ENGN3227 Analogue Electronics Problem Sets V1.0 Dr. Salman Durrani November 2006 c 2006 by Salman Durrani. Copyright Problem Set List 1. Op-amp Circuits 2. Differential Amplifiers 3. Comparator Circuits 4. Digital to Analog Converters 5. Op-amp Frequency Response 6. Active Filter Circuits 7. Filter Design 8. 555 Timer Circuits 9. Oscillator Circuits 10. Instrumentation Amplifier 11. Integrator and Differentiator 12. Matlab Circuits 3 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #01 Operational Amplifier Circuits Q1 Consider the circuit shown in Figure 1. Assume Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 10 kΩ, R2 = 220 kΩ, RL = 10 kΩ. (a) Derive an expression for closed-loop gain Acl . Express Acl in standard form. What is Acl for the special case when Aol → ∞. (b) Find the optimum value of Rc . (c) Find the values of closed loop gain, closed loop input resistance and closed loop output resistance. (d) Examine the stability of Acl if Aol increases from 2 × 105 to 5 × 105 . (e) Find the maximum frequency of a 0.1V peak sine-wave input that can be amplified without distortion. Rc vin vout R1 RL R2 Figure 1: The circuit for Question 1. Q2 Repeat Q1 (c),(e) for a buffer amplifier. Q3 Consider the circuit shown in Figure 2. Assume Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 1 kΩ, R2 = 100 kΩ, RL = 10 kΩ. (a) Derive an expression for closed-loop gain Acl . Express Acl in standard form. What is Acl for the special case when Aol → ∞. (b) Find the optimum value of Rc . (c) Find the values of closed loop gain, closed loop input resistance and closed loop output resistance. (d) Examine the stability of Acl if Aol increases from 2 × 105 to 5 × 105 . (e) Find the maximum frequency of a 0.1V peak sine-wave input that can be amplified without distortion. R2 vin R1 vout Rc RL Figure 2: The circuit for Question 3. Problem Set #01 page 1 ANU ENGN 3227 Q4 Consider the difference amplifier circuit shown in Figure 3. (a) Derive an expression for output voltage vout . What is vout for the special case when Aol → ∞. (b) Design the difference amplifier circuit to produce output vout = 0.5(v1 − v2 ). R2 v1 R1 v2 R1 vout R2 RL Figure 3: The circuit for Question 4. Problem Set #01 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #01 Solution Q1 Complete Solution The given circuit is Rc vin vout R1 R2 RL The given data is Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 1 kΩ, R2 = 100 kΩ, RL = 10 kΩ. (a) Acl Acl Acl Aol (R1 + R2 ) R1 + R2 + Aol R1 R2 (when Aol → ∞) = 1+ R1 Aol R1 = (where feedback ratio B = ) 1 + Aol B R1 + R2 = See Lecture 04 for complete derivation steps. (b) The optimum value of bias-current compensating resistor is Rc = R1 ||R2 = 220k||10k = 9.56 kΩ (c) The closed loop gain is given by B = = Acl = = R1 R1 + R2 10k 1 = 10k + 220k 23 Aol 1 + Aol B 2 × 105 5 = 22.997 1 + 2×10 23 Problem Set #01 page 3 ANU ENGN 3227 The closed loop input and output resistances are given by Rin(cl) = Rin (1 + BAol ) = 2 × 106 (1 + Ro(cl) = = 2 × 105 ) = 17.393 GΩ 23 Ro (1 + BAol ) 75 5 = 8.624 mΩ 1 + 2×10 23 (d) For A1(ol) = 2 × 105 , A1(cl) = 22.997. For A2(ol) = 5 × 105 , A1(cl) = 5×105 5 1+ 5×10 23 = 22.998. % change in Aol is A2(ol) − A1(ol) × 100 ∆Aol = A1(ol) 5 × 105 − 2 × 105 = × 100 = 150% 2 × 105 % change in Acl is A2(cl) − A1(cl) ∆Acl = × 100 A1(cl) 22.998 − 22.997 = × 100 = 0.00434% 22.997 We can see that a 150% increase in Aol results in only 0.00434% increase in Acl , i.e. Acl is a stable parameter. (e) From bandwidth limitation, the upper frequency limit is fc(cl) = = fT Acl 1 × 106 = 43.48 kHz 22.997 From slew rate limitation, the upper frequency limit is f = SR 2πVo(p) = 0.5 × 106 = 34.59 kHz 2π(22.997)(0.1) We see that in this case, the slew rate sets the upper limit. Hence the maximum frequency of 0.1V peak sine wave that can be amplified without distortion is fmax = 34.59 kHz. Problem Set #01 page 4 ANU ENGN 3227 Q2 Complete Solution The given circuit is vin vout The given data is Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, RL = 10 kΩ. (c) For buffer amplifier, the closed loop gain is given by B = 1 Acl Aol 1 + Aol B 2 × 105 = 0.999995 ≈ 1 1 + (2 × 105 )(1) = = The closed loop input and output resistances are given by = Rin (1 + BAol ) Rin(cl) = 26 [1 + (2 × 105 )(1)] = 400 GΩ Ro = (1 + BAol ) 75 = = 0.375 mΩ [1 + (2 × 105 )(1)] Ro(cl) Note: Compare these values with the answers in Q1(c) and Q3(c). We can see that buffer amplifier has highest input resistance and lowest output resistance of the three op-amp configurations. (e) For buffer amplifier, from bandwidth limitation, the upper frequency limit is fc(cl) = = fT Acl 1 × 106 = 1 MHz 1 From slew rate