H H H H H L : Low; H : High. ______________________________________________________________________________________ 10.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Chapter 11 Instructor Notes Chapter 11 introduces field-effect transistors. The material on transistors has been reorganized in this 4th Edition, and is now divided into two independent chapters, one on bipolar devices, and one on fieldeffect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect devices. Section 11.1 briefly reviews the classification and symbols for the major families of field-effect devices. Section 11.2 introduces the fundamental ideas behind the operation of N-channel field-effect enhancement-mode transistors, and illustrates the calculation of the state and operating point of basic fieldeffect transistor circuits. A brief explanation of P-channel devices is also presented in this section. Section 11.3 briefly outlines the operation of MOSFET amplifiers. Section 11.4 introduces the analysis of MOSFET switches and presents CMOS gates. The box Focus on Measurements: MOSFET bidirectional analog gate (pp. 572-573) presents ananalog application of CMOS technology. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. Learning Objectives 1. Understand the classification of field-effect transistors. Section 1. 2. Learn the basic operation of enhancement-mode MOSFETs by understanding their i-v curves and defining equations. Section 2. 3. Learn how enhancement-mode MOSFET circuits are biased. Section 2. 4. Understand the concept and operation of FET amplifiers. Section 3 5. Understand the concept and operation of FET switches. Section 4. 6. Analyze FET switches and digital gates. Section 4. 11.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Section 11.2: n-channel MOSFET Operation Problem 11.1 Solution: Known quantities: For the transistors shown in Figure P11.1, VT = 3 V . Find: The operating state of each transistor. Analysis: a) This is an n-channel enhancement MOSFET, with condition is: v DS < vGS − VT . VT = - 3 V. To operate in the triode region, the To operate in the saturation region, the condition is: v DS ≥ vGS − VT . To turn the transistor on, the condition is: vGS > VT . vGS = −2.5 V − VT = −2.5 + 3 = 0.5 V = 2.5 V > vGS − VT = 0.5 V . We can compute: vGS v DS v DS = 2.5 V Therefore, the transistor is in the saturation region. b) This is a p-channel enhancement MOSFET, with condition is: v DS > vGS − VT . VT = 3 V. To operate in the triode region, the To operate in the saturation region, the condition is: v DS ≤ vGS − VT . To turn the transistor on, the condition is: vGS < VT . We can compute: vGS = 2 V v DS = −1V vGS − VT = 2 − 3 = −1V v DS = −1V = vGS − VT = −1V . c) Therefore, the transistor is in the saturation region. This is a p-channel enhancement MOSFET, with VT condition is: v DS > vGS − VT . = - 3 V. To operate in the triode region, the To operate in the saturation region, the condition is: v DS ≤ vGS − VT . To turn the transistor on, the condition is: vGS < VT . We can compute: vGS = −5 V v DS = −1V vGS − VT = −5 + 3 = −2 V v DS = −1V > vGS − VT = −2 V . Therefore, the transistor is in the triode region. d) This is an n-channel enhancement MOSFET, with condition is: v DS < vGS − VT . VT = - 3 V. To operate in the triode region, the To operate in the saturation region, the condition is: v DS ≥ vGS − VT . To turn the transistor on, the condition is: vGS > VT . We have: vGS = −2V > VT . vGS − VT = −2 + 3 = 1V vDS = 6V > vGS − VT = 1V Therefore, the transistor is in the saturation region. ______________________________________________________________________________________ 11.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Problem 11.2 Solution: Known quantities: The potentials of an n-channel enhancement-mode MOSFET (4, 5, and 10 V respectively). Find: The circuit symbol, if the device is operating: a) In the ohmic state. b) In the active region. Analysis: a) To operate in the ohmic region, the condition is: v DS < vGS − VT and VT > 0, v DS > 0 . The circuit for operation in the ohmic region is shown below. - D + vGD = 10V vDS = 4V + G + vGS = 5V - b) To operate in the active region, the condition is: - S v DS ≥ vGS − VT and VT > 0, v DS > 0 . The circuit for operation in the active region is shown below. D - + vGD = 4V vDS = 10V + + G vGS = 5V - S ______________________________________________________________________________________ Problem 11.3 Solution: Known quantities: The threshold voltage, VT = 2 V, of an enhancement-type NMOS that has its source grounded and a 3 V DC source connected to the gate. 11.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Find: The operating state if: a) v D = 0.5 V . vD = 1V . c) v D = 5 V b) Analysis: v DS = v D = 0.5 V vGS − VT = 3 − 2 = 1V a) v DS < vGS − VT The transistor is in the triode region. v DS = v D = 1V vGS − VT = 3 − 2 = 1V b) v DS = vGS − VT The transistor is either in the triode or in the saturation region. v DS = v D = 5 V c) vGS − VT = 3 − 2 = 1V v DS > vGS − VT The transistor is in the saturation region. ______________________________________________________________________________________ Problem 11.4 Solution: Known quantities: The threshold voltage, VT = 2 Find: R and vD for id V, of the p-channel transistor shown in Figure P11.4. k = 10 mA/V 2 . = 0.4 mA . Analysis: The device shown is a p-channel enhancement mode MOSFET, with , operate in the saturation region we require: v DS Since ≥ vGS − VT . VT = 2 V and , VDG = 0 V. To v DG = v DS − vGS = 0 > −VT = −2 V , the transistor is in the saturation region. Knowing k = 10 mA/V 2 , we can write: 0.4 = 10 ⋅ (vGS − 2) and determine v D = v DS = vGS = 2.2 V . R can 2 be found as follows: R= 20 − v D 20 − 2.2 = = 44.5 kΩ iD 0.4 ⋅10 −3 ______________________________________________________________________________________ Problem 11.5 Solution: Known quantities: The threshold voltage, VT = 2 = 3 V. V, of an enhancement-type NMOS transistor. iD = 1 mA when vGS = vDS 11.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Find: The value of iD for vGS = 4 V. Analysis: Because v DS > vGS − VT , the transistor is in the saturation region: i D = k ⋅ (vGS − VT )2 = k ⋅ (3 − 2)2 = 0.001A k = 0.001 . For vGS = 4 V we have: i D = 0.001 ⋅ (4 − 2)2 = 4 mA . ______________________________________________________________________________________ Problem 11.6 Solution: Known quantities: Characteristics of an n-channel enhancement-mode MOSFET operated in the ohmic region: vDS = 0.4 V, VT = 3.2 V. Effective resistance of the channel, given by: RDS = Find: The value of iD when vGS = 5 V, RDS 500 Ω. (vGS − 3.2) = 500 Ω, and vGD = 4 V. Analysis: Since VDS = 0.4 < vGS − VT = 5 − 3.2 = 1.8 V the transistor is operating in the ohmic region. The effective resistance iD = is: RDS = 500 = 277.78Ω . (5 − 3.2) Since RDS = VDS , iD we have: VDS = 1.44 mA . RDS ______________________________________________________________________________________ Problem 11.7 Solution: Known quantities: The threshold voltage, VT = 2.5 V, of an enhancement-type NMOS that has its source grounded and a V DC source connected to the gate. 4 Find: The operating state if: a) v D = 0.5 V b) v D = 1.5 V . Analysis: a) vDS = 0.5 < vGS - VT = 4 – 2.5 = 1.5 V, therefore the transistor is in the triode region. b) vD = 1.5 V = vDS , therefore the transistor is at the border of the saturation and triode regions. ______________________________________________________________________________________ 11.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Problem 11.8 Solution: Known quantities: The threshold voltage, VT = 4 V, of an enhancement-type NMOS. iD = 1 mA when vGS = vDS = 6 V. Find: The value of iD when vGS = 5 V. Analysis: k(6 - 4)2, we have k = 0.25×10-3 For vGS = 5 V, and assuming active operation: iD = 0.25×10-3(5 - 4)2 = 0.25 mA. From 0.001 = ______________________________________________________________________________________ Problem 11.9 Solution: Known quantities: The threshold voltage, VT = 1.5 V, of the NMOS transistor shown in Figure P11.9. Find: The voltage levels of the pulse signal at the drain output, if vG is a pulse with 0 k = 0.4 mA/V 2 . V to 5 V. Analysis: = 1.5 V, with vG = 0 V, vGS < VT, the transistor is cut off. Therefore, vD = 5 V. When vG = 5 V, and assuming that the transistor is in the active region: Since VT iD = k (vGS - VT)2 = 0.4 (5 - 1.5)2 = 4.9 mA. Therefore, vD = 5 - 4.9×1 = 0.1 V. ______________________________________________________________________________________ 11.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Section 11.3: n-channel MOSFET Amplifiers Problem 11.10 Solution: Known quantities: The i-v characteristic of Figure P11.10(a), and the circuit in Figure P11.10(b): VGG = 7 V, VDD = 10 V, RD = 5 Ω Find: The current iDQ the voltage vDSQ, and the region of operation of the MOSFET. Analysis: The operating point can be determined using the load line method. iD = VDD v DS − = 2 − 0.2 v DS RD RD By superimposing the load line on Figure P11.10(a), and by noticing that VGS = VGG = 7 V , we obtain iDQ = 0.8 A, v DSQ = 6 V The MOSFET is in the saturation region. ______________________________________________________________________________________ Problem 11.11 Solution: Known quantities: The circuit in Figure P11.10(b): VGG = 7 V, VDD = 20 V, VT = 3 V, RD = 5 Ω, K = 50 mA/V 2 Find: The current iDQ the voltage vDSQ, and the region of operation of the MOSFET. Analysis: Assuming that the MOSFET is in the saturation region, the quiescent drain current is iDQ = K (vGSQ − VT ) 2 = 0.05(7 − 3) = 0.8 A 2 The drain-to-source voltage is vDSQ = VDD − RD iDQ = 20 − 5 ⋅ 0.8 = 16 V Since vDG = v DS − vGS = 9V > VT hypothesis was correct ______________________________________________________________________________________ Problem 11.12 Solution: Known quantities: The circuit in Figure 11.12: VDD = 36 V, VT = 4 V, RD = 10 kΩ, R1 = R2 = 2 MΩ, K = 0.1 mA/V 2 Find: The current iDQ, the voltage vDSQ, the resistance RS, and the operating region of the MOSFET. Analysis: Using Thevenin equivalent, 11.7 G. Rizzoni, Principles and Applications of Electrical Engineering VGG = Problem solutions, Chapter 11 R2 VDD = 18 V R1 + R2 We can write the equations VGG = vGSQ + RS iDQ = 18, VDD = ( RD + RS )iDQ + v DSQ = RD iDQ + 18 − vGSQ + v DSQ = 36 RD iDQ + v DSQ = 18 + vGSQ Assuming saturation conditions, the current iD can be written as iDQ = K (vGSQ − VT ) 2 vGSQ + RS K (vGSQ − VT ) 2 = 18 and RD K (vGSQ − VT ) 2 + v DSQ = 18 + vGSQ Notice that the problem has more unknown than equations; we can impose the vDSQ to ensure saturation conditions as vDSQ = VDD / 2 = 18 V 2 (vGSQ − VT ) 2 = vGSQ vGSQ − 9vGSQ + 16 = 0 vGSQ = 6.56 V Remark: The other solution of the algebraic equation is not acceptable because < VT. The resistance RS is given by RS = 18 − vGSQ K (vGSQ − VT ) 2 = 18 − 6.56 = 17.45 kΩ 2 0.1 ⋅ 10 −3 (6.56 − 4) and the drain current iDQ = K (vGSQ − VT ) 2 = 0.655 mA ______________________________________________________________________________________ Problem 11.13 Solution: Known quantities: n-channel MOSFET operating in saturation region with vDSQ = 5 V, vGSQ = 3 V, VT = 1 V, K = 0.1 mA/V 2 Find: The transconductance gm. Analysis: The drain current is iDQ = K (vGSQ − VT ) 2 = 0.4 mA The transconductance is g m = 2 KiDQ = 2 0.04 ⋅ 10 −3 = 0.4 mA/V 2 ______________________________________________________________________________________ Problem 11.14 Solution: Known quantities: The circuit in Figure 11.12: VDD = 12 V, VT = 1 V, RS = RD = 10 kΩ, R1 = R2 = 2 MΩ, K = 1 mA/V 2 11.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Find: The current iDQ, the voltage vDSQ, and the voltage vGSQ. Analysis: Using Thevenin, VGG = R2 VDD = 6 V R1 + R2 We can write the equations VGG = vGSQ + RS iDQ = 6, VDD = ( RD + RS )iDQ + v DSQ = 2 RD iDQ + v DSQ = 12 Assuming saturation conditions, the current iD can be written as 2 − 19vGSQ + 4 = 0 iDQ = K (vGSQ − VT ) 2 vGSQ + RS K (vGSQ − VT ) 2 = 6 10vGSQ vGSQ = 1.66 V The other solution is not acceptable because less then VT. It follows iDQ = 6 − vGSQ RS = 0.434 mA, vDSQ = 12 − 2 RD iDQ = 3.32 V ______________________________________________________________________________________ 11.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Section 11.4: MOSFET Switches Problem 11.15 Solution: Known quantities: The CMOS NAND gate of Figure 11.21. Find: Identify the state of each transistor for v1 = v2 = 5 V. Analysis: The two transistors at the top are cut off and the two at the bottom are on. ______________________________________________________________________________________ Problem 11.16 Solution: Known quantities: The CMOS NAND gate of Figure 11.21. Find: Identify the state of each transistor for v1=5V, v2=0V. Analysis: The transistor at the bottom and the first on the top are off, the other two are on. ______________________________________________________________________________________ Problem 11.17 Solution: Find: Draw the schematic diagram of a two-input CMOS OR gate. Analysis: The output of the circuit of Figure 11.18 is connected as an input to the circuit of Figure 11.14. VDD M1 v1 vout M2 v2 M3 M4 Figure 11.18 11.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 ______________________________________________________________________________________ Problem 11.18 Solution: Find: Draw the schematic diagram of a two-input CMOS AND gate. Analysis: The output of the circuit of Figure 11.21 is connected as an input to the circuit of Figure 11.14. VDD vout v1 VT = 1.5V v2 Figure 11.21 ______________________________________________________________________________________ Problem 11.19 Solution: Find: Draw the schematic diagram of a two-input CMOS NOR gate. Analysis: The circuit of Figure 11.18 11.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 VDD M1 v1 vout M2 v2 M3 M4 Figure 11.18 ______________________________________________________________________________________ Problem 11.20 Solution: Find: Draw the schematic diagram of a two-input CMOS NAND gate. Analysis: The circuit of Figure 11.21. VDD vout v1 VT = 1.5V v2 Figure 11.21 _________________________________________________ _____________________________________ 11.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 Problem 11.21 Solution: Known quantities: The circuit of Figure P11.21. Find: Show that the given circuit functions as a logic inverter. Analysis: Construct a state table: vin Q1 Q2 vout low resistive open high high open resistive low This table clearly describes an inverter. ______________________________________________________________________________________ Problem 11.22 Solution: Known quantities: The circuit of Figure P11.22. Find: Show that the given circuit functions as a NOR gate. Analysis: Construct a state table: v1 v2 Q1 Q2 vout 0 0 off off high 0 high off on low high 0 on off low high high on on low This table clearly describes a NOR gate. ______________________________________________________________________________________ Problem 11.23 Solution: Known quantities: The circuit of Figure P11.23. Find: Show that the given circuit functions as a NAND gate. Analysis: Construct a state table: v1 v2 Q1 Q2 vout 0 0 off off high 0 high off on high 11.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 11 high 0 on off high high high on on low This table clearly describes a NAND gate. ______________________________________________________________________________________ 11.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Chapter 12 Instructor Notes Chapter 12 introduces the subject of power electronics. The importance of power electronics cannot be overemphasized, considering the widespread industrial application of electric machines, and other high current loads in practical engineering applications. The chapter discusses the basic characteristics and limitations of power amplifiers, practical voltage regulators, inductive loads (such as electric motors), and SCRs. The aim is to give the student sufficient understanding of the device characteristics to be able to complete simple "order of magnitude" calculations to be able to size a device for a given application. This chapter is much more practically oriented than some of the others in the text, and may be used to accompany a course in electric power and machines based on Chapters 7, 16, 17 and 18. After Sections 12.1 and 12.2 present a classification of power electronic devices (Figure 12.1, p. 578) and circuits (Table 12.1, p. 579), the discussion is divided into the topics of Voltage Regulators (Section 12.3), Power Amplifiers and Transistor Switches (Section 12.4), Rectifiers and Controlled Rectifiers (Section 12.5), and Electric Motor Drives (Section 12.6) Homework problems are divided into three major sections. The first, on voltage regulators, includes problems two different voltage regulator circuits (12.2, 12.3) The second, on rectifiers and controlled , illustrates a battery charging circuit (12.7) and two simple motor speed control problems (12.9, 12.10). The last section, on drives, introduces choppers, and more advanced problems on DC motor supplies based on controlled rectifiers (12.23 and 12.24). Learning Objectives 1. Learn the classification of power electronic devices and circuits. Sections 1 and 2. 2. Analyze the operation of practical voltage regulators. Section 3. 3. Understand the principal limitations of transistor power amplifiers. Section 4. 4. Analyze the operation of single- and three-phase controlled rectifier circuits. Section 5. 5. Understand the operation of power converters used in electric motor control, and perform simplified analysis on DC-DC converters. Section 6. 12.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Section 12.3: Voltage Regulators Problem 12.1 Solution: Find: Repeat Example 12.1 for a 7-V Zener diode. Analysis: Calculating the collector and base currents according to: 7 − 1 .3 = 0.57 A 10 I I B = E = 51.8 mA 11 IE = We find: 20 − 7 = 0.277 A 47 I Z = I R − I B = 0.225 A IR = VCE = 20 − VL = 20 − 5.7 = 14.3 V > 0.6 V Thus, the transistor is in the active region. The Zener power is: I Z ⋅ VZ = 1.576 W . ______________________________________________________________________________________ Problem 12.2 Solution: Known quantities: The current regulator circuit shown in Figure P12.2. Find: The expression for RS. Analysis: Assuming that the Zener voltage is VZ , that VBE = Vγ = 0.6 V , and that the required current is I, we have: VZ + VBE VZ + 0.6 = . I I ______________________________________________________________________________________ RS = Problem 12.3 Solution: Known quantities: The shunt-type voltage regulator shown in Figure P12.3. Find: The expression for the output voltage, Vout . Analysis: If the Zener diode is to be in the regulator mode, the CB junction must be forward biased; in this case, both the CB and the BE junctions are forward biased, since a substantial base current will be generated through 12.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 the Zener diode (depending on the value of the shunt resistor in the output circuit). Thus, the collectorV − VCEsat emitter voltage is equal to: VCEsat ≈ 0.2 V , and the source current will be: I S = S = IC + I Z . RS The voltage across the shunt resistor will therefore be Vγ, and the output voltage is: Vout = VZ + Vγ . ______________________________________________________________________________________ 12.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Section 12.5: Rectifiers and Controlled Rectifiers (AC-DC Converters) Problem 12.4 Solution: Known quantities: The circuit shown in Figure 12.17. Find: If the LR load is replaced by a capacitor, draw the output waveform and label the values. Analysis: When the sinusoidal source voltage is in the positive half cycle, the series diode conducts, and the shunt diode is an open circuit; thus, the positive half cycle appears directly across the capacitor (assuming ideal diodes). During the negative half cycle, the series diode is open, and therefore the voltage across the capacitor remains zero, as shown in the sketches below. vAC vC ______________________________________________________________________________________ Problem 12.5 Solution: Known quantities: The circuit shown in Figure 12.17. Find: If the diode forward resistance is 50 Ω, the forward bias voltage is 0.7 V, and the load consists of a resistor R = 10Ω and an inductor L = 2 H , draw vL(t) and label the values for the given circuit. Analysis: To obtain exact numerical values, we assume a 111 Vrms source, R = 10Ω , and L = 2 H , then: v AC (t ) = A sin (ωt ) = 155.6 ⋅ sin (377t ) , and from Equation 12.6, the average load current is: 155.6 A A IL = = 4.95 A . Using the approximation: v L (t ) ≈ + sin (ωt ) we have: 2 2 πR A A v L (t ) ≈ + sin (ωt ) = 77.8 + 77.8 sin (377t ) . The waveform is shown below: 2 2 12.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 rectified load voltage 200 volts 150 100 50 0 0 0.02 0.04 0.06 0.08 0.1 time, s ______________________________________________________________________________________ Problem 12.6 Solution: Known quantities: For the circuit shown in Figure P12.6, vAC is a sinusoid with 11 V peak amplitude, R = 2 kΩ and the forward-conducting voltage of D is 0.7 V. Find: a. Sketch the waveform of vL(t). b. Find the average value of vL(t). Analysis: a. Assume v AC (t ) = 10 sin (ωt ) V . The output voltage is: v L (t ) = (10 − 0.7 )sin (ωt ) . The waveform is shown below: vL (V) 10 9.3 5 0 t -5 b. vL = 1 π 9.3 9.3 sin (ωt ) d(ωt ) = = 2.96 V . ³ π 2π 0 ______________________________________________________________________________________ Problem 12.7 Solution: Known quantities: The vehicle battery charge circuit shown in Figure P12.7. Find: Describe the circuit, and draw the output waveform (L1 and L2 represent the inductances of the windings of the alternator). Analysis: The positive half cycle from w1 is conducted by diode D1. Diode D2 does not conduct due to negative bias at w2. The first half cycle is passed through to the battery. 12.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 D 1 (on) + ~ vAC w1 w2 + Vehicle Battery D2 (off) - The second half cycle finds w2 positive and diode D2 conducts current to the battery while diode D1 is negatively biased and is off. D 1 (off) ~ vAC w1 w2 + + Vehicle Battery D2 (on) - The full-wave rectified output waveform is shown below. Max VAV 0 180 360 The average DC value VAV is 63% of the peak value. ______________________________________________________________________________________ Problem 12.8 Solution: Find: NOTE: Typo in problem statement, referring to the wrong example problem. Repeat Example 12.6 for α = π/3 and π/6. Analysis: 2π 1 § π · 120 2 π/3 we have: v L ¨ ¸ = = 105 V . 1 − + sin 3 3 2 ©3¹ 2 v The power is: P = L = 45.94 W . R π 1 § π · 120 2 b) For α = π/6 we have: v L ¨ ¸ = 1 − + sin = 110.6 V . 6 3 2 ©6¹ a) For α = 12.6 G. Rizzoni, Principles and Applications of Electrical Engineering The power is: P= Problem solutions, Chapter 12 2 vL = 50.97 W . R ______________________________________________________________________________________ Problem 12.9 Solution: Known quantities: o For the circuit shown in Figure P12.9, assume the thyristors are fired at α = 60 and that the motor current is 20 A and is ripple free. The supply is 111 VAC (rms). Find: a) Sketch the output voltage waveform, v0. b) Compute the power absorbed by the motor. c) Determine the volt-amperes generated by the supply. Analysis: a) α = 60 × π π = , Ra = 0.2 . 180 3 1 b) Vorms 120 2 ª 1 º2 = 1 − + sin 120 ° » = 105 V Æ Pm = I oVorms = (20 A )(105 V ) = 2.1 kW . 2 «¬ 3 ¼ c) PR = I o2 Ra = (20)2 (0.2 ) = 80 W ; PS = Pm + PR = 2180 W . ______________________________________________________________________________________ Problem 12.10 Solution: Known quantities: The circuit of Figure 12.2, replacing the resistive load with a DC motor. The motor operates at 110 V and absorbs 4 kW of power. The AC supply is 80 V, 60 Hz. Assume that the motor inductance is very large (i.e., the motor current is ripple free), and that the motor constant is 0.055 V/rev/min. 12.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Find: If the motor runs at 1,000 rev/min at rated current: a) Determine the firing angle of the converter. b) Determine the rms value of the supply current. Analysis: = k T × N = 0.055 × 1000 = 55 V P 4000 = = = 36.4 A 110 V Back emf I DC V = 110 V . Assume R = 1 . a) 36.4 = 1 π I DC = 1 πR [ 2V (cosα ) − VB (π − α )] [ 2 (110) cosα − 55(π − α )] . Solving yelds: α ≈ 127 π rad. b) With zero ripple, I rms = I DC = 36.4 A . ______________________________________________________________________________________ Problem 12.11 Solution: Find: NOTE: Typo in problem statement, referring to the wrong example problem. For the light dimmer circuit of Example 12.6, determine the load power at firing angles α = 0 0 , 30 0 , 60 0 , 90 0 ,120 0 ,150 0 ,180 0 , and plot the load power as a function of α. Analysis: Note that VLrms = and VLrms = α 120 1 − + sin 2α , for α ≤ 90 0 , π 2 α sin 2α 120 , 1− + π 2π 2 VLrms α 0° for a > 90 0 P= VL2rms RBULB 84.85 30 30 ° 111.60 50.98 60 ° 115.00 45.98 90° 60.00 15.00 120 ° 37.53 5.87 150 ° 14.41 0.87 180 ° 0 0 A sketch of power vs. firing angle is shown below: 12.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 ______________________________________________________________________________________ Problem 12.12 Solution: Known quantities: For the circuit shown in Figure P12.12: V L = 10 V V r = 10 % = 1 V I L = 650 mA vline = 170 cos ωt V rad ω = 2513 s Find: Determine the conduction angle of the diodes, if the diodes are fabricated form silicon. Analysis: 1 1 V m = V L + V r = 10 V + [ 1 V ] = 10.5 V 2 2 1 1 v L-min = V L - V r = 10 V - [ 1V ] = 9.5 V 2 2 D2 and D4 conduct during ωt2 < ωt < π. First, determine the amplitude of the source voltage. [The secondary of the transformer acts as a source.] Then use the same KVL to determine the angle at which the diodes start conducting. 12.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 KVL : + v D 4 + v s + v D 2 + v L = 0 At ωt = π : v D 2 = v D 4 = v D-on = 0.7 V [Si] vs = - V so vL = V m V so = v D-on + v D-on + V m = 0.7 V + 0.7 V + 10.5 V = 11.9 V At ωt = ωt 2 : v D 4 = v D 2 = v D-on = 0.7 V v s = V so cos ω t 2 v L = v L-min +v +v 0.7 V + 0.7 V + 9.5 V v = - 0.91597 cos ωt 2 = - D-on D-on L-min = 11.9 V V so ∴ φ = 180o - 156.3o = 23. 66o ωt 2 = 156. 3o ______________________________________________________________________________________ Problem 12.13 Solution: Known quantities: For the circuit shown in Figure P12.13, assume that the conduction angle of the diodes shown (which are Silicon) is: Φ = 23o v s1 (t ) = v s 2 (t ) = 8 cos(ωt ) V rad ω = 377 R L = 20 kΩ C = 0.5 µF s Find: The ripple voltage. Analysis: KVL : - v s1 (t ) + v D1 + v L (t ) = 0 At t = 0 : v L (0) = V m = V so - v D-on = 8 V - 0.7 V = 7.3 V π π - 23o 180o = 7.268 ms ωt 2 = π - Φ t 2 = rad 377 s t t v L (t ) = v L (∞ ) + (v L (0) − v L (∞ )) e - TC = 0 + [ V m - 0 ] e- R L C 7.286⋅10-3 v L (t 2 ) = v L-min = 7.3 ⋅ e - [ 20⋅103 ] [ 0.5⋅10-6 ] = 3.529 V V r = V m - v L-min = 7.3 - 3.529 = 3.771 V Note the ripple is quite large. This is primarily due to the very small (for this type circuit) value of the capacitance. Also the conduction angle assumed above is not correct for this circuit. ______________________________________________________________________________________ 12.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Problem 12.14 Solution: Known quantities: The diodes in the full-wave DC power supply shown are Silicon. If: I L = 85 ma rad vline = 156 cos ωt V s φ = Conduction Angle = 23. 90o V L = 5.3 V V r = 0.6 V C = 1023 µF ω = 377 Find: The value of the average and peak current through each diode. Analysis: Diodes D1 and D3 will conduct half of the load current and Diodes D2 and D4 will conduct the other half. i D-ave = Therefore: 1 1 I L = [ 85 mA ] = 42.5 mA 2 2 The waveforms of the diode currents are complex but can be roughly approximated as triangular [recall area of triangle = bh/2]: I L = [ i D1,3 + i D2,4 ] ave = i D- pk = 1 2π 1 φ i D- pk φ i D- pk [ + ] ³0 [ i D1,3 ( ωt) + i D2,4 ( ωt) ] d [ ωt] = 2π 2π 2 2 2π I L [ 2π rad ] [ 85 ma ] 2π I L = = = 1.280 A . 1 1 π rad φ o [ 23.90 ] [ ] Φ+ Φ 2 2 180o ______________________________________________________________________________________ Problem 12.15 Solution: Known quantities: The diodes in the full-wave DC power supply shown in Figure P12.13 are Silicon. If: I L = 600 mA V L = 50 V V r = 8 % = 4 V vline = 170 cos ωt V ω = 377 rad s Find: The value of the conduction angle for the diodes and the average and peak current through the diodes. The load voltage waveform is shown in Figure P12.15. Analysis: 1 1 v L-min = V L - V r = 50 - 4 = 48 V 2 2 1 1 V m = V L + V r = 50 + 4 = 52V 2 2 KVL : - v s1 (t ) + v D1 + v L (t ) = 0 At t = 0 : - V so cos(0 ) + v D-on + V m = 12.11 0 V so = 0.7 + 52 = 52.7 V G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 KVL : + v s 2 (t ) + v D 2 + v L (t ) = 0 At t = t 2 : + V so cos(ω t 2 ) + v D-on + v L-min = 0 0.7V + 48V v D-on + V L-min ] = cos- [ ] = 2.749 rad ωt 2 = cos- [ 52.7 V V so o Φ = π - ωt 2 = π - 2.749 rad = 392.1 mrad 180 = 22. 47o π rad The waveforms of the diode currents are complex but can be roughly approximated as triangular [recall the area of triangle = bh/2]. i D1-ave = i D2-ave = 1 1 I L = 600 mA = 300 mA 2 2 [ i D1 + i D 2 ] ave = I L 1 2π ³ [ i D1( ωt) + i D 2 ( ωt) ] d[ ωt] = I L ωT 0 1 1 1 [ Φ i D- pk + Φ i D- pk ] = I L ωT 2 2 ωT I L [ 2π rad ] [ 600 mA ] = = 9.615 A i D- pk = Φ 392.1 mrad ______________________________________________________________________________________ 12.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Section 12.6: Electric Motor Drives Problem 12.16 Solution: Known quantities: For the chopper of Figure P12.35, the supply voltage is 120 V, and the armature resistance of the motor is 0.15 Ω. The motor back emf constant is 0.05 V/rev/min and the chopper frequency is 250 Hz. Assume that the motor current is free of ripple and equal to 125 A at 120 rev/min. Find: a) The duty cycle of the chopper, δ, and the chopper-on time, t1. b) The power absorbed by the motor. c) The power generated by the supply. Analysis: a) vo = io Ra + Ea = (125)(0.15) + 6 = 24.75 V 24.75 = 0.2063 120 t § 1 · δ = 1 t1 = δT = (0.2063)¨ ¸ = 825µs T © 250 ¹ 24.75 = δ (120) δ = b) Pm = E a io = (6)(125) = 750 W PR = Ra io2 = (0.15)(125) 2 = 2.344 kW c) PS = Pm + PR = 3.094 kW or PS = δ ⋅ V S ⋅ io = (0.2063)(120)(125) = 3.094 kW . ______________________________________________________________________________________ Problem 12.17 Solution: Known quantities: For the circuit shown in Figure 12.39, the motor constant is 0.3 V/rev/min, the supply voltage is 600 V, and the armature resistance is Ra = 0.2Ω . Find: If the motor speed is 800 rev/min and the motor current is 300 a) The duty cycle of the chopper, δ. b) The power fed back to the supply. Analysis: a) vo = E a + io Ra = 240 + (−300)(0.2) = 180 V δ = A, determine: 180 = 0.300 . 600 Pm = E a io = (240)(−300) = −72.0 kW ½° ¾ PS = Pm + PR = −54.0 kW PR = Ra io2 = (0.2)(−300) 2 = 18 kW °¿ or PS = δ ⋅ VS ⋅ io = (0.300)(600)(−300) = −54.0 kW . ______________________________________________________________________________________ b) Problem 12.18 12.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Solution: Known quantities: For the two quadrant chopper of Figure 12.40, assume the thyristors S1 and S2 are turned on for time t1 and off for time T – t1 (T is the chopping period). Find: An expression for the average output voltage in terms of the supply voltage, VS, and the duty cycle, δ. Analysis: vo = t1 t 1 VS = 1 VS = VS = δ ⋅ VS . T t ⋅ t1 t ______________________________________________________________________________________ Problem 12.19 Solution: Known quantities: Supply voltage; chopper duty cycle. Find: Average and rms value of ideal switched supply voltage. Analysis: t1 1 The duty cycle is δ 0.4 The average value is therefore <Vsupply> = δ×Vsupply = 40 V. T 2.5 To compute the rms value we use the definition of eq. 4.24: V˜supply δ 0 2 Vsupply dt δ 100 2 63.25 V ______________________________________________________________________________________ Problem 12.20 Solution: Known quantities: Load resistance and inductance; ideal supply voltage; duty cycle. Find: Average values of current and voltage; power drawn from battery supply. Analysis: From the data given, δ = 0.333. Since the period of the waveform is 3 ms, we can calculate the switching frequency to be: 1 1 333.33 Hz f T 3 10 3 ω = 2, 094.4 rad/s L 10 -3 H = 2 ms . The time constant of the load impedance is τ = = 0.5 R The average load voltage is: <VL> = δ×Vsupply = 33.33 V. Vsupply 100 V δ 0.33 66.67 A . The average load current is I L 0.5 R To compute the power drawn from the battery supply (which is assumed equal to the load power if 2 ˜I L R; switching losses are held negligible), we really need to compute the rms load current, since PL however, this calculation cannot be completed without knowing exactly the shape of the load current. 12.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Without further analysis we can only state that the power drawn from the battery is greater than <PL> = <VL><IL> = 2.22 kW. ______________________________________________________________________________________ Problem 12.21 Solution: Known quantities: The converter of Problem 12.20 with a DC motor as load. Ra = 0.2Ω, La = 1 mH, Ea = 10 V, T = 3 ms, δ = 1 3 Find: The average load current and voltage. Analysis: The average load voltage is: <VL> = δ×Vsupply = 33.33 V. The average current is: < I L >= < VL > − Ea 33.33 − 10 = = 116.5 A 0 .2 Ra ______________________________________________________________________________________ Problem 12.22 Solution: Known quantities: Load resistance and inductance; load current; motor armature constant; DC supply and desired rpm range. Find: Range of duty cycles required. Analysis: Assume steady-state operation, so that the effects of the load inductance may be neglected. When the rpm is zero, the back emf is also 0, so VL Ia Ra 25 0.3 7.5 V . The motor emf constant is kaφ = 0.00167 V-s/rev, or 60*0.00167 = 0.1004 V-min/rev At 2000 rpm, the back emf is: Ea = kaφn = 200.4 V. Thus, the total load voltage is VL Ia Ra Ea 7.5 200.4 207.9 V From the range of voltages required by the motor for its proper operation, we conclude that the range of required duty cycles is: Vmin 7.5 δ min 0.0341 Vsupply 220 Vmax 207.9 δ max 0.943 220 Vsupply ______________________________________________________________________________________ Problem 12.23 Solution: Known quantities: Motor ratings; motor armature resistance and armature constant; power supply ratings. Find: Motor speed, power factor and efficiency for α = 0° and α = 20°. 12.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Analysis: The nominal torque from the rated data is Tm = 10000 P = = 95.49 Nm ω m 2π 1000 60 It follows that the DC current is Ia = Tm 95.49 = = 47.75 A 2 Ka The average load voltage for firing angle of 0° is < VL , 0, >= 2 2 2 2 VS = 240 = 216 V π π The speed at zero degree of firing angle is given by < VL , 0, >= Ea , 0, + Ra I a Ea , 0, = 216 − 0.42 ⋅ 47 .75 = 196 V E a , 0, ω m, 0 = , = Ka 196 = 97 .97 rad s 2 The efficiency is η= E a , 0, I a E a , 0, I a + R a I 2 a = E a , 0, = E a , 0, + R a I a 196 = 91 % 196 + 0 .42 * 47 . 75 The rms voltage is Vrms , L = VS = 240 The power factor can be calculate as follows pf (0 ) ≅ , Ea , 0 , I a + Ra I a2 Vrms , L I a = Ea , 0 , + Ra I a Vrms , L = 196 + 0.42 * 47.75 = 0 .9 240 ° In the case of firing angle of 20 , we have 2 2 V S cos 20 , = 203 V π =< V L , 20 , > − R a I a = 203 − 0 .42 * 47 .75 = 182 .95 V < V L , 20 , >= E a , 20 , ω m , 20 = , η= E a , 20 , Ka Ea , 20 , Ea , 20 , + Ra I a pf (20, ) ≅ = 182 .95 = 91 .47 rad s 2 = 182.95 = 90% 182.95 + 0.42 ⋅ 47.75 Ea , 20 , + Ra I a Vrms , L = 182.95 + 0.42 * 47.75 = 0.846 240 ______________________________________________________________________________________ Problem 12.24 12.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 12 Solution: Known quantities: Separately excited DC motor: P = 10 kW, V = 300 V, ω = 1000 rev min , Ra = 0.2Ω, K a = 1.38 V ⋅ s / rad Power supply: VS = 220 (rms)V, f = 60 Hz Three phase controlled bridge rectifier. Find: Speed, power factor, and efficiency for a firing angle of 30 deg. Assumptions: Load torque is constant, and the DC motor deliver the power P at 0 deg of firing angle. Additional inductance is present to ensure continuous conduction. Analysis: The average voltage supplied by the rectifier at 0 deg of firing angle is < VL , 0 , >= It follows 3 3 2VS = 514.6 V π < VL,0, >= Ra Ia + Ea, 0, = Ra P + Ea,0, Ea2,0, − < VL, 0, > Ea,0, + Ra P = 0 Ea2,0, − 514.6Ea,0, + 2000= 0 Ea,0, Ea,0, = 510.68 V Ia = Ia,0, = Ia,30, = P 10000 = = 19.58 A Ea,0, 510.68 The average voltage supplied by the rectifier at 30° of firing angle is < VL , 30, >=< VL , 0, > ⋅ cos 30, = 3 3 3 2VS = 445.65V π 2 The emf of the DC motor in this condition is given by Ea,30, =<VL,30, > −Ra Ia = 445.65− 0.2 ⋅19.58 = 441.74 V Finally, the speed is given by ωm = E a , 30 , = 320 . 1 Ka rad s If we assume that the inductance is big enough, then the current ripple is negligible. Under this conditions η= Ea , 30 , I a Ea , 30 , I a + R I 2 a a = 8649.27 = 99.1% 8725.94 The rms voltage supplied for a firing angle of 30° is VL , rms = 453.03 V The power factor is Ea , 30 , I a + Ra I a2 Ea , 30 , + Ra I a 441.74 + 3.92 Pout pf = ≅ = = = 0.984 VL , rms I L , rms VL , rms I a VL , rms 453.03 ______________________________________________________________________________________ 12.