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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.002 Circuits and Systems Midterm Review by Jason Kim 1. Primitive Elements Resistor: time domain: V = IR freq. domain: Z = R • Resistor Network N-unknowns & N-equations (most primitive approach) Node Analysis (KCL, KVL) *KCL: Sum of all currents entering and leaving a node is zero. *KVL: Sum of all voltages around a closed loop path is zero. (be careful with polarities) 1) Label(!) all current directions and voltage polarities. (Remember, currents flow from + to -) 2) write KCL & KVL Eq. 3) subst. and solve. Simplification (by inspection) • Thevenin Equivalent Circuit Model Norton Equivalent Circuit Model Rth Vth + Voc=Vth - + - same i-v characteristics at the ports IN RN + Isc=IN - * Three variables: Vth=Voc, Rth=RN, and IN=Isc . They are related by Voc=Isc*Rth . * Vth=Voc: Leave the port open(thus no current flow at the port) and solve for Voc. For a resistive network, this gives you a point on the V-axis of the i-v plot. * IN=Isc: Short the port(thus no voltage across the port) and solve for Isc flowing out of + and into -. For a resistive network, this gives you a point on the I-axis of the i-v plot. * Rth: Set all sources to zero, except dependent sources. Solve the resistive network. When setting sources to zero, V source becomes short and I source becomes open. Rth may also be found by attaching Itest and Vtest and setting Rth = (Vtest)/(Itest). If dependent sources are present, set only the independent sources to zero and attach Itest and Vtest. Use KCL and KVL to find the expression (Vtest)/(Itest)=Rth. Itest Original Circuit N + - Vtest note the direction of Itest and polarity of Vtest. V test R th = ----------I test * For both Thevenin & Norton E.C.M.’s, they have the same power consumption at the port as the original circuit, but not for its individual components. Power dissipated at Rth does not equal power dissipated at the resistors of the original circuit. Same holds for the power delivered by the sources in the Thevenin model and the original circuit. Thevenin and Norton E.C.M.’s are for terminal i-v characteristics only. Page 1 • Power = Real Power = IV = I2R = V2/R (energy dissipated as heat) T 1 • ⟨ Power⟩ = --- ∫ I ( t )V ( t ) dt T 0 • Superposition "In a linear network with a number of independent sources, the response can be found by summing the response to each independent sources acting alone, with all other independent sources set to zero." VR when only V1 is on. + Linear Network with n sources VR = VR VR R _ 1 V2 ∼ n = 0 + VR 2 V 1 ,V 3 ∼ n = 0 … + VR n V1 ∼ n – 1 = 0 1) Leave one source on and turn off all other sources. (Voltage source "off" = short & Current source "off" = open) 2) Find the effect from the "on" source. 3) Repeat for each sources. 4) Sum the effect from each sources to obtain the total effect. For cases where a dependent source is present along with multiple sources, DO NOT turn off the dependent source. Leave the dependent source on and carry it in your expressions. Tackle the dependent source term last by solving linear equations. Remember that the variable which the dependent source is depended on is affected by the individual independent sources that you are turning on and off. • Multiport Network i1 V1 + - + + V - 2 M M R11 i1 i2 R12i2 V1 R22 + - + - i2 + R21i1 V2 - - * By attaching V1, I1 and V2,I2 at the ports, V 1 = R 11 I 1 + R 12 I 2 V 2 = R 21 I 1 + R 