limitation, the upper frequency limit is fSR = SR 2πVo(p) = 0.5 × 106 = 34.59 kHz 2π(22.997)(0.1) We see that in this case, the slew rate sets the upper limit. Hence the maximum frequency of 0.1V peak sine wave that can be passed without distortion is fmax = 34.59 kHz. Problem Set #01 page 5 ANU ENGN 3227 Q3 Complete Solution The given circuit is R2 vin R1 vout Rc RL The given data is Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 1 kΩ, R2 = 100 kΩ, RL = 10 kΩ. (a) Acl Acl Acl Aol R2 R1 + R2 + Aol R1 R2 = − (when Aol → ∞) R1 R1 Aol (where feedback ratio B = ) = − 1 + Aol B R2 = − See Lecture 04 for complete derivation steps. (b) The optimum value of bias-current compensating resistor is Rc = R1 ||R2 = 1k||100k = 990.01 Ω (c) The closed loop gain is given by R1 B = R2 1k 1 = = 100k 100 Aol Acl = − 1 + Aol B 2 × 105 = − 5 = −99.95 1 + 2×10 100 The closed loop input and output resistances are given by Rin(cl) Ro(cl) = R1 = 1 kΩ Ro = (1 + BAol ) 75 = 5 = 37.48 mΩ 1 + 2×10 100 Problem Set #01 page 6 ANU ENGN 3227 (d) For A1(ol) = 2 × 105 , A1(cl) = −99.95. For A2(ol) = 5 × 105 , A1(cl) = − 5×105 5 1+ 5×10 100 = −99.98. % change in Aol is A2(ol) − A1(ol) × 100 ∆Aol = A1(ol) 5 × 105 − 2 × 105 = × 100 = 150% 2 × 105 % change in Acl is A2(cl) − A1(cl) ∆Acl = × 100 A1(cl) −99.98 + 99.95 = × 100 = 0.03% −99.95 We can see that a 150% increase in Aol results in only 0.03% increase in Acl , i.e. Acl is a stable parameter. (e) From bandwidth limitation, the upper frequency limit is fc(cl) = fT |Acl | = 1 × 106 = 10.005 kHz 99.95 From slew rate limitation, the upper frequency limit is f = SR 2πVo(p) = 0.5 × 106 = 7.96 kHz 2π(| − 99.95|)(0.1) We see that in this case, the slew rate sets the upper limit. Hence the maximum frequency of 0.1V peak sine wave that can be amplified without distortion is fmax = 7.96 kHz. Note: Compare these values with the answers in Q1(e). We see that higher the closed loop gain, more strict the frequency limitations. Problem Set #01 page 7 ANU ENGN 3227 Q4 Partial Solution The given circuit is R2 v1 R1 v2 R1 vout R2 RL (a) v+ v− vin vout R2 v2 R1 + R2 R2 R1 = v1 + vout R1 + R2 R1 + R2 = (v+ ) − (v− ) = Aol vin = The output voltage is vout = vout = Aol R2 (v2 − v1 ) R1 + R2 + Aol R1 R2 (v2 − v1 ) (when Aol → ∞) R1 (b) vout = 0.5(v2 − v1 ). RL = 10 kΩ (assume) R1 = 10 kΩ (assume). R2 = 5 kΩ (calculate from output voltage formula). Problem Set #01 page 8 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #02 Differential Amplifiers Q1 Consider the circuit shown in Figure 1. Assume RC = 2.2 kΩ, RE = 4.7 kΩ, +VCC = 10 V, −VEE = −10 V, β = 100, VBE = 0.7 V, VT = 26 mV. (a) Derive an expression for differential voltage gain Ad of the amplifier. (b) Determine the Q-point. (c) What is the maximum peak to peak output voltage without clipping. (d) Find the values of differential voltage gain, input resistance and output resistance. (e) Determine the output voltage vout (t) if vin1 = 5 × 10−3 sin(2π1000t) and vin2 = 2 × 10−3 sin(2π1000t). +VCC RC RC - + vo1 vo2 vin1 vin2 RE -VEE Figure 1: The circuit for Question 1. Q2 Consider the circuit shown in Figure 1. Assume RC = 1.5 kΩ, RE = 4.7 kΩ, +VCC = 15 V, −VEE = −15 V, β = 100, VBE = 0.7 V, VT = 26 mV. (a) Derive an expression for differential voltage gain Ad of the amplifier. (b) Determine the Q-point. (c) What is the maximum peak to peak output voltage without clipping. (d) Find the values of differential voltage gain, input resistance and output resistance. (e) Determine the output voltage vout (t) if vin1 = 5 × 10−3 sin(2π1000t) and vin2 = −5 × 10−3 sin(2π1000t). Problem Set #02 page 1 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #02 Solution Q1 Complete Solution The given circuit is +VCC RC RC - + vo1 vo2 vin1 vin2 RE -VEE The given data is Assume RC = 2.2 kΩ, RE = 4.7 kΩ, +VCC = 10 V, −VEE = −10 V, β = 100, VBE = 0.7 V, VT = 26 mV. (a) Ad RC re = See Lecture 05 for complete derivation steps. (b) The find the Q-point, we have to determine values of ICQ and VCEQ . ICQ = VCEQ VEE −VBE 2RE 10 − 0.7 2(4.7k) 0.989 mA VCC +VBE − RC ICQ 10 + 0.7 − (2.2k)(0.989m) 8.52 V = IE = = = = = Hence the Q-point is (0.989 mA, 8.52 V). Problem Set #02 page 2 ANU ENGN 3227 (c) Each BJT can swing to a maximum collector voltage of +VCC at cutoff and a minimum voltage of approximately 0 V (dc level of the bases) at saturation. The output voltage is the difference of the two collector voltages and the collector voltages moves in the opposite direction by an equal amount. The voltage drop across each collector resistor is VRC = RC IC = (2.2k)(0.989m) = 2.17 V < VCE This means that the maximum change in voltage across each collector resistor is ±(2.17) V or 4.35 V peak to peak. The maximum peak-to-peak output voltage without clipping is (2)(4.35) = 8.7 V. See PSPICE: P02_Q01.sch. (d) The differential voltage gain is re = = = Ad = = = VT ICQ 26m 0.989m 26.29 Ω RC re 2.2k 26.29 83.68 The input and output resistances are Rin1 = Rin2 Ro1 = Ro2 = = = = = 2(β + 1)re 2(101)(26.29) 5.31 kΩ RC 2.2 kΩ (e) The output voltage is vin1 (t) = 5 × 10−3 sin(2π1000t) vin2 (t) = 2 × 10−3 sin(2π1000t) vout (t) = Ad (vin1 − vin2 ) = 83.68(3 × 10−3 sin(2π1000t)) = 0.251 sin(2π1000t) Note: (i) The output is in phase with the differential input voltage. (ii) no clipping will take place. Problem Set #02 page 3 ANU ENGN 3227 Q2 Partial Solution The given circuit is +VCC RC RC - + vo1 vo2 vin1 vin2 RE -VEE The given data is Assume RC = 1.5 kΩ, RE = 4.7 kΩ, +VCC = 15 V, −VEE = −15 V, β = 100, VBE = 0.7 V, VT = 26 mV. (a) See Lecture 05 for complete derivation steps. (b) ICQ VCEQ = 1.52 mA = 13.42 V (c) Each BJT can swing to a maximum collector voltage of +VCC at cutoff and a minimum voltage of approximately 0 V (dc level of the bases) at saturation. The output voltage is the difference of the two collector voltages and the collector voltages moves in the opposite direction by an equal amount. The voltage drop across each collector resistor is VRC = RC IC = (1.5k)(1.52m) = 2.28 V < VCE This means that the maximum change in voltage across each collector resistor is ±(2.28) V or 4.56 V peak to peak. The maximum peak-to-peak output voltage without clipping is (2)(4.56) = 9.12 V. Verify using PSPICE. Problem Set #02 page 4 ANU ENGN 3227 (d) re Ad Rin1 = Rin2 Ro1 = Ro2 = = = = 17.1 Ω 87.72 3.45 kΩ 1.5 kΩ (e) vout (t) = 0.877 sin(2π1000t) Problem Set #02 page 5 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #03 Comparators Q1 Consider the circuit shown in Figure 1. Assume R1 = 50 kΩ, R2 = 50 kΩ, +VCC = 5 V, −VEE = −5 V, |Vout(max) | = power supply voltage. (a) Derive expressions for upper and lower trigger points of the comparator. (b) Sketch the output if vin = 5 sin(2π1000t). vin vout R2 R1 Figure 1: The circuit for Question 2. Q2 Consider the circuit shown in Figure 2. Assume R1 = 50 kΩ, R2 = 50 kΩ, R3 = 100 kΩ, +VCC = 10 V, −VEE = −10 V, Vre f = 5 V, |Vout(max) | = power supply voltage. (a) Derive expressions for upper and lower trigger points of the comparator. (b) Sketch the output if vin = 5 sin(2π1000t). vin vout Vref R2 R1 R3 Figure 2: The circuit for Question 3. Problem Set #03 page 1 ANU ENGN 3227 Q3 Consider the circuit shown in Figure 3. Assume R1 = 50 kΩ, R2 = 50 kΩ, +VCC = 15 V, −VEE = −15 V, VZ = 4.7 V, VD = 0.7 V. (a) Find the bounded maximum output voltages. (b) Find values of upper and lower trigger points of the comparator. (c) Sketch the output if vin = 6 sin(2π1000t). Z1 Z2 Rc vin vout R1 R2 Figure 3: The circuit for Question 3. Q4 Consider the circuit shown in Figure 3. Assume R1 = 10 kΩ, R2 = 47 kΩ, +VCC = 15 V, −VEE = −15 V, VZ = 4.7 V, VD = 0.7 V. (a) Find the bounded maximum output voltages. (b) Find values of upper and lower trigger points of the comparator. (c) Sketch the output if vin = 3 sin(2π1000t). Problem Set #03 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #03 Solution Q1 Partial Solution The given circuit data is R1 = 50 kΩ, R2 = 50 kΩ, +VCC = 5 V, −VEE = −5 V, |Vout(max) | = power supply voltage. vin vout R1 R2 (a) VUT P VLT P R1 V R1 + R2 out(max) R1 = − V R1 + R2 out(max) = See Lecture 06 for complete derivation steps. (b) VUT P VLT P = 2.5 V = −2.5 V Problem Set #03 page 3 ANU ENGN 3227 The sketch of output voltage is shown below: 6 Input Output 4 Voltage (V) 2 0 −2 −4 −6 0 0.5 1 Time (s) 1.5 2 −3 x 10 Figure 4: Comparator output voltage vout (t). Note: the trigger points where output changes state are highlighted by a black dot. SEE MATLAB: L06_Example03.m and PSPICE:L06_Example03.sch. Problem Set #03 page 4 ANU ENGN 3227 Q2 Solution The given circuit data is R1 = 50 kΩ, R2 = 50 kΩ, R3 = 100 kΩ, +VCC = 10 V, −VEE = −10 V, Vre f = 5 V, |Vout(max) | = power supply voltage. vin Vref vout R2 R1 R3 (a) VUT P = VLT P = R R Vre f + Vout(max) R2 R3 R R Vre f − Vout(max) R2 R3 where R = R1 ||R2 ||R3 . (b) R = 20 kΩ VUT P = 4 V VLT P = 0 V The sketch of output voltage is shown below: 10 Input Output Voltage (V) 5 0 −5 −10 0 0.5 1 Time (s) 1.5 2 −3 x 10 Figure 5: Comparator output voltage vout (t). Verify using PSPICE P03_Q02.sch. Problem Set #03 page 5 ANU ENGN 3227 Q3 Partial Solution The given circuit data is R1 = 50 kΩ, R2 = 50 kΩ, +VCC = 15 V, −VEE = −15 V, VZ = 4.3 V, VD = 0.7 V. Z1 Z2 Rc vin vout R1 R2 (a) The circuit is a double-bounded Schmitt Trigger. One zener is always forward biased when the other one is in