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Chapter 13 Instructor Notes Chapter 13 is a stand-alone chapter that does not require much more than a general introduction to the idea of analog and digital signals as a prerequisite. Thus, the chapter could be covered as early as desired. Some instructors may find it desirable to first introduce the basics of electronic switching circuits by covering the appropriate sections of Chapters 9-11. The first section introduces the ideas of analog and digital signals, and the concepts of sampling and quantization, in an intuitive fashion. Section 13.2 introduces the binary number system, and binary codes; the box Focus on Measurements: Digital Position Encoders (pp. 622-623) discusses optical position encoders of the type commonly encountered in many industrial applications (e.g., robotics). The third section presents the foundations of Boolean algebra, and defines the properties of logic gates; the box Focus on Measurements: Fail-safe Autopilot Logic (p. 629) illustrates a simple application of digital logic to motivate the content of the section from a more practical perspective. Combinational logic design through the use of Karnaugh maps is presented in section 13.4; the boxes Focus on Methodology: Sum-ofProducts Realizations (p. 639) and Focus on Methodology: Products-of-Sums Realizations (p. 644) summarize design procedures for sum-of-products and product-of-sums circuits. The box Focus on Measurements: Safety Circuit for the Operation of a Stamping Press (pp. 646-648) demonstrates of the usefulness of even the simplest logic circuits in an industrial setting. A brief survey of digital logic could stop here, if desired. Section 13.5 describes more advanced combinational logic modules; The box Focus on Measurements: EPROM-based Look-up Table for Automotive Fuel Injection System Control (pp. 654-655) is centered around the air-to-fuel ratio control problem in an internal combustion engine, and illustrates a truly wide-spread application of digital logic, since this type of circuit is present in virtually every modern automobile. The end-of-chapter problems are divided into four sections. The first contains a few simple exercises related to number systems; the second section on combinational logic offers a selection of simple problems that are extensions of the examples given in the text, and also includes 5 applied problems (13. 19-13.23) that demonstrate the use of Boolean logic in five every-day situations. The third section covers logic design, and also includes a couple of problems with an applied flavor (13.38, 13.53), in addition to a variety of more traditional design problems. Problems 13.48 - 13.52 are related to number codes. The fourth section contains a few problems related to combinational logic modules. Learning Objectives 1. Understand the concepts of analog and digital signals and of quantization. Section 1. 2. Convert between decimal and binary number system and use the hexadecimal system and BCD and Gray codes. Section 2. 3. Write truth tables, realize logic functions from truth tables using logic gates. Section 3. 4. Systematically design logic functions using Karnaugh maps. Section 4. 5. Study various combinational logic modules, including multiplexers, memory and decoder elements, and programmable logic arrays. Section 5. 13.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Section 13.2: The Binary Number System Problem 13.1 Solution: Known quantities: The base 10 representation of five numbers: 40110, 27310, 1510, 3810, 5610. Find: The hex and the binary representation for these numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) 19116, 1100100012 b) 11116, 1000100012 c) F16, 11112 d) 2616, 1001102 e) 3816, 1110002 ______________________________________________________________________________________ Problem 13.2 Solution: Known quantities: The hex representation of five numbers: A16, 6616, 4716, 2116, 1316. Find: The base 10 and the binary representation for these numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) 1010, 10102 b) 10210, 11001102 c) 7110, 10001112 d) 3310, 1000012 e) 1910, 100112 ______________________________________________________________________________________ Problem 13.3 Solution: Known quantities: The base 10 representation of four numbers: 271.2510, 53.37510, 37.3210, 54.2710. Find: The binary representation for these numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) 100001111.012 b) 110101.0112 c) 100101.010102 d) 110110.0100012 ______________________________________________________________________________________ Problem 13.4 Solution: Known quantities: The binary representation of six numbers: 11112, 10011012, 11001012, 10111002, 111012, 1010002. Find: The hex and the base 10 representation for these numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) F16, 1510 b) 4D16, 7710 c) 6516, 10110 d) 5C16, 9210 e) 1D16, 2910 f) 2816, 4010 ______________________________________________________________________________________ 13.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.5 Solution: Known quantities: Three couples of binary numbers. Find: The addition for each couple. Analysis: Using the methodologies introduced in paragraph 13.2: a) 11111010 b) 100010100 c) 110000100 ______________________________________________________________________________________ Problem 13.6 Solution: Known quantities: Three couples of binary numbers. Find: The subtraction for each couple. Analysis: Using the methodologies introduced in paragraph 13.2: a) 11100 b) 1101110 c) 1000 ______________________________________________________________________________________ Problem 13.7 Solution: Known quantities: Three eight-bit binary numbers in sign-magnitude form. Find: The decimal value of these numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) -120 b) -31 c) 121 ______________________________________________________________________________________ Problem 13.8 Solution: Known quantities: Three decimal numbers. Find: The sign-magnitude form binary representation. Analysis: Using the methodologies introduced in paragraph 13.2: a) 01111110 b) 11111110 c) 01101100 d) 11100010 ______________________________________________________________________________________ 13.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.9 Solution: Known quantities: Four binary numbers. Find: The two's complement of these four numbers. Analysis: Using the methodologies introduced in paragraph 13.2: a) 24 - 1111 = 10000 -1111 = 0001 b) 27 - 1001101 = 0110011 7 c) 2 - 1011100 = 0100100 d) 25 - 11101= 100000 - 11101 = 00011 ______________________________________________________________________________________ 13.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Section 13.3: Boolean Algebra Problem 13.10 Solution: Known quantities: The expression B = AB + A B . Find: The truth table that proves that the expression is true. Analysis: Using a truth table as explained in paragraph 13.3: A B A AB AB AB + A B 0 0 1 0 0 0 0 1 1 0 1 1 1 0 0 0 0 0 1 1 0 1 0 1 we prove that the expression is true. ______________________________________________________________________________________ Problem 13.11 Solution: Known quantities: The expression BC + BC + B A = A + B . Find: The truth table that proves that the expression is true. Analysis: Using a truth table as explained in paragraph 13.3: A B C BC BC BA BC + BC + B A A+B 1 1 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 0 1 0 0 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 we prove that the expression is true. ______________________________________________________________________________________ 13.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.12 Solution: Known quantities: ( ) The expression ( X + Y ) ⋅ X + X ⋅ Y = Y . Find: The proof that the expression is true using the perfect induction method. Analysis: Using a truth table as explained in paragraph 13.3: X Y X+Y X X ⋅Y X + X ⋅Y ( X + Y ) ⋅ (X + X ⋅ Y ) 0 0 0 1 0 1 0 0 1 1 1 0 1 1 1 0 1 0 0 0 0 1 1 1 0 1 1 1 we prove that the expression is true. ______________________________________________________________________________________ Problem 13.13 Solution: Known quantities: ( ) The logic function F ( X ,Y, Z ) = X ⋅ Y ⋅ Z + X ⋅ Y ⋅ Z + X ⋅ Y + Z . Find: The simplification of the expression using the rules of Boolean algebra and De Morgan's theorems. Analysis: Applying De Morgan's theorems, F ( X ,Y, Z ) = Applying the rules of Boolean algebra, X ⋅Y ⋅ Z + X ⋅Y ⋅ Z + X ⋅Y ⋅ Z F ( X ,Y, Z ) = X ⋅ Y ⋅ Z + X ⋅ Y ⋅ Z + X ⋅ Y ⋅ Z = Y ⋅ Z + X ⋅ Y ⋅ Z ______________________________________________________________________________________ Problem 13.14 Solution: Known quantities: The logic function f ( A, B , C , D ) = A ⋅ B ⋅ C + A ⋅ C ⋅ D + B ⋅ C ⋅ D . Find: The simplification of the expression using the rules of Boolean algebra and De Morgan's theorems. Analysis: f ( A, B, C , D) = ABC + ACD + BCD ( ) = ABC + CD A + B ← Distributive ______________________________________________________________________________________ 13.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.15 Solution: Known quantities: The logic function F ( A, B ,C ) = A ⋅ B ⋅ C + A ⋅ B ⋅ C + A ⋅ B ⋅ C + A ⋅ B ⋅ C . Find: The simplification of the expression using the rules of Boolean algebra. Analysis: Applying the rules of Boolean algebra, ( ) ( ) ( ) F ( A, B, C ) = A ⋅ B ⋅ C + C + A ⋅ B ⋅ C + C = A ⋅ B + A ⋅ B = A + A ⋅ B = B ______________________________________________________________________________________ Problem 13.16 Solution: Known quantities: The truth table in Figure P13.16 for a logic function. Find: The expression for the logic function. Analysis: A 0 BC 00 0 01 11 10 1 1 0 1 1 1 We first find the Karnaugh map for the function. 1 1 From the map, we can see that: F = A + C ______________________________________________________________________________________ Problem 13.17 Solution: Known quantities: The circuit of Figure P13.17. Find: The Boolean function describing the operation of the circuit. Analysis: F = AB ⋅ CD ⋅ E = AB + CD + E where the second expression is a result of applying DeMorgan’s theorem to the first. ______________________________________________________________________________________ Problem 13.18 Solution: Known quantities: The circuit of Figure P13.18. Find: The truth table of the circuit. 13.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Analysis: The truth table can be obtained from the circuit considering the steps A, B and F, as reported in the figure of the circuit. x3 x2 x1 A B F 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 x 3 x2 x1 A F B 1 1 1 0 0 0 ______________________________________________________________________________________ Problem 13.19 Solution: Known quantities: The rules that have to be followed in order to implement a strategy able to decide when the steal sign has to be given, for a baseball team. The steal sign has to be given if: a) there are no other runners, the pitcher is right-handed and the runner is fast, or b) there is no one other runner on third-base, and one of the runner is fast, or c) there is one other runner on second-base, the pitcher is left-handed, and both runners are fast. d) Under no circumstances should the steal sign be given if all three bases have runners. Find: The circuit that implements these rules. Analysis: The table below lists the 8 possible conditions under which the steal sign should be given, using the following logic notations: A runner on a base is 1, no runner is 0 A fast runner on a base is 1, a non-fast runner is 0 A right-handed pitcher is 1, a left-handed pitcher is 0 Base Runner Pitcher b1 b2 b3 f1 f2 f3 P 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 0 1 1 0 1 1 0 1 1 0 0 1 1 0 1 1 0 0 0 0 1 1 1 0 0 1 0 0 0 1 1 1 0 For all other conditions, the output is off. 13.8 Output Light On, y 1 1 1 1 1 1 1 1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Next, convert the truth table to a logical expression for the output: y = b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P y = b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 ⋅ P + (b ⋅ b 1 2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 + b1 ⋅ b2 ⋅ b3 ⋅ f1 ⋅ f 2 ⋅ f 3 )(P + P ) y = (b1 ⋅ b3 ⋅ f1 ⋅ f 3 )(b2 ⋅ f 2 ⋅ P + b2 ⋅ f 2 ⋅ P ) + (b1 ⋅ b2 ⋅ b3 )( f1 ⋅ f 2 ⋅ f 3 + f1 ⋅ f 2 ⋅ f 3 + f1 ⋅ f 2 ⋅ f 3 ) [ ] y = b1 (b3 ⋅ f1 ⋅ f 3 )(b2 ⋅ f 2 ⋅ P + b2 ⋅ f 2 ⋅ P ) + (b2 ⋅ b3 )( f1 ⋅ f 2 ⋅ f 3 + f1 ⋅ f 2 ⋅ f 3 + f1 ⋅ f 2 ⋅ f 3 ) From this final expression, a logic circuit diagram can be made to achieve the correct response: b2 b3 f1 f2 f3 p AND NOT AND NOT OR AND AND NOT AND b1 NOT AND y AND OR NOT AND NOT NOT AND AND NOT AND OR AND AND OR NOT AND AND NOT 13.9 AND G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Note: The circles in the figure are nodes, not inversions. ______________________________________________________________________________________ Problem 13.20 Solution: Known quantities: The rules to pass a law in a small county. Find: The logic circuit that takes three votes as inputs and lights either a green or red light to indicate whether or not a measure passed. Assumptions: Green is enlightened if the law passes. Analysis: The function that implements the rules is: A If f ( A, B ,C ) is equal to 1 then the green light is lighted, otherwise the red one. B f ( A, B ,C ) = AB + BC + AC C The circuit that implement this function is reported in the figure besides. ______________________________________________________________________________________ Problem 13.21 Solution: Known quantities: The set-up for a water purification plant. Find: The logic circuit that sounds an alarm if a dangerous situation occurs. Analysis: The four variables are the flow and the height of water in the tank A and in the tank B: Af, Ah, Bf, Bh. From the Karnaugh map reported below we can find the minimum expression: A f Ah (B f + Bh ) + B f Bh (A f + Ah ) and the following realization of the circuit. 13.10 G. Rizzoni, Principles and Applications of Electrical Engineering BfBh AfAh 00 01 11 10 00 0 0 0 0 01 0 0 1 0 11 0 1 0 1 10 0 0 1 0 Problem solutions, Chapter 13 Af Ag Bf Bh ______________________________________________________________________________________ Problem 13.22 Solution: Known quantities: The rules for an alert system designed for cars. Find: The logic circuit that implements these functions. Analysis: Let's design two different circuits for the buzzer and the starting conditions. For both circuits we find first the Karnaugh map and then the minimum expression and the circuit to implement it. A = Ignition key (1 if turned); B = Door (1 if closed); C = Seat belt (1 if fasten); D = Lights (1 if on); E = Park (1 if on); For the first circuit: Buzzer = ABC + AD 13.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 AB CD 00 01 11 10 00 0 0 0 1 A 01 1 1 0 1 D 11 1 1 0 0 B 10 0 0 0 0 C 01 11 10 Start = ABCE AB 00 CE 00 0 0 0 0 01 0 0 0 0 11 0 0 1 0 10 0 0 0 0 A B C E ______________________________________________________________________________________ Problem 13.23 Solution: Known quantities: The on/off strategy for the compressor motor of a large commercial air conditioning unit. Find: The logic diagram that incorporates the state of four devices (S, D, T and M) and produces the correct on/off condition for the motor startup. Analysis: From the Karnaugh map we determine the minimum expression: SDT + M that is implemented by the following logic diagram. 13.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 SD 00 01 11 10 00 0 0 0 0 01 1 1 1 1 11 1 1 1 1 10 0 0 1 0 T M S D T M ______________________________________________________________________________________ 13.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Section 13.4: Karnaugh Maps and Logic Design Focus on Methodology: Sum-of-Products Realizations 1. 2. 3. 4. Begin with isolated cells. These must be used as they are, since no simplification is possible. Find all cells that are adjacent to only one cell, forming two-cell subcubes. Find cells that form four-cell subcubes, eight-cell subcubes, and so forth. The minimal expression is formed by the collection of the smallest number of maximal subcubes. Focus on Methodology: Products-of Sums Realizations 1. 2. Solve for the 0s exactly as for the 1s in the sum-of-products expressions. Complement the resulting expression. Problem 13.24 Solution: Known quantities: The truth table of Figure P13.24. Find: The logic function corresponding to the truth table. Analysis: A From the map, we can see that: BC 00 01 11 10 0 1 0 0 0 1 1 0 1 1 F = A⋅ B + B ⋅C ______________________________________________________________________________________ Problem 13.25 Solution: Known quantities: The circuit shown if Figure P13.25. Find: The minimum expression for the output. Analysis: The logic circuit gives the following expression: ) ( )(( ( )( ) ( )(( )( = (C ⋅ D )(A ⋅ B ⋅ C + A ⋅ B ⋅ C ) = (C ⋅ D )(A ⋅ B ⋅ C + (A + B )⋅ C ) )( )) F = C + D A⋅ B ⊕ C = C ⋅ D A⋅ B + C ⋅ A⋅ B ⋅C = C ⋅ D A⋅ B + C ⋅ A⋅ B ⋅C = 13.14 G. Rizzoni, Principles and Applications of Electrical Engineering A AB Problem solutions, Chapter 13 AB⊕ C = (A+B)C + ABC B F(A,B,C,D) C ( C D ) ( (A+B)C + ABC ) D C+D = C D ______________________________________________________________________________________ Problem 13.26 Solution: Known quantities: The logic function f ( A, B ,C ) = A ⋅ B ⋅ C + A ⋅ B ⋅ C + A ⋅ B ⋅ C . Find: The minimum expression for the function. Analysis: Using a Karnaugh map (reported besides) we determine: f = AB + AC ______________________________________________________________________________________ Problem 13.27 Solution: Known quantities: The truth table of Figure P13.27. Find: a) The Karnaugh map for the logic function defined by the truth table. b) The minimum expression for the function. c) The realization of the function using AND, OR and NOT gates. Analysis: a) The Karnaugh map is shown below. 13.15 G. Rizzoni, Principles and Applications of Electrical Engineering CD Problem solutions, Chapter 13 AB 00 01 11 10 00 1 0 0 1 C B 01 1 0 0 1 B A B D A B C 11 0 1 0 1 10 1 1 0 1 b) The map leads to the expression: F f = B ⋅C + A⋅ B + A⋅ B ⋅C + B ⋅ D c) and to the gate realization shown above. ______________________________________________________________________________________ Problem 13.28 Solution: Known quantities: The truth table of Figure P13.28. Find: The Karnaugh map and the minimum expression for the function defined by the truth table. Analysis: The Karnaugh map is shown besides. The minimum expression is reported below. BC A 00 01 11 10 0 0 1 0 1 1 1 0 1 0 f ( A, B ,C ) = A ⋅ B ⋅ C + A ⋅ B ⋅ C + A ⋅ B ⋅ C + A ⋅ B ⋅ C ______________________________________________________________________________________ Problem 13.29 Solution: Known quantities: The rules that define a logic function. Find: The minimum expression for this function and a sketch of a circuit to implement this function using only AND and NOT gates. Analysis: The truth table for the function is reported below as well as the Karnaugh map the minimum expression for the function and its realization using AND and NOT gates. 13.16 G. Rizzoni, Principles and Applications of Electrical Engineering A B C D F 0 0 0 0 0 0 0 0 1 X 0 0 1 0 0 0 0 1 1 1 0 1 0 0 X 0 1 0 1 X 0 1 1 0 1 0 1 1 1 X 1 0 0 0 0 1 0 0 1 1 1 0 1 0 0 1 0 1 1 X 1 1 0 0 1 1 1 0 1 X 1 1 1 0 X 1 1 1 1 1 Problem solutions, Chapter 13 F = B+D ______________________________________________________________________________________ Problem 13.30 Solution: Known quantities: The truth table that defines a logic function shown in Figure P13.30. Find: The construction of a logic circuit describing the function using only two gates. Analysis: SOP: POS: F = AB + BC Ö 3 gates F = B + A ⋅ C ⇒ F = B( A + C ) Ö 2 gates 13.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 ______________________________________________________________________________________ Problem 13.31 Solution: Known quantities: The requirement for the circuit to be designed: to produce the one's complement of an 8-bit signed binary number. Find: The design of the logic circuit. Analysis: ______________________________________________________________________________________ Problem 13.32 Known quantities: The truth table of Figure P13.32. Find: The Karnaugh map and the minimum expression for the logic function. Analysis: The Karnaugh map and the expression for the function are reported below. F = B ⋅ D + A⋅ D + A⋅ B ⋅C + A⋅ B ⋅C ⋅ D CD 00 01 11 10 00 1 0 0 1 01 0 0 1 0 11 1 1 0 1 10 1 0 0 1 AB Solution: ______________________________________________________________________________________ Problem 13.33 Solution: Known quantities: The requirement for the circuit to be designed: to produce the two's complement of an 8-bit signed binary number. Find: The design of the logic circuit. Analysis: The two’s complement is the one’s complement plus one. 13.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 X CIN SUM COUT 0 0 0 0 SUM = X ⊕ C IN 0 1 1 0 COUT = X • C IN 1 0 1 0 1 1 0 1 ______________________________________________________________________________________ Problem 13.34 Solution: Known quantities: The circuit of Figure P13.34. Find: The minimum output expression for the circuit. Analysis: f = A ⋅ B + B + C = A ⋅ B ⋅ B + C = A ⋅ B ⋅ (B + C ) = A ⋅ B (1 + C ) = A ⋅ B ______________________________________________________________________________________ Problem 13.35 Solution: Known quantities: The combinational logic to be designed: the addition between two 4-bit binary numbers. Find: The circuit to implement this operation. Analysis: A one-bit adder truth table is as follows, and from this table we find the following expression, and the two circuits for those functions: 13.19 G. Rizzoni, Principles and Applications of Electrical Engineering CIN x y SUM COUT 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Problem solutions, Chapter 13 SUM = C IN ⊕ x ⊕ y COUT = C IN + xy The complete 4-bit adder can be constructed as shown below: Note that this circuit assumes a carry-in for the lsb. If this is not necessary, then the circuit can be reduced correspondingly. ______________________________________________________________________________________ Problem 13.36 Solution: Known quantities: The truth table of Figure P13.36. Find: The minimum output expression and the circuit for the function. Analysis: Reported below are the Karnaugh map, the expression for the function and circuit realization. BC 00 01 11 10 A C 0 1 1 1 0 A C 1 1 1 0 1 A B 13.20 F G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 F = B + A⋅C + A⋅C ______________________________________________________________________________________ Problem 13.37 Solution: Known quantities: The logic circuit of Figure P13.37. Find: The minimum output expression for the circuit. Analysis: AB = A + B A B F C BC + BC f(A,B,C) = ABC ______________________________________________________________________________________ Problem 13.38 Solution: Known quantities: The rules for blood donation. Find: The circuit which will approve or disapprove any particular transfusion between donator and receiver. Analysis: Let WX represent a 2-bit code for the donor blood type, and let yz represent a 2-bit code for the recipient blood type. Then WXYZ will represent a donor-recipient pair. Let F be true if a transfusion can be made. Blood type codes may be assigned as follows: WX YZ A 00 00 B 01 01 AB 10 10 O 11 11 The truth table is: 13.21 G. Rizzoni, Principles and Applications of Electrical Engineering Donor−Recipient WX YZ F A-A 00 00 1 A-B 00 01 0 A-AB 00 10 1 A-O 00 11 0 B-A 01 00 0 B-B 01 01 1 B-AB 01 10 1 B-O 01 11 0 AB-A 10 00 0 AB-B 10 01 0 AB-AB 10 10 1 AB-O 10 11 0 O-A 11 00 1 O-B 11 01 1 O-AB 11 10 1 O-O 11 11 1 Problem solutions, Chapter 13 And the Karnaugh map, then, is as follows: From the Karnaugh map, F = W ⋅ X ⋅ Z + X ⋅ Y ⋅ Z + WX + Y Z and the resulting circuit is shown above. ______________________________________________________________________________________ 13.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.39 Solution: Known quantities: The logic circuit of Figure P13.39. Find: The minimum expression for the logic function. Analysis: The minimum expression is: A A+B B F = A + B + B + C + D = ( A + B ) ⋅ (B + C ) ⋅ D B+C C = (B + A ⋅ C ) ⋅ D = B ⋅ D + A ⋅ C ⋅ D f D ______________________________________________________________________________________ Problem 13.40 Solution: Known quantities: The combinational logic that have to be followed. Find: The circuit that implements this logic. Analysis: The appropriate truth table can be constructed as follows: ABCD 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 F4 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 F3 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 F2 0 0 0 1 0 0 0 1 1 0 1 1 1 0 1 1 F1 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 F0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 Next, we construct a Karnaugh map for each bit of the output, and determine its corresponding function. 13.23 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 F4 = AD F3 = BC F2 = AB + AD F1 = B + D F0 = C D This completes the design. 13.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 ______________________________________________________________________________________ Problem 13.41 Solution: Known quantities: The truth table of Figure P13.41. Find: a) The Karnaugh map for the function defined in the truth table. b) The minimum expression for the function. c) The circuit using AND, OR, NOT gates. Analysis: The answers are reported in the following plots. BC A 0 00 01 11 10 1 1 1 0 1 1 1 0 A C 1 A C B f = B + AC + AC F ______________________________________________________________________________________ Problem 13.42 Solution: Known quantities: The truth table of Figure P13.42. Find: a) The Karnaugh map for the function defined in the truth table. b) The minimum expression for the function. 13.24 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: The answers are reported in the following plots. a) CD 00 01 11 10 00 1 0 0 1 01 0 0 1 0 11 1 1 0 1 10 1 0 0 1 AB Problem solutions, Chapter 13 b) F = B ⋅ D + A⋅ D + A⋅ B ⋅C + A⋅ B ⋅C ⋅ D ______________________________________________________________________________________ Problem 13.43 Solution: Known quantities: The truth table of Figure P13.43. Find: a) The Karnaugh map for the function defined in the truth table. b) The minimum expression for the function. c) The circuit implementation using only NAND gates. Analysis: The answers are reported in the following plots. a) b) CD AB 00 00 01 11 10 1 1 1 1 01 0 1 1 0 11 0 1 0 1 10 1 1 0 0 F = A⋅ B + A⋅ D + C ⋅ D + B ⋅C + A⋅ B ⋅C ⋅ D c) 13.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 A B A D C D F B C A B C D ______________________________________________________________________________________ Problem 13.44 Solution: Known quantities: The logic that defines a function. Find: a) The Karnaugh map for the function and the truth table. b) The minimum expression for the function. c) The circuit using only AND, OR, and NOT gates. Analysis: a) A3 A2 A1 A0 F 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 1 1 0 1 0 d 1 0 1 1 d 1 1 0 0 d 1 1 0 1 d 1 1 1 0 d 1 1 1 1 d b) From K-Map groupings, F = A3 ⋅ A0 + A2 ⋅ A1 ⋅ A0 + A1 ⋅ A0 ⋅ A2 c) The circuit for this function is: 13.26 A1 A0 A3 A2 00 01 11 10 00 01 0 0 1 0 0 0 0 1 11 10 d d d d 0 1 d d G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 A3 A 0 A2 A1 A0 F A0 A2 A1 ______________________________________________________________________________________ Problem 13.45 Solution: Known quantities: The Karnaugh map reported in Figure P13.45. Find: The simplified sum-of-products representation of the function. Analysis: AB CD 11 10 0 01 1 0 0 1 1 0 0 0 x 1 0 0 0 1 0 00 00 01 11 10 F = A ⋅ C ⋅ D + A ⋅ B ⋅C + A ⋅ B ⋅ C ______________________________________________________________________________________ Problem 13.46 Solution: Known quantities: The semplification for the circuit of Problem 13.40 if it is known that the input represents a BCD (binary-coded decimal) number. Find: The simplified circuit or the reason for which it is not possible to simplify the circuit. Analysis: For this problem, the truth table is as follows: 13.27 G. Rizzoni, Principles and Applications of Electrical Engineering ABCD 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 F4 0 0 0 0 0 0 0 0 0 1 X X X X X X F3 0 0 0 0 0 1 0 1 0 0 X X X X X X F2 0 0 0 1 0 0 0 1 1 0 X X X X X X F1 0 1 0 1 1 1 1 1 0 1 X X X X X X Problem solutions, Chapter 13 F0 0 0 1 0 0 0 1 0 0 0 X X X X X X Now, we construct the Karnaugh maps and determine the corresponding functions. F3 = BC F4 = AD F1 = B + D F2 = AB + AD 13.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 F0 = C D These expressions are identical to those obtained in Problem 12.36. Surprisingly, the presence of the don’t cares did not change (or simplify) the solution. _____________________________________________________________________________________________ Problem 13.47 Solution: Known quantities: The Karnaugh map shown in Figure P13.47. Find: The simplified sum-of-product representation of the function. Analysis: AB 00 CD 00 0 01 11 10 01 1 11 10 x 0 0 1 x 0 0 1 0 1 x x 1 0 F = A⋅ B ⋅C ⋅ D + B ⋅ D + A⋅ B ______________________________________________________________________________________ 13.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.48 Solution: Known quantities: The parity bit method to ensure reliability in data transmission systems. Find: The logic circuit that checks the nibble of data and transmits the proper parity bit for even-parity systems. Analysis: If we write the nibble in the form ABCD, the Karnaugh map for this problem is: CD 00 01 11 10 00 0 1 0 1 01 1 0 1 0 11 0 1 0 1 10 1 0 1 0 AB A B C D Hence the function that gives the parity bit can be written as: F = A BC D + ABC D + ABC D + ABC D + ABCD + ABCD + ABC D + ABC D The logic circuit is shown above. ______________________________________________________________________________________ Problem 13.49 Solution: Find: The logic circuit that check a nibble of data and its parity-bit. Analysis: If we write the nibble in the form ABCD and E is the parity-bit, the logic circuit is: A B C D E Parity-bit evaluation ______________________________________________________________________________________ 13.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.50 Solution: Find: A logic circuit that takes a 4-bit gray code input into a 4-bit nibble of BCD code. Analysis: A A1 B C D B1 C1 D1 _____________________________________________________________________________________________ Problem 13.51 Solution: Find: A logic circuit that takes a 4-bit gray-code Analysis: AB CD 00 01 11 10 00 0 0 0 1 01 0 1 1 0 11 0 0 0 0 A 10 1 0 0 1 C B D F F = BC D + ABC D + BC D ______________________________________________________________________________________ 13.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.52 Solution: Find: A logic circuit that takes a BCD nibble as input and converts it into its 4221 equivalent, and reports an error if the BCD value exceeds 1001. Analysis: A B A1 B1 C1 C D D1 ______________________________________________________________________________________ Problem 13.53 Solution: Find: A logic circuit that reports an error if the 4-bit outputs of two sensors differ by more than one part per 30 seconds period. Analysis: The output of the logic circuit will be one when a difference greater or equal to 2 is detected between the two sensor outputs. By defining with a1, a2, a3, a4 the four bits of the first sensor, and with b1, b2, b3, b4 the bits of the second sensor, the following Karnaugh maps can be derived. From the maps, considering the zeros, the following expression can be obtained y = (a1 + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + b1 + b2 + b3 ) (a (a (a (a (a (a (a 1 1 1 1 1 1 1 + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + b1 + b2 + b3 ) + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + b1 + b2 + b3 )(a1 + a2 + a3 + a4 + b1 + b2 + b4 ) + a2 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b4 ) + a2 + a3 + a4 + b1 + b2 + b4 )(a1 + a2 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a4 + b1 + b2 + b3 + b4 ) + a2 + a3 + a4 + b1 + b2 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 ) + a2 + a3 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 ) + a2 + a3 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 ) 13.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 The expression can be simplified as follows y = (a1 ⊕ b1 + a 2 + a3 + b2 + b3 )(a1 ⊕ b1 + a2 + a3 + b2 + b3 )(a1 ⊕ b1 + a2 + a3 + b2 + b3 ) (a (a (a 1 1 1 ⊕ b1 + a2 + a3 + b2 + b3 )(a1 ⊕ b1 + a2 + a3 + a4 + b2 + b4 )(a1 ⊕ b1 + a2 + a 4 + b2 + b3 + b4 ) ⊕ b1 + a2 + a4 + b2 + b3 + b4 )(a1 ⊕ b1 + a2 + a3 + a 4 + b2 + b4 )(a1 ⊕ b1 + a2 + a3 + a4 + b2 + b3 + b4 ) ⊕ b1 + a2 + a3 + a4 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 )(a1 + a2 + a3 + a4 + b1 + b2 + b3 + b4 ) Using De Morgan’s Theorem y = (a1 ⊗ b1 )(a2 a3b2b3 + a2 a3b2b3 + a2 a3b2b3 + a2 a3b2b3 + a2 a3a4b2b4 + a2 a4b2b3b4 + a2 a4b2b3b4 + 13.33 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 + a 2 a3 a 4 b2b4 + a 2 a3 a 4 b2b3b4 + a 2 a3 a 4 b2 b3b4 ) + a1a 2 a3 a 4b1b2b3b4 + a1a 2 a3 a 4 b1b2 b3b4 = (a1 ⊗ b1 )((a 2 ⊗ b2 )(a3 ⊗ b3 ) + (a 2 ⊗ b2 )(a3 a 4 b4 + a 4 b3b4 ) + a 2 a3 a 4 b2 b3b4 + a 2 a3 a 4b2 b3b4 ) + + a1a 2 a3 a 4 b1b2 b3b4 + a1a 2 a3 a 4b1b2b3b4 From the expression above, it is possible to derive the following logic circuit _____________________________________________________________________________________ 13.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Section 13.5: Combinational Logic Modules Problem 13.54 Solution: Known quantities: The truth table for a function. Find: a) The Karnaugh map for the function; b) Its minimum expression c) Realize the function using a 1-of-8 multiplexer Analysis: a) AB CD 00 01 11 10 00 1 0 1 0 01 0 1 0 1 11 1 0 1 0 10 1 0 1 0 b) ( )( ) ( F = AB D + ABC + ABC D + AB D + ABC + ABC D = AB + AB C + D + C D AB + AB ) c) A I1 I2 I3 I4 I5 I6 I7 I8 F B C D _____________________________________________________________________________________________ 13.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.55 Solution: Known quantities: The multiplexer circuit shown in Figure P13.55. Find: a) The truth table for the multiplexer b) The binary function performed by the multiplexer. Analysis: a) x y C S 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 b) Binary Addition - S is the sum, and C is the carry. ______________________________________________________________________________________ Problem 13.56 Solution: Known quantities: A circuit that can operate as a 4-to-16 decoder is shown in Figure P13.56. Find: The operation of the 4-to-16 decoder, and the role of the logic variable A. Analysis: Assuming that the enable input (EN) is active high, when EN is logic 0 ( A is logic 1), all decoder outputs of the first decoder are forced to logic 1 independent of the inputs. However, when EN is logic 1 ( A is logic 0), all decoder outputs of the second decoder are forced to logic 1 independent of the select inputs. Therefore, A functions as the fourth bit of the select inputs. Thus, the circuit operates as a 4 of 16 decoder. ______________________________________________________________________________________ Problem 13.57 Solution: Known quantities: The circuit of Figure P13.57. Find: The ability of the circuit of performing a conversion from 4-bit binary numbers to 4-bit Gray code. Analysis: We construct the truth table for this circuit as shown below: 13.15 G. Rizzoni, Principles and Applications of Electrical Engineering Binary Input B3 B2 B1 B0 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 G3 G2 G1 G0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 B2 ⊕ B3 B1 ⊕ B2 Problem solutions, Chapter 13 B0 ⊕ B1 The output is clearly a Gray code since each number only changes by one bit relative to the previous number. ______________________________________________________________________________________ Problem 13.58 Solution: Known quantities: Four Boolean expressions. Find: a) Show that these four expressions represent the conversion from 4-bit Gray code to 4-bit binary numbers. b) Draw the circuit which implements the conversion. Analysis: a) Note that: B3 = G3 B2 = G3 ⊕ G2 = B3 ⊕ G2 B1 = G3 ⊕ G2 ⊕ G1 = B2 ⊕ G1 B0 = G3 ⊕ G2 ⊕ G1 ⊕ G0 = B1 ⊕ G0 Then, the truth table is: B3 G3G2 G1G0 G3 0000 0001 0010 0011 0100 0101 0110 0 0 0 0 0 0 0 B2 B1 B0 0 0 0 0 1 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 G3 ⊕ G2 B2 ⊕ G1 B1 ⊕ G0 13.33 0111 1000 1001 1010 1011 1100 1101 1110 1111 0 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 0 1 0 1 0 1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 The table verifies that the claim is correct. b) The circuit is shown below: ______________________________________________________________________________________ Problem 13.59 Solution: Known quantities: The function f ( A, B ,C ) = ABC + A BC + AC Find: The inputs for a 4-input multiplexer to implement the function. Assumptions: The inputs I 0 , I 1 , I 2 , I 3 correspond to AB , AB , AB and AB respectively, and each input may be 0 , 1 , C or C . Analysis: f = ABC + ABC + AC From the truth table it is clear that: I0 = 0 I1 = C I2 = 1 and I3 = C ______________________________________________________________________________________ 13.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 13 Problem 13.60 Solution: Known quantities: The function f ( A, B ,C , D ) = ∑ (2,5,6,8,9,10,11,13,14) 10 . Find: The inputs for an 8-bit multiplexer to implement the function. Assumptions: The inputs I 0 through I 7 correspond to A BC , ABC , ABC , ABC , A BC , A BC , ABC and ABC respectively, and each input may be 0 , 1 , D or D . Analysis: From the truth table it is clear that: I0 = 0 I1 = D I2 = D I3 = D I4 = 1 I5 = 1 I6 = D and I7 = D ______________________________________________________________________________________ 13.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Chapter 14 Instructor Notes Chapter 14 logically follows the material on combinational digital logic circuits introduced in Chapter 13. Section 14.1 contains a discussion of sequential logic modules; the box Focus on Measurements: Digital Measurement of Angular Position and Velocity (pp. 679-681) illustrates the use of a counter to measure the speed of rotation of a slotted wheel; this is a very common measurement in mechanical systems; the box Focus on Measurements: Seven-Segment Display (pp. 682-683) draws attention to a very common logic module. Section 14.2 provides an elementary introduction to sequential logic design, and Section 14.3 introduces microcomputers. This, and the two following sections on microcomputer architecture and microcontrollers, have been extensively revised, and are designed to provide an introduction to the concept of mechatronic design (see box on p. 688), recognizing that it is impossible to include a serious coverage of microcontrollers in a single chapter. Thus, the treatment is focused on an overview of the organization of microcontrollers, including a brief, qualitative illustration of computer interface issues presented in Focus on Measurements: Reading Sensor Data By Using Interrupts (pp. 697-698). The last section, 14.6, is dedicated to an example (courtesy of Delphi – Delco Electronic Systems) that describes a current automotive engine microcontroller. The example is a qualitative description, but will permit the instructor who desires to do so to motivate further study of this topic. The Find-It-On-The-Web references will provide students and instructor with additional reference material on the subject of automotive engine control. In recent years mechanical and industrial engineering programs have seen a significant growth in courses related to mechatronics, or more specifically to microcontroller applications in industrial and mechanical systems. The objective of Chapter 14 is to serve as an introduction to such courses. The homework problems are mostly devoted to sequential logic circuits; the last 6 problems review simple concepts related to the architecture and functions of microcomputers. Learning Objectives 1. Analyze the operation of sequential logic circuits. Section 1. 