22 I 2 The first subscript denote the place(port) of voltage measurement. The second subscript denote the place(port) of current source. Thus, R12 means the resistance V1/I2 when voltage V1 is measured at open port 1 and I2 current source is placed at port 2. * If all the Z-parameters(Rxx’s) are identical, then the networks can’t be differentiated using only the i-v characteristics at the ports. * Reciprocity: R12=R21. * Using the definitions from above, R11-R12 i1 M R22-R21 + V1 i2 + + V2 R12=R21 - - M i1 G11+G12 V1 -G12=-G21 - Page 2 + G22+G21 V - Pi model T model i2 time domain: I = C dV Capacitor: dt 1 freq. domain: Z = ---------jwC High Frequency = Short & Low Frequency = Open C C C1 + C2 1 2 Series: ------------------ Parallel: C 1 + C 2 Vc(0-)=Vc(0+) except when the input source is an impulse. Vc(t) is continuous while Ic(t) may be discontinuous. "I C E" : Current(I) LEADS Voltage(EMF) by 90o. Energy Conservation: ∫ I dt = Q = CV . 1 2 Energy Stored: E = --- CV . (no energy dissipation) 2 R Note: Two identical capacitors, C1 and C2, in series may act like an open circuit after a long time where the total voltage across the capacitors are split between the two cap’s. For this case, Vc does not discharge to zero but VC1=VC2. Inductor: time domain: V = L dI dt + VC1 C2 + C1 _ VC2 _ VC1(0-) = VC2(0-) freq. domain: Z = jwL High Frequency = Open & Low Frequency = Short Series: L 1 + L 2 L L L1 + L2 1 2 Parallel: ----------------- IL(0-)=IL(0+) except when the input source is an impulse. IL(t) is continuous while VL(t) may be discontinuous. "E L I" : Voltage(EMF) LEADS Current(I) by 90o. Energy Conservation: ∫ V dt = λ = LI . 1 2 Energy Stored: E = --- LI . (no energy dissipation) 2 Note: Two identical inductors, L1 and L2, in parallel may act like a short circuit after a long time and have current flow through this loop indefinitely. For this case, iL is not zero but iL1=-iL2. 2. L1 i1 L2 i2 iL1(0-) = iL2(0-) First Order Circuits RC Network RL Network R Vi + - iL R C + Vc=Vo - Vi dV c V c V + -------- = -------idt RC RC + - L di L i L Vi + --------- = ----d t L L -- R Page 3 + Vo - R L Time Constant: τ = --- Time Constant: τ = RC R Homogeneous: V_i R Vi Vi iC VL t t t t iL VC V - _i R Particular: Vi VC V_i R iL Vi iC V_i R VL t t t t Total: V_i R Vi VC iL Vi iC VL t t t * Notice the Duality between Vc and IL & Ic and VL. Approach to First Order Circuits 1) Set up differential equations, using KCL and KVL. Use the constitutive laws for C and L. dV c V c V + -------- = -------idt RC RC 2) Find Homogeneous solution by setting input to zero and substituting Vh=Aest as a solution. 1 st st Ase + -------- Ae = 0 RC 1 ∴s = – -------RC 3) Find Particular solution. Remember the output follows the form of the input. V p = Vi ;t > 0 4) Combine Homogeneous and Particular solutions, and use initial conditions to find missing variables. (Be careful when input is an impulse!) V o = V h + V p = Ae –t -------RC + Vi V o = A + V i = 0 at t=0+. ∴ A = – V i –t Vo V_i R -------- RC = V i1 – e at t=0+, Vo=0 and t= ∞ , Vo=Vi. *After a long time, transients(homogeneous solution) die away, leaving only the particular solution. Thus, the output may look different from the input in the beginning but slowly catches up to follow the input. Page 4 t Impulse as Input ex. RC circuit: V i = Qδt 0 1 Q dV o + -------- V o dt = -------- δt dt ➔ RC RC 3. plus Q ) – 0 = -------- ➔ RC minus plus 0 1 dV o + -------RC ∫ 0 V o(0 V o(0 and minus ∴V o ( 0 