breakdown. We have Vout v− ∴ Vout = v− ± 5 (VZ +VD = 4.3 + 0.7 = 5) = v+ (summing point constraint due to feedback arrangement) = v+ ± 5 (1) Apply KCL to non-inverting pin, v+ −Vout v+ − 0 + R2 R1 = 0 Vout v+ = v+ = Vout 2 1 + RR21 (2) Substituting in (1) Vout 1 1− Vout 2 Vout = Vout ±5 2 = ±5 = ±5 = ±10 V 0.5 Hence Vout(max) = ±10 V. (b) The trigger points are VUT P VLT P R1 V =5V R1 + R2 out(max) R1 = − V = −5 V R1 + R2 out(max) = Problem Set #03 page 6 ANU ENGN 3227 (c) The sketch of output voltage is shown below: 10 Input Output Voltage (V) 5 0 −5 −10 0 0.5 1 Time (s) 1.5 2 −3 x 10 Figure 6: Output voltage vout (t). Problem Set #03 page 7 ANU ENGN 3227 Q4 Solution The given circuit data is R1 = 10 kΩ, R2 = 47 kΩ, +VCC = 15 V, −VEE = −15 V, VZ = 4.7 V, VD = 0.7 V. Z1 Z2 Rc vin vout R1 Vout(max) VUT P VLT P R2 = ±6.55 V = 1.15 V = −1.15 V The sketch of output voltage is shown below: 10 Input Output 6.55 Voltage (V) 5 1.15 0 −1.15 −5 −6.55 −10 0 0.5 1 Time (s) 1.5 2 −3 x 10 Figure 7: Output voltage vout (t). Problem Set #03 page 8 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #04 Digital to Analogue Converters Q1 Consider the DAC circuit shown in Figure 1. Assume RF = 10 kΩ, R = 5 kΩ, +VCC = 15 V, −VEE = −15 V, VH = 5 V, VL = 0 V. (a) Determine the resolution of the DAC. (b) Determine the full scale output voltage of the DAC. (c) Determine the output voltage if binary code (0101)2 is applied at the input. (d) Determine the output voltage if binary code (1001)2 is applied at the input. D0 RF 8R D1 4R D2 2R vout D3 R Figure 1: The circuit for Question 1. Q2 Consider the circuit shown in Figure 2. Assume R = 10 kΩ, +VCC = 18 V, −VEE = −18 V. Determine the output voltage. +10V +5V 2R 2R 2R R 2R 2R R RF=2R R vout Figure 2: The circuit for Question 2. Problem Set #04 page 1 ANU ENGN 3227 Q3 Consider the circuit shown in Figure 3. Assume R = 10 kΩ, +VCC = 18 V, −VEE = −18 V. Determine the output voltage. +5V 2R -5V 2R 2R 2R 2R R R RF=2R R vout Figure 3: The circuit for Question 3. Q4 Consider the DAC circuit shown in Figure 4. Assume R = 10 kΩ, +VCC = 15 V, −VEE = −15 V, VH = 5 V, VL = 0 V. (a) Determine the resolution of the DAC. (b) Determine the full scale output voltage of the DAC. (c) Using the DAC equation, determine the output voltage if binary code (0101)2 is applied at the input. (d) Using the DAC equation, determine the output voltage if binary code (1001)2 is applied at the input. D0 2R 2R 2R R D3 D2 D1 2R 2R R RF=2R R vout Figure 4: The circuit for Question 4. Problem Set #04 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #04 Solution Q1 Complete Solution The given circuit data is RF = 5 kΩ, R = 10 kΩ, +VCC = 15 V, −VEE = −15 V, VH = 5 V, VL = 0 V. D0 8R RF D1 4R D2 2R vout D3 R (a) To find the DAC resolution, we find the output voltage when input binary code is 00012 i.e. D0 = 5V and D1 = D2 = D3 = 0V. The equivalent circuit for this case is +5V 8R RF 4R 2R vout R Writing KCL equation, v− − 5 v− − vout + + i− 8R RF 0 − 5 0 − vout + 80k 5k vout = 0 = 0 = −0.3125 V Hence the resolution = −0.3125 V/LSB. Problem Set #04 page 3 ANU ENGN 3227 (b) To find the full scale output voltage of the DAC, we find the output voltage when input binary code is 11112 i.e. D0 = D1 = D2 = D3 = 5V. The equivalent circuit for this case is +5V 8R RF +5V 4R +5V 2R vout +5V R Writing KCL equation, v− − 5 v− − 5 v− − 5 v− − 5 v− − vout + + + + + i− 8R 4R 2R R RF 0 − 5 0 − 5 0 − 5 0 − 5 0 − vout + + + + 80k 40k 20k 10k 5k vout = 0 = 0 = −4.6875 V Hence Vo(FS) = −4.6875 V. Note: For DACs, once resolution is known, Vo(FS) can also be found using resolution = Vo(FS) 2n −1 . (c) Using the DAC equation (0101)2 vout = 510 = D = resolution × D = (−0.3125)(5) = −1.5625 V (d) Using the DAC equation (1001)2 vout = 910 = D = resolution × D = (−0.3125)(9) = −2.8125 V Modify PSPICE: L07_Example02.sch to check answer. Problem Set #04 page 4 ANU ENGN 3227 Q2 Complete Solution The given circuit data is Assume R = 10 kΩ, +VCC = 18 V, −VEE = −18 V. The equivalent circuit is +10V RF=2R +5V 2R R R R 2R 2R 2R vout 2R VT H The equivalent circuit for determining Thevenin equivalent voltage is Vx R R Vy VTH +5V 2R 2R 2R 2R Writing KCL equations, we have Vx − 5 Vx − 0 Vx −Vy + + 2R 2R R Vy − 0 Vy −Vx Vy −VT H + + 2R R R VT H − 0 VT H −Vy + 2R R = 0 = 0 = 0 Simplifying, 4Vx − 2Vy −2Vx + 5Vy − 2VT H −2Vy + 3VT H = 5 = 0 = 0 Converting to standard matrix form, 4 −2 0 Vx 5 −2 5 −2 Vy = 0 0 −2 3 VT H 0 Problem Set #04 page 5 ANU ENGN 3227 Using Cramer’s rule, 4 −2 0 ∆ = −2 5 −2 0 −2 3 −2 −2 5 −2 − (−2) = (4) 0 3 −2 3 4 −2 5 −2 5 0 0 −2 0 20 VT H = = = 0.625 V 32 32 + (0) −2 5 0 −2 = 32 RT H The equivalent circuit for determining