2. Understand the operation of digital counters. Section 1. 3. Design simple sequential circuits using state transition diagrams. Section 2. 4. Study the basic architecture of microprocessors and microcomputers. Sections 3,4, 5. 14.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Section 14.1: Sequential Logic Modules Section 14.2: Sequential Logic Design Problem 14.1 Solution: Known quantities: For the circuit shown in Figure P14.1, the input is a square wave having a period of 2 s, maximum value of 5 V, and minimum value of 0 V. Assume all flip-flops are initially in the RESET state. Find: a) Explain what the circuit does. b) Sketch the timing diagram, including the input and all four outputs. Analysis: 4 a) The device is called a MOD-16 ripple counter. It can count clock pulses from 0 to (2 -1). The 1 2 3 4 outputs divide the frequency by 2 , 2 , 2 , and 2 respectively. Therefore, you can use this circuit as a divide by N counter, where N is 2, 4, 8 and 16. b) ______________________________________________________________________________________ Problem 14.2 Solution: Find: a) How many flip-flops would be required to construct a binary pulse counter, which can count up to 10010, by interconnecting T-type flip-flops in an appropriate manner. b) Sketch the circuit needed to implement this counter. Analysis: a) 10010 = 1100100 2 7 flip flops required. b) This circuit could be modified with combinational logic if it is desired to have it reset at 10010 and start counting again from 00000002. ______________________________________________________________________________________ Problem 14.3 14.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Solution: Known quantities: The circuit shown in Figure P14.3. Find: Explain what the circuit does and how it works. Analysis: The basic operation of the circuit is to count up when X = 0, and to count down when X = 1. X clk T1 Output #1 Output #2 counting UP 00 01 10 11 counting DOWN 00 11 10 01 00 11 Clock X Output #2 T1 Output #1 ↑ 0 0 0 No change ↑ 1 1 0 No change ↑ 0 1 1 toggle ↑ 1 0 1 toggle ______________________________________________________________________________________ Problem 14.4 Solution: Known quantities: If a circuit is constructed from 3 D-type flip-flops, with: D0 = Q 2 D1 = Q2 ⊕ Q0 D2 = Q1 . Find: a) Draw the circuit diagram. b) Assuming the circuit starts with all flip-flops SET, sketch a timing diagram which shows the outputs of all three flip-flops. Analysis: a) 14.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 b) ______________________________________________________________________________________ Problem 14.5 Solution: Find: Assuming you have all the logic gates available, make a D flip-flop using a T flip-flop and some logic gate(s). Analysis: Q T Flip-Flop T D D Flip-Flop Assume that Q is logic 0 initially. If the input D is logic 0, the output of the gate is logic 0 which means that the output of the T flip-flop will not be toggled (i.e., will remain logic 0). When the input D is logic 1, the output of the gate will be logic 1. Therefore, the output of the flip-flop will be 1. Thus, the circuit will operate as a D flip-flop. ______________________________________________________________________________________ Problem 14.6 Solution: Known quantities: For the circuit shown in Figure P14.6, assume that all the initial values are 0. Note that all the flip-flops are negative edge-triggered. Find: Draw a timing diagram (four complete clock cycles) for A0, A1, and A2. 14.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Analysis: CLK A0 A1 J =1 K=1 J =K=0 J =1 K=1 J =K=1 J =1 K=0 J =0 K=0 J =K=1 J =K=0 J =1 J =1 J =0 J =1 K=0 K=0 K=1 K=0 ______________________________________________________________________________________ A2 Problem 14.7 Solution: Known quantities: Assume that the slotted encoder shown in Figure P14.7 has a length of 1 meter and a total of 1,000 slots (i.e., there is one slot per millimeter. Find: If a counter is incremented by 1 each time a slot goes past a sensor, design a digital counting system that determines the speed of the moving encoder (in meters per second). Analysis: Assuming a maximum speed of 10 m/s and a minimum speed of 1 mm/s, we can calculate the instantaneous speed of the slotted encoder by counting the number of clock pulses between slots using a fixed frequency clock. This resolution should be sufficient to measure the speed of the encoder over the range of interest. The figure depicts the arrangement: A 10 kHz clock increments a 16-bit binary counter. 16 = 65,536, will be the -4 maximum count between slots. At a speed of 10 m/s, the time for one slot to go by is 10 s, thus the number of counts would be 1 count; at the minimum speed of 1 mm/s the number of counts would be 4 10 . A 14-bit counter would be sufficient, but in practice it is easier to cascade two 8-bit counters; thus the The choice for a 16-bit counter is due to the maximum speed requirement: 2 choice of a 16-bit counter. The count is held by a latch, and then converted to BCD for use with sevensegment displays. The details of the seven-segment display encoders are not shown (see Focus on Measurements: Seven-Segment Display). If a decimal point is placed to the right of the second sevensegment display (starting from the left in the figure), the display will read the speed in m/s, up to a maximum of 10 m/s, with a resolution of 1 mm/s. 14.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 clock f = 10 kHz 1 mm signal from linear position encoder reset 16-bit binary counter CK 16-bit latch BCD output binary-todecimal converter 7-segment displays ______________________________________________________________________________________ Problem 14.8 Solution: Known quantities: The circuit shown in Figure P14.8. Find: The output Q for the given circuit. Analysis: Output 1 2 3 4 5 6 7 8 9 10 11 t ______________________________________________________________________________________ Problem 14.9 Solution: Find: Describe how the ripple counter works. Why is it so named? What disadvantages can you think of for this counter? Analysis: This is briefly discussed in the digital counters section (see Figure 14.12). ______________________________________________________________________________________ Problem 14.10 14.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Solution: Find: Write the truth table for an RS flip-flop with enable (E), preset (P), and clear (C) lines. Analysis: Knowing that an input to the R or S line will be effective only when the enable input is 1, and the outputs are initially 0, the truth table for an RS flip-flop with set and preset is as follows: S R P C Q 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 0 1 1 0 1 0 1 ______________________________________________________________________________________ Problem 14.11 Solution: Known quantities: The JK flip-flop shown in Figure P14.11, with a given input signal. Find: Assuming that Q is at logic 0 initially and the trailing edge triggering is effective, sketch the output Q. Analysis: Jn Kn 1 1 Qn+1 Q n (toggle ) Input Output ______________________________________________________________________________________ Problem 14.12 Solution: Known quantities: With reference to the JK flip-flop shown in figure P14.11, assume that the output at the Q terminal is made to serve as the input to a second JK flip-flop wired exactly as the first. Find: Sketch the output Q of the second flip-flop. 14.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 Analysis: Jn Kn 1 1 Qn+1 Q n (toggle ) Input Output ______________________________________________________________________________________ Problem 14.13 Solution: Known quantities: The flip-flop with the characteristic given in Figure P 14.13, where A and B are the inputs to the flip-flop and Q is the next state output. Find: Using necessary logic gates, make a T flip-flop from the given flip-flop. Analysis: T A B Q 0 1 0 1 0 0 q q T A B Q Q T Flip-Flop ______________________________________________________________________________________ Section 14.3: Microprocessors Section 14.4: Computer System Architecture Section 14.5: Microcontrollers Problem 14.14 Solution: Find: If a typical PC has 32 Mbytes of standard memory: a) How many words is this? 14.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 b) How many nibbles is this? c) How many bits is this? Analysis: 32 Mbytes = 32 × 2 20 bytes = 33554432 bytes 1 word a) 33554432 bytes × = 16777216 words 2 bytes 2 nibbles b) 33554432 bytes × = 67108864 nibbles 1 byte 8 bits c) 33554432 bytes × = 268435456 bits 1 byte ______________________________________________________________________________________ Problem 14.15 Solution: Find: Suppose a microprocessor has n registers. a) How many control lines do you need to connect each register to all other registers? b) How many control lines do you need if a bus is used? Analysis: a) n ⋅ (n − 1) . b) 2n . ______________________________________________________________________________________ Problem 14.16 Solution: Find: Suppose it is desired to implement a 4K 16-bit memory. a) How many bits are required for the memory address register? b) How many bits are required for the memory data register? Analysis: We need 2 = 4Kbytes = 4096 bits N = 12 Therefore, we need 12 bits for the memory address register. b) The data register must be at least as large as each word in memory. Therefore, the data register must be 16 bits in length. ______________________________________________________________________________________ N a) Problem 14.17 Solution: Find: What is the distinction between volatile and nonvolatile memory. Analysis: “Volatile” memory is memory whose contents are lost when the power is turned off. This is the RAM in a computer. "Nonvolatile" means that the information in the memory is not lost when the power is off. This is a magnetic disk, magnetic tape or ROM in a computer. 14.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 ______________________________________________________________________________________ Problem 14.18 Solution: Known quantities: Suppose that a particular magnetic tape can be formatted with 8 tracks per centimeter of tape width. The recording density is 200 bits/cm, and the transport mechanism moves the tape past the read heads at a velocity of 25 cm/s. Find: How many bytes/s can be read from a 2-cm-wide tape. Analysis: 8tracks × 2cm = 16tracks cm bits bits = 3200 16tracks × 200 cm cm bits 1byte bytes × = 400 3200 cm 8bits cm bytes cm bytes × 25 = 10000 400 cm s s ______________________________________________________________________________________ Problem 14.19 Solution: Find: Draw a block diagram of a circuit that will interface two interrupts, INT0 and INT1, to the INT input of a CPU so that INT1 has the higher priority and INT0 has the lower. In other words, a signal on INT1 is to be able to interrupt the CPU even when the CPU is currently handling an interrupt generated by INT0, but not vice versa. Analysis: There are two types of interrupts: non-maskable, and maskable. When a logic signal is applied to a maskable interrupt input (INT1 in this case), the microprocessor is immediately interrupted. When a logic signal is applied to a maskable input (INT0 in this case), the microprocessor is interrupted only if that particular input is enabled. Maskable interrupts are disabled or enabled under program control. 14.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 14 ______________________________________________________________________________________ 14.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Chapter 15 Instructor Notes Chapter 15 continues the discussion of integrated circuit electronics begun in Chapter 8with op-amps. The Chapter is extremely modular, and the degree of coverage can vary widely, depending on the requirements of each individual Instructor. The first two sections cover measurement systems and transducers, providing asummary in Table 15.1 (p. 712), and noise problems. The third covers the instrumentation amplifier (which had been introduced in Chapter 8) in greater depth. This section might be of interest by itself as an extension of Chapter 8, and can be covered immediately following Section 8.2. Section 15.3 also discusses practical active filters, focusing on Butterworth and Chebyshev designs, and isa logical continuation of Section 8.3 on active filters; the material in this section is fairly advanced, and will require the student to have had a rigorous introduction to the Laplace Transform and to systems concepts. Thus, this section will be appropriate for a second course on electronics and instrumentation. The active filter design material would be nicely complemented by laboratory exercises; the importance of filter design and analysis in instrumentation problems is not to be underestimated. This material could also be supplemented very effectively by a review of filter design and analysis procedures using computer aids (e.g., MATLAB® or Electronics Workbench®). The material in Section 15.4, on the subject of signal interface (A/D, D/A conversion and sample-and-hold amplifiers) is, on the other hand, much more quantitative, and can be presented separately, for example, in conjunction with Chapter 8, or with Chapters 13 and 14. There are excellent possibilities for very useful laboratory experiments in connection with this material. The emphasis is on illustrating the important parameters and performance limitations in the application of commercial ADCs, DACs, and sample-and-hold amplifiers. A commercial data acquisition board is also described in Focus on Measurements: Data Acquisition Card for Personal Computer (pp. 747-750). It is this Author's opinion that this material is of great practical importance to non-majors, many of whom will at some point make use of a microcomputer-based digital data acquisition system, regardless of their specialty. Section 15.5 can also be viewed as an extension of Chapter 8, and could be covered in a first course. Again, the section is independent of the other sections in the chapter. In this section, the op-amp comparator and the Schmitt trigger are introduced first. The section closes with a functional description of timing circuits including a one-shot IC (74123) and the NE 555 timer IC. Section 15.6 provides a summary of other IC instrumentation circuits, togive the student a flavor of the capabilitiesof modern integrated electronics. The box Focus on Measurements: Using the ADXL202 Accelerometer as a Multifunction Sensor in Car Alarms (pp. 762-766) provides a detailed look at an application note (Courtesy: Analog Devices). The aim of this box is to present Application Notes as a valuable design resource for practicing engineers. The last section, 15.7, is somewhat unusual for a textbook of this nature in covering the basic elements of digital data transmission. However, given the pervasive presence of digital instruments and microcomputers in engineering laboratories and in the field, a brief survey of the IEEE 488 and RS-232 standards can prove extremely useful in practice. New in the 4th Edition are a brief overviewof the USB and CAN bus standards. The homework problems present a variety of analysis and design problems on instrumentation amplifiers and active filters. Several design problems are also given to complement the section on timing circuits; a few of the problems require the student to explore the data sheets for the AD 625 instrumentation amplifier, the 555 timer, and the 74123 one-shot. Although these problems are fairly simple, they can be used to educate the student to search for design parameters in the data sheets. The data sheets are provided in the CD-ROM and on the website.. Also included is a series of problems on DAC and ADC analysis and design. Some emphasis is again placed on reading and understanding the device data sheets of commercial ADCs and DACs. Issues in sampling frequency selection and resolution are approached in a few applied problems, where practical measurement situations pertaining to the measurement of angular position (problem 15.59), torque (problem 15.61), and altitude (problem 15.62) are described. The chapter problems end with a few simple problems on data transmission and coding. Learning Objectives 1. Review the major classes of sensors. Section 1. 2. Learn how to properly ground circuits, and methods for noise shielding and reduction. Section 2. 3. Design signal conditioning amplifiers and filters. Section 3. 4. Understand A/D and D/A conversion, and select the specifications of a the appropriate conversion system for a given application. Section 4. 5. Analyze and design simple comparator and timing circuits using integrated circuits. Review other common instrumentation integrated circuits. Section 5, 6. 6. Learn basics of data transmission in digital instruments. Sections 6. 15.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Section 15.1: Measurement Systems and Transducers Problem 15.1 Solution: Find: Explain the differences between tachometers and speedometers. Analysis: Frequency - engine speed is normally measured right at the crankshaft prior to any gearing in rpm typically several thousands. The transducers used with speedometers measure speed at the axle in rpm typically much lower than at the engine output. Scale factors - the tachometer would require none, the speedometer requires a conversion factor from rpm of the axle (rotational) to mph (linear). ______________________________________________________________________________________ Problem 15.2 Solution: Find: Explain the differences between the engineering specifications you would write for a transducer to measure the frequency of an audible sound wave and a transducer to measure the frequency of a visible light wave. Analysis: Audio frequencies: 0 < f ≤ 15 kHz Visible frequencies: 3.9 × 10 Hz ≤ f ≤ 7.9 × 10 Hz Devices used to measure quantities at audio frequencies will be incapable of sensing or measuring accurately those same quantities at frequencies in the visual range. Various types of photocells are available for use as light sensors. For audio frequencies, more conventional devices like bridges may be used for measuring signals. ______________________________________________________________________________________ 14 14 Problem 15.3 Solution: Find: A simple schematic diagram of a circuit to compute the sum of a temperature and a percentage relative humidity signal. Analysis: Use a transducer that will convert temperature in degrees Fahrenheit to volts between the values of the 2sided supply voltage. Similarly measure the percentage relative humidity and convert the transducer output to the requisite voltage. 15.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Summing Amplifier RF Temp. Transducer RT V+ - Humidity Transducer Output (Note: Output will be inverted) + RH V- ______________________________________________________________________________________ Problem 15.4 Solution: Known quantities: 0.255 A F , where A is the cross-sectional d 2 area of the transducer plate (in ), and d the air-gap length (in). The capacity of a capacitive displacement transducer C= Find: The change in voltage (∆v0) when the air-gap changes from 0.01 in to 0.015 in. Analysis: i=C ∆vo ∆q , ∆q = C∆v o , = ∆vo ∆t C Assume no change in charge. C new = 2 3 C old ∆v o = V 3 2 ______________________________________________________________________________________ Problem 15.5 Solution: Known quantities: The circuit of Figure P15.5 in which i D H = 0.5 10 −6 µ Am 2 W . Find: a) Show that the output voltage varies linearly with H. H = 1,500 W m 2 , VD = 7.5 V , and an output voltage of 1 V is desired, determine an appropriate value for RL. b) If Analysis: a) Vout = i D R L Given a large enough value of VD, Vout = 0.5 10 −6 HRL ; hence, varies linearly with H. b) Vout = i D H R L 1 = 0.5 10 −6 (1500 )R L , RL = 1333Ω 15.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 ______________________________________________________________________________________ Problem 15.6 Solution: Known quantities: The value of the constant G for quartz in compressive stress, 0.055 fluoride in axial stress, 0.22 V⋅ m . N V⋅ m , and for polyvinylidene N 2 The quartz element of a force sensor is 0.25 in thick and has a rectangular cross section of 0.09 in . The fluoride film of a piezoelectric sensor is 30 µm thick, 1,5 cm wide, and 2.5 cm in the axial direction. Find: a) The output of the force sensor in V/N. b) The output of the load sensor in V/N. Analysis: a) Vout = 0.055 b) 0 V ⋅ m 1 in 1 V × = 8.66 N 0.0254 m 0.25 in N V N ______________________________________________________________________________________ Problem 15.7 Solution: Known quantities: The allowable levels of error in the measurement of K, m and ξ: ±5 %, ±2 %, ±10 %. The expression for b, b = 2ξ Km . Find: The percentage error limit for b. Analysis: 5+ 2· § error = ±¨10 + ¸ = ±13.5 % 2 ¹ © ______________________________________________________________________________________ Problem 15.8 Solution: Known quantities: The measurements taken by a sensor that measure the thickness of a wet pulp layer every 2 feet along the sheet: 8.2, 9.8, 9.92, 10.1, 9.98, 10.2, 10.2, 10.16, 10.0, 9.94, 9.9, 9.8, 10.1, 10.0, 10.2, 10.3, 9.94, 10.14, 10.22, 9.8. The roller speed is adjusted based on the last 20 measurements unless the probability that the mean thickness lies within ±2 % of the sample mean exceeds 0.99. Find: The adjustment of the speed roller. 15.4 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: mean = ¦ measurements = 9.945 ± 2 %, 20 average deviation = standard deviation = ¦ deviations 20 max . = 10.144 , min . = 9.746 = 0.225 ¦ deviations 20 Problem solutions, Chapter 15 2 = 0.427 Measurement #1 exceeds the standard deviation σ probability<0.99 roller speed will be adjusted ______________________________________________________________________________________ Problem 15.9 Solution: Find: Discuss: a) Measurement accuracy. b) Instrument accuracy. c) Measurement error. d) Precision. Analysis: a) This term and instrument accuracy are used interchangeably if only one instrument is involved and if the measurement method is appropriate. Basically, the accuracy of the measurement is given by the instrument’s specifications (ordinarily in terms of percent of indicated value or full scale value). b) See answer to part a). c) Measurement error can be synonymous with measurement accuracy, but can also refer to sloppy methods of data acquisition, use of multiple transducers and/or instruments whose individual errors combine, or simply a lack of reliable, multiple data points. d) Precision and resolution are interchangeable terms and refer to the smallest increment of measured quantity that can be detected by the instrument. ______________________________________________________________________________________ Problem 15.10 Solution: Known quantities: Four sets of measurements shown in Figure P15.10 taken on the same response variable of a process using four different sensors. Find: Rank these data sets with respect to: a) Precision. b) Accuracy. Analysis: (b) and (c) are precise, (a) and (d) are not. (a) and (c) are accurate, (b) and (d) are not. ______________________________________________________________________________________ 15.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Section 15.3: Signal Conditioning Problem 15.11 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure P15.11, R1 = 1 kΩ , R2 = 5 kΩ . Find: The gain of the input stage. Analysis: A = 1+ 2 R2 2(5 kΩ) = 1+ = 1 + 10 = 11 R1 1 kΩ ______________________________________________________________________________________ Problem 15.12 Solution: Known quantities: The resistance R1 = 1 kΩ for the instrumentation amplifier of Figure P15.11, and the value of the gain for the input stage, 50. Find: The resistance R2 to make that gain for the amplifier. Analysis: A = 1+ 2 R2 1 1 R2 = R1 ( A − 1) = (1 kΩ)(50 − 1) = 24.5 kΩ 2 2 R1 ______________________________________________________________________________________ Problem 15.13 Solution: Known quantities: The resistance R2 = 10 kΩ for the instrumentation amplifier of Figure P15.11, and the value of the gain for the input stage, 16. Find: The resistance R1 to make that gain for the amplifier. Analysis: A = 1+ 2 R2 2 R2 2(10 kΩ) R1 = = ≈ 1333 Ω A −1 16 − 1 R1 ______________________________________________________________________________________ Problem 15.14 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure 15.16, 15.6 R1 = 1 kΩ , R2 = 10 kΩ . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Find: The gain of the input stage. Analysis: A = 1+ 2 R2 2(10 kΩ) = 1+ = 1 + 20 = 21 R1 1 kΩ ______________________________________________________________________________________ Problem 15.15 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure 15.16, R1 = 1.5 kΩ , R2 = 80 kΩ . Find: The gain of the input stage. Analysis: A = 1+ 2 R2 2(80 kΩ) = 1+ = 1 + 106.7 = 107.7 R1 1.5 kΩ ______________________________________________________________________________________ Problem 15.16 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure 15.16, R1 = R' = R = 1 kΩ , R2 = 5 kΩ , R f = 10 kΩ . Find: The differential gain for the IA. Analysis: Adif = A § 2 R · R f § 2(5) · 10 = ¨1 + = ¨¨1 + 2 ¸¸ ¸ = 110 1 ¹ 1 R © R1 ¹ R © Rf ______________________________________________________________________________________ Problem 15.17 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure P15.11, ∆R = 2 % of R. Find: The CMMR of the IA. 15.7 R = 1 kΩ , R f = 200 kΩ , G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Analysis: Adif CMRRdB = 20 log 10 Adif = A Rf R , · Rf § R + Rf ¨ − 1¸ A R ¨© R f + R + ∆R ¸¹ ∆R = 0.02 ⋅ R = 0.02 ⋅ 1 kΩ = 20 Ω A Rf R R + Rf CMRRdB = 20 log 10 · Rf § ¨ ¸A R ¨© R f + R + ∆R ¸¹ 1 R + Rf = 20 log 10 R f + R + ∆R ≈ 80dB −1 ______________________________________________________________________________________ Problem 15.18 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure P15.11, ∆R = 2 % R = 1 kΩ , R f = 200 kΩ , of R. Find: The mismatch in gains for the differential components id dB. Analysis: Assume A = 10 . Then Adif = 20 log 10 Adif ≈ 66 dB Rf R A = 2000 CMRRdB = 80 dB (from Problem 15.17) 20 log10 Adif − CMRRdB = −14 dB ______________________________________________________________________________________ Problem 15.19 Solution: Known quantities: The resistances for the instrumentation amplifier of Figure 15.16, differential gain, 900. Find: The resistances R and R2 to achieve that gain. Analysis: Adif = R f § 2 R2 ¨1 + R ¨© R1 · 10 § 2 R2 · 10 ¸¸ = ¨1+ ¸ = (1+ R2 ) = 900 2 ¹ R ¹ R© 15.8 R1 = 2 kΩ , R f = 10 kΩ , and the G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 1+ R2 = 90 R Choose: R = 1 kΩ R2 = 89 kΩ Thus, ______________________________________________________________________________________ Problem 15.20 Solution: Known quantities: The cutoff specification of Example 15.3, ω c = 10 rad s . Find: The order of the filter required to achieve 40 dB attenuation at ω s = 24 rad s . Analysis: 20 log10 1 + ω s2 n ≤ 40 @ ω s = 25 rad s Solving the equation, we obtain n = 1.43 . Thus n = 2 is desired. ______________________________________________________________________________________ Problem 15.21 Solution: Known quantities: The circuit for a low-pass filter shown in Figure P15.21 with gain. Find: a) The relationship between output amplitude and input amplitude. b) The relationship between output phase angle and input phase angle. Analysis: This is an inverting amplifier circuit, with Vout = − ZF Vin , where: Z F = R F Z in Therefore, Vout = − 1 jω C F RF 1 = = and Z in = Rin . 1 jωC F 1 + jω R F C F RF + jω C F RF ⋅ RF 1 Vin Rin 1 + jωR F C F 1 a) Vout R = F Vin Rin b) ∠Vout − ∠Vin = π − tan −1 (ωR F C F ) rad, or ∠Vout − ∠Vin = 180° − tan −1 (ωR F C F ) deg. 1 + (ωR F C F ) 2 ______________________________________________________________________________________ Problem 15.22 Solution: Known quantities: The circuit for a low-pass filter shown in Figure P15.21 with gain, with C F = 100 pF , vin = 2 sin(2,000πt ) V . 15.9 Rin = 20 kΩ , RF = 100 kΩ , G. Rizzoni, Principles and Applications of Electrical Engineering Find: The expression of Problem solutions, Chapter 15 vout . Analysis: Vout R = F Vin Rin 1 1 + (wR F C F ) 2 = 4.99 ( ) ( ) ∠Vout − ∠Vin = π − tan −1 2000π ⋅ 100 ⋅ 10 3 ⋅ 100 ⋅ 10 −12 = π − tan −1 62.8 ⋅ 10 −3 = 3.079 rad vout (t ) = 9.98 sin(2000πt + 3.079) V ______________________________________________________________________________________ Problem 15.23 Solution: Known quantities: The circuit of the low-pass filter of Figure 15.22. Find: The frequency response of the filter. Analysis: The circuit is shown below: We have: RB Vout R A + RB V −V V − V AB i1 = s , i2 = , i3 = jωC1 (Vout − V ) , R1 R2 i4 = jωC 2V AB . From i2 = i4 , we have: V = V AB (1 + jωR2 C 2 ) From i1 + i3 = i 2 , we have: V AB = § RB § § ¨ ¨ jωC1 + Vout ¨ − jωC1 + ¨¨ ¨ R R + A B ©© © Vout ( jω ) = Vs § RB R1 ¨ − jωC1 + ¨ R A + RB © Therefore, the frequency response is: Vout ( jω ) = Vs where: i3 i1 vs R1 i2 v R2 C2 i4 vAB RB C1 + vout RA · 1 · Vs 1 1 · ¸ ¸¸(1 + jωR2 C 2 )¸ − + ¸ R ¸= R R1 R2 ¹ 2 ¹ 1 ¹ 1 · 1 · §§ 1 1 · ¸ ¸− ¨ ¨ jωC1 + ¸ 1 ( ω ) j R C + + 2 2 ¸ ¸ R ¸ ¨¨ R R 1 2 ¹ 2 ¹ ¹ ©© K (1 R1 R2 C1C 2 ) ª 1 º 1 1 1 (1 − K )» ( jω ) + + + R1 R2 C1C 2 ¬ R1C1 R2 C1 R2 C 2 ¼ ( j ω )2 + « K = 1+ RA RB ______________________________________________________________________________________ 15.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Problem 15.24 Solution: Known quantities: The circuit of the high-pass filter of Figure 15.22. Find: The frequency response of the filter. Analysis: The circuit is shown below: We have: i1 RB V AB = Vout R A + RB i1 = jωC1 (Vs − V ) , i2 = jωC 2 (V − V AB ) , i3 = i3 vs C1 R1 i2 v C 2 Vout − V V , i4 = AB . R1 R2 R2 + i4 vAB vout RA RB · § 1 ¸ i2 = i4 , we have: V = V AB ¨¨1 + jωR2 C 2 ¸¹ © From i1 + i3 = i 2 , we have: From °§ RB jωC1Vs = Vout ®¨¨ °̄© R A + RB º 1 ½° · ª§ 1 ·§ · 1 ¸¸ «¨¨ + jωC1 + jωC 2 ¸¸¨¨ + 1¸¸ − jωC 2 » − ¾ ¹ ¬© R1 ¹© jωC 2 R2 ¹ ¼ R1 °¿ Therefore, the frequency response is: Vout ( jω ) = Vs K ( jω ) 2 ª 1 º 1 (1 − K ) + 1 + 1 » ( jω ) + R2 C1 R2 C 2 ¼ R1 R2 C1C 2 ¬ R1C1 ( j ω )2 + « where: K = 1 + RA RB ______________________________________________________________________________________ Problem 15.25 Solution: Known quantities: The circuit of the band-pass filter of Figure 15.22. Find: The frequency response of the filter. Analysis: The circuit is shown below: We have: i1 RB V AB = Vout R A + RB V −V V −V i1 = s , i 2 = jωC1 (V − V AB ) , i3 = out . R1 R2 15.11 vs R1 v i2 C 1 R3 i3 R 2 C2 vAB RB + vout RA G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 · § 1 § C 1 · ¸¸ i2 = V AB ¨¨ + jωC 2 ¸¸ , we have: V = V AB ¨¨1 + 2 + © C1 jωR3C1 ¹ ¹ © R3 From i1 + i3 = i 2 , we have: From °§ R B Vs = Vout ®¨¨ R1 °̄© R A + RB · ª§ R1 + R2 ·§ C + C2 1 ¸¸ «¨¨ + jωC1 ¸¸¨¨ + 1 C1 ¹ «¬© R1 ¹© jωC1 R3 Therefore, the frequency response is: Vout ( jω ) = Vs Kjω º 1 ½° · ¸¸ − jωC1 » − ¾ »¼ R2 °¿ ¹ 1 R1C 2 º ª 1 1 1 1 (1 − K )» ( jω ) + R1 + R2 + + + R1 R2 R3C1C 2 ¼ ¬ R1C1 R1C 2 R2 C1 R2 C 2 ( j ω )2 + « where: K = 1 + RA RB ______________________________________________________________________________________ Problem 15.26 Solution: Known quantities: The circuit of Figure P15.21, where C F = 100 pF , and the desired cutoff frequency and gain magnitude are respectively 20 kHz and 5. Find: The appropriate values of RF and Rin. Analysis: f C = 20 kHz ω C = 2πf C = 40π k ωC = 1 RF C F RF = rad s 1 1 = = 79.6 kΩ 3 ω C C F 40π ⋅10 ⋅100 ⋅10 −12 RF R = 5 Rin = F = 15.9 kΩ Rin 5 ______________________________________________________________________________________ Problem 15.27 Solution: Known quantities: The cutoff frequency, 10 kHz, and the DC gain, 10, for a second-order Butterworth high pass filter, with Q = 5 and Vs = ±15 V . Find: The design of the filter. Analysis: K = 1+ RA = 10 ; choosing R A = 9 kΩ RB = 1 kΩ RB 15.12 G. Rizzoni, Principles and Applications of Electrical Engineering f = 1 2π R1C1 R2 C 2 Problem solutions, Chapter 15 = 10 kHz ª R2 C 2 R1C 2 RC º + − 9 1 1 » = 0 .2 « R2 C1 R2 C 2 »¼ «¬ R1C1 Choose C1 = C2 = 0.01 µF and solve for R1 = 540 Ω and R2 = 4.7 kΩ. Then, substitute the values thus obtained in the high-pass filter of Figure 15.22. ______________________________________________________________________________________ Problem 15.28 Solution: Known quantities: The cutoff frequency, 25 kHz, and the DC gain, 15, for a second-order Butterworth high pass filter, with Q = 10 and Vs = ±15 V . Find: The design of the filter. Analysis: RA = 15 ; choosing R A = 14 kΩ RB = 1 kΩ RB 1 f = = 25 kHz 2π R1C1 R2 C 2 K = 1+ ª R2 C 2 R1C 2 RC º + − 14 1 1 » = 0.1 « R2 C1 R2 C 2 »¼ «¬ R1C1 Choose C1 = C2 = 1 µF and solve for R1 = 1.8 Ω and R2 = 23 Ω. Then, substitute the values thus obtained in the high-pass filter of Figure 15.22. ______________________________________________________________________________________ Problem 15.29 Solution: Known quantities: The circuit of Figure P15.29. Find: The characteristic of the filter. Analysis: Note that Vb ≈ Vout . Therefore, I= and Vb § ¨ ¨ 1 ¨ C ¨ jω 2 © · ¸ ¸ ¸ ¸ ¹ = jω C 2 Vout 15.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 § · § C ωRC · ¸¸Vout V a = RI + Vb = R¨¨ jω Vout ¸¸ + V out = ¨¨1 + j 2 2 ¹ © ¹ © Now, writing a KCL equation at node a, V − Vout Va − Vin =0 +I+ a R § 1 · ¨ ¸ ¨ jω 2C ¸ © ¹ ωRC · ωRC · § § ¨1 + j ¸Vout − Vin ¨1 + j ¸Vout − Vout C 2 ¹ 2 ¹ © © + jω Vout + =0 R § · 2 1 ¨ ¸ ¨ jω 2C ¸ © ¹ ª§ ωRC · ωRC · º § « ¨1 + j − 1¸ » ¸ ¨1 + j C © 1 2 2 ¹ «© ¹ »V = ª 1 + j ωC + j ωC − ω 2 RC 2 ºV Vin = « + jω + out «R » out R R § · » 2 2 2 1 ¬ ¼ « ¨ ¸ » ¨ jω 2C ¸ » «¬ © ¹ ¼ ( ) Vin = 1 + j 2ωRC − ω 2 R 2 C 2 Vout Vout 1 = 2 Vin 1 − (ωRC ) + j 2ωRC Vout 1 1 = = 2 4 2 4 Vin 1 − 2(ωRC ) + (ωRC ) + 2(ωRC ) 1 + (ωRC ) which is a second-order Butterworth low-pass function with cutoff frequency ωC = 1 RC ______________________________________________________________________________________ Problem 15.30 Solution: Known quantities: The cutoff frequency, 15 kHz, and the DC gain, 15, for a second-order Butterworth low pass filter, with Q = 5 and Vs = ±15 V . Find: The design of the filter. Analysis: RA = 15 ; choosing R A = 14 kΩ RB = 1 kΩ RB 1 f = = 15 kHz 2π R1C1 R2 C 2 K = 1+ ª R2 C 2 + « «¬ R1C1 R1C 2 RC º − 14 1 1 » = 0.2 R2 C1 R2 C 2 »¼ 15.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Choose C1 = C2 = 1 µF and solve for R1 = 2.6 Ω and R2 = 43.9 Ω. Then, substitute the values thus obtained in the high-pass filter of Figure 15.22. ______________________________________________________________________________________ Problem 15.31 Solution: Known quantities: The low cutoff frequency, 200 Hz, the high cutoff frequency, 1 kHz, and the pass band gain, 4. Find: The value of Q for the filter, and the approximate frequency response of this filter. Analysis: Q= fH fL = 0.6 fH − fL R K = 1 + A = 2 ; choosing R A = 1 kΩ RB = 1 kΩ RB 1 f = = 1 kHz 2π R1C1 R2 C 2 Choose R1 = R2 and C1 = C2 = 1 µF R1 = R2 = 160 Ω. Then, substitute the obtained values in the high-pass filter of Figure 15.22. R K = 1 + A = 2 ; choosing R A = 1 kΩ RB = 1 kΩ RB 1 f = = 200 Hz 2π R1C1 R2 C 2 Choose R1 = R2 and C1 = C2 = 1 µF R1 = R2 = 800 Ω. Then, substitute the obtained values in the high-pass filter of Figure 15.22. By connecting the output of the high-pass filter to the input of the low-pass filter, we obtain the desired filter. ______________________________________________________________________________________ Problem 15.32 Solution: Known quantities: The circuit of Figure P15.29 and the cutoff frequency, 10 filter. Find: The design of the filter. Analysis: f C = 10 Hz = 1 2πRC ωC = Choose R = 20 kΩ . Then, Hz, for a second-order Butterworth low pass 1 rad = 2π ⋅ 10 = 20π RC s 1 = 20π (20 kΩ)C C= and the two capacitors have values given by 15.15 1 = 796 nF 20 ⋅ 10 3 ⋅ 20π G. Rizzoni, Principles and Applications of Electrical Engineering 2C = 1.125 µF and C 2 Problem solutions, Chapter 15 = 563 nF . ______________________________________________________________________________________ Problem 15.33 Solution: Known quantities: The low pass Sallen Key filter of Figure P15.33. Find: Vout as a function of frequency and generate its Bode magnitude plot. Show that the Vin R 1 and that the low frequency gain is 4 . cutoff frequency is 2πRC R3 The voltage gain Analysis: We have: V AB = R4 Vout R3 + R4 Vs − V V − V AB , i3 = . R R From i 2 = jωCV AB , we have: V = V AB (1 + jωRC ) From i1 + i3 = i 2 , we have: i1 = jωC (Vout − V ) , i 2 = °§ R4 Vs = Vout ®¨¨ °̄© R3 + R4 ½° · ¸¸[(2 + jωRC )(1 + jωRC ) − 1] − jωRC ¾ °¿ ¹ Therefore, the frequency response is: Vout ( jω ) = Vs 1 (RC )2 ª 1 (3 − K )º» ( jω ) + 1 2 +« (RC ) ¬ RC ¼ K ( jω )2 where: K = 1 + R3 R4 ______________________________________________________________________________________ Problem 15.34 Solution: Known quantities: The circuit shown in Figure P15.34 Find: The transfer functions relating each of the three outputs to the input Analysis: 15.16 Vin . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 1 V2 = − V1 s 1 1 V3 = − V2 = 2 V1 s s and 1 V4 = −V2 = V1 s Using Laplace transforms, note that Writing a KCL equation at the inverting input to the leftmost op-amp, a b − KVin − V1 − aV4 − bV3 = 0 or − KVin − V1 − V1 − 2 V1 = 0 s s V1 Ks 2 =− 2 Vin s + as + b which is a second-order high-pass function. Also, V2 1V Ks =− 1 = 2 Vin s Vin s + as + b which is a bandpass function, and V3 K 1 V = 2 1 =− 2 Vin s Vin s + as + b which is a low-pass function. ______________________________________________________________________________________ Problem 15.35 Solution: Known quantities: The filter shown in Figure P15.35. Find: Verify that the filter's frequency response has the following expression: H ( jω ) = ( jω )2 − (1 R3 R2 C1C 2 ) R3 R1 § 1 1 1 · 1 ¸¸ jω + + ¨¨ + + R3 R2 C1C 2 © R1C1 R2 C1 R3C1 ¹ Analysis: The circuit is shown below: We have: Vs − V Vout − V , i2 = jωC 2Vout , i3 = , i4 = jωC1V , R1 R3 i1 R2 V v vs i5 = . R2 R1 i i5 From i5 = −i2 , we have: V = − jωR2 C 2Vout C1 4 From i1 + i3 = i5 + i 4 , we have: Vs jω C 2 R 2 V jωC 2 R2 2 + Vout + out + Vout = −( jω ) C1C 2 R2Vout − jωC 2Vout R1 R1 R3 R3 i1 = 15.17 R3 i3 i2 + - C2 vout G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Therefore, the frequency response is: H ( jω ) = ( jω )2 − (1 R3 R2 C1C 2 ) R3 R1 § 1 1 1 · 1 ¸¸ jω + + ¨¨ + + R3 R2 C1C 2 © R1C1 R2 C1 R3C1 ¹ ______________________________________________________________________________________ Problem 15.36 Solution: Known quantities: The filter shown in Figure P15.36. Find: Verify that the filter's frequency response has the following expression: H ( jω ) = jωK R1C1 § 1 R1 + R2 1 1 1− K · ¸¸ jω + + + + R1 R2 R3C1C 2 © R1C1 R3 C 2 R3C1 R2 C1 ¹ ( jω )2 + ¨¨ Analysis: The circuit is shown below: We have: i1 V R V AB = Vout = out R + R (K − 1) K V −V V , i2 = jωC 2 (V − V A ) = A , i1 = i R1 R3 V −V i3 = out , i4 = jωC1V . R2 vi v R1 C1 C2 i4 i3 R3 i2 i2 = jωC 2 (V − V A ) = From i1 + i3 = i 2 + i4 , we have: + vo vA R · Vout § 1 VA ¸= , we have: V = V A ¨1 + ¨ R3 jωR3C 2 ¸¹ K © From R2 R(K-1) · § 1 ¸ ¨1 + ¨ jωR3C 2 ¸¹ © ·V § ·V § jωC1 ·V Vi § V V 1 1 − ¨¨ + 1¸¸ out + out − ¨¨ + 1¸¸ out = out + ¨¨ + jωC1 ¸¸ out R1 © jωC 2 R3 ¹ KR1 R2 © jωC 2 R3 ¹ KR2 KR3 © jωC 2 R3 ¹ K Therefore, the frequency response is: H ( jω ) = jωK R1C1 § 1 R1 + R2 1 1 1− K · ¸¸ jω + + + + R1 R2 R3C1C 2 © R1C1 R3 C 2 R3C1 R2 C1 ¹ ( jω )2 + ¨¨ where: K = 1 + RA RB ______________________________________________________________________________________ 15.