plus ) = 0 plus ∫ 0 V dV o V o + -------- = -------i- . RC RC dt and 0 Q V o dt = -------RC minus plus ∫ 0 δt dt ➔ V o ( 0 plus ) – V o(0 minus Q ) = -------RC minus Q ) = -------RC Digital Abstraction Boolean Logic *become comfortable expressing in different forms: truth table ↔ formula ↔ gates ↔ transistor level. A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 OUT 1 0 1 0 1 0 0 0 ↔ OUT = ( A ⋅ B ) + C A B ↔ ↔ OUT C OUT A C B *Note that in the transistor-level implementation, transistors in series form an AND expression and those in parallel form an OR expression. The final output is always inverted, equivalent to adding a NOT(inverter) to the output. Primitive Rules A + (B + C) = (A + B) + C A+B=B+A A + (B ⋅ C) = (A + B) ⋅ (A + C) A+1=1 A+0=A A+A=1 A+A+A A + (A ⋅ B) = A A + (A ⋅ B) = A + B A A A A A A A A A ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ (B ⋅ C) = (A ⋅ B) ⋅ C (associative) B=B ⋅ A (communative) (B + C) = (A ⋅ B) + (A ⋅ C) (distributive) 0 =0 1=A A=0 (complement) A=A (idempotence) (A + B) = A (absorption) (A + B) = A ⋅ B (absorption) DeMorgan’s Laws 1) A ⋅ B = A + B Gate Equivalent: 2) A + B = A ⋅ B Gate Equivalent: A B A B C = A C = A B C B Minimum Sum of Products(MSP): (1)first level AND’s, (2) combine with OR’s. Minimum Product of Sums(MPS): (1)first level OR’s, (2) combine with AND’s. Page 5 C ex) (A ⋅ B) + (B ⋅ C) ex) (A+B) ⋅ (B+C) 4. MOSFET Transistor Large Signal Model S Model: SR Model: D D Off G D On G Ron S VGS ≥ VT VGS < VT When (VDS ≥ VGS - VT), When (VDS < VGS - VT), SCS Model (Saturation): SVR Model (Linear): D D iDS= Off On G S VGS ≥ VT VGS < VT G Off G S S D K(VGS-VT)2 2 On G D D Off G On G Ron= S S S VGS ≥ VT iDS S VGS ≥ VT VGS < VT VDS = VGS - VT VGS < VT Load Line iDS VS RL VGS5 Operating Point -1 RL VGS5 Lin ear Reg ion Saturation Region 1 K(VGS-VT) VDS < VGS - VT VGS4 VGS4 VGS3 VGS2 VGS1 VGS3 VGS2 VGS1 VO VDS VDS > VGS - VT MOSFET characteristics 5. VDS VS Load Line superimposed on MOS characteristics for operating point analysis of MOSFET Amplifier MOSFET Amplifier VS VS vO RL VI + _ VI+vi + _ vO vO vO iD= Linear for small vi id=K(vI-vT)vi K(VI-vT)2 2 vO (VI,VO) RL vi + _ vi MOSFET Amplifier Small Signal gain: Large Signal Model(SCS) Small Signal Model vo ----- = – K ( V I – v T )R L = – g m R L vi vI Operating Point@VI *Note that the output is a function of both VI(DC) and vi(AC). Page 6 Parameter 6. Value (for small signal model) Parameter Value (for small signal model) Input Resistance ∞ Current Gain Ri K ( V I – v T ) ( R L || R O ) -------- ; = ∞ if Ri= ∞ RO Output Resistance RL Power Gain 2 Ri [ K ( V I – v T ) ( R L || R O ) ] -------- ; = ∞ if Ri= ∞ RO Noise Margins (1) for large noise margins VO need high gain Noise Margin "Low" VOH (2) VOL Valid Low VIH Valid High Forbidden Zone VOL VIL VIL Noise Margin "High" VIH VOH Noise Margins VI Inverter characteristic curve To improve noise margins: (1) increase VIL and lower VIH, and/or (2) increase VOH and lower VOL. (1) : increase vT of the MOSFET. This would make the transistor to turn on at a higher threshold voltage, increasing VIL and lowering VIH. (2): R R on + R L L W on - and R on ∝ ----- . increase load resistance RL or (W/L) ratio of the MOSFET since V O ∝ --------------------- References Agarwal, Anant and Jeff Lang. 6.002 Notes (1998) Leeb, Steve. 6.002 Recitation Notes (fall ’97) Schlecht, Martin. 6.002 Lecture Notes (fall ’97) Searle, Campbell. 6.002 Notes (1991) Good Luck! 10. 27. 1998. JK. Page 7