Thevenin equivalent resistance R R 2R RTH 2R 2R 2R Combining the resistances starting from far end (i.e. series and parallel combinations respectively) and moving back towards the Thevenin terminals, RT H = R Overall Equivalent Circuit The overall equivalent circuit for determining vout is +10V 2R R RF=2R v− VTH RTH vout Using KCL, v− −VT H v− − 10 v− − vout + + + i− RT H + R 2R 2R vout = 0 = −10.625 V SEE PSPICE: P04_Q02.sch. Problem Set #04 page 6 ANU ENGN 3227 Q3 Partial Solution The given circuit data is Assume R = 10 kΩ, +VCC = 18 V, −VEE = −18 V. +5V 2R -5V 2R 2R R RF=2R 2R 2R R R vout The resistors at the far end (left of +5V) can be combined to get Req = (2R||2R) + R = 2R The equivalent circuit is +5V -5V 2R R RF=2R 2R 2R R vout Req=2R +5V VT H Vx R VTH 2R Req=2R 2R -5V Vx − 5 Vx − 0 Vx −VT H + + 2R 2R R VT H − (−5) VT H −Vx + 2R R Problem Set #04 = 0 = 0 page 7 ANU ENGN 3227 4 −2 −2 3 ∆ = VT H = Vx VT H = 5 −5 4 −2 = 16 −2 3 4 5 −2 −5 −10 = = −0.625 V ∆ 16 RT H R 2R RTH Req=2R RT H 2R = R Overall Equivalent Circuit RF=2R R v− VTH RTH 2R vout vout = 1.25 V SEE PSPICE: P04_Q03.sch. Problem Set #04 page 8 ANU ENGN 3227 Q4 Solution The given circuit data R = 10 kΩ, +VCC = 15 V, −VEE = −15 V, VH = 5 V, VL = 0 V. D0 2R 2R 2R R D3 D2 D1 2R 2R R RF=2R R vout (a) resolution = -0.625 V/LSB. See Lecture 07 for complete derivation steps. (b) Using the DAC equation n = 4 Vo(FS) resolution = 2n − 1 Vo(FS) = (−0.625)(24 − 1) = −9.375 V (c) Using the DAC equation (0101)2 vout = 510 = D = resolution × D = (−0.625)(5) = −3.125 V (d) Using the DAC equation (1001)2 vout Problem Set #04 = 910 = D = resolution × D = (−0.625)(9) = −5.625 V page 9 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #05 Op-Amp Frequency Response Q1 (a) Derive an expression for the op-amp open loop transfer function Aol ( f ). (b) Derive an expression for the op-amp closed loop transfer function Acl ( f ). Q2 Consider the circuit shown in Figure 1. Assume Aol(mid) = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 10 kΩ, R2 = 220 kΩ, RL = 10 kΩ. (The data is same as in ProblemSet01: Q1) (a) Find the closed loop transfer function Acl ( f ) (b) Find the closed loop cut-off frequency fc(cl) . (c) Find Acl ( f ) for f = 0, 10, 100, 1k, 30k, fc(cl) , fT respectively. (d) Find output voltage vout (t) if vin (t) = 0.1 sin(2π30000t). Rc vin vout R1 RL R2 Figure 1: The circuit for Question 2. Q3 Consider the circuit shown in Figure 2. Assume Aol(mid) = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 1 kΩ, R2 = 100 kΩ, RL = 10 kΩ. (The data is same as in ProblemSet01: Q2) (a) Find the closed loop transfer function Acl ( f ) (b) Find the closed loop cut-off frequency fc(cl) . (c) Find Acl ( f ) for f = 0, 10, 100, 1k, fc(cl) , fT respectively. (d) Find output voltage vout (t) if vin (t) = 0.1 sin(2π1000t). R2 vin R1 vout Rc RL Figure 2: The circuit for Question 3. Problem Set #05 page 1 ANU ENGN 3227 Q4 Consider the circuit shown in Figure 3. Assume Aol(mid) = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs. (a) Find the overall closed loop transfer function for the cascaded op-amps. (b) Find output voltage vout if vin = 0.1 sin(2π10000t + 30◦ ). R4=2 kΩ vout1 vin vin2 R3=1 kΩ vout R1=1 kΩ R2=4 kΩ Figure 3: The circuit for Question 4. Problem Set #05 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #05 Solution Q1 Solution (a) Aol ( f ) = Aol(mid) 1+ j f f c(ol) See Lecture 09 for complete derivation steps. (b) Acl ( f ) = Aol(mid) 1 + BAol(mid) + j f f c(ol) See Lecture 09 for complete derivation steps. Problem Set #05 page 3 ANU ENGN 3227 Q2 Complete Solution The given circuit data is Aol(mid) = 2×105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 10 kΩ, R2 = 220 kΩ, RL = 10 kΩ. Rc vin vout R1 R2 RL (a) The feedback ratio is B = = R1 R1 + R2 1 10k = 10k + 220k 23 The open loop cut-off frequency is fc(ol) = = fT Aol 106 = 5 Hz 2 × 105 The closed loop transfer function is Acl ( f ) = = = Aol(mid) 1 + BAol(mid) + j f f c(ol) 2 × 105 1 1 + (2 × 105 )( 23 ) + j 5f 2 × 105 8696.65 + j0.2 f The above equation shows how closed loop gain varies with frequency. At f = 0 Hz (i.e. DC), we have Acl (0) = = = Aol(mid) 1 + BAol(mid) + j f 0 c(ol) Aol(mid) 1 + BAol(mid) 2 × 105 5 1 + 2×10 23 = 22.997 Compare with ProblemSet01: Q1. Problem Set #05 page 4 ANU ENGN 3227 (b) The closed loop cut-off frequency can be calculated in two ways. Using unity gain relationship and Acl (0), we have fc(cl) = = fT Acl 1 × 106 = 43.48 kHz 22.997 Using the open loop cut-off frequency result, we have fc(cl) = fc(ol) (1 + BAol(mid) ) = (5)(1 + 2 × 105 ) = 43.48 kHz 23 Compare with ProblemSet01: Q1. (c) Acl ( f ) = Acl (0) = Acl (10) = Acl (100) = Acl (1k) = Acl ( fc(cl) ) = Acl ( fT ) = 2 × 105 8696.65 + j0.2 f 2 × 105 = 22.997 = 22.997∠0◦ 8696.65 2 × 105 = 22.997 − j0.0053 = 22.997∠−0.013◦ 8696.65 + j2 2 × 105 = 22.997 − j0.053 = 22.997∠−0.13◦ 8696.65 + j20 2 × 105 = 22.997 − j0.53 = 22.991∠−1.32◦ 8696.65 + j200 2 × 105 = 11.5 − j11.5 = 16.26∠−45◦ 8696.65 + j8696.65 2 × 105 = 0.043 − j0.9981 = 0.999∠−87.51◦ 8696.65 + j2 × 105 We can see that at closed loop cut-off frequency, |Acl ( fc(cl) )| = 16.26 = Acl √(0) . 