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Problem 15.37 Solution: Known quantities: The filter shown in Figure P15.35. Find: Verify that for the filter: C § 1 1 1 1 · ¸ = R2 R3 2 ¨¨ + + Q C1 © R1 R2 R3 ¸¹ Analysis: From the expression, we can see that ωc = 1 C 1C 2 R 2 R 3 and ωc 1 1 1 = + + Q C1 R1 C1 R2 C1 R3 or § 1 1 1 1 · ¸¸ = C1C 2 R2 R3 ¨¨ + + Q C R C R C R 1 1 1 2 1 3 ¹ © or C § 1 1 1 1 · ¸ = R2 R3 2 ¨¨ + + C1 © R1 R2 R3 ¸¹ Q ______________________________________________________________________________________ 15.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Section 15.4: Analog-to-Digital and Digital-to-Analog Conversion Problem 15.38 Solution: Find: List two advantages of digital signal processing over analog signal processing. Analysis: 1. Digital signals are less subject to noise, since one only needs to discriminate between two voltages. 2. Digital signals are directly compatible with.digital computers, and can therefore be easily stored on a disk, or exchanged between computers. Thus, digital signals are intrinsically more portable than analog signals. ______________________________________________________________________________________ Problem 15.39 Solution: Find: Discuss the role of a multiplexer in a data acquisition system. Analysis: It sequentially switches a set of analog inputs to the system input. ______________________________________________________________________________________ Problem 15.40 Solution: Known quantities: The circuit shown in Figure P15.40. Find: Explain the operation of the circuit. Analysis: Op-amp #1 is an input buffer. The JFET behaves as a low-leakage diode which enables and disables the RC holding circuit, and op-amp #2 is a voltage-follower whose purpose is to isolate the circuit from the load. ______________________________________________________________________________________ Problem 15.41 Solution: Known quantities: The circuit shown in Figure P15.40. The input is a 1 kHz sinusoidal signal with 0° phase angle, 0 V DC offset and 20 V peak to peak amplitude. VG is a rectangular pulse train with 10 µs, and a period 100 µs, with a leading edge of the first pulse at t=0. Find: a) Sketch Vout if the RC circuit has a time constant equal to 20 µs. b) Sketch Vout if the RC circuit has a time constant equal to 1 ms. Analysis: a) 15.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 b) ______________________________________________________________________________________ Problem 15.42 Solution: Known quantities: The input to a four-digit DAC, 1210, given that RF = R0 15 , logic 0 corresponds to 0 V, and logic 1 corresponds to 4.5 V. Find: a) The output of the DAC. b) The maximum voltage that can be outputted from the DAC. c) The resolution over the range 0 to 4.5 V. d) The number of bits required if an improved resolution of 20 Analysis: [ RF 3 2 b3 + 2 2 b2 + 21 b1 + b0 R0 1 a) Va = −4.5 [12] = −3.6 V 15 1 (15) = −4.5 V b) (Va )max = −4.5 15 1 c) δVa = 4.5 = 0 .3 V 15 Va = −4.5 15.21 ] mV is desired. G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 § (Va )max − (Va )max · log ¨ + 1¸ ¨ ¸ (δVa )req © ¹ = 7.82 d) n ≥ log 2 Therefore, we choose n = 8. ______________________________________________________________________________________ Problem 15.43 Solution: Known quantities: The input to a eight-digit DAC, 21510, given that RF = R0 255 , logic 0 corresponds to 0 V, and logic 1 corresponds to 10 V. Find: a) The output of the DAC. b) The maximum voltage that can be outputted from the DAC. c) The resolution over the range 0 to 10 V. d) The number of bits required if an improved resolution of 3 mV is desired. Analysis: 21510 = 110101112 R Va = −10 F 2 7 b7 + 2 6 b6 + + 21 b1 + b0 R0 1 [215] = −8.341 V a) Va = −10 255 1 [255] = −10 V b) (Va )max = −10 255 1 = 39.2 mV c) δVa = 10 255 § (Va )max − (Va )max · log ¨ + 1¸ ¨ ¸ (δVa )req © ¹ = 11.703 d) n ≥ log 2 Therefore, we choose n = 12. [ ] ______________________________________________________________________________________ Problem 15.44 Solution: Known quantities: The circuit of Figure P15.45, a simple 4-bit DAC. In the circuit if the bit is 1, the corresponding switch is up, if the bit is 0 the switch is down. Find: If the digital number is represented by b3b2b1b0, determine an expression relating v0 to the binary input bits.. Analysis: This circuit is just a summing amplifier, with 15.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 R2 R R R RV§ b b b · b3V − 2 b2V − 2 b1V − 2 b0V = − 2 ¨ b3 + 2 + 1 + 0 ¸ 2 R1 4 R1 8 R1 R1 © 2 4 8¹ R1 RV = − 2 (8b3 + 4b2 + 2b1 + b0 ) 8 R1 vO = − ______________________________________________________________________________________ Problem 15.45 Solution: Known quantities: The input to a eight-digit DAC, 9810, given that RF = R0 255 , logic 0 corresponds to 0 V, and logic 1 corresponds to 4.5 V. Find: a) The output of the DAC. b) The maximum voltage that can be outputted from the DAC. c) The resolution over the range 0 to 4.5 V. d) The number of bits required if an improved resolution of 0.5 mV is desired. Analysis: 1 [98] = 1.729 V 255 1 b) (Va )max = 4.5 [255] = 4.5 V 255 1 = 17.6 mV c) δVa = 4.5 255 § (Va )max − (Va )max · log ¨ + 1¸ ¨ ¸ (δVa )req © ¹ = 13.136 d) n ≥ log 2 Therefore, we choose n = 14. a) Va = 4.5 ______________________________________________________________________________________ Problem 15.46 Solution: Known quantities: The four-digit DAC of Figure P15.46, with an output range − 10 ≤ VO ≤ 0 V . Find: The value of RF that will give that output. Assumptions: The logic 0 = 0 V and the logic 1 = 5 V. Analysis: For the circuit of Figure P15.46, Therefore, (Va )max RO = 1 kΩ and n = 4. Rf = −10 V , and − 10 = −5 15 , or R f = 133.3 Ω . 1000 ______________________________________________________________________________________ 15.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 § (Va )max − (Va )max · log ¨ + 1¸ ¨ ¸ (δVa )req © ¹ = 14.288 d) n ≥ log 2 Therefore, we choose a 15-bit ADC. ______________________________________________________________________________________ Problem 15.50 Solution: Known quantities: The four-digit DAC of Figure P15.46, with an output range − 15 ≤ VO ≤ 0 V . Find: The value of RF that will give that output. Assumptions: The logic 0 = 0 V and the logic 1 = 5 V. Analysis: Here, − 15 ≤ VO ≤ 0 V . For the circuit of Figure P15.46, RO = 1 kΩ and n = 4. Rf Therefore, (Va )max = −15 V , and − 15 = −5 15 , or R f = 200 Ω . 1000 ______________________________________________________________________________________ Problem 15.51 Solution: Known quantities: The circuit of Figure P15.44, and the desired output of the 4-bit DAC, V0 = − 1 (8b3 + 4b2 + 2b1 + b0 ) V . 10 Find: The design of the DAC. Analysis: From the results of Problem 15.44, we see that we must choose V0, R2 and R1 such that: R2 ⋅ V0 1 = 8 ⋅ R1 10 One possible choice is to let V = 15 V , R1 = 30 kΩ and R2 = 1.6 kΩ . ______________________________________________________________________________________ Problem 15.52 Solution: Known quantities: The range, ±15 V, and the resolution, 0.01 Fin The number of bits of the DAC. Analysis: V, of a data acquisition system. § 15 − (− 15) · n ≥ log 2 ¨ + 1¸ = 11.55 . Choose n = 12. © 0.01 ¹ 15.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 ______________________________________________________________________________________ Problem 15.53 Solution: Known quantities: The range, ±10 V, and the resolution, 0.04 V, of a data acquisition system. Find: The number of bits of the DAC. Analysis: § 10 − (− 10 ) · n ≥ log 2 ¨ + 1¸ = 8.96 . Choose n = 9. © 0.04 ¹ ______________________________________________________________________________________ Problem 15.54 Solution: Known quantities: The range, −10 − 15 V, and the resolution, 0.004 V, of a data acquisition system. Find: The number of bits of the DAC. Analysis: § 15 − (− 10 ) · n ≥ log 2 ¨ + 1¸ = 12.6 . Choose n = 13. © 0.004 ¹ ______________________________________________________________________________________ Problem 15.55 Solution: Known quantities: The range, 0 − 2,500 rev/min, and the resolution, 1 commands to a motor. Find: The number of bits of the DAC. Analysis: rev/min, of a DAC used to deliver velocity § 2500 − (0 ) · n ≥ log 2 ¨ + 1¸ = 11.29 . Choose n = 12. 1 © ¹ With this choice we compute the following resolution: res = 2500 = 0.61 rev/min 212 ______________________________________________________________________________________ Problem 15.56 Solution: Known quantities: The range, 0 − 10 V, for an ADC. 15.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Find: a) The resolution if this is a 3-bit device. b) The resolution if this is a 8-bit device. c) A general comment about the relationship between the number of bits and the resolution. Analysis: res = 2 −3 ⋅ 10 V = 1.25 V . −8 b) res = 2 ⋅ 10 V = 39.0625 mV . a) c) more bits give better resolution. ______________________________________________________________________________________ Problem 15.57 Solution: Known quantities: The range, -5 − 15 V, and the resolution required, 0.05 %, for a DAC. Find: The number of bits required. Analysis: The range is 15 - (-5) = 20 V 20 § · + 1¸ = 10.97 . Choose n = 11. © 20 ⋅ 0.0005 ¹ Thus, n ≥ log 2 ¨ ______________________________________________________________________________________ Problem 15.58 Solution: Known quantities: The number of channels of an ADC, eight, the time required for ADC conversion, 100 µs, the time required for computation and output time for four of the channels, 500 µs, and for the other four, 250 µs. Find: The number of bits required. Analysis: We assume a data acquisition system of the type shown in Figure 15.32. Therefore, each channel will be 1 of the external clock rate and the slowest channels will determine the rate. 8 Thus, sampling rate = 8(100 µs + 500 µs) = 4.8 ms. f 1 Thus, f s = = 208.3 Hz and f max = s = 104.15 Hz −3 2 4.8 ⋅ 10 sampled at ______________________________________________________________________________________ Problem 15.59 Solution: Known quantities: The range of the potentiometer, 270° and 10 V, and the maximum displacement to be measured, 180°. Find: a) The voltage to be resolved by an ADC to resolve an angular displacement of 0.5°, and the number of bits to do that. 15.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 b) If the ADC requires a 10 V input voltage find the optimum amplifier gain to take advantage of the full range of the ADC. Analysis: For a dynamic range of 10 V for 270° of rotation, we compute the following resolution: 10 0.5 = 18.52 mV 270 a) Finding the range for 180° rotation it is possible to determine the bits requirement for the ADC. § · § 180 · ¨ 10¨ ¸ ¸ 270 ¹ © ¨ n = log 2 + 1¸ = 8.50 , and we choose n = 9. ¨ 18.52 ⋅ 10 −3 ¸ ¨ ¸ © ¹ res = b) The voltage gain of the amplifier is gain = 270 = 1 .5 180 ______________________________________________________________________________________ Problem 15.60 Solution: Known quantities: The maximum frequency of the signal to be digitized, 250 kHz, and the number of bits, 10, of the successive approximation ADC used. Find: The maximum permissible conversion time. Analysis: The conversion time should be no more than 10% of the signal period. For this case, the signal period is T= 1 = 4 µs 250 ⋅ 10 3 Therefore, the conversion time should be no longer than 400 ns. ______________________________________________________________________________________ Problem 15.61 Solution: Known quantities: The maximum frequency of the signal to be digitized, a torque signal from a torque sensor mouted on a farm tractor engine, is twice the shaft rotation frequency; the rotational speed of the crankshaft is 800 rpm. Find: The minimum sampling period according to Nyquist criterion. Analysis: Shaft rotation frequency is 800 rpm or 13.33 rev/sec; therefore, the fluctuation frequency is 26.67 Ts = 1 = 18.75 ms 26.67 ⋅ 2 Hz: ______________________________________________________________________________________ 15.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Problem 15.62 Solution: Known quantities: The range of an aircraft altimeter, from 0 sensing, 10 m. V at 0 m to 10 V at 10000m, and the allowable error in Find: The minimum number of bits for the ADC. Analysis: The allowable error is ±10 m. Therefore, an equivalent 20 m step size is allowable and § 10000 · n = log 2 ¨ + 1¸ = 8.97 © 20 ¹ Thus, a 9-bit ADC is required. ______________________________________________________________________________________ Problem 15.63 Solution: Known quantities: The characteristic of the circuit needed, a circuit that generates interrupts at fixed time intervals. Find: The design of the circuit using a square wave that has a period equal to the desired time interval between interrupts. Analysis: T MICROPROCESSOR INT Clock SET Q CK INT D 5V INTERRUPT REQUEST FROM THE ADC (CONVERSION COMPLETE) CLEAR ______________________________________________________________________________________ Problem 15.64 Solution: Find: The minimum number of bits required to digitize an analog signal with a resolution of: a) 5%. b) 2%. c) 1%. Analysis: 5% 2 − n ≤ 0.05 n = 5 −n b) 2% 2 ≤ 0.02 n = 6 −n c) 1% 2 ≤ 0.01 n = 7 a) 15.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Section 15.5: Comparator and Timing Circuits Problem 15.65 Solution: Known quantities: The window comparator circuit of Figure P15.65. Find: Show that Vout = 0 whenever Vlow ≤ Vin ≤ Vhigh , and that Vout = Vin otherwise. Analysis: If Vin ≤ Vlow , If Vlow ≤ Vin ≤ Vhigh , If Vin ≥ Vhigh , V1 = 0 ½ Vout = V + +¾ V2 = V ¿ V1 = 0 ½ ¾ Vout = 0 V2 = 0¿ V1 = V + ½ + ¾ Vout = V V2 = 0 ¿ ______________________________________________________________________________________ Problem 15.66 Solution: Known quantities: The noise peak amplitude, ±150 mV, the reference value around which the circuit is to switch, -1 V, and the characteristic of the op-amp, ±10 V supplies (Vsat = 8.5 V). Find: The design of the circuit. Analysis: This is very similar to Example 15.14, Therefore, v+ = R2 R2 v s+ vout + R2 + R3 R2 + R1 Since the required noise protection level is ±150 mV, R1 and R2 can be computed from: R2 R2 ∆v = 8.5 = 0.15 V where ∆v = 300 mV v sat = 2 R2 + R1 R2 + R1 Assuming R1 = 100 kΩ , R2 can be calculated to be approximately 1.8 kΩ. Since the required reference voltage is -1 V, we can find R3 by solving the equation R2 R2 10 = 1 V to obtain: R3 = 16.2 kΩ . v s− = R2 + R3 R2 + R3 ______________________________________________________________________________________ Problem 15.67 Solution: Known quantities: The circuit of Figure P15.67; R1 = 100 Ω , R2 = 56 kΩ , Ri = R1 R2 , and vin is a 1 V peak to peak sine wave. Find: The threshold voltages and the output waveform. 15.32 G. Rizzoni, Principles and Applications of Electrical Engineering Assumptions: The supply voltages are ±15 Problem solutions, Chapter 15 V. Analysis: Applying KCL: V Vo R = 1 + 2 . Therefore, o = 561 Vin Vin R1 ______________________________________________________________________________________ Problem 15.68 Solution: Known quantities: The circuit of Figure P15.68. Find: Explain the operation of the circuit. Analysis: Operation of this circuit depends on the magnitude of the input voltage, Vin , being much larger than zero. When Vin >> 0 , the op-amp will be saturated, and Vout will be at its saturation limit, VR (positive rail). When the op-amp is saturated, the approximation V+ = V− is no longer valid. Writing KCL at the non- inverting input, we have § 1 V+ − Vin V+ − Vout 1 · 1 1 ¸¸ = + = 0 or V+ ¨¨ + Vin + Vout Rin RF RF © Rin RF ¹ Rin The output changes from positive rail (+ V R ) to negative rail (− V R ) when V+ = V− = 0 . When Vin is §R V+ ≈ V− = 0 , and Vout = −¨¨ F © Rin §R · = VR , then: Vin = −¨¨ in ¸¸VR © RF ¹ small, then If Vout What this means is the following: As · ¸¸Vin ¹ Vin drops from a large positive value, through zero, to a negative value, the output of the op-amp “switches” (from §R · + VR to − VR ) at the point Vin = −¨¨ in ¸¸VR . © RF ¹ 15.33 G. Rizzoni, Principles and Applications of Electrical Engineering Similarly, as Problem solutions, Chapter 15 Vin increases in the opposite direction, the “switch” from − VR to + VR occurs at §R · Vin = +¨¨ in ¸¸VR . © RF ¹ This is a form of voltage hysteresis, as shown in the figure below. ______________________________________________________________________________________ Problem 15.69 Solution: Known quantities: The circuit of Figure P15.68. The op-amp is a LM741 with ±15 V, RF = 104 kΩ , and Vin is a 1 kHz sinusoidal signal with 1 V amplitude. Find: a) The appropriate value for Rin if the output is to be high whenever Vin ≥ 0.25 V . b) Sketch the input and the output waveforms. Analysis: a) We know that V R ≈ 13V when the LM741 op-amp is used with ± 15V bias supplies. Then, from the discussion in the answer to Problem 15.69, §V §R · Vin = −¨¨ in ¸¸Vout Rin = ¨¨ in © RF ¹ © Vout · 0.25 · ¸¸ R F = §¨ ¸(104 kΩ) = 2 kΩ © 13 ¹ ¹ b) The input and output waveforms are sketched below. 15.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 ______________________________________________________________________________________ Problem 15.70 Solution: Known quantities: The circuit of Figure P15.70. Find: a) The output waveform for vin a 4 b) The output waveform for vin a 4 V peak to peak sine wave at 100 Hz and Vref = 2 V. V peak to peak sine wave at 100 Hz and Vref = -2 V. Analysis: a) b) 15.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 ______________________________________________________________________________________ Problem 15.71 Solution: Known quantities: The go-no go detector application circuit of Figure P15.71. Find: a) Explain how the circuit works. b) Design a circuit such that the green LED will turn on when Vin exceeds 5 whenever Vin is less than 5 V. V, and the LED will be on Assumptions: Only 15 V supplies are available. Analysis: a) Define V2 = R1 V as the voltage at the inverting input of the op-amp. Then: R2 + R1 Vin > V2 the output of the op-amp will be positive and the green LED will turn on (go). When Vin < V2 the output of the op-amp will be negative and the red LED will turn on (no go) b) For this design, V2 = 5 V and V = 15 V . R1 R1 1 15 V = 5 V = R2 + R1 3 R2 + R1 R2 R2 =2 or +1 = 3 R1 R1 Choose R1 = 10 kΩ and R2 = 20 kΩ to complete the design. When ______________________________________________________________________________________ Problem 15.72 Solution: Known quantities: The circuit of Figure P15.72, where vin is a 100 D2 are 6.2 V Zener diodes. mV peak sine wave at 5 kHz, R = 10 kΩ, and D1 and Find: Draw the output voltage waveform. 15.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Analysis: ______________________________________________________________________________________ 15.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 Section 15.7: Data Transmission in Digital Instruments Problem 15.73 Solution: Known quantities: An ASCII (hex) encoded message. Find: Decode the message. Analysis: ASCII decoding is easy! ______________________________________________________________________________________ Problem 15.74 Solution: Known quantities: An ASCII (binary) encoded message. Find: Decode the message. Analysis: This is a time-consuming problem. ______________________________________________________________________________________ Problem 15.75 Solution: Known quantities: Some decimal numbers. Find: The ASCII form for the numbers. Analysis: Decimal ASCII 12 345.2 43.5 31 32 33 34 35 2E 32 34 33 2E 35 ______________________________________________________________________________________ Problem 15.76 Solution: Known quantities: Some words. Find: The ASCII form for the words. Analysis: a) 44 69 67 69 74 61 6C b) 43 6F 6D 70 75 74 65 c) 41 73 63 69 69 72 15.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 15 d) 41 53 43 49 49 ______________________________________________________________________________________ Problem 15.77 Solution: Find: Explain why data transmission over long distances is usually done via a serial scheme rather than parallel. Analysis: Serial data transmission requires only a single data path. Parallel requires 16 (or more, depending on word length), and would, therefore, be much more expensive. ______________________________________________________________________________________ Problem 15.78 Solution: Known quantities: The on-board memory of an automated data-logging, 16 K-words, that samples the variable of interest once every 5 min. Find: How often must the data be downloaded and the memory cleared in order to avoid losing any data. Analysis: Longest possible delay = 16 ⋅ 1024 ⋅ 5 = 81920 min = 1hr 1day = 81920min × × ≈ 56.9days 60min 24hr ______________________________________________________________________________________ Problem 15.79 Solution: Find: Explain why three wires are required for the handshaking technique employed by IEEE 488 bus system. Analysis: Three lines are used for handshaking in the IEEE 488 bus to accomplish the following functions: One line is used to declare the bus ready to accept data; another line to declare that data has been accepted, and a third one to declare that the data was indeed valid. ______________________________________________________________________________________ Problem 15.80 Solution: Known quantities: The information held in a CD-ROM, 650 MB. The CD-ROM are packaged 50 per box, and 100 boxes are shipped. The distance for the trip is 2,500 miles and the airplane speed is 400 mi/hr. Find: The transmission rate between the two cities in bits/s. Analysis: 15.39 G. Rizzoni, Principles and Applications of Electrical Engineering 650 ⋅ 1024 ⋅ 1024 Problem solutions, Chapter 15 bytes CDs mi ⋅ 50 ⋅ 100 boxes ⋅ 400 CD box hr = 5.4525952 ⋅ 1011 bytes 2500 mi hr bytes bits bits 1hr ⋅8 ⋅ ≈ 1.21 ⋅ 10 9 hr byte 3600 sec s Gbits . or approximately 1.13 s 5.4525952 ⋅ 1011 ______________________________________________________________________________________ 15.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Chapter 16 Instructor Notes The last part of the book presents an introduction to electro-magneto-mechanical systems. Some of the foundations needed for this material (AC power) were discussed in Chapter 7; the polyphase AC power material in Chapter 7 may be introduced prior to covering Chapter 17, or together with the AC machine material of Chapter 17. The emphasis in this chapter (and the next two) is on preparing the student for the use of electromagneto-mechanical systems as practical actuators for industrial applications. Thus, more emphasis is placed on describing the performance characteristics of linear motion actuators and of rotating machines than on a description of their construction details. The material in Chapters 16-18 has been used by several instructors over the last several years in a second (quarter-length) course in system dynamics (System Dynamics and Electromechanics) designed for mechanical engineering juniors. Section 16.1 16 reviews basic laws of electricity and magnetism, which should already be familiar to the student from an earlier Physics course. The box Focus on Measurements: Linear Variable Differential Transformer (pp. 795-796) presents an example related to sensors with a discussion of the LVDT as a position transducer. Section 16.2 discusses approximate linear magnetic circuits and the idea of reluctance, and introduces magnetic structures with air gaps and simple electro-magnets. The box Focus on Methodology: Magnetic Structures and Magnetic Equivalent Circuits (p. 804) summarizes the analysis methods used in this section. A magnetic reluctance position sensor is presented in Focus on Measurements: Magnetic Reluctance Position Sensor (pp. 811-812) and Focus on Measurements: Voltage Calculation in Magnetic Reluctance Position Sensor (pp. 812-814). The non-ideal properties of magnetic materials are presented in Section 16.3, where hysteresis, saturation, and eddy currents are discussed qualitatively. Section 16.4 introduces simple models for transformers; more advanced topics are presented in the homework problems. Section 16.5 is devoted to the analysis of forces and motion in electro-magneto-mechanical structures characterized by linear motion. The boxes Focus on Methodology: Analysis of Moving-Iron Electromechanical Transducers (pp. 823-824) and Focus on Methodology: Analysis of Moving-Coil Electromechanical Transducers (p. 836) summarize the analysis methods used in this section. The author has found that it is pedagogically advantageous to introduce the Bli and Blu laws for linear motion devices before covering these concepts for rotating machines: the student can often visualize these ideas more clearly in the context of a loudspeaker or of a vibration shaker. Example 16.9 (pp. 824-825) analyzes the forces in a simple electromagnet, and Examples 16.10 (pp. 825-827) and 16.12 (p.831) extend this concept to a solenoid and a relay. Example 16.11 (p. 827-828) ties the material presented in this chapter to the transient analysis topics of Chapter 5. Example 16.13 (pp. 836-839) performs a dynamic analysis of a loudspeaker, showing how the frequency response of a loudspeaker can be computed from an electromechanical analysis of its dynamics. Finally, The box Focus on Measurements: Seismic Transducer (pp.839-840) presents the dynamic analysis of an electromechanical seismic transducer. The homework problems are divided into four sections. The first reviews basic concepts in electricity and magnetism; the second presents basic and more advanced problems related to the concept of magnetic reluctance; the third offers some problems related to transformers. Section 4 contains a variety of applied problems related to electromechanical transducers; some of these problems emphasize dynamic analysis (16.40, 16.41, 16.42, 16.47-50) and are aimed at a somewhat more advanced audience. Learning Objectives 1. Review the basic principles of electricity and magnetism. Section 1. 2. Use the concepts of reluctance and magnetic circuit equivalents to compute magnetic flux and currents in simple magnetic structures. Section 2. 3. Understand the properties of magnetic materials and their effects on magnetic circuit models. Section 3. 4. Use magnetic circuit models to analyze transformers. Section 4. 5. Model and analyze force generation in electro-magneto-mechanical systems. Analyze moving iron transducers (electromagnets, solenoids, relays), and moving-coil transducers (electro-dynamic shakers, loudspeakers, seismic transducers. Section 5. 16.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Section 16.1: Electricity and Magnetism Problem 16.1 Solution: Known quantities: As shown in Figure P16.1. Find: a) The flux density in the core. b) Sketch the magnetic flux lines and indicate their direction. c) The north and south poles of the magnet. Assumptions: None. Analysis: a) B= φ 4 × 10 −4 = = 0.04 T A 0.01 b) Viewed from the top: c) See above. _____________________________________________________________________________________________ Problem 16.2 Solution: Known quantities: As shown in Figure P16.2. Find: If there is a resultant force on the single coil? If so, in what direction? Why? Assumptions: None. Analysis: Yes, the resultant force on the single coil is in the downward direction. If the coils are thought of as electromagnets, there is a north pole from the lower coil attracting a south pole from the upper coil. _____________________________________________________________________________________________ Problem 16.3 Solution: Known quantities: A LVDT is connected to a resistive load R L . Find: The LVDT equations. 16.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Assumptions: None. Analysis: Assume both secondary windings have resistance RS and inductance L S , if M S is the mutual coupling, we have: di di di V1 = M 1 − ( R S i L + L L ) + M S L dt dt dt di L di di + ( RS i L + L )− MS L V2 = M 2 dt dt dt ∴Vout = V1 − V2 = (M 1 − M 2 ) di di di − 2( RS i L + L L ) + 2M S L dt dt dt V i L = out RL R ∴(2( L − M S ) s + 1 + 2 S )Vout = ( M 1 − M 2 ) sI L RL Therefore, the transfer function is: Vout (M 1 − M 2 )s = R IL 2( L − M S ) s + 1 + 2 S RL ______________________________________________________________________________________ Problem 16.4 Solution: Known quantities: Equations of "Focus on Measurements: Linear Variable Differential Transformer", and the results of Problem 16.3. Find: The frequency response of the LVDT and the range of frequencies for which the device will have maximum sensitivity for a given excitation. Assumptions: None. Analysis: di We have Vex = L p + R pi dt ∴Vex ( s ) = ( L p s + R p ) I ( s ) R Vout ( s )(2( L − M ) s + 1 + 2 s ) = M s I ( s ) RL where M = M 1 − M 2 . Therefore, Vout ( s ) = Vex ( s ) Ms R ( L p s + R p ) 2( L − M ) s + 1 + 2 s RL To determine the maximum sensitivity of the output voltage to the excitation we could compute the V (s) ∂H ( s ) derivative of H ( s ) = out with respect to s, set = 0 , and solve for s . By setting s = jω , this ∂s Vex ( s ) 16.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 procedure will yield the excitation frequency for which the sensitivity of the output is maximum. It may, however, be more useful to compute the frequency response H ( jω ) numerically, to visualize the range of frequencies over which the sensitivity is acceptable. _____________________________________________________________________________________________ Problem 16.5 Solution: Known quantities: λ i= 0.5 + λ a. λ = 1V ⋅ s b. R = 1Ω i (t ) = 0.625 + 0.01 sin 400t A Find: a) The energy, coenergy, and incremental inductance. b) The voltage across the terminals on the inductor. Assumptions: None. Analysis: a) For λ = 1 V ⋅ s : The current is: λ i= = 0.667 A 0.5 + λ The energy is : § λ · ³0 ¨© 0 .5 + λ ¸¹ d λ = 1 .0 W m= 1 .0 ³ ¨© 1 − 0 .5 + λ ¸¹d λ = (λ − 0 .5 ln 0 .5 + λ ) 0 .5 § · 0 1 .0 0 W m = 1 .0 − 0 .5 ln 0 .5 + 1 .0 − 0 + 0 .5 ln 0 .5 = 0 .4507 J The coenergy is: W m' = i λ − W m = 0 . 2163 J The incremental inductance is: dλ | λ =0.625 di di 0 .5 + λ − λ = dλ (0 .5 + λ )2 L∆ = d λ (0 .5 + λ ) = di 0 .5 2 (0.5 + λ ) | dλ |λ =0.625 = λ =0.625 = 4.5H di 0.5 2 L∆ = b) To compute the voltage, we must add the contribution of the voltage across the resistive part of the inductor plus that generated by the inductance: V L (t ) = Ri (t ) + L∆ di dt di = 4 cos(400t ) dt 16.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 VL (t ) = (1Ω)(0.625 + 0.01sin(400t )) + (4.5H )(4 cos(400t )) VL (t ) = 0.625 + 18 sin(400t + 90°) It is important to observe that this is the inductor terminal voltage only for values of flux linkage in the neighborhood of 1V ⋅ s . ______________________________________________________________________________________ Problem 16.6 Solution: Known quantities: λ2 i= 0.5 + λ2 a) λ = 1 V ⋅ s b) R = 1 Ω i (t ) = 0.625 + 0.01 sin 400t A Find: a) The energy, coenergy, and incremental inductance. b) The voltage across the terminals on the inductor. Assumptions: None. Analysis: a) For λ = 1 V ⋅ s : The current is: λ2 i= = 0.667 A 0.5 + λ2 The energy is : § λ2 ³0 ¨¨© 0 .5 + λ 2 1 .0 W m= Integral of form: ³a 2 · ¸¸ d λ = ¹ 1 .0 0 .5 § ³ ¨© 1 − 0 .5 + λ 0 2 · ¸d λ ¹ du 1 u = tan −1 + c 2 a a +u 1 .0 λ ·· ¸¸ ¸ 0 . 5 ¹ ¸¹ 0 § § 0 .5 · 0 .5 § W m= ³ ¨1 − d λ = ¨¨ λ − tan −1 ¨¨ 2 ¸ 0 .5 + λ ¹ 0 .5 © © 0 © 1 .0 W m = 1 .0 − § 1 · § 0 · 0 .5 tan −1 ¨¨ ¸¸ − 0 + tan −1 ¨¨ ¸¸ = 0 .3245 J 0 .5 0 .5 © 0 .5 ¹ © 0 .5 ¹ 0 .5 The coenergy is: W m' = i λ − W m = 0 . 3425 J The incremental inductance is: dλ | λ =0.625 di di 2 λ (0 .5 + λ 2 ) − 2 λ (λ 2 ) = dλ (0 .5 + λ 2 )2 L∆ = ( dλ 0.5 + λ2 L∆ = |λ =0.625 = λ di ) ( dλ 0 .5 + λ 2 = λ di 2 |λ =0.625 = 2.25H 16.5 ) 2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 b) To compute the voltage, we must add the contribution of the voltage across the resistive part of the inductor plus that generated by the inductance: V L (t ) = Ri (t ) + L∆ di dt di = 4 cos(400t ) dt VL (t ) = (1Ω)(0.625 + 0.01sin(400t )) + (2.25H )(4 cos(400t )) VL (t ) = 0.625 + 9 sin(400t + 89.9°) It is important to observe that this is the inductor terminal voltage only for values of flux linkage in the neighborhood of 1V ⋅ s . _____________________________________________________________________________________________ Problem 16.7 Solution: Known quantities: Characteristic plot shown in Figure P16.7. Find: a) The energy and the incremental inductance for i = 1.0 A. b) The voltage across the terminals of the inductor when R = 2 Ω, i (t ) = 0.5 sin 2πt. Assumptions: None. Analysis: a) The diagram is shown below: (V S) 4 3 2 0 wm 0.5 1.0 1.5 Wm is the area at the left of the curve as shown. 1 1 Wm = × 0.5 × 2 + 0.5 × 1 + × 0.5 × 1 = 1.25 J 2 2 The incremental inductance is: ∆λ 2 L∆ = |i =1 = = 2 H ∆i 1 b) For i = 0.5 sin 2πt and R = 2 Ω : dλ V = Ri + dt 16.6 i(A) G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 For i < 0.5, λ = 4i : V L (t ) = sin( 2πt ) + 4 × 0.5 × 2π cos(2πt ) = sin(2πt ) + 4π cos(2πt ) ______________________________________________________________________________________ Problem 16.8 Solution: Known quantities: Structure of Figure 16.12 A = 0.1m 2 µ r = 2000 Find: The reluctance of the structure. Assumptions: Each leg is 0.1 m in length Mean magnetic path runs through the exact center of the structure Analysis: Calculation of mean path: Using the assumption that the mean magnetic path runs through the exact center of the structure, and since the structure is square, the mean path is determined using the following figure: .08m 0.1m .09m .1m l c = 4 × 0.09m = 0.36m Calculation of Reluctance: ℜ= lc lc 0.36 = = = 1432 A − turns / Wb µA µ r µ o A 2000 × 4π × 10 −7 × 0.1 ____________________________________________________________________________________ 16.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Section 16.2: Magnetic Circuits Focus on Methodology: Magnetic Structures and Equivalent Magnetic Circuits Direct Problem Given – The structure geometry and the coil parameters. Calculate – The magnetic flux in the structure. 1. Compute the mmf. 2. Determine the length and cross section of the magnetic path for each continuous leg or section of the path. 3. Calculate the equivalent reluctance of each leg. 4. Generate the equivalent magnetic circuit diagram, and calculate the total equivalent reluctance. 5. Calculate the flux, flux density, and magnetic field intensity, as needed. Direct Problem Given – The desired flux or flux density and structure geometry. Calculate – The necessary coil current and number of turns. 1. Calculate the total equivalent reluctance of the structure from the desired flux. 2. Generate the equivalent magnetic circuit diagram. 3. Determine the mmf required to establish the desired flux. 4. Choose the coil current and number of turns required to establish the desired mmf. Problem 16.9 Solution: Known quantities: a) φ = 4.2 × 10 −4 Wb, mmf = 400 A ⋅ t . b) l = 6in. Find: a) The reluctance of a magnetic circuit. b) The magnetizing force in SI units. Assumptions: None. Analysis: a) ℜ= F A⋅t 400 = = 9.52 × 10 5 −4 φ 4.2 × 10 Wb b) A⋅t = 2625 m m 6 × 0.0254 in _____________________________________________________________________________________________ H= F 16.8 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.10 Solution: Known quantities: As shown in Figure P16.10. Find: a) The reluctance values and show the magnetic circuit when µ = 3000 µ 0 . b) The inductance of the device. c) The new value of inductance when a gap of 0.1 mm is cut in the arm of length d) The limiting value of inductance when the gap is increased in size (length). Assumptions: Neglect leakage flux and fringing effects. Analysis: a) l1 µA1 l2 µA2 1 2 3 1 0.3 7.96 103 H 1 (3000)4π 10 7 (0.01) 0.1 10.671 103 H 7 4 (4π 10 )3000(25 10 ) 7.96 10 3 H l3 . 1 1 The circuit is shown below: R1 + R3 R2 F ℜT = ℜ1 + ℜ 2ℜ3 = 12.51 × 10 3 H −1 ℜ2 + ℜ3 b) L= N2 100 2 = = 0.8 H ℜ T 12.51 × 10 3 c) We have ℜg = 0.0001 = 7.96 × 10 3 H −1 −7 −4 (4π × 10 )(100 × 10 ) ℜ g is in series with ℜ 3 and thus: ℜT = ℜ1 + L= ℜ 2 (ℜ 3 + ℜ g ) ℜ 2 + ℜ3 + ℜ g = 14.33 × 10 3 H −1 N2 = 0.7 H ℜT d) As the gaps get longer, ℜ g will get larger and as an extreme case the circuit is made of ℜ1 and ℜ 2 in series, therefore: 16.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 ℜT = 18.57 × 10 3 H −1 N2 = 0.54 H ℜT _____________________________________________________________________________________________ L= Problem 16.11 Solution: Known quantities: N = 1000 turns, i = 0.2 A, l g1 = 0.02 cm, l g 2 = 0.04cm. Find: The flux and flux density in each of the legs of the magnetic circuit. Assumptions: Neglect fringing at the air gaps and any leakage fields. Assume the reluctance of the magnetic core to be negligible. Analysis: Calculate Reluctance in each air gap ℜ g1 = 0.0002 4π × 10 −7 × (0.01) 2 = 1.59 × 10 6 ℜ g 2 = 2ℜ g1 = 3.18 × 10 6 Assume the reluctance of the material can be neglected when compared to the reluctance of the air gaps; the analogous circuit is shown below: 1 + R g1 φ1 = - F 2 R g2 Ni = 1.26 × 10 −4 Wb ℜ g1 φ1 = 1.26Wb m 2 A 1 φ 2 = φ1 = 0.63 × 10 −4 Wb 2 1 B2 = B1 = 0.63Wb m 2 2 B1 = _____________________________________________________________________________________________ Problem 16.12 Solution: Known quantities: φ = 3 × 10 −4 Wb, l iron = l steel = 0.3 m, A = 5 × 10 −4 m 2 , N = 100 turns. Find: The current needed to establish the flux. 16.