2 Also at unity gain frequency, |Acl ( fT )| = 0.999 ≈ 1 (which follows from definition of unity gain frequency). See MATLAB: P05_Q01.m Problem Set #05 page 5 ANU ENGN 3227 (d) Given that vin (t) = 0.1 sin(2π30000t) f = 30 kHz Converting to phasors, we have, → − V in = 0.1∠0◦ = 0.1 Using definition of closed loop transfer function → − V out Acl ( f ) = → − V in Substituting values → − V out = (Acl (30000))(0.1∠0◦ ) = (0.1)(15.58 − j10.75) = 1.558 − j1.075 = 1.893∠−34.60◦ Converting back to time domain, vout (t) = 1.893 sin(2π1000t − 34.6◦ ) This is the equation of the output voltage waveform. See PSPICE: P05_Q02.sch In PSPICE, add trace 1.893*SIN(2*3.14*30000*TIME-34.6*3.14/180) to compare simulation and predicted result. Problem Set #05 page 6 ANU ENGN 3227 Q3 Solution The given circuit data is Aol(mid) = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs, R1 = 1 kΩ, R2 = 100 kΩ, RL = 10 kΩ. R2 vin R1 vout Rc RL (a) B = fc(ol) = Acl ( f ) = R1 1 = R2 100 fT = 5 Hz Aol −Aol(mid) 1 + BAol(mid) + j f f = −2 × 105 2001 + j0.2 f c(ol) (b) fc(cl) = fc(cl) = fT = 10.005 kHz Acl (0) fc(ol) (1 + BAol(mid) ) = 10.005 kHz (c) Acl (0) = Acl (10) = Acl (100) = Acl (1k) = Acl ( fc(cl) ) = Acl ( fT ) = −2 × 105 = −99.95∠0◦ 2001 −2 × 105 = −(99.95 − j0.099) = −(99.95∠−0.057◦ ) 2001 + j2 −2 × 105 = −(99.94 − j0.998) = −(99.94∠−0.57◦ ) 2001 + j20 −2 × 105 = −(98.96 − j9.98) = −(99.45∠−5.71◦ ) 2001 + j200 −2 × 105 = −(49.975 − j49.975) = −(70.67∠−45◦ ) 2001 + j2001 −2 × 105 = −(0.01 − j0.999) = −(0.999∠−89.42◦ ) 2001 + j2 × 105 (d) vin (t) = 0.1 sin(2π1000t) vout (t) = −9.945 sin(2π1000t − 5.71◦ ) See PSPICE: P05_Q03.sch In PSPICE, add trace -0.9945*SIN(2*3.14*1000*TIME-5.71*3.14/180) to compare simulation and predicted result. Problem Set #05 page 7 ANU ENGN 3227 Q4 Partial Solution The given circuit is R4=2 kΩ vout1 vin vin2 R3=1 kΩ vout R1=1 kΩ R2=4 kΩ (a) This is a cascaded two-stage op-amp circuit. For 1st op-amp, B = = fc1(ol) Acl1 ( f ) = R1 1 = R1 + R2 5 fT = 5 Hz Aol −Aol(mid) 1 + BAol(mid) + j f f = 2 × 105 40001 + j0.2 f = −2 × 105 100001 + j0.2 f c1(ol) For 2nd op-amp, B = = fc2(ol) Acl2 ( f ) = R3 1 = R4 2 fT = 5 Hz Aol −Aol(mid) 1 + BAol(mid) + j f f c2(ol) The overall closed loop transfer function is the product of the two individual closed loop transfer functions. Acl ( f ) = Acl1 ( f )Acl2 ( f ) −4 × 1010 (4 × 109 − 0.04 f 2 ) + j14002 f Acl ( f ) = (b) f = 10 kHz Acl (10k) = −4 × 1010 = −(10∠−2◦ ) 3.996 × 109 + j140.02 × 106 The output voltage is vin (t) = 0.1 sin(2π10000t + 30◦ ) vout (t) = − sin(2π10000t + 28◦ ) See PSPICE: P05_Q04.sch Problem Set #05 page 8 ANU ENGN 3227 Appendix A: Useful Formulas Notation → − V = a + jb (rectangular form) → − → − ◦ V = | V |∠θ ((polar form) p → − 2 2 |V | = a +b b θ = tan−1 ± 180◦ a (1) (2) (3) (4) Note: arctangent function is multivalued so an adjustment is needed obtain correct value of θ. Scientific calculators can, however, convert complex numbers from polar to rectangular and vice versa in single operation and give correct value of θ automatically. Practice with your scientific calculator to become proficient in using complex, Pol and Rec modes. Complex Conjugate → −∗ V = a − jb (5) i.e the conjugate of the complex number is formed by reversing the sign of the imaginary part. In polar form → −∗ → − V = | V |∠−θ◦ . Multiplication by Complex Conjugate → −→ − V V∗ = (a + jb)(a − jb) = a2 − jab + jab − j2 b2 = a2 + b2 (6) Multiplication in Polar form → − → − → − → − If V 1 = | V 1 |∠θ◦1 and V 2 = | V 2 |∠θ◦2 → − → − V 1V 2 → − → − → − → − = (| V 1 |∠θ◦1 )(| V 2 |∠θ◦2 ) = (| V 1 || V 2 |)∠(θ1 + θ2 )◦ (7) i.e to multiply numbers in polar form, we multiply the magnitudes and add the angles. Division in Polar form → − → − If V 1 = |V1 |∠θ◦1 and V 2 = |V2 |∠θ◦2 → − V1 → − V2 → − → − | V 1 |∠θ◦1 | V 1| ◦ = → − → − ∠(θ1 − θ2 ) ◦ | V 2 |∠θ2 | V 2| = (8) i.e to divide numbers in polar form, we divide the magnitudes and subtract the angles. Useful Identities j 1 j j2 = √ −1 (9) = −j (10) = −1 (11) Problem Set #05 page 9 ANU ENGN 3227 Appendix B: Complex Number Calculation Examples Addition and Substraction → − V1 → − V2 → − → − V 1+V 2 → − → − V 1−V 2 = 8 + j16 = 12 − j3 = 20 + j13 (using