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Assumptions: None Analysis: Reluctance for each material is calculated as follows: ℜ= l l = µA µ r µ o The total reluctance of the structure is the sum of the reluctances for each material. From Table 16.1: Cast Iron: µ r = 5195 Cast Steel: µ r = 1000 Cast Iron: ℜ CI = l CI 0.3m = = 9.1909 × 10 4 A − turns / Wb 2 −7 −4 µ r µ o A (5195) 4π × 10 5 × 10 m ( )( ) Cast Steel: ℜ CS = lCS 0.3m = = 4.7746 × 10 5 A − turns / Wb 2 −7 −4 µ r µ o A (1000) 4π × 10 5 × 10 m ( )( ) Note: Cast Steel is less permeable than cast iron Total Reluctance: ℜ T = ℜ CI + ℜ CS = 9.1909 × 10 4 A − turns / Wb + 4.7746 × 10 5 A − turns / Wb = ℜ T = 5.6937 × 10 5 A − turns / Wb From φ= Ni , we can compute the current. ℜT φRT (3 × 10 −4 Wb)(5.6937 × 10 5 A − turns / Wb) i= = = 1.71 A N 100turns ______________________________________________________________________________________ Problem 16.13 Solution: Known quantities: As shown in Figure P16.13. Find: The magnetic flux φ . Assumptions: None. 16.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Analysis: I = 2 A, r = 0.08 m, N = 100, Across = 0.009 m 2 , µ r = 1000 l = 2πr = 0.50265 m, µ = µ r µ 0 R= 1 µAcross = 4.44444 × 10 4 A Wb mmf = I ⋅ N mmf = 0.0045Wb R ______________________________________________________________________________________ ∴φ = Problem 16.14 Solution: Known quantities: φ = 2.4 × 10 −4 Wb, l ab = l ef = 0.05 m, l af = lbe = 0.02 m, lbc = l dc ,A = 2 × 10 −4 m 2 . The material is sheet steel. Find: a) The current required to establish the flux. b) Compare the mmf drop across the air gap to that across the rest of the magnetic circuit and discuss your results using the value of µ for each material. Assumptions: None. Analysis: a) Assume the material is cast steel. B= 2.4 × 10 −4 = 12 T 2 × 10 −4 A⋅t H CS = 1400 m A⋅t H AG = 9.55 × 10 5 m 100 I = 1400(0.1 + 0.02 + 0.017) + (9.55 × 10 5 )(0.003) I = 30.6 A b) FCS = 191.8 A ⋅ t F AG = 2865 A ⋅ t Note: µ CS = 0.0009 µ AG = 0.00000126 µ CS ≈ 700 µ AG 16.12 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 ______________________________________________________________________________________ Problem 16.15 Solution: Known quantities: Magnet of Figure P16.15, φ = 2 × 10 −4 Wb, l ab = l bg = l gh = l ha = 0.2 m, l bc = l fg = 0.1 m, l cd = l ef = 0.099 m. The material is sheet steel. Find: The value of I required to establish the flux. Assumptions: None. Analysis: A1 = 2 × 10 −4 m 2 , A2 = 5 × 10 −4 m 2 µ r = 4000 l gap = l ha − 2l cd = 0.002 m R gap = l gap = 3.1831 × 10 6 A Wb µ 0 A2 mmf gap = φ1 R gap = 636.61977 A Ref = Rcd = l cd A = 3.93908 × 10 4 µA2 Wb mmf cd = φ1 Rcd = 7.87817 A l A R fg = Rbc = bc = 3.97887 × 10 4 µA2 Wb mmf bc = φ1 Rbc = 7.95775 A To find the mmf in the rightmost leg of the magnetic circuit, l A Ref = Rcd = cd = 3.93908 ×10 4 µA2 Wb l A R ab = ab = 7.95775 × 10 4 µA2 Wb A Wb mmf series = φ T R series = 848.69641 A mmf total = mmf series + mmf parallel = 1.51799 × 10 3 A R series = 3R ab = 2.38732 × 10 5 mmf total = 7.59 A N mmf parallel = mmf gap + 2mmf cd + 2mmf bc = 66829161 A ∴i = Rbg = φ2 = l bg = 1.98944 × 10 5 µA1 mmf parallel Rbg A Wb = 0.00336 Wb φT = φ1 + φ 2 = 0.00356 Wb ______________________________________________________________________________________ 16.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.16 Solution: Known quantities: Actuator of Figure P16.16, N = 2000 turns , g = 10 mm, B = 1.2 T , the air gap is fixed. Find: a) The coil current. b) The energy stored in the air gaps. c) The energy stored in the steel. Assumptions: None. Analysis: a) The equivalent circuit is: where: From Table 16.1, the relative permeability of sheet steel is 4000. ℜ L = ℜR = ℜC = 80 × 10 −3 = 25,984 A ⋅ t Wb 4000(4π × 10 −7 )(35 × 17.5) × 10 −6 80 × 10 −3 = 12,992 A ⋅ t Wb 4000(4π × 10 −7 )(35 × 35) × 10 −6 ℜ B1 = ℜ B 2 = ℜ g1 = 61.25 × 10 −3 = 19,894 A ⋅ t Wb 4000(4π × 10 −7 )(35 × 17.5) × 10 −6 10 × 10 −3 = 6,496,120 A ⋅ t Wb (4π × 10 −7 )(35 × 35) × 10 −6 10 × 10 −3 ℜg2 = = 12,955,225 A ⋅ t Wb (4π × 10 −7 )(35 × 17.5) × 10 −6 ℜ R + ℜ g 2 + ℜ B2 ℜT = ℜ C + ℜ g1 + = 13.01 × 10 6 A ⋅ t Wb 2 φT = BA = 1.2(35 × 35) × 10 −6 = 1.47 × 10 −3 Wb 2000 I = ℜ T φT I = 9.56 A b) 16.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 1.2 (10 × 10 −3 ) = 9549.3 A ⋅ t −7 4π × 10 1 wg1 = (1.47 × 10 −3 )(9549.3) = 7.02 J 2 1 1.47 × 10 −3 wg 2 = ( )(9549.3) = 3.51 J 2 2 wg = w g1 + 2wg 2 = 14.04 j H lg = c) 1 (1.47 × 10 − 3 )(2000)(9.56) = 14.05 J 2 = wT − w g = 0.01 J wT = wST ______________________________________________________________________________________ Problem 16.17 Solution: Known quantities: µ r = 2000, N = 100 . Find: a) The current needed to produce φ = 0.4 Wb m 2 in the center leg. b) The current needed to produce φ = 0.8Wb m 2 in the center leg. Assumptions: None. Analysis: With l1 = 34 cm, l2 = l3 = 90 cm and A = (8 × 10 −2 ) 2 cm 2 , we compute: ℜ1 = 0.34 2000 × 4π × 10 4 ℜ 2 = 5.595 × 10 H ℜT = ℜ1 + −7 × (8 × 10 −1 = ℜ3 −2 2 ) = 2.114 × 10 4 H −1 ℜ 2ℜ3 = 4.91 × 10 4 H −1 ℜ 2 + ℜ3 φT = 0.4 × (0.08) 2 = 2.56 × 10 −3 Wb BART Ni From φT = BA = , we have i = . ℜT N a) 2.56 × 10 −3 × 4.91 × 10 4 125.7 = 100 N = 1.257 A i= b) Since the current is directly proportional to B, the current will have to be doubled. ______________________________________________________________________________________ 16.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Section 16.4: Transformers Problem 16.18 Solution: Known quantities: N = 1000 turns, l 1 = 16 cm , A1 = 4 cm 2 ,l 2 = 22 cm, A2 = 4 cm 2 , l 3 = 5 cm ,l3 = 2 cm 2 , µ r = 1500. Find: a) Construct the equivalent magnetic circuit and find the reluctance associated with each part of the circuit. b) The self-inductance and mutual-inductance for the pair of coils. Assumptions: None. Analysis: a) The analogous circuit is shown below: R1 R2 + + R3 F1 - F2 - The individual reluctances are: 16 × 10 −2 4π × 10 −7 × 1500 × 4 × 10 − 4 = 2.12 × 10 5 H −1 ℜ1 = 22 × 10 − 2 ℜ2 = 4π × 10 −7 × 1500 × 4 × 10 − 4 = 2.92 × 10 5 H −1 5 × 10 − 2 4π × 10 − 7 × 1500 × 2 × 10 − 4 = 1.33 × 10 5 H −1 ℜ3 = b) The inductance can be computed as follows: Lm1 = N2 = 4.72 H ℜ1 Lm 2 = N2 = 3.43 H ℜ2 Lm3 = N2 = 7.54 H ℜ3 let LT = Lm1 + Lm 2 + Lm3 = 15.68 H L L Lm = m1 m 2 = 1.03 H = L12 = L21 = M LT 16.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 L L L1 = m1 m3 = 2.27 H LT L L L2 = m 2 m3 = 1.65 H LT L11 = L1 + Lm = 3.3 H L22 = L2 + Lm = 2.68 H ______________________________________________________________________________________ Problem 16.19 Solution: Known quantities: A 300 Ω resistive load referred to the primary is 7500 Ω . r1 = 20 Ω, L1 = 1.0 mH , Lm = 25 mH , r2 = 20 Ω, L2 = 1.0 mH . Find: a) The turns ratio. b) The input voltage, current, and power and the efficiency when this transformer is delivering 12 W to the 300 Ω load at a frequency f = 10,000 2π Hz . Assumptions: Core losses are negligible. Analysis: The equivalent circuit is: a) From 7500 = N 2 × 300 , we have N = 5 . b) X L1 = 2πfL1 = 10 = X L 2 X Lm = 250 From I L2 R L = 12 W , I L2 = 0.04 , we have: I L = 0.2∠0 , A V L = 60∠0 , V 16.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 V2 = I L ( R L + r2 + jX L 2 ) = 64 + j 2 = 64.03∠1.79 , V I m = 0.256∠ − 88.21, = 0.008 − j 0.2559 A I1' = I m + I L = 0.33∠ − 50.9 , A V1' = I1' (r1 + jX L1 ) + V2 = 70.72 − j1.04 = 70.72∠ − 0.84 , V V1 = NV1' = 353.6∠ − 0.84 , V 1 ' I1 = 0.066∠ − 50.9 , A N Pin = V1 I1 cos θ = 14.98 W I1 = P ∴efficiency = η = out = 80.1% Pin ______________________________________________________________________________________ Problem 16.20 Solution: Known quantities: A 220 20 V transformer has 50 turns on its low-voltage side. Find: a) b) The number of turns on its high side. The turns ratio α when it is used as a step-down transformer. The turns ratio α when it is used as a step-up transformer. c) Assumptions: None. Analysis: 220 α= = 11 20 a) The primary has N P = 50 × 11 = 550 turns b) α = 11 is a step-down transformer. c) 1 α= is a step-up transformer. 11 ______________________________________________________________________________________ Problem 16.21 Solution: Known quantities: The high-voltage side of the transformer has 750 turns, and the low-voltage side 50 turns. The high side is connected to a rated voltage of 120 V . A rated load of 40A is connected to the low side. Find: a) The turns ratio. b) The secondary voltage. 16.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 c) The resistance of the load. Assumptions: No internal transformer impedance voltage drops Analysis: a) 750 α= = 15 50 b) 1 120 V2 = V1 = = 8V 15 α c) 8 RL = = 0.2 Ω 40 ______________________________________________________________________________________ Problem 16.22 Solution: Known quantities: A transformer is used to match an 8 Ω loudspeaker to a 500 Ω audio line. Find: a) The turns ratio of the transformer. The voltages at the primary and secondary terminals when 10 W of audio power is delivered to the speaker. Assumptions: The speaker is a resistive load and the transformer is ideal. Analysis: a) b) From α 2 R L = 500 , we have α = 7.91 b) From 10 = V22 , we have V2 = 8.94 V RL c) V1 = αV2 = 70.7 V ______________________________________________________________________________________ Problem 16.23 Solution: Known quantities: It is a step-down transformer. The high-voltage and low-voltage sides have 800 turns and 100 turns respectively. 240 VAC voltage is applied to the high side. The impedance of the low side is 3 Ω . Find: a) The secondary voltage and current. b) The primary current. c) The primary input impedance from the ratio of primary voltage and current. d) The primary input impedance. Assumptions: None. 16.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Analysis: We have α = N1 =8. N2 a) 1 V1 = 30 V α V I 2 = 2 = 10 A RL V2 = b) I1 = c) 1 I 2 = 1.25 A α Z in = 240 = 192 1.25 d) Z in = α 2 R L = 192 ______________________________________________________________________________________ Problem 16.24 Solution: Known quantities: It is a step-up transformer. All the others are the same as Problem 16.23. Find: The transformer ration of the transformer. Assumptions: None. Analysis: 100 1 α= = 800 8 ______________________________________________________________________________________ Problem 16.25 Solution: Known quantities: It is a 2,300 240 − V , 60 − Hz, 4.6 − kVA transformer. It has an induced emf of 2.5V turn . Find: N and low-side turns l . I b) The rated current of the high-voltage side h . c) The transformer ratio when the device is used as a step-up transformer. Assumptions: It is an ideal transformer. Analysis: a) a) The number of high-side turns Nh 16.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 2300 = 920 turns 2 .5 240 = 96 turns Nl = 2.5 Nh = b) Ih = 4.6 × 10 3 =2A 2300 c) Nl = 0.1044 Nh ______________________________________________________________________________________ α= 16.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Section 16.5: Electromechanical Transducers a) Moving-Iron Transducers Focus on Methodology Analysis of moving-iron electromechanical transducers a. Calculation of current required to generate a given force 1. 2. Derive an expression for the reluctance of the structure, as a function of air gap displacement: R(x). Express the magnetic flux in the structure, as a function of the mmf (i.e., of the current, I) and of the reluctance, R(x): 3. F (i) R(x) Compute an expression for the force using the known expressions for the flux and for the 2 reluctance: f 4. dR(x) 2 dx Solve the expression in 3. for the unknown current, i. b. Calculation of force generated by a given transducer geometry and mmf Repeat steps 1-3 above, substituting the known current to solve for the force, f. Problem 16.26 Solution: Known quantities: Electromagnet of Example 16-9 (Figure 16.38) N = 700turns; f restore = 8,900 N ; Agap = 0.01m 2 ; L = 1m; µ r = 1000 Find: a) The current required to keep the bar in place b) Initial current to lift the magnet if the bar is initially 0.1 m away from the electromagnet Assumptions: a) Air gap becomes zero and the iron reluctance cannot be neglected b) Neglect the iron reluctance Analysis: a) To compute the current we need to derive an expression for the force in the air gap. Without neglecting the iron reluctance, we can write the expression for the reluctance as follows: L 2x (x) r 0A 0A where L is the total length of the iron magnetic path (excluding the air gap). Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current: Ni A Ni r 0 (x) L 2 rx Then, the magnitude of the force in the air gap is given by the expression 2 2 2 d ( x) Ni r 0 A Ni r 1 2 0A f 2 2 2 2 2 L 4L r x 4 r x 2 dx L 4 L r x 4 r2 x 2 0A As x approaches zero, we can calculate the force to be: 16.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 (Niµ r )2 µ 0 A f ( x = 0) = L2 and the current required to maintain the given force is: (Niµ r )2 µ 0 A f ( x = 0) = i=± L2 L2 f ( x = 0) ( N µ r )2 µ 0 A =± L Nµ r f ( x = 0) µ0 A Assuming that the total length of the magnetic path is L=1 m and that µr = 1,000, we can calculate a value for the current to be 1.20 A. b) Since the bar is initially 0.1 m away from the structure, the reluctance of the air dominates the reluctance of the structure. The reluctance is calculated as: 2x µ0 A ℜ( x ) = Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current: φ= µ A Ni = Ni 0 ℜ( x ) 2x Then, the magnitude of the force in the air gap is given by the expression 2 (Ni )2 µ 0 A φ 2 dℜ( x) 1 ( Niµ 0 A) 2 = = f = 2 dx 2 4x 2 µ0 A 4x 2 Finally, the current required is: i=± 2x N f µ0 A =± 2(0.1m ) 8900 N = ±240.3 A 700 4π × 10 −7 × 0.01m 2 Note that the holding current from part a is significantly smaller than the current required to lift the bar from the initial distance of 0.1 m. ______________________________________________________________________________________ Problem 16.27 Solution: Known quantities: ℜ( x) = 7 × 10 8 (0.002 + x) H −1 , N = 980 t , R = 30 Ω, Va = 120 V Find: a) The energy stored in the magnetic field for x = 0.005 m . b) The magnetic force for x = 0.005 m . Assumptions: None. Analysis: a) We have L( x) = N2 . ℜ( x ) Wm' = Wm = L( x)i 2 2 The current is: 16.23 G. Rizzoni, Principles and Applications of Electrical Engineering I DC = 120 = 4A 30 ∴Wm = b) f =− Problem solutions, Chapter 16 980 2 × 4 2 2 × 7 × 10 8 × 0.007 = 1.568 J i 2 N 2 dℜ( x) = −224 N 2 (ℜ( x)) 2 dx The minus sign indicates that the force f is in a direction opposite to that indicated in the figure. _____________________________________________________________________________________________ Problem 16.28 Solution: Known quantities: Solenoid of Example 16.10 (Figure 16.40) Find: The best combination of current magnitude and wire diameter to reduce the volume of the solenoid coil. Will this minimum volume result in the lowest possible resistance? How does the power dissipation of the coil change with the wire gauge and current value? Assumptions: Use of Copper wire in solenoid Analysis: In order to access the effects of the wire diameter and current magnitude to the volume, resistance, and power dissipated, mathematical expressions need to be developed for each variable. Volume: Vcoil = l coil Acoil The length of the coil is given by the circumference: l coil = πNd coil The area of the coil: π 2 d coil 4 π 2 Nd coil 3 π 2 = πNd coil d coil = 4 4 Acoil = Vcoil From Example 16.10, the relationship between the current and the number of turns is given as: Ni = 56.4 A − turns Hence: Vcoil = (56.4)π 2 d coil 3 4i Resistance: The resistance of the wire is given as: R= ρl coil Acoil where ρ is the resistivity of the wire, which is assumed to be copper with a Using the previous derivations: 16.24 ρ = 1.725 × 10 −8 Ω / m G. Rizzoni, Principles and Applications of Electrical Engineering R= Problem solutions, Chapter 16 ρπNd coil 4 ρN (4)(56.4)ρ = = π 2 d coil id coil d coil 4 Power: The dissipated power is given by: P = i2R = (4)(56.4)iρ d coil Using a chart for AWG wire gauge and current carrying capacity, a table can be developed relating the wire gauge and current carrying capacity to the volume, resistance, and power dissipated. AWG Gauge Wire Diameter [in] Wire Diameter [m] Current Capacity [A] 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 1.019E-01 8.081E-02 6.408E-02 5.082E-02 4.030E-02 3.196E-02 2.530E-02 2.010E-02 1.590E-02 1.260E-02 1.000E-02 8.000E-03 6.300E-03 5.000E-03 4.000E-03 3.100E-03 2.500E-03 2.000E-03 1.600E-03 1.200E-03 1.000E-03 2.588E-02 2.053E-02 1.628E-02 1.291E-02 1.024E-02 8.118E-03 6.426E-03 5.105E-03 4.039E-03 3.200E-03 2.540E-03 2.032E-03 1.600E-03 1.270E-03 1.016E-03 7.874E-04 6.350E-04 5.080E-04 4.064E-04 3.048E-04 2.540E-04 30.0 20.0 15.0 10.0 5.0 3.3 2.5 1.25 0.83 0.63 0.31 0.21 0.16 7.81E-02 5.16E-02 3.91E-02 1.95E-02 1.29E-02 9.77E-03 4.88E-03 3.22E-03 Required Number of Turns [turns] 2 3 4 6 11 17 23 45 68 90 180 273 361 722 1094 1444 2888 4375 5775 11551 17501 Coil Volume [m3] Resistance [Ω ] Power Dissipated [W] 8.041E-05 6.017E-05 4.001E-05 2.993E-05 2.986E-05 2.256E-05 1.477E-05 1.481E-05 1.111E-05 7.299E-06 7.297E-06 5.661E-06 3.649E-06 3.649E-06 2.831E-06 1.739E-06 1.824E-06 1.415E-06 9.565E-07 8.070E-07 7.076E-07 5.012E-06 9.480E-06 1.594E-05 3.015E-05 7.603E-05 1.453E-04 2.422E-04 6.098E-04 1.168E-03 1.946E-03 4.903E-03 9.286E-03 1.556E-02 3.922E-02 7.428E-02 1.265E-01 3.138E-01 5.943E-01 9.806E-01 2.615E+00 4.754E+00 4.511E-03 3.792E-03 3.586E-03 3.015E-03 1.901E-03 1.582E-03 1.514E-03 9.528E-04 7.950E-04 7.600E-04 4.788E-04 3.950E-04 3.800E-04 2.394E-04 1.975E-04 1.931E-04 1.197E-04 9.875E-05 9.351E-05 6.234E-05 4.938E-05 As the wire gauge increases, the wire diameter and current carrying capacity both decrease, and in turn the number of turns required increases, and the coil volume decreases. However, the resistance also increases with an increase in wire gauge. Hence, the minimum volume will not result in the lowest possible resistance. Finally, the power dissipation decreases with the increase of wire gauge due to decrease in current capacity. _____________________________________________________________________________________ Problem 16.29 Solution: Known quantities: Solenoid of Example 16.10 (Figure 16.40) a = 0.01m , l gap = 0.001m , K = 10 N / m Find: f , mmf using equation 16.46 and equation 16.30 16.25 G. Rizzoni, Principles and Applications of Electrical Engineering Assumptions: The reluctance of the iron is negligible, Neglect fringing At x = 0 , the plunger is in the gap by an infinitesimal displacement Problem solutions, Chapter 16 ε Analysis: From Example 16.10: ℜ( x ) = 2l gap µ o ax Compute inductance if the magnetic circuit as a function of reluctance (equation 16.30): L= N 2 µ o ax N2 = 2l gap ℜ( x ) Compute stored magnetic energy: Wm = 1 2 1 N 2 i 2 µ o ax Li = 2 2 2l gap Finally, use equation 16.46 to write the expression for the magnetic force: dWm N 2i 2 µ o a =− fe = − dx 4l gap This matches the derivation from example 16.10 using the relationship between the magnetic flux, the reluctance of the structure, and the magnetic force. The calculation for the required mmf is identical to example 16.10: f gap = kx = ka = (10 N / m ) × (0.01m ) = 0.1N _____________________________________________________________________________________________ Problem 16.30 Solution: Known quantities: Solenoid of Example 16.11 (Figure 16.40) a = 0.01m , l gap = 0.001m , K = 10 N / m , N = 1000turns , V = 12V , Rcoil = 5Ω Find: Current and magnetic force response as a function of time using equation 16.46 and equation 16.30 in the derivation Assumptions: The reluctance of the iron is negligible, Neglect fringing The inductance of the solenoid is approximately constant and is equal to the midrange value (plunger displacement equal to a / 2 ). Analysis: From Example 16.11: ℜ gap ( x ) = 2l gap µ o ax Compute inductance if the magnetic circuit as a function of reluctance (equation 16.30): L= N2 ℜ gap ( x ) = N 2 µ o ax 2l gap Compute stored magnetic energy: 1 N 2 i (t ) 2 µ o ax 1 2 Wm = Li (t ) = 2 2l gap 2 16.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Finally, use equation 16.46 to write the expression for the magnetic force: dWm N 2i (t ) 2 µ o a (1000) × 4π × 10 −7 × 0.01 2 f gap = − i (t ) = πi (t ) 2 =− = dx 4l gap 4 × 0.001 With the assumption that the inductance is constant with x = a / 2 : N 2 µ o a 2 (1000)2 × 4π × 10 −7 × (0.01) 2 N2 L≈ = = = 31.4mH ℜ gap ( x ) 4l gap 4 × 0.001 2 From Example 16.11: i (t ) = f gap − t V § 12 § − Rt · 6 .3×10 − 3 L 1 e 1 e − = − ¨ ¸ ¨ ¹ 5 © R© − t ª12 § − 3 ·º = π « ¨1 − e 6.3×10 ¸» ¹¼ ¬5 © · ¸A ¹ 2 This matches the derivation from example 16.11 using the relationship between the magnetic flux, the reluctance of the structure, and the magnetic force. The response curves are shown below: ______________________________________________________________________________________ Problem 16.31 Solution: Known quantities: Solenoid of Example 16.11 (Figure 16.40) a = 0.01m , l gap = 0.001m , K = 10 N / m , N = 1000turns , V = 12V , Rcoil = 5Ω 16.27 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 m = 0.5kg Find: Generate a simulation program that accounts for the fact that the solenoid inductance is not constant, but is a function of plunger position Compare graphically the current and force step responses of this system to the step response obtained in Example 16.11 Assumptions: The reluctance of the iron is negligible, Neglect fringing Neglect damping on the plunger Analysis: From Example 16.11: ℜ gap ( x ) = 2l gap µ o ax The inductance is now a function of plunger position N 2 µ o ax L= = 2l gap ℜ gap ( x ) N2 The differential equation for the current using the equivalent circuit shown below: R V L V (t ) = Ri (t ) + L di dt To find the equation of the force, compute stored magnetic energy as a function of current: Wm = 1 1 N 2 i (t ) 2 µ o ax Li (t ) 2 = 2 2 2l gap Finally, use equation 16.46 to write the expression for the magnetic force: f gap dWm N 2 i(t ) 2 µ o a (1000)2 × 4π × 10 −7 × 0.01 2 =− =− = i (t ) = πi (t ) 2 dx 4l gap 4 × 0.001 This matches the derivation from example 16.11 using the relationship between the magnetic flux, the reluctance of the structure, and the magnetic force. Simulink Block Diagram: 16.28 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Comparison of Step Responses for Constant Inductance and Variable Inductance Systems: Note the quicker response of the variable inductance system due to the smaller inductance initially. The larger constant inductance results in a delayed response. ______________________________________________________________________________________ 16.29 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.32 Solution: Known quantities: Relay of Example 16.12 (Figure 16.46) N = 10,000turns; f restore = 5 N ; Agap = (0.01m ) ; x = 0.05mµ r = 1000 2 Find: Required holding current to keep relay closed Assumptions: Air gap becomes zero and the iron reluctance cannot be neglected Analysis: To compute the current we need to derive an expression for the force in the air gap. Without neglecting the iron reluctance, we can write the expression for the reluctance as follows: L 2x (x) r 0A 0A where L is the total length of the iron magnetic path (excluding the air gap). Knowing the reluctance we can calculate the magnetic flux in the structure as a function of the coil current: Ni A Ni r 0 (x) L 2 rx Then, the magnitude of the force in the air gap is given by the expression 2 2 2 d ( x) Ni r 0 A Ni r 1 2 0A f 2 2 2 2 2 L 4L r x 4 r x 2 dx L 4 L r x 4 r2 x 2 0A As x approaches zero, we can calculate the force to be: f ( x = 0) = (Niµ r )2 µ 0 A L2 and the current required to maintain the given force is: f ( x = 0) = i=± (Niµ r )2 µ 0 A L2 L2 f ( x = 0) ( N µ r )2 µ 0 A =± f ( x = 0) L Nµ r µ0 A The total length of the magnetic path: L = 0.05m + 0.05m + 0.10m + 0.10 = 0.30m The current: i=± (0.30) (10000)(1000) (5) (4π ×10 )(0.01) −7 2 = ±0.0060 A = ±6.0mA _____________________________________________________________________________________________ Problem 16.33 Solution: Known quantities: Relay Circuit shown in Figure P16.33 N = 500turns; Agap = 0.001m 2 ; L = 0.02m; k = 1000 N / m; R = 18Ω Find: Minimum DC supply voltage v for which the relay will make contact when the electrical switch is closed 16.30 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Assumptions: Neglect the iron reluctance Analysis: The reluctance of the gap: 2x µ 0 Agap ℜ gap ( x ) = dℜ gap ( x ) dx = 2 µ 0 Agap Magnetic Flux: φ= Niµ 0 Agap Ni = 2x ℜ( x ) Magnetic force: (Ni ) µ 0 Agap φ 2 dℜ( x ) § Niµ 0 Agap · 1 2 ¸¸ = = ¨¨ = 2 dx 2x 4x 2 © ¹ 2 µ 0 Agap 2 f gap 2 The force that must be overcome is the spring force, fk: f gap = f k = kx Equating the two force equations and solving for the current: i=± 2 N kx 3 µ 0 Agap The voltage is determined using Ohm’s law, and x = L: (1000)(0.02) = 182.0V 2R 2(18) kx 3 =± N µ 0 Agap (500) 4π × 10 −7 (0.001) 3 v = iR = ± ______________________________________________________________________________________ Problem 16.34 Solution: Known quantities: The simplified representation of a surface roughness sensor shown in Figure P16.34 Find: Derive an expression for the displacement x as a function of the various parameters of the magnetic circuit and of the measured emf e Assumptions: The flux φ = β / ℜ( x ) where β is a known constant Frictionless contact between the moving plunger and the magnetic structure The plunger is restrained to vertical motion only 16.31 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Analysis: e N d ; dt ; (x) L x (x) r 0A (x) e N d dt x 1 L 0A r 0A r 0A ; r 1 x L N dx x dt N r 0A L 1x r 1 dx 1 x dt r 2 r ______________________________________________________________________________________ Problem 16.35 Solution: Known quantities: As shown in Figure P16.35. The air gap between the shell and the plunger is uniform and equal to 1 mm . The diameter is d = 25 mm . The exciting current is 7.5 A . N = 200 . Find: The force acting on the plunger when x = 2 mm . Assumptions: None. Analysis: The cross-section area A is: 25 × 10 −3 2 2 ) m 2 From this expression we can compute the variable reluctance of the air gap: x ℜx = = 1621 × 10 6 x −3 2 π ( 25 × 10 ) 4π × 10 −7 × 4 and the resulting force is: A = π( f = i 2 N 2 dR x 2 R x2 dx 7 .5 2 200 2 × 1621 × 10 6 | x =2×10−3 2 (1621 × 10 6 x) 2 = 173.5 N ______________________________________________________________________________________ = Problem 16.36 Solution: Known quantities: As shown in Figure P16.36, the double-excited electromechanical system moves horizontally. The cross section of the structure is w × w . 16.32 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Find: a) The reluctance of the magnetic circuit. b) The magnetic energy stored in the air gap. c) The force on the movable part as a function of its position. Assumptions: Resistance, magnetic leakage and fringing are negligible. The permeability of the core is very large. Analysis: a) The reluctance of the magnetic circuit is: ℜ= lg x + 2 µ 0 w( w − x) µ0 w b) The magnetic energy stored in the air gap is: Wm = ( N1 + N 2 ) 2 i 2 2ℜ c) The magnetic force is: lg · x 2§ ¨¨ ¸ ( N + N ) + 1 2 2 2 ¸ µ w µ w ( w − x ) i2 0 © 0 ¹ f = 2 2 lg § x · ¨¨ ¸¸ + 2 µ w ( w − x ) µ w 0 © 0 ¹ ______________________________________________________________________________________ Problem 16.37 Solution: Known quantities: The flux density in the cast steel pathway is 1.1T . The diameter of the plunger is 10 mm . Find: The force f between the faces of the poles. Assumptions: Air gap is negligible between walls and plunger Since the pathway is cast steel, µ r = 1000 (from Table 16.1). Analysis: Using Equation 16.50: f = φ 2 dℜ( x ) 2 dx The flux is determined from the flux density and the area: φ = BA Since the plunger is cylindrical and the air gap between the plunger and the coil is negligible, the reluctance is calculated as: ℜ( x ) = x µA where x is the gap between the plunger and back wall of the solenoid. The reluctance of the cast steel pathway can be neglected due to the low reluctance of the structure. The derivative of the reluctance is: 16.33 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 dℜ( x ) 1 = dx µA The area is the cross-sectional area of the plunger: A= π 2 d 4 Combining all of the equations into the force equation: φ 2 dℜ( x ) (BA)2 1 πd 2 B 2 1 π (0.010 )2 (1.1)2 f = = = = = 37.8 N 2 dx 2 µA (2 )(4 ) µ o 8(4π × 10 −7 ) ______________________________________________________________________________________ Problem 16.38 Solution: Known quantities: A force of 10,000 N is required to support the weight. The cross-sectional area of the magnetic core is The coil has 1000 turns . 0.01 m 2 . Find: The minimum current that can keep the weight from falling for x = 1.0 mm . Assumptions: Negligible reluctance for the steel parts and negligible fringing in the air gaps. Analysis: The variable reluctance is given by: 2x ℜT ( x) = 2ℜ( x) = 4π × 10 −7 (0.01) = 159.15 × 10 6 x The force is related to the reluctance by: N 2 i 2 dℜT ( x) f = −10000 = − 2 RT2 ( x) dx Therefore, i = 3.18 = 1.784 A ______________________________________________________________________________________ Problem 16.39 Solution: Known quantities: The 12 − VDC control relay is made of sheet steel. Average length of the magnetic circuit is 12 cm . The average cross section of the magnetic circuit is 0.60 cm 2 . The coil has 250 turns and carries 50 mA . Find: a. The flux density B in the magnetic circuit of the relay when the coil is energized. b. The force F exerted on the armature to close it when the coil is energized. Assumptions: None. Analysis: a. 16.34 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 NI = Hl 250(250 × 10 −3 ) = H (12 × 10 −2 ) H = 104.2 from the curve, B = 0.75 T . b. F= 1 AB 2 1 (0.6 × 10 −4 )(0.75) 2 = = 13.4 N 2 µ0 2 4π × 10 −7 ______________________________________________________________________________________ Problem 16.40 Solution: Known quantities: As shown in Figure P16.40. Find: The differential equations describing the system. Assumptions: None. Analysis: The equation for the electrical system is: di v = iR + L( x) dt where: N 2µ0 A N2 = ℜT ( x) 2x The equation for the mechanical system is: L( x) = Fm = m where Fm d 2x + kx dt 2 is the magnetic pull force. To calculate this force we use the following equation: dWm dx is the energy stored in the magnetic field. Fm = − where Wm Let F and ℜ be the magnetomotive force acting on the structure and its reluctance, respectively; then: Wm = φ 2 ℜ( x) F2 N 2 i 2 µA = = 2 2ℜ( x) 4x dWm N 2 i 2 µA = dx 4x 2 Finally, the differential equations governing the system are: Fm = − v = iR + m d 2x N 2 µ 0 A di 2 x dt + kx = N 2 i 2 µA dt 2 4x 2 This system of equations could be solved using a numerical simulation, since it is nonlinear. ______________________________________________________________________________________ 16.35 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.41 Solution: Known quantities: As shown in Figure P16.41. The solenoid has a cross section of 5 cm 2 . Find: a. The force exerted on the plunger when the distance x is 2 cm and the current is 5 A . N = 100 turns . Relative permeabilities of the magnetic material and the nonmagnetic sleeve are 2,000 and 1 . b. Develop a set of differential equations governing the behavior of the solenoid. Assumptions: The fringing and leakage effects are negligible. Analysis: a. ℜx = x µ 0 Ax ℜg = lg 0.005 = 7.96 × 10 6 H −1 −7 −4 (4π × 10 )(5 × 10 ) = µ 0 Ag l1 0.335 = = 2.67 × 10 5 H −1 −7 −4 µ r µ 0 Ag 2000(4π × 10 )(5 × 10 ) ℜ1 = ℜ 2 = ℜ3 = l3 0.055 = = 2.19 × 10 4 H −1 µ r µ 0 Ax 2000(4π × 10 −7 )(10 × 10 − 4 ) ℜm = lx lx = = (3.98 × 10 5 )(0.095 − x) H −1 µ r µ 0 Ax 2000(4π × 10 −7 )(10 × 10 −4 ) The circuit is shown below: Rg Rg Rx Rm R2 F ℜ eq = R3 Rx R1 R3 + - ℜ g + ℜ1 2 Rm + - F = lg 2 µ 0 Ag + R eq l1 = 4.11 × 10 6 H −1 2µ r µ 0 A ℜ T = ℜ 3 + ℜ x + ℜ m + ℜ eq = lg l3 l1 x (0.095 − x) + + + + µ r µ 0 Ax µ 0 Ax µ r µ 0 Ax 2 µ 0 Ag 2 µ r µ 0 Ag Since Ax = 2Ag: 16.36 G. Rizzoni, Principles and Applications of Electrical Engineering ℜT = = l 3 + xµ r + (0.095 − x) + l g µ r + l1 2 µ r µ 0 Ag Problem solutions, Chapter 16 = 0.055 + 0.095 − x + 2000 x + (0.005)(2000 ) + 0.335 10.485 + 1999 x = = 2(2000 ) 4π × 10 −7 5 × 10 −4 2.51 × 10 −6 ( )( ) = 4.18 × 10 6 + 7.96 × 10 8 x 2 N 2 i 2 (2 µ r µ 0 Ag ) (µ r − 1) = N 2 i 2 dℜ T φ 2 ∂ℜ T ( x) =− =− fm = − 2 2 2 ∂x 2 RT dx 2(l3 + xµ r + (0.095 − x) + l g µ r + l1 ) 2 µ r µ 0 Ag =− (l µ r µ 0 Ag N 2 i 2 (µ r − 1) + xµ r + (0.095 − x) + l g µ r + l1 ) 2 3 = 2 2 2000 )(4π × 10 −7 )(5 × 10 −4 )(100 ) (5) (2000 − 1) ( =− (0.055 + 0.095 − x + 2000 x + (0.005)(2000 ) + 0.335) 2 For x = 0.02, f = −0.25N . b. Electrical subsystem: v s (t ) = L( x) di (t ) + Ri (t ) dt Mechanical subsystem: mx(t ) = f m ( x) − d dx(t ) k − x(t ) l dt Reluctance: ℜT = Flux: φ= l 3 + xµ r + (0.095 − x) + l g µ r + l1 2 µ r µ 0 Ag 2 µ r µ 0 Ag Ni (t ) Ni (t ) = ℜ m ( x) l 3 + xµ r + (0.095 − x) + l g µ r + l1 Magnetic force: 2 N 2 i 2 (2 µ r µ 0 Ag ) (µ r − 1) φ 2 ∂ℜ T ( x) N 2 i 2 dℜ T = =− = f m (x ) = 2 2 2 ∂x 2 RT dx 2(l 3 + xµ r + (0.095 − x) + l g µ r + l1 ) 2 µ r µ 0 Ag = (l µ r µ 0 Ag N 2 i 2 (µ r − 1) + xµ r + (0.095 − x) + l g µ r + l1 ) 2 3 Inductance: L( x ) = N 2 2 µ r µ 0 Ag N2 = ℜ T ( x) l3 + xµ r + (0.095 − x) + l g µ r + l1 Substituting the expressions for fm and L(x) in the two differential equations, we have the final answer. 16.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Electrical subsystem: § · di (t ) N 2 2µ r µ 0 Ag ¨ ¸ v s (t ) = + Ri (t ) ¨ l 3 + xµ r + (0.095 − x) + l g µ r + l1 ¸ dt © ¹ Mechanical subsystem: mx(t ) = (l µ r µ 0 Ag N 2 i 2 (µ r − 1) + xµ r + (0.095 − x) + l g µ r + l1 ) 2 3 −d dx(t ) k − x(t ) dt l Note that these equations are very nonlinear! ______________________________________________________________________________________ Problem 16.42 Solution: Known quantities: The relay shown in Figure P16.42 Find: Derive the differential equations (electrical and mechanical) for the relay Assumptions: The inductance is a function of x The iron reluctance is negligible Analysis: Electrical subsystem: v(t) L(x) di(t) dt Ri(t) Mechanical subsystem: mx(t ) = f m ( x) − b dx(t ) − kx(t ) dt Next, we calculate the magnetic force and inductance as functions of x. Reluctance: ( x) Flux: 2x A Ni( t ) ( x) Ni(t ) A 2x Magnetic force: φ 2 ∂ℜ( x) N 2 i 2 µA f m (x ) = = 2 ∂x 4x 2 Inductance: L( x ) = N2 N 2 µA = ℜ( x ) 2x Substituting the expressions for fm and L(x) in the two differential equations, we have the final answer. 16.38 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Electrical subsystem: v(t ) = N 2 µA di (t ) + Ri (t ) 2 x(t ) dt Mechanical subsystem: mx(t ) = N 2 i 2 µA dx(t ) −b − kx(t ) 2 dt 4 x(t ) Note that these equations are very nonlinear! ______________________________________________________________________________________ b) Moving-Coil Transducers Focus on Methodology: Analysis of moving-coil electromechanical transducers 1. 2. 3. Apply KVL to write the differential equation for the electrical subsystem, including the back emf, e = Blu, term. Apply Newton’s Second Law to write the differential equation for the mechanical subsystem, including the magnetic force f = Bli, term. Laplace transform the two coupled differential equations to formulate a system of linear algebraic equations, and solve for the desired mechanical and electrical variables. Problem 16.43 Solution: Known quantities: Length of the wire is 20 cm ; Flux density is 0.1T ; The position of the wire is x(t ) = 0.1sin 10t m Find: The induced emf across the length of the wire as a function of time. Assumptions: None. Analysis: From e = Blu , we have e(t ) = Bl dx = 0.02 cos(10t ) V dt ______________________________________________________________________________________ 16.39 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.44 Solution: Known quantities: Emf: e1 (t ) = 0.02 cos 10t ; the length of the second wire: 0.1 m ; the position of the second wire: x(t ) = 1 − 0.1sin 10t . Find: The induced emf e(t ) defined by the difference in emf's e1 (t ) and e2 (t ) . Assumptions: None. Analysis: We have e(t ) = e1 (t ) − e2 (t ) e2 (t ) = (0.1)(0.1)(−1 cos10t ) V e(t ) = 0.02 cos(10t ) + 0.01 cos(10t ) = 0.03 cos(10t ) V ______________________________________________________________________________________ Problem 16.45 Solution: Known quantities: I = 4 A, B = 0.3Wb m 2 . Find: The magnitude and direction of the force induced on the conducting bar. Assumptions: None. Analysis: f = Bli = 0.3 × l × 4 = 1.2l N Force will be to the left if current flows upward. ______________________________________________________________________________________ Problem 16.46 Solution: Known quantities: B = 0.3Wb m 2 . Find: The magnitude and direction of the induced voltage in the wire. Assumptions: None. Analysis: e = Blu cos 45 , = 2.83V ______________________________________________________________________________________ 16.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.47 Solution: Known quantities: The electrodynamic shaker shown in Figure P16.47 Shaker mass, m; air gap dimension, d; number of coil turns, N; spring parameter, k; armature resistance, R and inductance, L Find: a. Reluctance of the structure and Flux density B b. Dynamic equations of the shaker c. Transfer function and frequency response of the shaker velocity to input voltage Vs. Assumptions: Iron reluctance is negligible. Neglect fringing. Assume no damping in this system. Analysis: a. Reluctance of structure: 2d 0A Compute flux density: B b. A NIf A NI f 0 2d Electrical Subsystem, using KVL: Vs Ri L di dt Bl dx dt Mechanical Subsystem: c. d2x m 2 dt Bli kx Laplace Transform: Vs (s) (R Ls)I(s) BlsX(s) 2 0 BlI(s) (ms k)X(s) Vs (s) X(s) Vs (s) (R Ls)(ms 2 Bl mLs 3 mRs 2 k) Bls X(s) Bl (kL 2 2 B l )s Rk U ( s ) SX ( s ) Bls = = 3 2 Vs ( s ) Vs ( s ) mLs + mRs + (kL + B 2 l 2 ) s + Rk Frequency Response: s = jϖ U ( jϖ ) Bljϖ = 3 Vs ( jϖ ) mL( jϖ ) + mR( jϖ )2 + (kL + B 2 l 2 ) jϖ + Rk U ( jϖ ) Bljϖ = 2 Vs ( jϖ ) R(k − mϖ ) + j (kL + B 2 l 2 )ϖ − mLw 3 [ ] ______________________________________________________________________________________ 16.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.48 Solution: Known quantities: The electrodynamic shaker shown in Figure P16.47 is used to perform vibration testing of an electrical connector. B 1, 000 Wb/m 2; l 5 m; k 1000 N/m; m 1 kg; b 5 N - s/m; L 0.8 H; R 0.5 ; The test consists of shaking the connector at the frequency ϖ = 2π × 100 rad / s Find: The peak amplitude of sinusoidal voltage Vs required to generate an acceleration of 5g(49m/s2) under the stated test conditions Assumptions: Connector has negligible mass when compared to the platform. Analysis: Applying KVL around the coil circuit: di dx L Ri Bl VS dt dt Next, we apply Newton’s Second Law: d 2x dx Bli m 2 b kx 0 dt dt To derive the frequency response, Laplace transform the two equations to obtain: sL R I (s) BlsX (s) VS (s) 2 BlI(s) (ms bs k) X(s) 0 We can write the above equations in matrix form and resort to Cramer’s rule to solve for U(s) as a function of V(s): sL R Bls I (s) V (s) 2 bs k) X (s) (ms Bl 0 with solution det X(s) det Ls R V (s) Bl 0 Ls R Bl BlV (s) 2 Ls R (ms Bls 2 (ms bs k) Bl 2 s bs k) To obtain the acceleration response, we multiply the numerator by s2: s 2 X ( s ) X ( s ) Bls 2 = = V (s) V ( s ) (Lm )s 3 + (bL + Rm )s 2 + bR + kL + (Bl )2 s + (kR ) X ( jω ) − Blω 2 = V ( jω ) kR − (bL + Rm )ω 2 + j bR + kL + (Bl )2 ω − (Lm )ω 3 ( [ ] [( ) ] ) The magnitude of this complex number, evaluated at ω=2π×100 is 0.62. Thus, to obtain the desired acceleration (peak) of 49 m/s2, we wish to have a peak voltage amplitude |VS| = 49/0.62 ≈ 78 V. ______________________________________________________________________________________ 16.42 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 Problem 16.49 Solution: Known quantities: As described in Example 16.13. Find: Derive and sketch the frequency response of the loudspeaker in the following two cases. Describe qualitatively how the loudspeaker frequency response changes as the spring stiffness k increases and decreases. Find the limit of the frequency response and the kind of the speaker as k approaches zero. a. k = 50,000 N m . b. k = 5 × 10 6 N m . Assumptions: None. Analysis: a. For k = 50,000 N m , the transfer function is: U 1.478 × 10 5 s = 3 V s + 3075s 2 + 9 × 10 6 s + 4 × 10 9 The magnitude frequency response is plotted below: This response would correspond to a midrange speaker. b. For k = 5 × 10 6 N m , the transfer function is : 16.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 U 1.478 × 10 5 s = 3 V s + 3075s 2 + 5.04 × 10 8 s + 4 × 1011 The magnitude frequency response is plotted below: It should be apparent that this response would enhance the treble range, and is the response of a "tweeter". c. For k = 0 , the transfer function is : 1.478 × 10 5 U = 2 V s + 3075s + 4 × 10 6 The magnitude frequency response is plotted below: 16.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 In this case, the speaker acts as a "woofer", emphasizing the low frequency range. In practice, k cannot be identically zero, so the actual response of a woofer would resemble that of a midrange speaker, shifted towards the lower frequencies. ______________________________________________________________________________________ Problem 16.50 Solution: Known quantities: Loudspeaker of Example 16.13 (Figure 16.52) rcoil = 0.05m; L = 10mH ; R = 8Ω; N = 47; B = 1T ; m = 0.01kg ; b = 22.75 N − s / m; k = 5 × 10 4 N / m Find: Modify the parameters of the loudspeaker (mass, damping, and spring rate), so as to obtain a loudspeaker with a bass response centered on 400 Hz. Demonstrate that your design accomplishes the intended task, using frequency response plots. Assumptions: None. Analysis: From Example 16.13, the system transfer function is: U (s ) Bls = V (s ) (Lm )s 3 + (Rm + Lb )s 2 + Rb + kL + (Bl )2 s + kR The frequency response is: [ ] 16.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 16 U ( jϖ ) jBlϖ = V ( jϖ ) kR − (Rm + Lb )ϖ 2 + j Rb + kL + (Bl )2 ϖ − (Lm )ϖ 3 where: l = 2πNrcoil {[ ] } Using MATLAB, it is easy to adjust the mechanical parameters one-by-one in order to see each parameters’ effect on the system frequency response. Mass: Increasing the mass decreases the frequency response center frequency. Damping: Increasing the damping also decreases the center frequency, but also widens the response bandwidth. Spring Rate: Decreasing the spring rate also decreases the center frequency. There are many possible combinations of mechanical parameters that could be substituted to generate the desired response. One such set of parameters is: m = 0.05kg ; b = 30 N − s / m; k = 5 × 10 N / m The frequency response for this system is given below, with the response centered over 400 Hz: 3 _____________________________________________________________________________________________ 16.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Chapter 17 Instructor Notes The objective of Chapter 17 is to introduce the foundations for the analysis of rotating electric machines. In Section 17.1, rotating electric machines are classified on the basis of their energy conversion characteristics and of the nature of the electric power they absorb (or generate). Section 17.2 reviews the physical structure of a DC machine and presents a simple general circuit model that is valid for both motors and generators, including dynamic equations. Section 17.3 contains a brief discussion of DC generators. Section 17.4 describes the characteristics of the various configurations of DC motors, both of the wound stator and permanent magnet types. The torque speed characteristics of the different configurations are compared, and the dynamic equations are given for each type of motor. The section ends with a brief qualitative discussion of speed control in DC motors. The second half of the chapter is devoted to the analysis of AC machines. In Section 17.5, we introduce the concept of a rotating magnetic field. The next two sections describe synchronous generators and motors; the discussion is brief, but includes the analysis of circuit models of synchronous machines and a few examples. Circuit models for the induction motor, as well as general performance characteristics of this class of machines are discussed in Section 17.8, including a brief treatment of AC machine speed and torque control. Although the discussion of the AC machines is not particularly detailed, all of the important concepts that a non-electrical engineer would be interested in to evaluate the performance characteristics of these machines are introduced in the chapter, and reinforced in the homework problem set. The homework problems include a mix of traditional electric machinery problems based on circuit models and of more system-oriented problems. Problems 17.24-36 deal with the performance and dynamics of systems including DC motors. These problems are derived from the author’s experience in teaching a Mechanical Engineering System Dynamics course with emphasis on electromechanics, and are somewhat unusual (although relevant and useful for non–electrical engineers) in this type of textbook. These problems are well suited to a more mature audience that has already been exposed to a first course in system dynamics. Problem 17.39 provides a link to the power electronics topics covered in Chapter 12. All other problems are based on the content of the chapter. Learning Objectives 1. Understand the basic principles of operation of rotating electric machines, their classification, and basic efficiency and performance characteristics. Section 1. 2. Understand the operation and basic configurations of separately-excited, permanentmagnet, shunt and series DC machines. Section 2. 3. Analyze DC generators at steady-state. Section 3. 4. Analyze DC motors under steady-state and dynamic operation . Section 4. 5. Understand the operation and basic configuration of AC machines , including the synchronous motor and generator, and the induction machine. Sections 5, 6, 7, 8. 17.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.1: Rotating Electric Machines Problem 17.1 Solution: Known quantities: The relationship of the power rating and the ambient temperature is shown in the table. A motor with Pe = 10 kW is rated up to 85, C . Find: The actual power for the following conditions. 50, C . , b) Ambient temperature is 25 C . a) Ambient temperature is Assumptions: None. Analysis: a) The power at ambient temperature 50, C : Pe' = 10 − 10 × 0.125 = 8.75 kW b) The power at ambient temperature 30, C : Pe' = 10 + 10 × 0.08 = 10.8 kW ______________________________________________________________________________________ Problem 17.2 Solution: Known quantities: The speed-torque characteristic of an induction motor is shown in the table. The load requires a starting torque of 4 N ⋅ m and increase linearly with speed to 8 N ⋅ m at 1500 rev min . Find: a) The steady state operating point of the motor. b) The change in voltage if the load torque increases to Assumptions: None. Analysis: The characteristic is shown below: 17.2 10 N ⋅ m . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Speed 1600 1400 xx x x 1200 x 1000 x 800 x 600 400 x 200 x 10 20 30 Torque a) The operating point is: nm = 1425 rev min, T = 7N ⋅m b) From the following equation: Tnew Vs , new = Told Vs , old 10 KVs = 7 VS 2 2 = K2 K = 1.195 ∴Vs , new = 1.195Vs , old ___________________________________________________________________________________ 17.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.2: Direct Current Machines Problem 17.3 Solution: Known quantities: Each conductor of the DC motor is −4 6 in. long. The current is 90 A . The field density is 5.2 × 10 Wb in . 2 Find: The force exerted by each conductor on the armature. Assumptions: None. Analysis: Wb in 2 0.0254 m F = BI × l = 5.2 × 10 × × 90 × 6 in × 2 2 in in (0.0254 m) = 11.06 Nt −4 ______________________________________________________________________________________ Problem 17.4 Solution: Known quantities: The air-gap flux density of the DC machine is 4 Wb m 2 . The area of the pole face is 2 cm × 4 cm . Find: The flux per pole in the machine. Assumptions: None. Analysis: With B = 4 kG = 0.4 T = 0.4 Wb m 2 , we can compute the flux to be: φ = BA = 0.4 × 0.02 × 0.04 = 0.32 mWb ______________________________________________________________________________ 17.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.3: Direct Current Generators Problem 17.5 Solution: Known quantities: A 120V , 10 A shunt motor. The armature resistance is 0.6 Ω . The shunt field current is 2 A . Find: The LVDT equations. Assumptions: None. Analysis: VL at full load is 120 V and E b = 120 + ( 2 + 10) × 0.6 = 127.2 V Rf = 120 = 60 Ω 2 Eb to be constant, we have: 127.2 = 2 .1 A ia = i f = 0.6 + 60 Assuming Therefore: VL = 127.2 − 2.1 × 0.6 = 125.9 V Voltage reg. = 125.9 − 120 = 0.049 = 4.9% 120 __________________________________________________________________________ Problem 17.6 Solution: Known quantities: A 20 kW , 230V separately excited generator. The armature resistance is 0.2 Ω . The load current is 100 A . Find: a) The generated voltage when the terminal voltage is 230 V . b) The output power. Assumptions: None. Analysis: If we assume rated output voltage, that is VL = 230V , we have a) The generated voltage is 230 V . b) The output power is 23 kW . If we assume rated output power, that is Pout = 20 kW , we have 17.5 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 a) The generated voltage is 200V . b) The output power is 20 kW . Eb = 230V , and compute the output voltage to be: VL = 230 − 100 × 0.2 = 210V If we assume We have: a) The generated voltage is 210V . b) The output power is 21 kW . ______________________________________________________________________________________ Problem 17.7 Solution: Known quantities: A 10 kW , 120VDC series generator. The armature resistance is 0.1 Ω and a series field resistance is 0.05 Ω . Find: a) The armature current. b) The generated voltage. Assumptions: The generator is delivering rated current at rated speed. Analysis: The circuit is shown below: RS + Ra va - + - LS + ia RL Eb vL - a) ia = P 10 × 103 = = 83.33 A VL 120 b) Va = 120 + ia RS = 124.17 V ______________________________________________________________________________________ 17.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.8 Solution: Known quantities: A 30 kW , 440V shunt generator. The armature resistance is 0.1 Ω and a series field resistance is 200 Ω . Find: a) The power developed at rated load. b) The load, field, and armature currents. c) The electrical power loss. Assumptions: None. Analysis: The circuit is shown below: iL if Rf + Ra ia + - Eb Lf RL vL - 30 × 10 3 iL = = 68.2 A 440 440 if = = 2 .2 A 200 ia = 70.4 A a) Eb = VL + ia Ra = 440 + 70.4 × 0.1 = 447.04 V P = Ebia = 31.471 kW b) iL = 62.8 A i f = 2 .2 A ia = 70.4 A c) Ploss = ia2 Ra + i 2f R f = 1464 W ______________________________________________________________________________________ 17.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.9 Solution: Known quantities: A four-pole 450 kW , 4.6 kV shunt generator. The armature resistance is 2 Ω and a series field resistance is 333 Ω . The generator is operating at the rated speed of 3600 rev min . Find: The no-load voltage of the generator and terminal voltage at half load. Assumptions: None. Analysis: For n = 3600 rev min , ω m = 377 rad sec : 450 × 10 3 = 97.8 A 4.6 × 10 3 4.6 × 10 3 if = = 13.8 A 333 ia = i f + i L = 111.6 A iL = Using the relation: Eb = VL + ia Ra = 4823.2V At no-load, VL = Eb − ia Ra = 4820.4V At half load, iL = 48.9 A ia = i f + i L = 62.7 A VL = Eb − ia Ra = 4810.7 V ______________________________________________________________________________________ Problem 17.10 Solution: Known quantities: A 30 kW , 240V generator is running at half load at 1800 rev Find: The total losses and input power. Assumptions: None. Analysis: 1 rated load = 15 kW 2 At an efficiency of 0.85 , the input power can be computed to be: 15 × 10 3 Pin = = 17.647 kW 0.85 Pout = The total loss is: 17.8 min with efficiency of 85 percent. G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Ploss = Pin − Pout = 2.647 kW ______________________________________________________________________________________ Problem 17.11 Solution: Known quantities: 200 rev min , it delivers 20 A to a 100 V line. The armature resistance is 1.0 Ω and a series field resistance is 100 Ω . The magnetization characteristic is shown in A self excited DC shunt generator. At Figure P17.11. When the generator is disconnected from the line, the drive motor speed up to 220 rad s . Find: The terminal voltage. Assumptions: None. Analysis: From the figure, for I f > 0.5 A, ω = 200 rad sec Eb = 40 I f + 100 For ω = 220 rad sec , we have: 100 + 40 I f Eb' = 220 200 Therefore, 220 (100 + 40 I f ) = 110 + 44 I f 200 For no load, I a = I f . Therefore, Eb' = 110 + 44 I f = 101I f ∴I f = 1.93 A The terminal voltage is: V = I f R f = 193V ______________________________________________________________________________________ 17.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.4: Direct Current Motors Problem 17.12 Solution: Known quantities: A 220 V shunt motor. The armature resistance is 0.32 Ω . A field resistance is 110 Ω . At no load the armature current is 6 A and the speed is 1800 rpm . Find: a) The speed of the motor when the line current is 62 A . b) The speed regulation of the motor. Assumptions: The flux does not vary with load. Assume a 8 N ⋅ m brush drop. Analysis: a) 1800 = 220 − 2 − 6(0.32) K aφ K aφ = 0.12 ∴n = 220 − 2 − 6(0.32) = 1657 rpm K aφ b) % reg = 1800 − 1657 × 100 = 8.65% 1657 _____________________________________________________________________________________ Problem 17.13 Solution: Known quantities: A 50 hp, 550 volt shunt generator. The armature resistance including brushes is 0.36 Ω . Operating at rated load and speed, the armature current is 75 A . Find: What resistance should be inserted in the armature circuit to get a 20 percent speed reduction when the motor is developing 70 percent of rated torque. Assumptions: There is no flux change. Analysis: T = K aφI a I a = 0.7(75) = 52.5 A nR = 550 − 75(0.36) K aφ nR = 523 K aφ 0.8nR = 550 − 52.5 RT 0.8 × 523 = 550 − 52.5RT K aφ ∴RT = 2.51Ω Radd = 2.51 − 0.36 = 2.15 Ω 17.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 ______________________________________________________________________________________ Problem 17.14 Solution: Known quantities: 100 kW , 440V shunt DC motor. The armature resistance is 0.2 Ω and a series field resistance is 400 Ω . The generator is operating at the rated speed of 1200 rev min . The full-load efficiency is 90 A percent. Find: a) The motor line current. b) The field and armature currents. c) The counter emf at rated speed. d) The output torque. Assumptions: None. Analysis: At n = 1200 rev min , ω m = 125.7 rad sec , the output power is 100 hp = 74.6 kW . 0.9 , we have: 74.6 Pin = = 82.9 kW 0 .9 From full-load efficiency of a) From Pin = iSVS = 82.9 kW , we have: 82.9 × 103 iS = = 188.4 A 440 b) 440 = 1 .1 A 400 ia = 187.3 A if = c) Eb = VL − ia Ra = 402.5 A d) Tout = Pout = 593.5 N ⋅ m ωm ______________________________________________________________________________________ Problem 17.15 Solution: Known quantities: A 240V series motor. The armature resistance is 0.42 Ω and a series field resistance is 0.18 Ω . The speed is 500 rev min when the current is 36 A . Find: What is the motor speed when the load reduces the line current to 17.11 21 A . G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Assumptions: A 3 volts brush drop and the flux is proportional to the current. Analysis: 500 = n= 240 − 3 − 36(0.6) K aφ = 0.431 K aφ 240 − 3 − 21(0.6) = 893 rpm 21 ( )(0.431) 36 ______________________________________________________________________________________ Problem 17.16 Solution: Known quantities: A 220 VDC shunt motor. The armature resistance is 0.2 Ω . The rated armature current is 50 A . Find: a) The voltage generated in the armature. b) The power developed. Assumptions: None. Analysis: a) Eb = VL − ia Ra = 220 − 50 × 0.2 = 210V b) P = Ebia = 210 × 50 = 10.5 kW = 14.07 hp ______________________________________________________________________________________ Problem 17.17 Solution: Known quantities: A 550V series motor. The armature resistance is 0.15 Ω . The speed is current is 112 A and the load is 75 hp . Find: The horsepower output of the motor when the current drops to Assumptions: The flux is reduced by 15 percent. 17.12 84 A . 820 rev min when the G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: 2π ⋅ n ⋅ T 33,000 2π (820)T 75 = T = 480.4 lb ⋅ ft 33,000 T = K aφI a 480.4 = K aφ (112) K aφ = 4.29 HP = Tn = 4.29(0.85)(84) = 306.2 lb ⋅ ft 550 − 84(0.15) = 973 rpm 0.85(0.65) 2π (973)(306.2) HPn = = 56.7 hp 33,000 nn = ______________________________________________________________________________________ Problem 17.18 Solution: Known quantities: 220 VDC shunt motor. The armature resistance is 0.1 Ω and a series field resistance is 100 Ω . The speed is 1100 rev min when the current is 4 A and there is no load. A Find: E and the rotational losses at 1100 rev min . Assumptions: The stray-load losses can be neglected. Analysis: Since n = 1100 rev min corresponds to ω = 115.2 rad sec , we have: iS = 4 A 200 =2A 100 ia = i S − i f = 2 A if = Also, Eb = 200 − 2 × 0.1 = 199.8V The power developed by the motor is: P = Pin − Pcopper _ loss = 200 × 4 − (22 × 100 + 22 × 0.1) = 399.6W 17.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 ______________________________________________________________________________________ Problem 17.19 Solution: Known quantities: 230 VDC shunt motor. The armature resistance is 0.5 Ω and a series field resistance is 75 Ω . At 1100 rev min , Prot = 500 W . When loaded, the current is 46 A . A Find: a) The speed Pdev and Tsh . b) ia (t ) and ω m (t ) if L f = 25 H , La = 0.008 H and the terminal voltage has a 115V change. Assumptions: None. Analysis: 230 = 3.07 A 75 ia = i L − i f = 42.93 A if = ω m = 117.3 rad sec 230 At no load, 117.3 = , therefore, K aφ K aφ = 1.96 At full load, ωm = 230 − 0.5 × 42.93 K aφ The back emf is: Eb = 230 − 0.5 × 42.93 = 208.5V The power developed is: Pdev = Eb I a = 8.952 kW The power available at the shaft is: Po = Pdev − Prot = 8952 − 500 = 8452 W The torque available at the shaft is: Tsh = Po = 72.1 N ⋅ m ωm ______________________________________________________________________________________ 17.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.20 Solution: Known quantities: 200 VDC shunt motor. The armature resistance is 0.1 Ω and a series field resistance is 100 Ω . At 955 rev min with no load, Prot = 500 W , the line current is 5 A . A Find: The motor speed, the motor efficiency, total losses, and the load torque when the motor draws the line. Assumptions: Rotational power losses are proportional to the square of shaft speed. Analysis: iS = 5 A 200 =2A 100 ia = i S − i f = 3 A if = The copper loss is: Pcopper = i 2f R f + ia2 Ra = 400.9 W The input power is: Pin = 5 × 200 = 1 kW Therefore, Prot + PSL = 1000 − 409 = 599.1W 955 at ω m = 2π = 100 rad sec . 60 Also, E b at no load is: Eb = 200 − 3 × 0.1 = 199.7 V K aφ = 1.997 When iS = 40 A with i f = 2 A and ia = 38 A , Eb = 200 − 38 × 0.1 = 196.2 V ωm = Eb = 98.25 rad sec = 938.2 rev min K aφ The power developed is: P = Eb I a = 196.2 × 38 = 7456 W The copper loss is: Pcopper = i 2f R f + ia2 Ra = 544.4 W The input power is: Pin = 40 × 200 = 8 kW And 17.15 40 A from G. Rizzoni, Principles and Applications of Electrical Engineering PSH = 7456 − TSH = Problem solutions, Chapter 17 98.25 × 599.1 = 6867.4 W 100 6867.4 = 69.9 N ⋅ m 98.25 Finally, the efficiency is: eff = PSH = 85.84% Pin ______________________________________________________________________________________ Problem 17.21 Solution: Known quantities: 50 hp, 230V shunt motor operates at full load when the line current is 181 A at 1350 rev min . The field resistance is 17.7 Ω . To increase the speed to 1600 rev min , a resistance of 5.3 Ω is cut in via the field rheostat. The line current is increased to 190 A . A Find: a) The power loss in the field and its percentage of the total power input for the 1350 rev b) The power losses in the field and the field rheostat for the 1600 rev min speed. c) The percent losses in the field and in the field rheostat at min speed. 1600 rev min speed. Assumptions: None. Analysis: a) 230 = 13.0 A 17.7 Pf = ( 230)(13.0) = 2988.7 W If = Pf Pm = 2988.7 = 0.072 = 7.2% ( 230)(181) b) If = 230 = 10 A (17.7 + 5.3) Pf = 102 (17.7) = 1770W PR = 102 (5.3) = 530 W c) Pin = ( 230)(190) = 43,700 W 1770 × 100 = 4.05% 43700 530 × 100 = 1.21% % PR = 43700 % Pf = ______________________________________________________________________________________ 17.16 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.22 Solution: Known quantities: 10 hp, 230 V shunt-wound motor. The armature resistance is 0.26 Ω and a series field resistance is 225 Ω . The rated speed is 1000 rev min . The full-load efficiency is 86 percent. A Find: The effect on counter emf, armature current and torque when the motor is operating under rated load and the field flux is very quickly reduced to 50 percent of its normal value. The effect on the operation of the motor and its speed when stable operating conditions have been regained. Assumptions: None. Analysis: EC = Kφ n ; counter emf will decrease. V − EC ; armature current will increase. ra T = KφI a ; effect on torque is indeterminate. Ia = Operation of a dc motor under weakened field conditions is frequently done when speed control is an important factor and where decreased efficiency and less than rated torque output are lesser considerations. V − I a ra V − I a ra 1000 = Kφ Kφ V − I a ra nnew = 0 .5 K φ n= Assume small change in the steady-state value of I a . Then: 1000 0.5 = nnew = 2000 rpm nnew 1 ______________________________________________________________________________________ Problem 17.23 Solution: Known quantities: The machine is the same as that in Example 17.7. The circuit is shown in Figure P17.23. The armature resistance is 0.2 Ω and the field resistance is negligible. n = 120 rev min , I a = 8 A . In the operating region, φ = kI f , k = 200 . Find: a) The number of field winding turns necessary for full-load operation. b) The torque output for the following speeds: 1. n , = 2n n, = n 2 , 4. n = n 4 3. 2. n = 3n c) Plot the speed-torque characteristic for the conditions of part b. Assumptions: None. , 17.17 G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: in Example 17.7, i f = 0.6 A , the mmf F is: F = 200 × 0.6 = 120 At For a series field winding with: a) iseries = ia = 8 A , we have: N series = b) 120 = 15 turns 8 nm = 120 rev min ω m = 12.57 rad sec Eb = VS − ia ( Ra + RS ) Neglecting RS , we have Eb = 7.2 − 8 × 0.2 = 5.6 V = k aφω m 5 .6 k aφ = = 0.446 12.57 From T = k T φ ia and k T = k a , T = 0.446 × 8 = 3.56 N ⋅ m By using φ = kia , we have: Eb = k a kiaω m = VS − ia Ra T = k T ( kia )ia = k a kia2 VS 5 .6 , where K = k a k = = 0.056 From ia = Ra + Kω m 8 × 12.57 1 ) 2 , we have: And T ∝ ( Ra + Kω m Ra + ωm Ra + Kω m 2 TX K =( ) =( )2 Ra T Ra + Kω X + ωX K R a = 3.59 K 3.59 + ω m 2 ∴TX = 3.56( ) 3.59 + ω X 1. at ω X = 2ω m = 25.12 rad sec , TX = 1.13 N ⋅ m ω X = 3ω m = 37.71 rad sec , TX = 0.55 N ⋅ m 2. at 3. at ω X = 0.5ω m = 6.28 rad sec , 17.18 Problem solutions, Chapter 17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 TX = 9.54 N ⋅ m ω X = 0.25ω m = 3.14 rad sec , TX = 20.53 N ⋅ m 4. at c) The diagram is shown below: ______________________________________________________________________________________ Problem 17.24 Solution: Known quantities: PM DC motor circuit model; mechanical load model. Example 17.9. Find: Voltage-step response of motor. Assumptions: None. Analysis: Applying KVL and equation 17.47 to the electrical circuit we obtain: dI (t ) Eb (t ) 0 VL (t) Ra I a (t ) La a dt or dIa (t) Ra I a (t) K a PM m (t ) VL (t) dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: d (t) T(t) TLoad (t ) b J dt or La d (t ) b (t) 0 dt since the load torque is assumed to be zero. To derive the transfer function from voltage to speed, we use the result of Example 17.9 with Tload = 0: KT PM VL (s) m (s) sL a Ra (sJ b) Ka PM KT PM K TPM I a (t) J The step response of the system can be computed by assuming a unit step input in voltage: KT PM 1 m (s) sL a Ra (sJ b) Ka PM KT PM s 17.19 G. Rizzoni, Principles and Applications of Electrical Engineering F ( s) = KTPM 1 ( sLa + Ra )( sJ + b) + K aPM KTPM s F ( s) = KTPM s[ JLa s + ( La b + Ra J ) s + Ra b + K a KT ] F ( s) = KTPM ª (R b + K a K T ) º ( L b + Ra J ) s «s 2 + a s+ a » JLa JLa ¬ ¼ Set: m= For F(s) 2 ( La b + Ra J ) JLa F (s) = Problem solutions, Chapter 17 n= ( Ra b + K a KT ) JLa KTPM KTPM KTPM = = 2 s[ s + ms + n ] ª§ 2 ª§ m2 · m2 º m· § m 2 ·º s «¨¨ s + ms + ¸¸ + n − ¸» » s «¨ s + ¸ + ¨¨ n − 4 ¹ 4 ¼ 2¹ © 4 ¸¹¼» ¬© ¬«© 2 1 s[(s a)2 where : = tan-1 Therefore: 2 b ] b and bo a m ( L b + Ra J ) a= = a 2 2 JLa 1 bo2 f (t) b2 1 e bbo at sin(bt a2 2 ª ( L b + Ra J ) º R b + K a KT § La b + Ra J + a − ¨¨ b =« a » JLa JLa ¬ 2 JLa ¼ © 2 o φ = tan −1 2 Ra b + K a KT § La b + Ra J · 1 ¸¸ − ¨¨ JLa JLa © ¹ 4 m2 = b= n− 4 Ra b + K a KT § La b + Ra J − ¨¨ JLa JLa © ( L b + Ra J ) − a 2 JLa ) 2 · 1 Ra b + K a KT ¸¸ = JLa ¹ 4 2 · 1 ¸¸ ¹ 4 Thus giving the step response: 17.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 JLa 1 Ω m (t ) = + 2 Ra b + K a K T § · ¨ Ra b + K a K T §¨ La b + Ra J ·¸ 1 ¸§¨ Ra b + K a K T −¨ ¸ 4 ¸¸¨ ¨¨ JLa JLa JLa © ¹ ¹© © ª « «§¨ Ra b + K a K T § La b + Ra J − ¨¨ sin «¨ JL JLa ¨ a © «© « «¬ 2 · 1 ·¸ ¸¸ ¸t − tan −1 ¹ 4 ¸¹ e ª ( L b + Ra J ) º −« a »t 2 JLa ¬ ¼ * · ¸ ¸ ¹ Ra b + K a K T § La b + Ra J − ¨¨ JLa JLa © ( L b + Ra J ) − a 2 JLa 2 · 1 º» ¸¸ ¹ 4» » » » »¼ Expressions for the natural frequency and damping ratio of the second-order system may be derived by comparing the motor voltage-speed transfer function to a standard second-order system transfer function: KS H(s) . 2 s 2ζω ns ω n2 The motor transfer function is: KT PM KT PM m (s) VL (s) Ra (sJ b ) K a PM KT PM sL a JLa s 2 JRa bL a s Ra b K a PM KT PM KT PM s JRa 2 JLa Ra b bL a s Ka PM K T PM JLa JLa Comparing terms, we determine that: Ra b K a PM K T PM ωn2 JLa 2ζωn JRa bLa JLa or ωn ζ Ra b K a PM KT PM JLa 1 JRa bLa JLa 2 JLa Rab KaPM KT PM From these expressions, we can see that both natural frequency and damping ratio are affected by each of the parameters of the system, and that one cannot predict the nature of the damping without knowing numerical values of the parameters. ______________________________________________________________________________________ Problem 17.25 Solution: Known quantities: Torque-speed curves of motor and load: Tm = aω + b (motor), TL = cω2 + d (load), Find: Equilibrium speeds and their stability. 17.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Assumptions: All coefficients of torque-speed curve functions are positive constants Analysis: The first consideration is that the motor static torque, T0,m = b must exceed the load static torque, T0,L = d; thus, the first condition is b>d. The next step is to determine the steady state speed of the motor-load pair. If we set the motor torque equal to the load torque, the resulting angular velocities will be the desired solutions. 2 Tm TL aω b cω d resulting in the quadratic equation 2 cω aω d b 0 with solution ω a a2 4c b d . 2c Both solutions are positive, and therefore physically acceptable. The question of stability can be addressed by considering the following sketch. Torque U S Speed In the figure, we see that the intersection of a line with a quadratic function when both solutions are positive leads to two possible situations: the line intersecting the parabola when the rate of change of both curves is positive, and the line intersecting the parabola when the rate of change of torque w.r. to speed of the latter is negative. The first case leads to an unstable operating point; the second case to a stable operating point (you can argue each case qualitatively by assuming a small increase in load torque and evaluating the consequences). We can state this condition mathematically by requiring that the following steady-state stability condition hold: dT L dTm d d Evaluating this for our case, we see that dT L dTm 2c a. d d From the expression we obtained earlier, a a2 4c b d 2c it is clear that 2c ω>a always olds, since the term under the square root is a positive constant. Thus, this motor-load pair always leads to stable solutions. To verify this conclusion intuitively, you might wish to dT L dTm plot the motor and load torque-speed curves and confirm that the condition is always satisfied d d (note that the sketch above is not an accurate graphical representation of the two curves). ______________________________________________________________________________________ Problem 17.26 Solution: Known quantities: Expression for friction and windage torque, TFW, functional form of motor torque, T, or load torque, TL. Find: Sketch torque-speed curve 17.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Assumptions: None. Analysis: The sketches are shown below. T II braking T I motoring TL TFW TL TL TFW TL TL T FW T0 -T 0 TFW T0 TL TL -T 0 IV braking III reverse motoring Torque-speed curve for constant load torque Torque-speed curves for variable load torque ______________________________________________________________________________________ Problem 17.27 Solution: Known Quantities: A PM DC motor and parameters when 1) in steady-state, no load conditions, and 2) connected to a pump Find: a) A damping coefficient, sketch of the motor, the dynamic equations, the transfer function, and 3 dB bandwidth. b) A sketch of the motor, dynamic equations, transfer function, and 3 dB bandwidth. Assumptions: kt ka k PM Analysis: a) The magnetic torque balanced the damping torque gives s: k t *ia b* or b k PM * ia m m N m *.15 A A 3350 rev 2 rad 1 min * * min 60 sec rev 7*10 3 2.993 *10 6 N - m - sec Sketch: PM DC Motor-Load System ra Vs La TM J ωm TL b 17.23 G. Rizzoni, Principles and Applications of Electrical Engineering Dynamic equations: dia dt kPM * La * Vs Eb J* d Tm Eb ra * ia k PM * m di La * a dt Vs ra * ia m Tm dt kPM * ia b* TL m m d m b * m k PM * ia dt To get the transfer function: TL J* k PM º ª ia º ª Vs º ª( La s + ra ) = « −k ( Js + b)»¼ «¬ω m »¼ «¬ − TL »¼ PM ¬ ª( L s + ra ) Vs º det « a − TL »¼ ¬ − k PM ωm ( s) = k PM º ª( L s + ra ) det « a ( Js + b)»¼ ¬ − k PM ωm ( s) Vs T = L =0 k PM 2 ( J * La ) s + ( ra * J + La * b) s + ra * b + k PM 2 b) Sketch: ra La Vs TM ωm JL J TL b bL Dynamic Equations: Vs Eb dia ra * ia dt k PM * m La * Eb dia ra * ia k PM * m dt d m (J J L )* Tm TL (b bL ) * dt Tm kPM *ia d m TL (J J L ) * (b bL ) * m dt Vs La * m k PM * ia 17.24 Problem solutions, Chapter 17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 To get the transfer function: k PM ª( La s + ra ) º ª ia º ª Vs º = « −k (( J + J L ) s + (b + bL ))»¼ «¬ω m »¼ «¬ − TL »¼ PM ¬ ª( L s + ra ) Vs º det « a − k PM − TL »¼ ¬ ω m (s) = k PM ª( L s + ra ) º det « a (( J + J L ) s + (b + bL ))»¼ ¬ − k PM ωm k PM (s) = 2 2 Vs (( J + J L ) * La ) s + (ra * ( J + J L ) + La * (b + bL )) s + ra * (b + bL ) + k PM T =0 L Frequency Response: ______________________________________________________________________________________ Problem 17.28 Solution: Known Quantities: A PM DC motor is used to power a pump Find: The dynamic equations of the system and the transfer function between the motor voltage and pressure. Assumptions: The inertia and damping of the motor and pump can be lumped together. 17.25 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: Sketch: Motor Pump Circuit ra La C V s TL TM ωm kpu mp J R b dia dt di La a dt Vs La Vs J d m Tm dt TL J d Eb ra ia k PM TL m dt ra ia b m b m m kPM m p 0 dp Cacc R dt dp p Cacc kp m 0 dt R kp m To get the transfer function: ª La s + ra « « − k PM «¬ 0 º ª ia º ªVs º k PM 0 »« » « » Js + b kp » «ω m » = « 0 » − Rk p RCacc s + 1»¼ «¬ p »¼ «¬ 0 »¼ ª La s + ra k PM Vs º « » det « − k PM Js + b 0 » «¬ 0 − Rk p 0 »¼ p( s ) = ª La s + ra k PM º 0 « » det « − k PM Js + b kp » «¬ 0 − Rk p RCacc s + 1»¼ k PM Rk p p ( s) = 2 Vs ( La s + ra )( Js + b)( RCacc s + 1) + k PM ( RCacc s + 1) + Rk p2 ( La s + ra ) ______________________________________________________________________________________ 17.26 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.29 Solution: Known quantities: Motor circuit shown in Figure P17.29 and magnetization parameters, load parameters. Operating point. Note correction to the operating point: Ia0 = 186.67 A; Note correction to the parameter: kf = 0.12 V-s/A-rad Find: a) System differential equations in symbolic form b) Linearized equations Assumptions: The dynamics of the field circuit are negligible. Analysis: a) Differential equations Applying KVL and equation 17.47 to the electrical circuit we obtain: dI f (t) Lf R f I f (t) VS (t) field circuit dt or dIa (t) Ra I a (t) k f I f (t ) m (t) VS (t) armature circuit dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: d m (t) J Tm(t) TL (t ) b m dt or d m (t) k f I f (t)I a (t) J b m (t) TL (t ) dt Since the dynamics of the field circuit are much faster than those of the armature circuit (time constant Lf La VS ), we can write I f and the system of equations is now: Rf Ra Rf dI (t) V (t) La a Ra I a (t) k f S m (t ) VS (t ) dt Rf La VS (t) d m (t ) Ia (t) J b m (t) Rf dt b) Linearization Define perturbation variables: Ia (t) Ia I a (t) kf m(t) m TL (t) m (t) VS (t) VS VS (t) Next, we write the steady-state equations (all derivatives equal to zero): V Ra Ia k f S m VS Rf VS I a b m TL Rf These equations must be satisfied at the operating point. We can verify this using numerical values: kf 17.27 G. Rizzoni, Principles and Applications of Electrical Engineering 0.75 0.12 Ia 150 200 60 Problem solutions, Chapter 17 200 resulting in 1 150 200 0.12 200 186.67 0.75 60 150 0.12 186.67 0.6 200 T L 60 resulting in Ia TL 56 120 64 N - m Now, given that the system is operating at the stated operating point, we can linearize the differential equation for the perturbation variables around the operating point. To linearize the equation we recognize V (t) VS (t) I a (t ). the nonlinear terms: k f S m (t) and k f Rf Rf To linearize these terms, we use the first-order term in the Taylor series expansion: VS (t) m (t ) VS (t) m(t) kf kf V (t) kf S VS (t) m (t) m (t) Rf Rf VS Rf m VS , kf Rf m VS (t ) VS VS (t) I a (t) Rf kf VS , m m m (t) kf VS (t)I a (t ) Rf VS VS (t ) VS , I a kf VS (t)I a (t ) Rf Ia I a (t) VS , I a kf Ia VS (t ) VS Ia (t) Rf Now we can write the linearized differential equations in the perturbation variables: kf kf d Ia (t) La Ra I a (t ) VS m (t) VS (t ) m VS (t) dt Rf Rf kf Rf VS I a (t ) J d m (t) dt b m (t ) I a VS (t) TL (t ) This set of equations is now linear, and numerical values can be substituted to obtain a numerical answer, valid in the neighborhood of the operating point, for given voltage and load torque inputs to the system. ______________________________________________________________________________________ Problem 17.30 Solution: Known Quantities: TL = 5 + 0.05 + 0.001 kTPM k APM R a 0.2 VS 2 2.42 50V Find: What will the speed of rotation be of the fan? Assumptions: The fan is operating at constant speed. Analysis: Applying KVL for a PM DC motor (note at constant current short inductor) 17.28 G. Rizzoni, Principles and Applications of Electrical Engineering VS Problem solutions, Chapter 17 Eb (eqn. 17.48) ia Ra T ia k TPM Eb k aPM m T Vs Ra kaPM kTPM TL 5 0.05 m T 5 0.05 Vs m 0.001 0.001 m 2 m 2 m k TPM Ra kaPM Plug in the known variables and solving for 50V 5 0.05 8.26x10 2 5 m m 0.001 2 m 2.42 4.13x10 m m. 0.02 2.42 N m Amp 3 m 2.42 V sec rad ( 50 .413) m 0 20.5, or - 29318 rad/sec 20.5 rad/sec m m 20.5 rad/sec Nm 60 sec min rev 2 196RPM ______________________________________________________________________________________ Problem 17.31 Solution: Known Quantities: A separately excited DC motor Ra = 0.1Ω, R f = 100Ω, La = 0.2 H , L f = 0.02 H , K a = 0.8, K f = 0.9 J = 0.5kg − m 2 , b = 2 N − m − rad / s Find: a) A sketch of the system and its three differential equations b) Sketch a simulation block diagram c) Put the diagram into Simulink d) Run the simulation with Armature Control with a constant field voltage V f = 100V Plot the current and angular speed responses Run the simulation with Field Control with a constant armature voltage V f = 100V Plot the current and angular speed responses Assumptions: No external load torque is applied Analysis: a) Sketch: Separately Excited DC Motor 17.29 G. Rizzoni, Principles and Applications of Electrical Engineering Ia If Vf Lf La Ra VS Rf Problem solutions, Chapter 17 m Eb J Tm TL b The three dynamic equations are: Vf Lf dt dI La a dt Va J b) dI f d dt b Rf I f Ra Ia Tm Eb J d dt b Simulink block diagram kaia J d dt Simulink DC Motor Subsystem 17.30 b k a Iak f I f G. Rizzoni, Principles and Applications of Electrical Engineering d) Simulink Responses: Armature Control: Field Control: 17.31 Problem solutions, Chapter 17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 ______________________________________________________________________________________ Problem 17.32 Solution: Known quantities: Ra, La, ka = kT, Jm, bm, J, b, TL. Find: Transfer functions from armature voltage to angular velocity and from load torque to angular velocity. Schematics, diagrams, circuits and given data. See equations 17.16-18 and Figure 17.20. Assumptions: Analysis: Applying KVL and equation 17.47 to the electrical circuit we obtain: dI (t ) Va (t) Ra I a (t) L a a Eb (t) 0 dt or dIa (t) Ra I a (t) k a m (t) Va (t) dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: La 17.32 G. Rizzoni, Principles and Applications of Electrical Engineering Jm d (t ) dt J Tm (t) TL (t) bm Problem solutions, Chapter 17 b or d (t ) bm b (t ) TL (t) dt To derive the transfer function, we Laplace transform the two equations to obtain: sLa Ra I a (s ) k a (s) Va (s) k T Ia (t) Jm k a I a (s ) s Jm J J bm (s) b TL (s) We can write the above equations in matrix form and resort to Cramer’s rule to solve for Ωm(s) as a function of Va(s) and TL(s). sL a Ra ka I a (s) Va (s) ka s Jm J bm b T L (s) (s) m with solution det m (s) det sLa Va (s) Ra sL a T L (s) ka Ra ka ka s Jm J bm b or m (s) Ra sLa sL a Ra s J m J bm k a2 b T L (s) sL a R a s J m ka J bm b k a2 Va (s) and finally m (s) T L (s) Va (s) 0 sL a Ra s J m m (s) Va (s) Ra sLa J 2 bm b ka bm b k a2 ka TL (s ) 0 sLa Ra s J m J ______________________________________________________________________________________ Problem 17.33 Solution: Known Quantities: A PM DC motor that is coupled to a pump with a long shaft. Find: The dynamic equations of the system and the transfer function from input voltage to load inertia speed. Assumptions: The energy conversion is ideal. Analysis: Sketch: PM DC Motor-Load Coupling 17.33 G. Rizzoni, Principles and Applications of Electrical Engineering Ra Problem solutions, Chapter 17 La Vs TM k J ωm JL ωL TL bL b Knowing: d dt 1 s dt Then the three dynamic equations for the system are: Va dia dt La Ra ia d m bm dt d L bL JL dt Jm m L Ea K K La m s L s dia dt k aia K Putting them in matrix form: ªL s + R a « a − k « a « 0 ¬« ωL (s) Va T ka 2 J m s + bm s + K s −K s = L =0 Ra ia ka L 0 K m s s m TL ºª i º ª V º »« a » « a » » «ω m » = « 0 » s » J L s 2 + bL s + K » «¬ω L »¼ «¬− TL »¼ s¼ 0 −K ka K (La s + Ra )(J m s + bm + K s )(J L s + bL + K s ) s − −K ( ) (L s + R ) + (J s + b s 2 a a L L +K ______________________________________________________________________________________ Problem 17.34 Solution: Known quantities: Field and armature circuit parameters; magnetization and armature constants; motor and load inertia and damping coefficients. Find: a) sketch system diagrams for shunt and series configuration b) write expression for torque-speed curves for each configuration c) write the differential equations for each configuration d) determine whether equations are linear or nonlinear and how they could be linearized 17.34 s )(− k )(k ) a a G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Assumptions: Analysis: a) System diagrams IS Lf If Ia VS Ia La Lf Ra Rf Rf J m Ra VS Tm Eb TL Eb La m J Tm TL b b Series motor-load system Shunt motor-load system b) Write expressions for the torque-speed curves Shunt configuration Applying KVL and Newton’s Second Law for the steady-state system we write: VS R f I f field circuit and VS Ra Ia kf If m armature circuit or VS Ra Ia VS Rf kf m VS I a b m TL Rf To obtain the torque-speed curve of the motor (there will also be a load torque-speed equation, but we do not have any information on the nature of the load), we write: Tm k f I f Ia Tm k f I f Ia kf TmR f Tm k f If k f VS and substitute the expression for Ia in the electrical circuit equation: Ra R f V V VS Ra Ia k f S m Tm k f S m Rf k f VS Rf Ia or Tm k f VS Ra R f VS kf VS Rf m k f VS2 k 2f VS2 Ra R f Ra R 2f m Series configuration Applying KVL and Newton’s Second Law for the steady-state system we write: 17.35 G. Rizzoni, Principles and Applications of Electrical Engineering VS Ra R f Ia k f Ia2 Tm 2 k f Ia b TL m Problem solutions, Chapter 17 m To obtain the torque-speed curve of the motor (there will also be a load torque-speed equation, but we do not have any information on the nature of the load), we write: Tm Ia kf VS Ra R f Ia k f Ia2 m = Ra Rf Tm kf Tm m which leads to a quadratic equation in Tm and ωm. c) Write the differential equations Shunt configuration Applying KVL and equation 17.47 to the electrical circuit we obtain: dI f (t) Lf R f I f (t) VS (t) field circuit dt or dIa (t) Ra I a (t) k f I f (t ) m (t) VS (t) armature circuit dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: d m (t) J Tm(t) TL (t ) b m dt or d m (t) b m (t) TL (t ) k f I f (t)I a (t) J dt Note that we have three differential equations that must be solved simultaneously. If the dynamics of the Lf La field circuit are much faster than those of the armature circuit (time constant , as is often the Rf Ra case) one can assume that the field current varies instantaneously with the supply voltage, leading to VS If and to the equations: Rf dI (t) V (t) La a Ra I a (t) k f S ω m (t ) VS (t ) dt Rf La VS (t) dω m (t ) Ia (t) J bωm (t) TL (t) Rf dt Series configuration Applying KVL and equation 17.47 to the electrical circuit we obtain: dI a (t) Ra R f Ia (t) k f Ia (t)ω m (t) VS (t) La L f dt Applying Newton’s Second Law and equation 17.46 to the load inertia, we obtain: dωm (t) J Tm(t) TL (t ) bω m dt or dω m (t ) 2 k f Ia (t) J bωm (t) TL (t) dt d) Determine whether the equations are nonlinear Both systems of equations are nonlinear. In the shunt case, we have product terms in If and ωµ, and in If and Ia (or in VS and ω, and in VS and Ia if we use the simplified system of two equations). In the series case, 2 we have a quadratic term in Ia and a product term in If and ωm. In either case, no simple assumption leads kf 17.36 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 to a linear set of equations; thus either linearization or nonlinear solution methods (e.g.: numerical simulation) must be employed. ______________________________________________________________________ Problem 17.35 Solution: Known quantities: A shunt-connected DC motor shown in Figure P17.35 Motor parameters: k a , kT = armature and torque reluctance constant and k f = field flux constant Find: Derive the differential equations describing the electrical and mechanical dynamics of the motor Draw a simulation block diagram of the system Assumptions: None Analysis: Electrical subsystem dI f (t) R f I f (t) field dt dI (t ) dI (t ) VS (t ) = La a + Ra I a (t ) + k aφω m (t ) = La a + Ra I a (t ) + k a k f I f (t )ω m (t )armature dt dt VS (t) Lf Mechanical subsystem J dω m (t ) = Tm (t ) − TL (t ) − bω m (t ) = k aφI a (t ) − TL (t ) − bω m (t ) = k a k f I f (t ) I a (t ) − TL (t ) − bω m (t ) dt Simulation block diagram: 17.37 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 ______________________________________________________________________________________ Problem 17.36 Solution: Known quantities: A series-connected DC motor shown in Figure P17.36 Motor parameters: k a , kT = armature and torque reluctance constant and k f = field flux constant Find: Derive the differential equations describing the electrical and mechanical dynamics of the motor Draw a simulation block diagram of the system Assumptions: None Analysis: Electrical subsystem 17.38 G. Rizzoni, Principles and Applications of Electrical Engineering VS (t) La Lf dI a (t) dt Ra R f I a (t) k a Problem solutions, Chapter 17 m (t ) dI a (t) RI a (t) ka k f I a (t ) m (t) dt ka k f R dI a (t) 1 VS (t) I a (t) I (t ) m (t ) dt L L L a Mechanical subsystem d m (t) J T m (t) T L (t ) b m (t) kT I a (t) T L (t ) b m (t ) kT k f I a2 (t) T L (t) b m (t ) dt 1 b d m (t) kT k f 2 I a (t) T L (t ) (t) J J J m dt Simulation block diagram: VS (t) L ______________________________________________________________________________________ 17.39 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.6: The Alternator (Synchronous Generator) Problem 17.37 Solution: Known quantities: 550V ⋅ A, 20 V rated automotive alternator. At rated V ⋅ A , the power factor is 0.85 . The resistance per phase is 0.05 Ω . The field takes 2 A at 12 V . The friction and windage loss is 25W and core loss is 30 W . A Find: The percent efficiency under rated conditions. Assumptions: None. Analysis: 500 = 25 A 20 Pa = I a2 Ra = 31.25W Ia = Pout = 500(0.85) = 425W Pf = 2(12) = 24 W Pin = Pout + Pa + 25 + 30 + 24 = 535.25W %= 425 × 100 = 79.4% 535.25 ______________________________________________________________________________________ Problem 17.38 Solution: Known quantities: A three-phase 2300 V , 500 kV ⋅ A synchronous generator. X S = 8.0 Ω, ra = 0.1 Ω . The machine is operating at rated load and voltage at a power factor of 0.867 lagging. Find: The generated voltage per phase and the torque angle. Assumptions: None. 17.40 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: 500k = 125.5 A 3 (2300) 2300 E= ∠0 , + 125.5∠ − 30, (0.1 + j 0.8) 3 = 1327.9 + 101.2∠52.9, = 1389 + j80.7 I= = 1391.3∠3.3, V ∴ E = 1391.3V δ = 3.3, ______________________________________________________________________________________ Problem 17.39 Solution: Known quantities: As shown in Figure P17.39. Find: Explain the function of Q , D , Z , and SCR . Assumptions: None. Analysis: Q : The setting of R1 determines the biasing of Q . When Q conducts, the SCR will fire, energizing the alternator’s field. D : This diode serves as a “free-wheeling” element, allowing the field current to circulate without interfering with the commutation of the SCR . Z : The Zener diode provides a fixed reference voltage at the emitter of transistor Q ; i.e., determination of when Q conducts is controlled solely by the setting of R1 . SCR : The SCR acts as a half-wave rectifier, providing field excitation for the alternator. Without the field, of course, the alternator cannot generate. ______________________________________________________________________________________ 17.41 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.7: The Synchronous Motor Problem 17.40 Solution: Known quantities: A non-salient pole, Y-connected, three phase, two-pole synchronous machine. The synchronous reactance is 7 Ω and the resistance and rotational losses are negligible. One point on the open-circuit characteristic = 400V (phase voltage) for a field current of 3.32 A . The machine operates as a motor, with a terminal voltage of 400 V (phase voltage). The armature current is 50 A , with power factor 0.85 is given by V0 leading. Find: Eb , field current, torque developed, and power angle Assumptions: None. Analysis: The per phase circuit is shown below: δ . XL + Eb IS + - VS - Since the power factor is 0.85 , we have: θ = 31.79 , 2π 3600 = 377 rad sec ωm = 60 From VOC = 400 V , we have E b = 400 V (open circuit ) = kω m i f 400 = 0.3196 Therefore k = 377 × 3.32 Eb = 400∠0, − 50∠31.79 , × 7∠90, = 400 + 184.38 − j 297.49 = 655.74∠ − 26.98, V Eb = 5.44 A 120.48 θ T = 31.79 , + 26.98, = 58.77 , if = The torque developed is: T= 3 Eb I S cos θ T = 135.27 N ⋅ m 377 17.42 G. Rizzoni, Principles and Applications of Electrical Engineering δ is the angle from V to δ = −26.98 Problem solutions, Chapter 17 Eb : , The phase diagram is shown below: IS T V Eb ______________________________________________________________________________________ Problem 17.41 Solution: Known quantities: A factory load of 900 kW at 0.6 power factor lagging is increased by adding a 450 kW synchronous motor. Find: The power factor this motor operates at and the KVA input if the overall power factor is 0.9 lagging. Assumptions: None. Analysis: Pold = 900 kW Qold = 1200 kVAR Pm = 450 kW PT = 1350 kW QT = 653.8 kVAR Q m = 653.8 − 1200 = −546.2 kVAR pf m = cos(tan −1 Sm = Qm ) = 0.636 leading Pm Pm = 708 kVA pf m ______________________________________________________________________________________ Problem 17.42 Solution: Known quantities: A non-salient pole, Y-connected, three phase, two-pole synchronous generator is connected to a 400 V (line to line), 60 Hz , three-phase line. The stator impedance is 0.5 + j1.6 (per phase). The generator is delivering rated current 36 A at unity power factor to the line. Find: The power angle for this load and the value of E b for this condition. Sketch the phasor diagram, showing Eb , I S , and VS . 17.43 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Assumptions: None. Analysis: 400 , ∠0 = 230.9∠0, V 3 I L = 36∠0, A VL = Z S = 0.5 + j1.6 = 1.676∠72.65, Ω Eb = VL + I L Z S = 248.9 + j57.6 = 255.5∠13.03, V , The power angle is 13.03 . Eb I Vs s _____________________________________________________________________________________ Problem 17.43 Solution: Known quantities: A non-salient pole, three phase, two-pole synchronous generator is connected in parallel with a three-phase, Y-connected load. The equivalent circuit is shown in Figure P17.43. The parallel combination is connected to a 220V (line to line), , three-phase line. The load current is 25 A at a power factor of 0.866 inductive. X S = 2 Ω . The motor is operating with I f = 1 A, T = 50 N ⋅ m at a power angle of − 30, . Find: I S , Pin (to the motor), the overall power factor and the total power drawn from the line. Assumptions: Neglect all losses for the motor. Analysis: The phasor per-phase voltage is: VS = 127∠0, V Tdev = 50 N ⋅ m = − 3 E b VS sin δ 377 X S Therefore, Eb = − 50(377)2 = 197.9 V 3(127) sin( −30, ) Eb = 197.9∠ − 30, V For i f = 1 A , I S = 49.47 + j 22.2 = 54.23∠24.16 , A The load current is: 17.44 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 I L = 25∠ − cos −1 0.866 = 21.65 − j12.5 and I 1 = I L + I S = 71.12 + j 9.7 = 71.78∠7.77 , A Pin _ motor = 3 × 54.23 × 127 × cos 24.16, = 18.85 kW Pin _ total = 3 × 71.78 × 127 × cos 7.77 , = 27.10 kW The power factor is: pf = cos 7.77 , = 0.991 leading ______________________________________________________________________________________ Problem 17.44 Solution: Known quantities: A non-salient pole, Y-connected, three phase, four-pole synchronous machine. The synchronous reactance 10 Ω . It is connected to a 230 3 V (line to line), 60 Hz , three-phase line. The load requires a torque of Tshaft = 30 N ⋅ m . The line current is 15 A leading the phase voltage. is Find: The power angle δ and E for this condition. The line current when the load is removed. Is it leading or lagging the voltage. Assumptions: All losses can be neglected. Analysis: At ω m = 188.5 rad sec , we can calculate Pout = 30 × 188.5 = 5655W Since Pin = Pout and Pin' ( per phase ) = 1885W = 230 × 15 cos θ , we calculate θ = cos −1 0.5464 = 56.88, , , Since VS = 230∠0 V , I S = 15∠56.88 A Eb = 355.6 − j81.96 = 364.92 ∠ − 12.98 , V The power angle is: δ = −12.98, 0, and from 364.92 ∠0, = 230∠0, − 10∠90, If the load is removed, the power angle is I = 13.495∠90, A The current is leading the voltage. ______________________________________________________________________________________ 17.45 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.45 Solution: Known quantities: 10 hp, 230V , 60 Hz Y-connected, three phase synchronous motor delivers full load at a power factor of 0.8 leading. The synchronous reactance is 6 Ω . The rotational loss is 230 W , and the field loss is 50 W . A Find: a) The armature current. b) The motor efficiency. c) The power angle. Assumptions: Neglect the stator winding resistance. Analysis: Pout = 10 hp = 7460 W Pin = Pout + Pr + Pcopper = 7740 ∴Pin ( per phase) = 2580 = VS I S 0.8 230 = 132.8V 3 2580 ∴I S = = 24.3 A 132.8 × 0.8 VS = That is: VS = 132.8∠0, V , I S = 24.3∠36.87 , A Eb = VS − I S (6 ∠9 0, ) = 249.2 ∠ − 27.9 , V a) I S = 24.3∠36.87 , A b) efficiency = 7460 = 0.964 = 96.4% 7740 c) power angle = −27.9, ______________________________________________________________________________________ Problem 17.46 Solution: Known quantities: 2300 V , 60 Hz , 30 poles, 2000 hp , unity power factor synchronous motor. X S = 1.95 Ω per phase. A three-phase Find: The maximum power and torque. Assumptions: Neglect all losses. 17.46 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: 3600 = 240 rev min 15 ω S = 25.13 rad sec nS = At full load, Pin = 746 × 2000 = 1.492 MW VS = 2300 = 1327.9∠0 , V 3 For unity power factor, I S = 374.5∠0, A Eb = VS − I S jX S = 1327.9 − j 730.3 = 1515.5∠ − 28.2, V The maximum power and torque are: Pmax = 3 Tmax = E b VS = 3.096 MW XS Pmax = 123.2 kN ⋅ m ωS ______________________________________________________________________________________ Problem 17.47 Solution: Known quantities: 1200 V Y-connected, three phase synchronous motor takes 110 kW when operated under a certain load at 1200 rev min . The back emf of the motor is 2000 V . The synchronous reactance is 10 Ω per A phase. Find: The line current and the torque developed by the motor. Assumptions: Winding resistance is negligible. Analysis: VS = 1200 = 692.8∠0, 3 The input power per phase is: L= 100 2 N2 = = 0.8 H ℜ T 12.51 × 10 3 The power developed is: P = −3 E b VS sin δ XS ∴sin δ = −0.2646 δ = −15.34 , The torque developed is: 17.47 G. Rizzoni, Principles and Applications of Electrical Engineering T= Problem solutions, Chapter 17 P = 875.1 N ⋅ m ωS ______________________________________________________________________________________ Problem 17.48 Solution: Known quantities: 600V Y-connected, three phase synchronous motor takes 24 kW at a leading power factor of 0.707 . The per-phase impedance is 5 + j50 Ω . A Find: The induced voltage and the power angle of the motor. Assumptions: None. Analysis: 600 = 346.4∠0, V 3 Z S = 5 + j50 = 50.25∠84.29 , Ω VS = From pf = 0.707 , we have θ = 45, . From Pin = 3VS I S cos θ , we have 24 × 103 × 0.707 = 32.67 A 3 × 346.4 I S = 32.67∠45, A IS = Eb = VS − I S Z S = 1385 − j1270.6 = 1880.3∠ − 42.51, V The power angle is: δ = −42.51, The power developed and the copper loss are: Pdev = 3 Eb I S cos 87.51, = 8.006 kW 2 Ploss = 3 I S RS = 16.01 kW ______________________________________________________________________________________ 17.48 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Section 17.8: The Induction Motor Problem 17.49 Solution: Known quantities: A 74.6 kW three-phase, parameters are: 440 V (line to line), four-pole, 60 Hz induction motor. The equivalent circuit RS = 0.06 Ω RR = 0.08 Ω X S = 0.3 Ω X R = 0.3 Ω The no-load power input is X m = 5Ω 3240 W at a current of 45 A . Find: The line current, the input power, the developed torque, the shaft torque, and the efficiency at s = 0.02 . Assumptions: None. Analysis: 400 = 254∠0, V 3 j5( 4 + j 0.3) Z in = 0.06 + j 0.3 + 4 + j5.3 VS = = 2.328 + j 2.294 = 3.268∠44.59, Ω I S = 77.7∠ − 44.59 A Pin = 3 × 254 × 77.7 cos( −44.59, ) = 42.16 kW I2 = j5 I S = 58.51∠ − 7.55, A 4 + j5.3 The total power transferred to the rotor is: RS 2 I 2 = 41.1 kW S Pm = PT − Pcopper _ loss _ in _ rotor PT = 3 = 41.1 × 103 (1 − s ) = 40.25 kW ω m = (1 − s )ω S = 0.98 × 188.5 = 184.7 rad sec Therefore, the torque developed is: Pm = 218 N ⋅ m 184.7 = 1880.3∠ − 42.51, V Tdev = The rotational power and torque losses are: 17.49 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Prot = 3240 − 3 × 452 × 0.06 = 2875.5W Trot = 15.56 N ⋅ m The shaft torque is: Tsh = 218 − 15.56 = 202.4 N ⋅ m Efficiency is: Tsh = Pout 202.4 × 184.7 = = 0.887 Pin 42.16 × 10 3 ______________________________________________________________________________________ Problem 17.50 Solution: Known quantities: 60 Hz , four-pole, Y-connected induction motor is connected to a three-phase, 400V (line to line), 60 Hz line. The equivalent circuit parameters are: A RS = 0.2 Ω RR = 0.1 Ω X S = 0.5 Ω X R = 0.2 Ω When the machine is running at X m = 20 Ω 1755 rev min , the total rotational and stray-load losses are 800 W . Find: The slip, input current, total input power, mechanical power developed, shaft torque and efficiency. Assumptions: None. Analysis: From n S = 1800 rev min ¸ we have s = 0.025 RR =4 s Z in = 0.2 + j 0.5 + j 20( 4 + j 0.2) 4 + j 20.2 = 3.972 + j1.444 = 4.226∠19.98, Ω Therefore, 17.50 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 I S = 54.6∠ − 19.98, A Pin = 3(54.6)( 400 cos( −19.98, )) = 35.6 kW 3 2 Pt = Pin = 3 I S RS = 35.6 × 103 − 3(54.6) 2 × 0.2 = 33.81 kW Pm = (1 − s ) Pt = 32.97 kW Psh = Pout = Pm − 800 = 32.17 kW ω m = 183.8 rad sec Tsh = 175 N ⋅ m 32.17 efficiency = = 0.904 35.6 ______________________________________________________________________________________ Problem 17.51 Solution: Known quantities: A three-phase, 60 Hz , eight-pole induction motor operates with a slip of 0.05 for a certain load. Find: a) The speed of the rotor with respect to the stator. b) The speed of the rotor with respect to the stator magnetic field. c) The speed of the rotor magnetic field with respect to the rotor. d) The speed of the rotor magnetic field with respect to the stator magnetic field. Assumptions: None. Analysis: n S = 900 rev min , ω S = 94.25 rad sec a) nm = (1 − s )n S = 855 rev min b) The speed of the stator field is 900 rev min , the rotor speed relative to the stator field is − 45 rev min . c) 45 rev min d) 0 rev min ______________________________________________________________________________________ 17.51 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.52 Solution: Known quantities: 60 Hz , 400 V (per phase), two-pole induction motor develops Pm = 37 kW at a certain A three-phase, speed. The rotational loss at this speed is 800 W . Find: a) The slip and the output torque if the total power transferred to the rotor is b) 40 kW . I S and the power factor if Pm = 45 kW , RS = 0.5 Ω . Assumptions: Stray-load loss is negligible. Analysis: a) Pm = 3(1 − s ) Pt = 37 kW 1 − s = 0.925 s = 0.075 ω S = 377 rad sec ω m = (1 − s )ω S = 348.7 rad sec Psh = Pout = 37 − 0.8 = 36.2 kW P Tsh = sh = 103.8 N ⋅ m 348.7 n S = 3600 rev min , b) 2 Pin = 3 I S RS + Pt 2 3 I S RS = 5 kW ∴ I S = 57.7 A Pin = 3VS I S cos θ = 45 kW The power factor is: cos θ = 0.65 lagging ______________________________________________________________________________________ Problem 17.53 Solution: Known quantities: The nameplate speed of a 25 Hz induction motor is 720 rev min . The speed at no load is 745 rev min . Find: a) The slip. b) The percent regulation. Assumptions: None. 17.52 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: a) 120(25) = 4.17 p = 4 720 120( 25) n sync = = 750 rpm 4 750 − 720 slip = = 0.04 = 4% 750 p≈ b) reg = 745 − 720 = 0.035 = 3.5% 720 ______________________________________________________________________________________ Problem 17.54 Solution: Known quantities: The name plate of a squirrel-cage four-pole induction motor has 25 hp, 220 V , 60 Hz, 830 rev min , 64 A , three-phase line current. The motor draws when operating at a full load. Find: a) slip. b) Percent regulation if the no-load speed is 895 rpm . c) Power factor. d) Torque. e) Efficiency. Assumptions: None. Analysis: a) 20,800 W n sync = 900 rpm slip = 900 − 830 = 0.078 = 7.8% 900 reg = 895 − 830 = 0.078 = 7.8% 830 b) c) pf = d) T= 20,800 = 0.853 lagging 3 (220)(64) 7.04( 25 × 746) = 158.2 lb ⋅ ft 830 e) eff = 25 × 746 = 0.897 = 89.7% 20,800 ______________________________________________________________________________________ 17.53 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.55 Solution: Known quantities: 60 Hz , four-pole, Y-connected induction motor is connected to a 200 V (line to line), three-phase, 60 Hz line. The equivalent circuit parameters are: RS = 0.48 Ω Rotational loss torque = 3.5 N ⋅ m A X S = 0.8 Ω RR = 0.42 Ω ( referred to the stator ) X m = 30 Ω X R = 0.8 Ω ( referred to the stator ) The motor is operating at slip s = 0.04 . Find: The input current, input power, mechanical power, and shaft torque. Assumptions: Stray-load losses are negligible. Analysis: VS = 115.5V ω m = (1 − s )188.5 = 181 rad sec j 30(10.5 + j 0.8) Z in = 0.48 + j 0.8 + 10.5 + j 30.8 = 9.404 + j 4.63 = 10.48∠26.2 , ∴ I S = 11.02∠ − 26.2 , A Pin ( per phase) = 115.5 × 11.02 × cos( −26.2, ) Pin (total ) = 3426W 2 Pt = Pin (total ) − 3RS I S = 3251W ∴ Pm = (1 − s ) Pt = 3121W Tsh = 3121 = 17.24 N ⋅ m 181 ______________________________________________________________________________________ Problem 17.56 Solution: Known quantities: a) A three-phase, 220 V , 60 Hz induction motor runs at 1140 rev min . b) A three-phase squirrel-cage induction motor is started by reducing the line voltage to VS reduce the starting current. Find: a) The number of poles (for minimum slip), the slip, and the frequency of the rotor currents. b) The factor the starting torque and the starting current reduced. Assumptions: None. 17.54 2 in order to G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Analysis: a) For minimum slip, the synchronous speed, therefore, nS = 1200, 3600 , should be as close as possible to 1140 rev min , p 2 p = 6 poles 1200 − 1140 = 0.05 1200 f rotor = 3 Hz s= b) If the line voltage is reduced to half, the starting current is reduced by a factor of is proportional to IS 2 2 . The developed torque . Therefore, the starting torque is reduced by a factor of 4 . ______________________________________________________________________________________ Problem 17.57 Solution: Known quantities: A six-pole induction machine has a 50 kW rating and is 85 percent efficient. If the supply is 220V at 60 Hz . 6 poles 60 Hz 50 kW 85% efficient 220 Volt 4% slip Find: The motor speed and torque at a slip s = 0.04. Assumptions: None. Analysis: a) ns 120 f p 120 60 1200rev / min 6 @ slip of 4% n ns 1 s 1200 rev/ min 1 0.04 1152 rev/ min Pout = Pin × efficiency = (50kW )(0.85) = 42.5 kW b) Tout = Pout 1 rev 60 sec 42500 W = = 352.3 N - m * ω 1152 rev/min 2π rad min ______________________________________________________________________________________ Problem 17.58 Solution: Known quantities: 6 poles 17.55 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 60 Hz 240 Volt rms 10% slip Torque = 60 N-m Find: a) The speed and the slip of the induction machine if a load torque of 50 N-m opposes the motor. b) The rms current when the induction machine is operating under the load conditions of part a. Assumptions: The speed torque curve is linear in the region of our interests. Analysis: a) 120 f p ns 120 60 6 1200rev / min @ slip of 4% ns 1 s n 1200 rev/ min 1 0.04 1152 rev/ min T or que [ N- m] 60 50 T= m * n + b ns torque T m n b m 60 0 N m 1080 1200 rev/min b T N m rev/min N m 0.5 1080 rev/min rev/min 0.5 60 N m m n 600 N - m motor speed (@ 50 N-m) n T rev/min 50 N - m .5 N m b m 600 N - m 1000 rev/min n slip (@ 50 N-m) s ns n ns 1200 1100 1200 .0833 b) Output Power Pout Tω 50 N - m 1100 8.33% rev 2π rad 1min min rev 60sec Input Power Pin Pout efficiency Irms Vrms 17.56 5760 W G. Rizzoni, Principles and Applications of Electrical Engineering Current Irms Pout efficiency Vrms 5760 W 0.92 240 volts Problem solutions, Chapter 17 26.1 amps ______________________________________________________________________________________ Problem 17.59 Solution: Known quantities: A three-phase, 5 hp , 220 V , 60 Hz induction motor. V = 8V , I = 18 A, P = 610 W . Find: a) The equivalent stator resistance per phase, RS . b) The equivalent rotor resistance per phase, RR . c) The equivalent blocked-rotor reactance per phase, XR . Assumptions: None. Analysis: a) RS = 1 PBR = 0.314 Ω 2 2 3I BR b) RR = 0.314 Ω c) ZS = VBR 3 48 3 = = 1.54 Ω I BR 18 X R = Z S2 − R 2 = (1.54) 2 − (0.628) 2 = 1.4 Ω ______________________________________________________________________________________ Problem 17.60 Solution: Known quantities: The starting torque equation is: T= RR m ⋅ VS2 ⋅ 2 ωe ( RR + R S ) + ( X R + X S ) 2 Find: a) The starting torque when it is started at 220V . b) The starting torque when it is started at 110V . Assumptions: None. Analysis: a) 17.57 G. Rizzoni, Principles and Applications of Electrical Engineering T= Problem solutions, Chapter 17 1 q1V12 ( RR s ) ; s =1 ωS R2 + X 2 ∴T = 1 3(127) 2 (0.314) = 17.1 N ⋅ m 377 (0.628) 2 + (1.4) 2 b) 1 3(63.5) 2 (0.314) T= = 4.28 N ⋅ m 377 (0.628) 2 + (1.4) 2 ______________________________________________________________________________________ Problem 17.61 Solution: Known quantities: A four-pole, three-phase induction motor drives a turbine load with torque-speed characteristic given by TL = 20 + 0.006ϖ 2 At a certain operating point, the machine has 4% slip and 87% efficiency. Find: Torque at the motor-turbine shaft Total power delivered to the turbine Total power consumed by the motor Assumptions: Motor run by 60-Hz power supply Analysis: Synchronous speed of four-pole induction motor at 60-Hz: 120 f 60 s / min× 60r / s = = 1800r / min 4/2 P 2πrad / rev = 188.5rad / s ϖ s = 1800rev / min× 60 s / min ns = Rotor mechanical speed at 4% slip: ϖ m = (1 − s )ϖ s = (1 − 0.04)(188.5rad / s ) = 181.0rad / s Load torque at the shaft: TL = 20 + 0.006(181.0rad / s ) 2 = 216 N − m Total power delivered to the turbine: P = TLϖ m = (216 N − m )(181.0rad / s ) = 39.1kW Total power consumed by the motor: Pm = P 39.1kw = = 44.9kW 0.87 η ___________________________________________________________________________________ Problem 17.62 Solution: Known quantities: A four-pole, three-phase induction motor rotates at 1700 r/min when the load is 100 N-m. The motor is 88% efficient. 17.58 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Find: a) Slip b) For a constant-power, 10-kW load, the operating speed of the machine c) Total power consumed by the motor d) Sketch the motor and load torque-speed curves on the same graph. Show numerical values. Assumptions: Motor run by 60-Hz power supply Analysis: a) Synchronous speed of four-pole induction motor at 60-Hz: 120 f 60 s / min× 60r / s = = 1800r / min 4/2 P ns = Slip: s= b) ns − n 1800r / min − 1700r / min = = 0.056 = 5.6% ns 1800r / min Operating speed of machine for a constant-power load of 10-kW ϖ= 10000W P = = 100rad / sec TL 100 N − m n =ϖ c) 60s / min = 955r / min 2πrad / rev Total power consumed by the motor Pm = P 10kW = = 11.4kW 0.88 η d) Sketch of motor and load torque-speed curves on the same graph, with the operating point at the first intersection: 17.59 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 ___________________________________________________________________________________ Problem 17.63 Solution: Known quantities: A six-pole, three-phase motor. Find: The speed of the rotating field when the motor is connected to: a) a 60 Hz line. b) a 50 Hz line. Assumptions: None. Analysis: a) 4πf = 125.7 rad sec , P For 60 Hz , ω m = b) For 50 Hz , ω m = 104.72 rad sec , nm = 1000 rev min nm = 1200 rev min ______________________________________________________________________________________ 17.60 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.64 Solution: Known quantities: A six-pole, three-phase, 440V , 60 Hz induction motor. The model impedances are: RS = 0.8 Ω X S = 0.7 Ω RR = 0.3 Ω X R = 0.7 Ω X m = 35 Ω Find: The input current and power factor of the motor for a speed of 1200 rev min . Assumptions: None. Analysis: 440 = 254∠0, V 3 For n m = n S = 1200 rev min , s = 0 ( no load ) . Z in = RS + j ( X S + X m ) = 0.8 + j 35.7 VS = = 35.71∠88.7 , Ω I S = 7.11∠ − 88.7 , A The power factor is: cos 88.7, = 0.0224 lagging Pin = 3 I S VS cos θ = 121.4 W ______________________________________________________________________________________ Problem 17.65 Solution: Known quantities: A eight-pole, three-phase, RS = 0.78 Ω RR = 0.28 Ω 220 V , 60 Hz induction motor. The model impedances are: X S = 0.56 Ω X R = 0.84 Ω X m = 32 Ω Find: The input current and power factor of the motor for s = 0.02 . Assumptions: None. Analysis: For 8 poles, 17.61 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 3600 = 900 rev min 4 ω S = 94.25 rad sec nS = ω m = (1 − s )ω S = 92.4 rad sec By using the equivalent circuit, we have: Z in = 0.78 + j 0.56 + 0.28 + j 0.84) 0.02 14 + j 32.84 j 32( = 12.03 + j 6.17 = 13.52∠27.15, Ω VS = 127∠0, V I S = 9.39∠ − 27.15, A pf = cos( −27.15, ) = 0.8898 lagging ______________________________________________________________________________________ Problem 17.66 Solution: Known quantities: The nameplate is as given in Example 17.2. Find: The rated torque, rated volt amperes, and maximum continuous output power for this motor. Assumptions: None. Analysis: The speed is: nm = 3565 rev min 2π × 3565 = 373.3 rad sec 60 The rated volt ⋅ amperes is: ωm = 3 × ( 230 V ) × (106 A) = 42.23 kVA or 3 × ( 460 V ) × (53 A) = 42.23 kVA The maximum continuous output power is: PO = 40 × 746 = 29840 W The rated output torque is: T= PO = 79.93 N ⋅ m ωm ______________________________________________________________________________________ 17.62 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 17 Problem 17.67 Solution: Known quantities: At rated voltage and frequency, the 3-phase induction machine has a starting torque of maximum torque of 210 percent of full-load torque. 140 percent and a Find: a) The slip at full load. b) The slip at maximum torque. c) The rotor current at starting as a percent of a full-load rotor current. Assumptions: Neglect stator resistance and rotational losses. Assume constant rotor resistance. Analysis: a) TR = KV 2 ( R2 s R ) ( R2 s R ) 2 + X 2 TST = 1.4TR s ST = 1.0 TMT = 2.1TR s MT = R2 X The above leads to 3 equations in 3 unknowns: (1) 4.2 XR2 = 1.4 R22 + 1.4 X 2 R22 + s R X 2 = 1.4 R22 + 1.4 X 2 sR XR2 R (3) 4.2 = ( 2 )2 + X 2 sR sR (2) Solving the equations, we have: R2 = 0.382 X s R = 0.097 b) s MT = RT = 0.382 X c) IR = KV , 4.06 I ST = KV 1.07 I ST × 100 = 379% IR ______________________________________________________________________________________ 17.63 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Chapter 18 Instructor Notes The content of Chapter 18 is somewhat unusual for a textbook of this nature. The intent of this chapter is to provide a reasonably quantitative overview of the operation of small electric machines (mostly motors). In many practical industrial applications, ranging from servos for robots, to drug delivery systems, to actuation devices for control systems, to manufacturing equipment, to fluid power systems, small motors find widespread application. The first section discusses the brushless DC motor, including the basics of the electronic circuits that make its operation possible. The second section introduces the stepper motor and its drive. In section 18.3, the switched reluctance machine is introduced; this is a new section in the third edition, motivated by the increasing interest in this family of machines for industrial applications. Next, single phase AC motors are discussed in Section 18.4, starting with the universal motor, and continuing with a classification of single phase induction motors, which includes split-phase, capacitor-type and shaded-pole motors. The presentation detail is sufficient to permit quantitative analysis of these motors using circuit models. The final section, 18.5, on motor selection and application, introduces some of the basic ideas behind motor selection and performance calculations. This section, which describes calculations related to reflected load inertias in the presence of mechanical gear reductions, and calculations of acceleration, torque, efficiency, and thermal loading, could be covered at any point in Chapter 17 or 18, even if the material in Sections 18.1-18.4 is not covered. The examples given in the chapter are supplemented by over thirty homework problems, some of which are extensions of the examples presented in the text. Problems 18.5 and 18.9 require some background in digital logic circuits (Chapter 13); problems 18.7, 18.36 and 18.37 require some background in system dynamics; all remaining problems can be solved strictly on the basis of the material covered in the chapter. Learning Objectives 1. Understand the basic principles of operation of brushless DC motors, and the trade-offs between these and brush-type DC motors. Section 1. 2. Understand the operation and basic configurations of step motors, and understand step sequences for the different classes of step motors. Section 2. 3. Understand the operating principles of switched-reluctance machines. Section 3. 4. Classify and analyze single-phase AC motors, including the universal motor and various types of single-phase induction motors, using simple circuit models. Section 4. 5. Outline the selection process for an electric machine given an application; perform calculations related to load inertia, acceleration, efficiency, and thermal characteristics. Section 5. 18.1 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Section 18.1: Brushless DC Motors Problem 18.1 Solution: Known quantities: A permanent magnet six-pole two-phase synchronous machine. λ m = 0.1V ⋅ s . Find: The amplitude of the open-circuit phase voltage measured when the rotor is turned at n = 60 rev sec . Assumptions: None. Analysis: We know that λ m = 0.1V ⋅ s p=6 m=2 ω m = 60 rev s = 2π × 60 rad s Let flux linkage λ = λm sin ω t , where p ω = ω m = 3 × 60 = 180 rev s 2 = 2π × 180 rad s = 360π rad s Then, the generated voltage: d (λ m sin ω t ) dλ V =e= = = ωλ m cos ω t dt dt = Vm cos ω t Vm = ωλ m = 360π × 0.1 = 113.1V ______________________________________________________________________________________ Problem 18.2 Solution: Known quantities: A four-pole two-phase brushless dc motor. n = 3600 rev min . The open-circuit voltage across one of the phases is 50 V . Find: a) λ . b) The no-load rotor speed ω in rad/s when the mechanical source is removed and Va = 2 25 cos θ , Vb = 2 25 sin θ , where θ = ω e t . Assumptions: None. Analysis: We know that 18.2 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 p=4 m=2 ω m = 3600 rev min Vn = 50 V a) let e = V = 2V sin θ = 2V sin ω t ω= p ω m ⋅ 2π rad min = 2 × 60 × 2π rad s 2 λ = ³0t edt = ³0t 2V sin ω tdt = = 2 V cos ω t ω 2 50 cos 240πt = 0.094 cos ω t 240π b) Symmetric voltages in symmetric windings produce a rotational field in voltage with frequency f s . Let f s = 3600 rev min . Then, the rotor speed is ω 3600 2π ωm = = = 1800 rev min = × 1800 rad s = 60π rad s p 2 2 60 ______________________________________________________________________________________ Problem 18.3 Solution: Known quantities: T1 = T3 = 1 s (see Figure 18.7); maximum motor rpm, nmax = 1,800 rev/min. Find: T2. Assumptions: The motor covers 0.5 m in 100 revolutions. Analysis: The maximum rotational velocity of the motor is: v = nmax/60 = 1,800/60 = 30 rev/s. Using the expression derived in Example 18.2, we know that the maximum motor rotational velocity is: d v 1 1 T1 T2 T3 2 2 and we can calculate T2 as follows: d 1 1 100 rev T2 T1 T3 1 s 2.33 s v 2 2 30 rev/s Thus, the total trapezoidal profile has been shortened by 2/3 s. ______________________________________________________________________________________ Problem 18.4 Solution: Known quantities: Desired load motion profile (Figure P18.4). The motor covers 0.5 m (100 revolutions) in 3 s. Find: Maximum motor speed, acceleration and deceleration times. Assumptions: Assume a triangular speed profile.. 18.3 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Analysis: To simplify the analysis, choose a symmetrical speed profile; thus, the motor will accelerate for 1.5 s and decelerate for 1.5 s, or T1 = T2 = 1.5 s. Using the results of Example 18.2, if we set the flat portion of the speed profile (T2 in Example 18.2) to zero, we can write an expression for the total motor travel. 1 1 d v T1 T2 2 2 and calculate the maximum motor speed to be: d 100 rev v 66.67 rev/s 1 1 1 1 T1 T2 1.5 1.5 s 2 2 2 2 which corresponds to nmax = 66.67×60 = 4,000 rev/min. ______________________________________________________________________________________ 18.4 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Section 18.2: Stepping Motors Problem 18.5 Solution: Known quantities: Variable-reluctance step motor of Example 18.4 (Figure 18.11) Find: Design a logic circuit to achieve the step sequence given in Table 18.4 (see below) Assumptions: Hint: Use a counter and logic gates Analysis: Table 18.4: Current Excitation Sequence for VR Step Motor Rotor SA SB SC SD Position 1 0 0 0 0° 1 1 0 0 45° 0 1 0 0 90° 0 1 1 0 135° 0 0 1 0 180° 0 0 1 1 225° 0 0 0 1 270° 1 0 0 1 315° 1 0 0 0 360° There are eight possible configurations for the motor. Hence, a 3-bit binary counter was chosen that pulses every 45°. The table below lists the corresponding logic for the binary counter. Binary Counter Step Motor Response b0 b1 b2 SA SB SC SD 0 1 0 1 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 1 1 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 1 1 1 0 Next, convert the truth table to a logical expression for each of the outputs: S A = b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 = b1 ⋅ b2 (b0 + b0 ) + b0 ⋅ b1 ⋅ b2 S A = b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 S B = b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 = b2 (b0 ⋅ b1 + b0 ⋅ b1 + b0 ⋅ b1 ) S C = b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 = b1 ⋅ b2 (b0 + b0 ) + b0 ⋅ b1 ⋅ b2 S C = b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 S D = b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 + b0 ⋅ b1 ⋅ b2 = b2 (b0 ⋅ b1 + b0 ⋅ b1 + b0 ⋅ b1 ) 18.5 Rotor Position 0° 45° 90° 135° 180° 225° 270° 315° 360° G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 From these four expressions, a logic circuit diagram can be made for each individual output, and tied together to achieve the desired step response. 3-B it B inary Cou nter b0 b1 b2 NO T A ND NO T OR Sa A ND A ND A ND OR NO T A ND OR NO T Sb A ND A ND NO T NO T A ND NO T OR Sc A ND A ND A ND OR NO T A ND OR NO T Sd A ND A ND ______________________________________________________________________________________ 18.6 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.6 Solution: Known quantities: PM stepper motor with 6 poles, bipolar supply, Find: Smallest achievable step size. Assumptions: None Analysis: With reference to Example 18.3, we see that the half-step sequence for the 2-phase 4-pole motor leads to 45-degree steps. The addition of two poles will reduce the step size by 50%, resulting in 30-degree steps. ______________________________________________________________________________________ Problem 18.7 Solution: Known quantities: J m , J L , D, T f . Find: The dynamic equation for a stepping motor coupled to a load. Assumptions: None. Analysis: The equation will have the following form: di V = Ri + L + K E ω dt dω + Dω + TF + TL T = KT i = ( J m + J L ) dt ______________________________________________________________________________________ Problem 18.8 Solution: Known quantities: A hybrid stepper motor capable of 18 , steps. Find: Sketch the rotor-stator configuration of the motor. Assumptions: None. Analysis: The rotor and stator configuration is shown below: 18.7 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 The motor has 5 rotor teeth and 4 stator teeth (two phases). ______________________________________________________________________________________ Problem 18.9 Solution: Known quantities: Shown in Check Your understanding on page 928. Find: A binary counter and logic gates to implement the stepping motor binary sequence. Assumptions: None. Analysis: ______________________________________________________________________________________ Problem 18.10 Solution: Known quantities: A two-phase permanent magnet stepper motor has 50 rotor teeth. When driven at measured open circuit phase peak-to-peak voltage is 25V . Find: a) Calculate λ b) Express the developed torque when Assumptions: The winding resistance is ia = 1 A and ib = 0 . 0 .1 Ω . Analysis: a) We know that p = 50 m=2 ω m = 100 rad s 18.8 ω = 100 rad s , the G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Strikes per rev.: N = p ⋅ m = 100 2π 2π 2π = = N 100 50 π t V0 50 dθ λ = ³ edt = ³ 0 0 ω m ∆θ = = V0 π 12.5 π ⋅ = ⋅ = 0.00785V ⋅ s ω m 50 100 50 b) Let the winding resistance be represented by Rw , then V = k aω m + Rw I T = kT I where k a = kT = V − Rw I ωm V − Rw I N ⋅m ωm ______________________________________________________________________________________ Then T = k a I = Problem 18.11 Solution: Known quantities: The schematic diagram of a four-phase, two-pole PM stepper motor is shown in Figure P18.11. The phase coils are excited in sequence by means of a logic circuit. Find: The no-load voltage of the generator and terminal voltage at half load. Assumptions: a) The logic schedule for full-stepping of this motor. b) The displacement angle of the full step . Analysis: a) For full step clockwise rotation is: Phase 1 → phase 4 → phase 3 → phase 2 → phase 1 b) 18.9 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 , The displacement angle of the full step sequence is 90 . ______________________________________________________________________________________ Problem 18.12 Solution: Known quantities: A PM stepper motor provides a full-step angle of 15 , . Find: The number of stator and rotor poles. Assumptions: None. Analysis: The motor will require 24 stator teeth and 2 rotor teeth. ______________________________________________________________________________________ Problem 18.13 Solution: Known quantities: A bridge driver scheme for a two-phase stepping motor is as shown in Figure P18.13. Find: The excitation sequences of the bridge operation. Assumptions: None. Analysis: CK R 1 2 3 4 5 6 7 8 S1 1 1 0 0 0 0 0 1 1 S2 0 0 0 1 1 1 0 0 0 S3 0 0 0 1 1 1 0 0 0 S4 1 1 0 0 0 0 0 1 1 S5 0 1 1 1 0 0 0 0 0 S6 0 0 0 0 0 1 1 1 0 S7 0 0 0 0 0 1 1 1 0 S8 0 1 1 1 0 0 0 0 0 Where "1" means switch is closed. ______________________________________________________________________________________ 18.10 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.14 Solution: Known quantities: A PM stepper motor provides a full-step angle of 15 , . It is used to directly drive a 0.100 in. lead screw. Find: a) The resolution of the stepper motor in steps/revolution. b) The distance the screw travels in inches for each step 15 , of motor. The number of full 15 , steps required to move the lead screw and the stepper motor shaft through 17.5 revolution. d) The shaft speed (in rev/min) when the stepping frequency is 220 pps . c) Assumptions: None. Analysis: a) steps revolution = 360 = 24 15 b) d = 0.1'' × 15 , 360 , = 0.0042 '' c) steps = 175 rev × 24 d) steps = 420 steps rev # pulses =# steps 1 rev 60 s steps × × = 550 rpm s 24 steps min ______________________________________________________________________________________ n SH = 220 18.11 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Section 18.4: Single-Phase AC Motors Problem 18.15 Solution: Known quantities: The motor data are the following: 3 hp, 900 rev min 4 1 b) 1 hp, 3600 rev min 2 3 c) hp,1800 rev min 4 1 d) 1 hp, 6000 rev min 2 Find: Whether the following motors are integral- or fractional-horse power motors. Assumptions: None. Analysis: a) a) The power is 0.75 1800 = 1.5 hp . 900 Integral. b) The power is 1.5 1800 = 0.75 hp . 3600 Fractional. c) The power is 0.75 1800 = 0.75 hp . 1800 Fractional. d) The power is 1.5 1800 = 0.45 hp . 6000 Fractional. ______________________________________________________________________________________ Problem 18.16 Solution: Known quantities: F1 = F1( peak ) cos θ , Find: The expression for are present. F1( peak ) = F1(max) cos θ , . F1 and verify that for a single-phase winding, both forward and backward components 18.12 G. Rizzoni, Principles and Applications of Electrical Engineering Assumptions: None. Analysis: The stator mmf Problem solutions, Chapter 18 F1 can be expressed as: F1 = F1max cos(ω t ) cos θ 1 1 = F1max cos θ cos(ω t ) − F1max cosθ cos(ω t ) 2 2 1 1 + F1max cos θ cos(ω t ) + F1max cos θ cos(ω t ) 2 2 = FCW + FCCW where: FCW is a clockwise-rotating mmf. FCCW is a counter clockwise-rotating magnetic mmf. ______________________________________________________________________________________ Problem 18.17 Solution: Known quantities: A 200 V , 60 Hz , 10 hp single-phase induction motor operates at an efficiency of 0.86 a power factor of 0 .9 . Find: The capacitor that should be placed in parallel with the motor so that the feeder supplying the motor will operate at unity power factor. Assumptions: None. Analysis: We have: Pout = 746 × 10 = 7460 W Pout = 8674.4 W eff . From Pin = VS I S cos θ S = 8674.4 W , we have: Pin = I S = 48.2 A, θ = 25.84 , lagging Therefore, I S = 48.2∠ − 25.84 , = 43.38 − j 21.01 A To get unit power factor, I C = j 21.01 −200 , ω = 377 rad s , we have: From j 21.01 = 1 −j ωC C = 278.6 µF ______________________________________________________________________________________ 18.13 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.18 Solution: Known quantities: A 230 V , 50 Hz single-phase two-pole induction motor operates at 3 percent slip. Find: The slip in the opposite direction of rotation. Find the speed of the motor in the normal direction of rotation. Assumptions: None. Analysis: For a 2-pole machine, the synchronous speed is 3000 rev min for an excitation frequency of 50 Hz . From s = 0.03 , the slip in the opposite direction of rotation is 0.97 , the motor speed is 2910 rev min = 304.7 rad s . ______________________________________________________________________________________ Problem 18.19 Solution: Known quantities: A stepper motor with a 15, step angle operates in one-phase excitation mode. Find: The amount of time to take for the motor to rotate through 28 rev when the pulse rate is 180 pps . Assumptions: None. Analysis: ω = 180 × 15 = 2700 , / s = 7.5 rev s 28 rev t= = 3.73 s 7.5 rev s ______________________________________________________________________________________ Problem 18.20 Solution: Known quantities: 1 A hp,110V , 60 Hz , four-pole capacitor-start motor has the following parameters: 4 R S = 2.02 Ω X S = 2.8 Ω R R = 4.12 Ω X R = 2.12 Ω X m = 66.8 Ω s = 0.05 Find: a) The stator current. b) The mechanical power. c) The rotor speed. Assumptions: None. 18.14 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Analysis: a) 0.5 Zb = 0.991 + j1.057 = 0.5( Rb + jX b ) 0.5 Z f = 15.93 + j 20.07 = 0.5( R f + jX f ) Z in = 18.94 + j 23.93 = 30.52∠51.64 , ∴ I1 = V1 = 3.60∠ − 51.64 , A Z in b) P f = I12 (0.5 R f ) = 207.0 W Pb = I12 (0.5Rb ) = 12.84 W Pmech = (1 − s )( P f − Pb ) = 184.45 W c) For a 4-pole machine, ω s = 188.5 rad s , ω m = 179.1 rad s Thus, the rotor speed is: 179.1 rad s = 1710 rev min ______________________________________________________________________________________ Problem 18.21 Solution: Known quantities: 1 A hp,110V , 60 Hz , four-pole, single-phase induction motor has the following parameters: 4 R S = 1.86 Ω X S = 2.56 Ω R R = 3.56 Ω X R = 2.56 Ω X m = 53.5 Ω s = 0.05 Find: The mechanical power output. Assumptions: None. Analysis: 0.5 Zb = 0.830 + j1.248 0.5 Z f = 12.41 + j16.98 Z in = 15.1 + j 20.79 = 25.7∠54.0 , I1 = V1 = 4.28∠ − 54.0 , A Z in P f = I12 (0.5R f ) = 227.5W Pb = I12 (0.5Rb ) = 15.2W Pmech = (1 − s )( P f − Pb ) = 201.68 W ______________________________________________________________________________________ 18.15 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.22 Solution: Known quantities: A 115 V , 60 Hz , four-pole, one-phase induction motor has the following parameters: R S = 0.5 Ω X S = 0.4 Ω R R = 0.25 Ω X R = 0.4 Ω X m = 35 Ω Find: The input current and developed torque when the motor speed is 1,730 rev min . Assumptions: None. Analysis: The synchronous speed is 1,800 rev min for 1,730 rev min . (1 − s ) = 0.961 , therefore the slip s = 0.039 . We have: 0.5 Zb = 0.064 + j 0.2 0.5 Z f = 3.034 + j 0.747 Z in = 3.842∠20.52 , I S = 29.9∠ − 20.52 , A P f = 2717.5W Pb = 57.22 W Pmech = 2660.3W ω m = 181.2 rad s The torque developed is: 2551.6 Tdev = = 14.68 N ⋅ m 181.2 ______________________________________________________________________________________ Problem 18.23 Solution: Known quantities: No-load test of a single-phase induction motor at rated voltage and rated frequency. Find: The equivalent circuit of a single-phase induction motor for the no-load test. Assumptions: None. Analysis: At no load, s ≈ 0 . The circuit model is shown below: 18.16 G. Rizzoni, Principles and Applications of Electrical Engineering R1 X1 I1 0.5 X m Problem solutions, Chapter 18 + V1 0.5X'2 R'2 4 0.5 X m - ______________________________________________________________________________________ Problem 18.24 Solution: Known quantities: The locked-rotor test of the single-phase induction motor. Find: The equivalent circuit. Assumptions: Neglect the magnetizing current. Analysis: For locked rotor, ω m = 0, s = 1 . The circuit is shown below: R1 X1 X'2 + V1 R'2 I1 ______________________________________________________________________________________ Problem 18.25 Solution: Known quantities: 1 A hp,115 V two-pole universal motor has the effective resistances of the armature and series field as 8 4 Ω and 6 Ω . The output torque is 0.17N ⋅ m when the motor is drawing rated current of 1.5 A at a power factor of 0.88 at rated speed. Find: a) The full-load efficiency. b) The rated speed. c) The full-load copper losses. d) The combined windage, friction, and iron losses. e) The motor speed when the rms current is 0.5 A . Assumptions: Phase differences and saturation. 18.17 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Analysis: From P f = 0.88 , we have θ = −28.4 , and I S = 1.5∠ − 28.4 , A . The rated speed is: 1 8 × 746 = 548.53 rad s 0.17 a) Pin = 151.8W and the efficiency is: eff = 93.25 = 61.43% 151.8 b) The speed is 5238.1 rev min . c) d) 2 The copper loss is: 1.5 × 10 = 22.5W Other loss will be: 151.8 − 22.5 − 93.25 = 36.05 W e) Pin = (115)(0.5 + (0.5 2 × 10) + Pout )(0.88) = 12.05W 2 Assume T is proportional to I . 0 .5 Tnew = ( ) 2 (0.17) = 0.019 N ⋅ m 1 .5 12.05 = 637.9 rad s ω= 0.019 n = 6091.9 rpm ______________________________________________________________________________________ Problem 18.26 Solution: Known quantities: 240 V , 60 Hz , two-pole universal motor operates at 12,000 rev min on full load and draws a current of 6.5 A at a power factor of 0.94 lagging. The series field-winding impedance is 4.55 + j 3.2 Ω and the armature circuit impedance is 6.15 + j9.4 Ω A Find: a) The back emf of the motor. b) The mechanical power developed by the motor. c) The power output if the rotational loss is 65 W . d) The efficiency of the motor. Assumptions: None. Analysis: From P f = 0.94 (lagging ) , we have θ = −19.95 , and I S = 6.5∠ − 19.95 , A . a) Eb = 146.68 − j 53.25 = 156.05∠ − 19.95, V b) Pdev = I S Eb cos 0 , = 1014.3W c) Pout = 1014.3 − 65 = 949.3W d) 18.18 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Pin = 1466.4 W eff = 64.7% ______________________________________________________________________________________ Problem 18.27 Solution: Known quantities: A single-phase motor is drawing 20 A from a 400 V , 50 Hz supply. The power factor is 0.8 lagging. Find: The value of capacitor connected across the circuit to raise the power factor to unity. Assumptions: None. Analysis: VS = 400∠0 , ,θ = −36.9 , , therefore, I S = 20∠36.9 , = 16 − j12 For a unity power factor, I C = j12 . We have 12 = 400ωC , ω = 314.16 rad s . Therefore, C = 95.5 µF ______________________________________________________________________________________ Problem 18.28 Solution: Known quantities: A way of operating a three-phase induction motor as a single-phase source is shows in Figure P18.28. Find: Whether the motor will work. Explain why or why not. Assumptions: None. Analysis: It will work. The b and c windings will produce a magnetic field similar to a single phase machine, that is, two components rotating in opposite directions and the a winding would act as a starting winding. The phase shift provided by the capacitor is needed to provide a nonzero starting torque. ______________________________________________________________________________________ Problem 18.29 Solution: Known quantities: A 1 hp capacitor-start motor with its output adjusted to rated value. 4 E = 115 volts; I = 3.8 A; P = 310 W ; rev min = 1725. Find: a) Efficiency. b) Power factor. c) Torque in pound-inches. 18.19 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Assumptions: None. Analysis: a) 1 × 746 = 186.5W 4 P eff = out = 0.602 = 60.2% Pin Pout = b) pf = P 310 = = 0.709 lagging VI (115)(3.8) c) T = 7.04 Pout in 186.5 = 7.04 = 0.761lb ⋅ ft × 12 nR ft 1725 = 9.13 lb ⋅ in ______________________________________________________________________________________ 18.20 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Section 18.5: Motor Selection and Applications Problem 18.30 Solution: Known quantities: The tasks are the following: a) Vacuum clearner b) Refrigerator. c) Air conditioner compressor. d) Air conditioner fan. Variable-speed sewing machine. e) f) Clock. g) Electric drill. h) Tape drive. i) X-Y plotter. Find: The type of motor that can perform the above tasks. Assumptions: None. Analysis: The universal motor speed is easily controlled and thus it would be used for variable speed, that is, (e) and (g). The vacuum cleaner motors are often universal motors. This motor could also be used for the fan motors. A single-phase induction motor is used for (b) and (c). The clock should use a single-phase synchronous motor. The tape drive would be a single-phase synchronous motor also. An X-Y plotter uses a stepper motor. ______________________________________________________________________________________ Problem 18.31 Solution: Known quantities: A 5 hp,1150 rev min shunt motor. The speed control by means of a tapped field resistor is shown Figure P18.31. Find: The speed of the motor and the torque available at the maximum permissible load with the tap at position 3. Assumptions: None. Analysis: n = 230% × 1150 = 2645 rpm T = 40% × Trated = (0.4) (33,000)(5) = 9.13 lb ⋅ ft ( 2π )(1150) ______________________________________________________________________________________ 18.21 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.32 Solution: Known quantities: The applications are the following: a) Inexpensive analog electric clock. b) Bathroom ventilator fan. c) Escalator which must start under all load conditions. d) Kitchen blender. Table model circular saw operating at about 3,500 rev min . e) f) Hand-held circular saw operating at 15,000 rev min . g) Water pump. Find: The single-phase motor that can apply to the above cases. Assumptions: None. Analysis: a) reluctance b) shaded-pole c) capacitor-start d) universal e) permanent split capacitor f) universal g) permanent split capacitor ______________________________________________________________________________________ Problem 18.33 Solution: Known quantities: The power required to drive a fan varies as a cube of the speed. The motor driving a shaft-mounted fan is loaded to 100 percent of its horsepower rating on the top speed connection. Find: The horsepower output in percent of rating at the following speed reduction. a) 20 percent. b) 30 percent. c) 50 percent. Assumptions: None. Analysis: a) HP = (0.8) 3 = 0.512 = 51.2% of rated b) HP = (0.7) 3 = 0.343 = 34.3% of rated c) HP = (0.5) 3 = 0.125 = 12.5% of rated ______________________________________________________________________________________ 18.22 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Problem 18.34 Solution: Known quantities: An industrial plant has a load of 800 kW at a power factor of 0.8 lagging. A synchronous motor will be used to change them to 200 kW and 0.92 . Find: The KVA input rating and the power factor at which the motor operates. Assumptions: The motor has an efficiency of 91 percent. Analysis: 200 K = 219.8 KW 0.91 = 800 K + 219.8 K = 1019.8 KW Pm = Pnew Qnew = Pnew tan(cos −1 0.92) = 434.4 KVAR Qold = 800 K tan(cos −1 0.8) = 600 KVAR Qm = Qold − Qnew = 165.6 KVAR S m = Pm2 + Qm2 = 274.8 KVA pf m = cos tan −1 ( Qm ) = 0.8 leading Pm ______________________________________________________________________________________ Problem 18.35 Solution: Known quantities: An electric machine is controlled so that its torque-speed characteristics exhibit a constant-torque region and a constant-power region as shown in Figure P18.35. Average efficiency of the electric drive is 87%. Machine torque is constant at 150 N-m from 0 to 2500 rpm. The constant power region is from 2500 to 6000 rpm. Machine drives a constant torque load requiring 75 N-m. Find: a) Operating speed of the machine b) Electric power needed to operate the machine Assumptions: None Analysis: a) Since the load requires less than the starting torque, the motor should be operating in the constant power regime. The constant power is determined from the starting torque and the base speed: P = ϖ baseTstart = (2500r / min ) (2πrad / rev ) (150 N − m) = 39.3kW (60s / min ) The operating speed is determined from the load torque: 18.23 G. Rizzoni, Principles and Applications of Electrical Engineering ϖ= Problem solutions, Chapter 18 P (39300W ) = = 523.6rad / s TL (75 N − m ) n = (523.6rad / s ) (60s / min ) (2πrad / rev ) = 5000r / min b) The electric power needed to operate the machine: Pe = P 39.3kW = = 45.1kW η 0.87 ______________________________________________________________________________________ Problem 18.36 Solution: Known quantities: The PM synchronous (brushless DC) motor in Figure P18.36 Electrical subsystem parameters: RS, LS, k. (motor constant), VS(t), IS(t). Mechanical subsystem parameters: inertia and damping coefficient, J, b. Thermal subsystem parameters: thermal resistance, specific heat, mass, Rt, c, m. Find: Write the differential equations describing the electrothermomechanical dynamics of the systems Assumptions: All heat is generated in the stator by the stator current (i.e. the heat generated in the rotor is negligible) The rotor and stator are at the same temperature and specific heat c The stator is highly thermally conductive The dominant heat-transfer term is convection Overall thermal resistance Rt from stator to air The motor generates torque according to the equation The back emf is equal to Tm = kI S Eb = kw Analysis: Mechanical subsystem J dω (t ) = Tm − bω (t ) = kI S (t ) − bω (t ) dt Electrical subsystem VS (t ) = LS dI S (t ) dI (t ) + RS I S (t ) + Eb = LS S + RS I S (t ) + kω (t ) dt dt Thermal subsystem 2 RS IS (t) S (t) Rt air mc d S (t) dt ______________________________________________________________________________________ Problem 18.37 Solution: Known quantities: The wound separately excited motor in Figure P18.37 18.24 G. Rizzoni, Principles and Applications of Electrical Engineering Problem solutions, Chapter 18 Electrical subsystem parameters: Rf, Lf, Ra, La, (motor field and armature electrical parameters), kf. ka, kT (motor field and armature constants), VS(t), Vf(t), Ia(t), If(t). Mechanical subsystem parameters: load inertia, damping coefficient, load torque, J, b, TL. Thermal subsystem parameters: Ct-rotor, ht-rotor, Arotor(rotor thermal capacitance, film coefficient of heat transfer from rotor surface to air and from air to stator inner surface, rotor and inner stator surface area (assumed equal). Ct-stator, ht-stator, Astator(stator thermal capacitance, film coefficient of heat transfer from stator outer surface to air, stator outer surface area. Find: Write the differential equations describing the electrothermomechanical dynamics of the systems Assumptions: Heat is generated in the stator and the rotor by the respective currents The stator and rotor are highly thermally conductive The dominant heat-transfer term is convection through the air gap and to ambient Heat storage in the air gap is negligible, and the air gap is infinitely thin The motor generates torque according to the equation Tm = kT I a The back emf is equal to Eb = k a w The stator and rotor each act as a lumped thermal mass Analysis: Mechanical subsystem dω (t ) = Tm − bω (t ) − TL = kT I a (t ) − bω (t ) − TL dt kT = k aφ , φ = k f I f (t ), kT = k a k f I f (t ) J J dω (t ) = k a k f I f (t )I a (t ) − bω (t ) − TL dt Electrical subsystem Armature Circuit: Va (t ) = La dI (t ) dI a (t ) + Ra I a (t ) + Eb = La a + Ra I a (t ) + k aω (t ) dt dt Field Circuit: V f (t ) = L f dI f (t ) dt + R f I f (t ) Thermal subsystem Rotor: Ra I a2 (t ) − hrotor Arotor (θ R (t ) − θ S (t )) = mcrotor dθ R (t ) dt Stator: R f I 2f (t ) − hstator Astator (θ S (t ) − θ air ) + hrotor Arotor (θ R (t ) − θ S (t )) = mcstator dθ S (t ) dt ______________________________________________________________________________________ 18.25 Problem 19.1 Solution: Known quantities: 1 for | t |≤ ητ , τ = 1 sec , η = 30% 300 − 1 for | t |< τ The square wave signal: x (t ) =  Find: a. The Fourier series coefficients for the square wave signal. b. Frequency spectrum of the signal for the numerical values. Analysis: a. Fourier series coefficients: From Fourier series theory we know that the Fourier series representation of any signal is given as: f (t ) = a0 + ∑n =1 a n cos(nω 0 t ) + ∑n =1 bn sin( nω 0 t ) ∞ ∞ a0= 1 x (t )dt T ∫T an = 2 x (t ) cos(nω 0 t )dt T ∫T bn = 2 x (t ) sin(nω 0 t )dt T ∫T where x(t) is the signal, T is the period of the signal, and ω 0 is the natural frequency. The square wave signal is an odd function, so we need to compute only the bn Fourier coefficients. The interval of integration 0 ≤ t ≤ τ would be convenient.  τ τ  2  a = 1  ∫ dt − ∫ dt  = 0 0 τ τ  0  2   τ τ 2   b = 2  ∫ sin n 2π tdt − ∫ sin n 2π tdt  n τ τ τ τ  0 2     − cos n 2π t cos n 2π / 2 τ τ τ τ 2   | | b = + 0 n τ  n 2π t n 2π τ / 2    τ τ b = 1 − 2 cos nπ + 1 cos n 2π = 2 [1 − 2 cos nπ ] n ≠ 0 n nπ nπ nπ nπ  4 n : odd b =  nπ n 0 n : even ∴ x (t ) = 4 sin 2πt + 1 sin 3πt + " π τ τ 3 [ 19.2 ] b. Frequency spectrum of the signal for 19.xx η = 50% is shown in Figure 19.xx and for η = 30% is shown in Figure 1 0.9 0.8 0.7 Magnitude 0.6 0.5 0.4 0.3 0.2 0.1 −5000 −4000 −3000 −2000 −1000 0 Freq (Hz) 1000 2000 3000 4000 5000 −4000 −3000 −2000 −1000 0 Freq (Hz) 1000 2000 3000 4000 5000 1 0.9 0.8 0.7 Magnitude 0.6 0.5 0.4 0.3 0.2 0.1 −5000 19.2 Problem 19.2 Solution: Known quantities: Functional form of a full-wave rectified sinusoidal signal of time period T sec, x (t ) = | sin(ω o t ) | , and natural frequency ω 0 = 200π rad . s Find: a. The Fourier series coefficients. b. Frequency spectrum of the signal. Analysis: The rectified sine wave signal is an even function. Hence, we need to compute only the a n coefficients of the Fourier series. a0 = a0 = 1 π 2π ∫ | sin(ω o t ) | d (ω o t ) = 0 2 π π ∫ sin(ω t )d (ω t ) o o 0 4 π 2 2π 4 π π 4 1 a1 = ∫ | sin(ω o t ) | cos(ω o t )d (ω o t ) = ∫ sin(ω o t ) cos(ω o t )d (ω o t ) = ∫ sin(2ω o t )d (ω o t ) π 0 π 0 π 02 a1 = 0 a2 = 2 π 2π ∫ | sin(ω o t ) | cos(2ω o t )d (ω o t ) = 0 4 π π ∫ sin(ω t ) cos(2ω t )d (ω t ) o o o 0 π 4 1 = ∫ [sin(3ω o t ) − sin(ω o t )]d (ω o t ) π 02 42 π 3 a3 = 0 a2 = 4 2 π 15 2 4  a n =  π ( n − 1)( n + 1) 0  a4 = − n : even n : odd bn = 0  2  x (t ) = 4 1 + cos(2ω 0 t ) − 2 cos(4ω 0 t ) + 2 cos(6ω 0 t ) " π 3 65 15  b. The Frequency spectrum for the full wave rectified sinusoid of frequency Figure 19.xx. 19.2 ω 0 = 200π rad is shown in s 12 10 Magnitude 8 6 4 2 −800 −600 −400 −200 0 Freq (Hz) 200 400 600 800 Problem 19.3 Solution: Known quantities: Functional form of a full-wave rectified cosine wave of time period T sec, x (t ) = | sin(ω o t ) | , and natural frequency ω 0 = 150π rad . s Find: a. The Fourier series coefficients. b. Frequency spectrum of the signal. Analysis: π rad . The Fourier series 2 π 3π coefficients are to be found over a period of π rad. If we consider the period from rad , the analysis is to 2 2 The rectified cosine wave signal is same as the rectified sine wave with a phase shift of same as a full wave rectified sinusoid. Hence, the Fourier series coefficients are the same as computed in Problem 19.2. a0 = 1 π 2π 2 3π 2 ∫ | cos(ω t ) | d (ω t ) = π π∫ sin(ω t )d (ω t ) o o o 0 2 a0 = 19.2 4 π o a1 = 2 π 2π ∫ | cos(ω o t ) | cos(ω o t )d (ω o t ) = 0 4 π 3π 2 ∫ sin(ω o t ) cos(ω o t )d (ω o t ) = π π 4 1 sin(ω o t )d (ω o t ) π ∫0 2 2 a1 = 0 a2 = 2 π 2π 4 3π 2 ∫ | cos(ω t ) | cos(2ω t )d (ω t ) = π π∫ sin(ω t ) cos(2ω t )d (ω t ) o o o o o o 0 2 = 4 π 3π 2 1 ∫ [sin(3ω t ) − sin(ω t )]d (ω t ) π 2 o o o 2 42 π 3 2 4  a n =  π ( n − 1)( n + 1) 0  a2 = n : even n : odd bn = 0  2  x (t ) = 4 1 + cos(2ω 0 t ) − 2 cos(4ω 0 t ) + 2 cos(6ω 0 t ) " π 3 15 65  b. The Frequency spectrum for the full wave rectified cosine wave of frequency ω 0 = 150π Figure 19.xx. 12 10 Magnitude 8 6 4 2 −800 19.2 −600 −400 −200 0 Freq (Hz) 200 400 600 800 rad is shown in s Problem 19.4 Solution: Known quantities: Functional form of a cosine burst as shown in Fig. 19.xx and mathematically defined as: π x (t ) = cos( t ) T Find: Fourier series coefficients for the cosine burst. Analysis: The Fourier series coefficients can be calculated as follows: 1 1 a 0 = ∫ x (t )dt = TT T a0 = π 1 ∫T cos( T t )dt = T − T π 2 T t) sin(  π T  − T 2 2 2 π 2 an = T = T 2 T 2 π ∫ cos( T t ) cos( − T 2 21 T 2 T 2   ( 2n + 1)π T ∫ cos − T 2 n 2π t )dt T   ( 2n − 1)π t  + cos T    t  dt  T 1 = T  T T  ( 2n − 1)π  2  ( 2n + 1)π  t t  sin sin +     ( 2n + 1) T T  − T   ( 2n − 1)   2 π π  1 1 1   sin ( 2n + 1)  + sin ( 2n − 1)   π  ( 2n + 1)  2  ( 2n − 1)  2  2 a n = ( −1) n +1 ( 2n + 1)( 2n − 1)π Since the cosine burst is an even signal the b n coefficients are all zero. Hence b n = 0 . an = The cosine burst signal can be written as: x (t ) = 2 π + 2  2π cos 3π  T 2   4π t − cos  15π  T  t +"  Problem 19.5 Solution: Known quantities: Functional form of a triangular pulse signal as shown in Fig. 19.xx and mathematically defined as: 19.2  | t | x (t ) = A1 −  (u(t + T ) − u(t − T ) ) T  Find: a. Fourier transform of the function. b. Plot the frequency spectrum of the triangular pulse of period, T = 0.01 sec and amplitude, A = 0.5 . Analysis: The mathematical equation for the triangular pulse can be split into a function defined over different periods as follows:   t  A1 + T  for − T ≤ t < 0      t x (t ) =  A1 −  for 0 ≤ t ≤ T   T 0 elsewhere   ∞ The Fourier transform is defined as: X ( f ) = ∫ x(t ) exp( − j 2πft )dt and can be computed for the triangular −∞ pulse as follows: T X( f ) =  | t | ∫ A1 − T  exp( − j 2πft )dt −T T 0 t t   = ∫ A1 +  exp( − j 2πft )dt + ∫ A1 −  exp( − j 2πft )dt T  T −T  0  j exp( jπfT )  exp( − jπfT ) − exp( jπfT )  j exp( − jπfT )  exp( jπfT ) − exp( − jπfT )   = A 2 2   + 2π 2 f 2T   π f T j 2 2 2j       j exp( jπfT ) [− sin(πfT )] + j exp(2− j2πfT ) [sin(πfT )] = A 2 2 2π f T  2π f T  =  − exp( − jπfT ) + exp( jπfT )  A sin(πfT )  2  π f T 2j   = AT sin(πfT ) sin(πfT ) (πfT ) 2 2 X ( f ) = ATsinc 2 ( fT ) b. 19.2 where sinc(fT) = sin(πfT) πfT The Frequency spectrum for the signal as computed from Matlab is shown in Figure 19.xx. −3 5 x 10 4.5 4 3.5 Magnitude 3 2.5 2 1.5 1 0.5 −800 −600 −400 −200 0 Freq (Hz) 200 400 600 800 Problem 19.6 Solution: Known quantities: Functional form of a exponential pulse signal as shown in Fig. 19.xx and mathematically defined as: (exp) − at ,  x (t ) = 0, - (exp) at ,  for t > 0 for t = 0 for t < 0 Find: Fourier transform for the exponential pulse. Analysis: We can formulate a compact notation for the pulse signal by using “signum function” which equals +1 for positive time and –1 for negative time. This function is defined as: 1,  sgn(t ) = 0, - 1,  for t > 0 for t = 0 for t < 0 The signal x (t ) can be written as: x (t ) = exp( − a | t |) sgn(t ) The Fourier transform is now calculated as follows: ∞ X( f ) = ∫ exp( −a | t |) sgn(t ) exp( − j 2πft )dt −∞ 19.2 = 0 ∞ −∞ 0 ∫ − exp((a − j 2πf )t )dt + ∫ exp( −(a + j 2πf )t )dt 1 1 + a − j 2πf a + j 2πf − j 4πf X( f ) = 2 a + 4π 2 f 2 =− b. The Frequency spectrum for the signal is shown in Figure 19.xx. 1 0.9 0.8 0.7 Magnitude 0.6 0.5 0.4 0.3 0.2 0.1 −80 −60 −40 −20 0 Freq (Hz) 20 40 60 80 Problem 19.7 Solution: Known quantities: x (t ) = exp(-at) cos(2πf c t )u(t ) Functional form of a damped sinusoid signal as shown in Fig. 19.xx and mathematically defined as: Find: Fourier transform for the damped sinusoid. 19.2 Analysis: ∞ ∫ x(t ) exp( − j 2πft )dt X( f ) = −∞ ∞ = ∫ exp( −at ) cos(2πf t )u(t ) exp( − j 2πft )dt c −∞ ∞ = ∫ exp( −at ) 0 1 [exp( j 2πf c t ) + exp( − j 2πf c t )]exp( − j 2πft )dt 2 ∞ = X( f ) = b. 1 {exp( −[a + j 2π ( f − f c )]t ) + exp( −[a + j 2π ( f + f c )]t )}dt 2 ∫0  1 1 1 +  2  a + j 2π ( f − f c ) a + j 2π ( f + f c )  The Frequency spectrum from Matlab is shown in Figure 19.xx. 0.5 0.45 0.4 0.35 Magnitude 0.3 0.25 0.2 0.15 0.1 0.05 −200 −150 −100 −50 0 Freq (Hz) 50 100 150 200 Problem 19.8 Solution: Known quantities: Functional form of an ideal sampling function of frequency mathematical equation: δ T0 = 19.2 ∞ ∑ δ (t − mT ) m = −∞ 0 1 T 0 Hz as shown in Figure 19.xx and having Find: a. The Fourier transform for the periodic signal. b. Frequency spectrum of the signal for To = 0.01 sec . Analysis: In a limiting sense, Fourier transforms can be defined for periodic signals. Therefore, it is reasonable to represent that a periodic signal can be represented in terms of a Fourier transform, provided that this transform is permitted to include delta functions. An ideal sampling function consists of an infinite sequence of uniformly spaced delta functions. We observe that the generating function for the ideal sampling function is simply a delta function δ (t ) . The periodic signal can be represented in terms of the complex exponential Fourier series: δ ∞ = T 0 ∑c n exp( j 2πnf 0 t ) m = −∞ where cn is the complex Fourier series coefficients defined as: 1 cn = T0 ∞ ∫ δ (t ) exp( − j 2πnf t )dt 0 −∞ = f 0G ( nf 0 ) where G ( nf 0 ) is the Fourier transform of δ (t ) evaluated at the frequency nf 0 . For the delta function: G ( nf 0 ) = 1 for all n Therefore, using the relation for Fourier transform pair for a periodic signal gT0 (t ) with a generating function g (t ) and period T0 : ∞ ∞ ∑ g (t − mT ) ⇔ f ∑ G(nf 0 m = −∞ 0 n = −∞ 0 )δ ( f − nf 0 ) We get the Fourier transform pair for the ideal sampling function as: ∞ ∑ δ (t − mT0 ) ⇔ f 0 m = −∞ ∞ ∑ δ ( f − nf n = −∞ 0 ) We can see that the Fourier transform of a periodic train of delta functions, spaced T0 seconds apart, consists of another set of delta functions weighted by a factor f 0 = 1 and regularly spaced f 0 Hz apart along the frequency T0 axis. Problem 19.9 Solution: Known quantities: Functional form of the modulating signal m(t ) , the carrier signal c(t ) , and the modulation index c(t ) = Ac cos(2πf c t ) m(t ) = Am cos(2πf m t ) Find: The average power delivered to a 1-ohm resistor. Analysis: The AM signal is given by: s (t ) = Ac [1 + µ cos(2πf m t )] cos(2πf c t ) 19.2 µ. Expressing the product of two cosines as the sum of sinusoidal waves, we get: s (t ) = Ac cos(2πf c t ) + 1 1 µAc cos[2π ( f c + f m )t ] + µAc cos[2π ( f c − f m )t ] 2 2 The Fourier transform of s (t ) is therefore: 1 Ac [δ ( f − f c ) + δ ( f + f c )] 2 1 + µAc [δ ( f − f c − f m ) + δ ( f + f c + f m )] 4 1 + µAc [δ ( f − f c + f m ) + δ ( f + f c − f m )] 4 S( f ) = Thus the spectrum of an AM wave, for sinusoidal modulation, consists of delta functions at ± f c , f c ± f m ,− f c ± f m as seen from its Fourier transform. In practice, the AM wave s (t ) is a voltage or current wave. In either case, the average power delivered to a 1-ohm resistor by s (t ) is comprised of three components: Average power = 1 x 2 (t )dt T ∫T Using Parsevals energy relation we can find average power in frequency domain as: Average power = |X(f)|2 at f = 0 Hence the carrier frequency, upper side-frequency and lower side-frequency power is: 1 Carrier Power = Ac [δ ( − f c ) + δ ( f c )] 2 1 2 = Ac 2 2 1 Ac [δ ( − f c − f m ) + δ ( f c + f m )] 4 1 2 = µ 2 Ac 8 2 Upper side - frequency power = 2 1 Lower side - frequency power = Ac [δ ( − f c + f m ) + δ ( f c − f m )] 4 1 2 = µ 2 Ac 8 For a load resistor R different from 1-ohm, which is usually the case in practice, the expression for carrier power, upper side-frequency power, and lower side-frequency power are merely scaled by the factor 1 or R , depending R on whether the modulated wave s (t ) is a voltage or current, respectively. 19.2 Problem 19.10 Solution: Known quantities: Carrier signal frequency, f c = 0.82 MHz , upper side-band frequency components at frequencies f s1 = 0.825 MHz, f s 2 = 0.83 MHz, f s 3 = 0.84 MHz , their amplitudes and the modulation index µ = 1 . Find: a. Modulating signal equation. b. Plot spectrum of the modulating signal. c. Plot the spectrum of the AM signal including the lower side-band. Analysis: a. We know from the theory for AM that the upper side-band frequency has frequency components at frequencies: f sn = f c + f mn where n is the number of frequency components in the modulating signal and, their amplitudes in the AM signal are 1 times the original amplitude of the modulating signal for a 2 modulation index µ = 1 . Hence, we can find the modulating signal components to be: m1 (t ) = 0.8 sin(2π 5000t ) m2 (t ) = 0.4 sin(2π 10000t ) m3 (t ) = 0.5 sin(2π 20000t ) Hence the modulating signal is: m(t ) = 0.8 sin(2π 5000t ) + 0.4 sin(2π 10000t ) + 0.5 sin(2π 20000t ) b. c. The spectrum for the modulating signal is shown in Figure 19.xx The spectrum for the AM modulated signal with the lower side-band is shown in Figure 19.xx Problem 19.11 Known quantities: AM frequency spectrum from 525 kHz to 1.7 MHz , bandwidth for each channel is 10 kHz Find: a. Number of channels that can be transmitted in the given frequency range b. The maximum modulating frequency that can be transmitted without overlap. Analysis: Assume: No guardband between channels. a. The frequency range allocated for AM broadcast is f R = 1700 − 525 = 1175 kHz This range is partitioned to allow 10 kHz of separation between each channel; therefore, the total number of channels, N is 1175 ≈ 118 channels N= 10 b. The carriers of two separate channels are separated by 10 kHz . If we let the maximum frequency of the message signal increase, the outer edges of both sidebands move away from the carrier frequency and into each other, thereby increasing the bandwidth of each AM channel. The maximum allowable message frequency will occur at the midpoint of the spacing between the carriers. Hence, the maximum message frequency is half the frequency spacing between the carriers. f m (max) = 5 kHz 19.2 19.2

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