rectangular form) = −4 + j19 (using rectangular form) Note: If numbers to be added or subtracted are given in polar form, they must be first converted to rectangular form. Multiplication → − V1 → − V2 → − → − V 1V 2 → − V1 → − V2 → − → − V 1V 2 = 8 + j10 = 5 − j4 = 40 − j32 + j50 + 40 = 80 + j18 (using rectangular form) = 8 + j10 = 12.81∠51.34◦ = 5 − j4 = 6.4∠−38.66◦ = (12.81∠51.34◦ )(6.4∠−38.66◦ ) = 82∠12.68◦ = 80 + j18 (using polar form) Division → − V1 → − V2 → − V1 → − V2 → − V1 → − V2 → − V1 → − V2 = 6 + j3 = 3− j 6 + j3 3− j 6 + j3 3 + j = (multiply and divide by conjugate of denominator) 3− j 3+ j 18 + j6 + j9 − 3 = 9+1 = 1.5 + j1.5 (using rectangular form) = = 6 + j3 = 6.71∠26.57◦ = 3 − j = 3.16∠−18.43◦ = Problem Set #05 6.71∠26.57◦ = 2.12∠45◦ = 1.5 + j1.5 3.16∠−18.43◦ (using polar form) page 10 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #06 Active Filters Q1 Consider the circuit shown in Figure 1. Assume R1 = 1 kΩ, R2 = 2 kΩ, C1 = 1 µF. (a) Find the s-domain transfer function of the circuit in standard form. (b) Write the set of M ATLAB commands (4 lines expected) to obtain the Bode plot. vout(t) vin(t) C1 R2 R1 Figure 1: The circuit for Question 1. Q2 Consider the circuit shown in Figure 2. Assume R1 = 1 kΩ, C1 = 1 µF, C2 = 10 µF. (a) Find the s-domain transfer function of the circuit in standard form. (b) Write the set of M ATLAB commands (4 lines expected) to obtain the Bode plot. vout(t) vin(t) C2 R1 C1 Figure 2: The circuit for Question 2. Problem Set #06 page 1 ANU ENGN 3227 Q3 Consider the Sallen-Key low-pass filter circuit shown in Figure 3. Find the s-domain transfer function of the circuit in standard form. C1 R1 R2 v+ vout (t) vx vin(t) RF C2 v− RA Figure 3: The circuit for Question 3. Q4 Consider the Sallen-Key high-pass filter circuit shown in Figure 4. Find the s-domain transfer function of the circuit in standard form. R1 C1 vin(t) C2 v+ vout (t) vx R2 RF v− RA Figure 4: The circuit for Question 4. Problem Set #06 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #06 Solution Q1 Complete Solution The given circuit data is R1 = 1 kΩ, R2 = 2 kΩ, C1 = 1 µF. Re-drawing the circuit in s-domain, we have Vout(s) Vin(s) 1/sC1 R2 R1 (a) Applying KCL at -ve pin, V− (s) − 0 V− (s) −Vout (s) V− (s) −Vout (s) + + 1 R2 R1 sC = 0 1 Solving Vout (s) V− (s) = sC1 + R11 + R12 sC1 + R11 From circuit, Vin (s) = V+ (s). Applying op-amp assumption, V+ (s) = V− (s). Hence Vout (s) Vin (s) = sC1 + R11 + R12 sC1 + R11 Converting to standard form 1 1 1 s + + C1 R1 R2 Vout (s) = 1 Vin (s) s + R1C1 (a) Substituting the values, the transfer function is H(s) = s + 1500 s + 1000 Problem Set #06 page 3 ANU ENGN 3227 The M ATLAB commands are: >> >> >> >> num=[1 1500]; den=[1 1000]; H=tf(num,den); bode(H); From the shape of the magnitude bode plot, the given circuit provides 3.5 dB gain to low frequencies and allows high frequencies to pass unchanged (0dB gain). Q2 Partial Solution The given circuit data is R1 = 1 kΩ, C1 = 1 µF, C2 = 10 µF. The given circuit is vout(t) vin(t) C2 R1 C1 (a) In standard form H(s) = 1 R1C2 s + R11C1 (b) H(s) = 100 s + 1000 The M ATLAB commands are: >> >> >> >> num=[100]; den=[1 1000]; H=tf(num,den); bode(H); From the shape of the magnitude bode plot, the given circuit is a low-pass filter with roll-off -20 dB/decade. Problem Set #06 page 4 ANU ENGN 3227 Q3 Partial Solution Re-drawing the circuit in s-domain, we have 1/sC1 R1 R2 V+(s) Vout (s) Vx(s) Vin(s) 1/sC2 RF V−(s) RA Let G = 1+ RA + RF RF = RA RA Step 1: Apply KCL at -ve pin V− (s) −Vout (s) V− (s) − 0 + RF RA = 0 Solving, we have V− (s) = = RA Vout (s) RA + RF Vout (s) G Applying op-amp assumption, V+ (s) = V− (s). Hence V+ (s) = Vout (s) G Step 2: Apply KCL at +ve pin V+ (s) −Vx (s) V+ (s) − 0 + 1 R2 sC = 0 2 Substituting the value V+ (s) = Vx (s) = (sC2 R2 + 1) Vout (s) G and solving, we get Vout (s) G Step 3: Apply KCL at node x Vx (s) −V+ (s) Vx (s) −Vout (s) Vx (s) −Vin (s) + + 1 R2 R1 sC = 0 Vx (s) − VoutG(s) Vx (s) −Vout (s) Vx (s) −Vin (s) + + 1 R2 R1 sC = 0 1 1 Problem Set #06 page 5 ANU ENGN 3227 Simplifying, we have 1 1 sC1 + + Vx (s) = R1 R2 1 1 Vin (s) + sC1Vout (s) + Vout (s) R1 GR2 R1 (sC1 R1 R2 + R2 + R1 )Vx (s) = R2Vin (s) + sC1 R1 R2Vout (s) + Vout (s) G R1 Vout (s) = R2Vin (s) + sC1 R1 R2Vout (s) + Vout (s) (sC1 R1 R2 + R2 + R1 )(sC2 R2 + 1) G G (sC1 R1 R2 + R2 + R1 )(sC2 R2 + 1)Vout (s) = GR2Vin (s) + sGC1 R1 R2Vout (s) + R1Vout (s) (sC1 R1 R2 + R2 + R1 )(sC2 R2 + 1)Vout (s) − sGC1 R1 R2Vout (s) − R1Vout (s) 2 (sC1C2 R1 R2 + sC1 R1 R2 + sC2 R22 + R2 + sC2 R1 R2 + R1 )Vout (s) − sGC1 R1 R2Vout (s) − R1Vout (s) (s2C1C2 R1 R22 + sC1 R1 R2 (1 − G) + sC2 R1 R2 + sC2 R22 + R2 )Vout (s) (s2C1C2 R1 R2 + sC1 R1 (1 − G) + sC2 R1 + sC2 R2 + 1)Vout (s) = = = = GR2Vin (s) GR2Vin (s) GR2Vin (s) GVin (s) Hence Vout (s) Vin (s) G s2C1C2 R1 R2 + sC1 R1 (1 − G) + sC2 R1 + sC2 R2 + 1 G = 2 s C1C2 R1 R2 + [C1 R1 (1 − G) +C2 R1 +C2 R2 ]s + 1 Step 4: Convert transfer function to standard form. = Dividing numerator and denominator by C1C2 R1 R2 Vout (s) Vin (s) G = s2 + h 1−G C2 R2 1 C1C2 R1 R2 i + C11R2 + C11R1 s + C1C21R1 R2 Hence G H(s) = s2 + h 1−G C2 R2 1 C1C2 R1 R2 i + C11R2 + C11R1 s + C1C21R1 R2 Q4 Solution The given circuit is R1 C1 vin(t) C2 v+ vout (t) vx R2 RF v− RA Gs2 H(s) = s2 + Problem Set #06 h 1−G C1 R1 i + C11R2 + C21R2 s + C1C21R1 R2 page 6 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #07 Filter Design Q1 Consider the Sallen-Key biquad circuit shown in Figure 1. (a) Design a second-order (n = 2) Butterworth lowpass filter, with cut-off frequency fc = 4 kHz using this circuit. (b) Write the set of M ATLAB commands (6 lines expected) to obtain the Magnitude Bode plot vs. frequency for the filter. (c) Design a second-order (n = 2) Butterworth highpass filter, with cut-off frequency fc = 4 kHz. C1 R1 vin(t) R2 v+ vout (t) vx C2 RF v− RA Figure 1: The circuit for Question 1. Q2 Design a fourth-order (n = 4) Butterworth lowpass filter, with cut-off frequency fc = 4 kHz, using Sallen-Key biquads. Q3 Consider the Sallen-Key biquad circuit shown in Figure 1. (a) Design a second-order (n = 2) Bessel lowpass filter, with cut-off frequency fc = 4 kHz using this circuit. (b) Write the set of M ATLAB commands (6 lines expected) to obtain the Phase (in degrees) plot vs. frequency for the filter. Problem Set #07 page 1 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #07 Solution Q1 Complete Solution (a) The given design specifications are: n = 2, Butterworth filter, fc = 4 kHz. C1 R1 R2 v+ vout (t) vx vin(t) C2 RF v− RA The Sallen-Key lowpass filter transfer function in standard form is G H(s) = s2 + h 1−G C2 R2 1 C1C2 R1 R2 i + C11R2 + C11R1 s + C1C21R1 R2 where G = 1+ RF RA + RF = RA RA Assuming C1 = C2 = C and R1 = R2 = R, we have G 21 2 3−GRC s2 + CR s + R21C2 H(s) = The n = 2 lowpass Butterworth transfer function in standard form is K √ s2 + 2ωc s + ω2c H(s) = Comparing the constant term in the denominators, we have ω2c = ωc = fc = 1 R2C2 1 RC 1 2πRC Problem Set #07 page 2 ANU ENGN 3227 Comparing the coefficient of s terms in the denominators, we have √ 2ωc 3−G RC √ 1 3−G 2 = RC RC√ G = 3− 2 √ RF 1+ = 3− 2 RA √ RF = 2 − 2 = 0.5857 RA = Design Let C = 0.01 µF. Then fc = R = 1 2πRC 1 2πC fc 1 2π(0.01 × 10−6 )(4 × 103 ) = 3978.87 Ω = 3.9 kΩ (nearest standard value) = Let RA = 3.9 kΩ. Then RF RA RF = 0.5857 = 2.284 kΩ = 2.2 kΩ (nearest standard value) The filter implementation is C1=0.01µF R1=3.9kΩ R2=3.9kΩ vout (t) vin(t) C2=0.01µF RF =2.2kΩ RA =3.9kΩ See PSPICE: P07_Q01.sch Problem Set #07 page 3 ANU ENGN 3227 (b) Substituting the component values, H(s) = 1028338306.4448 s2 + 336817.883s + 657462195.9 The M ATLAB commands to generate Bode magnitude plot vs. frequency are: >> >> >> >> >> >> num=[1028338306.4448]; den=[1 336817.883 657462195.9]; [h,w] = freqs(num,den); f = w/(2*pi); mag = 20*log10(abs(h)); semilogx(f,mag); See also Matlab: P07_Q01.m (c) The position of Rs and Cs can be reversed in the lowpass biquad to create highpass biquad. Thus a second-order (n = 2) Butterworth highpass filter, with cut-off frequency fc = 4 kHz using Sallen-Key biquad is R 1=3.9kΩ C1=0.01µF C2=0.01µF vout (t) vin(t) R2=3.9kΩ RF =2.2kΩ RA =3.9kΩ See PSPICE: P07_Q01_hp.sch Problem Set #07 page 4 ANU ENGN 3227 Q2 Partial Solution The n = 4 lowpass Butterworth transfer function in standard form is K s4 + 2.61313ωc s3 + 3.14142ω2c s2 + 2.61313ω3c s + ω4c K1 K2 = s2 + 1.848ωc s + ω2c s2 + 0.765ωc s + ω2c H(s) = Thus two lowpass Sallen-Key biquads are required. 1st Stage Design K1 s2 + 1.848ωc s + ω2c H1 (s) = Assuming C1 = C2 = C and R1 = R2 = R, G 21 2 3−GRC s2 + CR s + R21C2 H(s) = Comparing the constant term in the denominators, we have ω2c = ωc = fc = 1 R2C2 1 RC 1 2πRC Comparing the coefficient of s terms in the denominators, we have 3−G RC G = 1.152 RF = 0.152 RA 1.848ωc = Let C = 0.01 µF. Then 1 2πC fc = 3978.87 Ω = 3.9 kΩ (nearest standard value) R = Let RA = 10 kΩ. Then RF = 1.52 kΩ = 1.5 kΩ Problem Set #07 (nearest standard value) page 5 ANU ENGN 3227 2nd Stage Design K2 2 s + 0.765ω H1 (s) = 2 c s + ωc Comparing the coefficient of s terms in the denominators, we have 3−G RC G = 2.235 RF = 1.235 RA = 0.765ωc C R RA RF = = = = 0.01 µF 3.9 kΩ 22 kΩ 27 kΩ (nearest standard value) (nearest standard value) The filter implementation is C1=0.01µF C1=0.01µF R1=3.9kΩ R2=3.9kΩ R1=3.9kΩ R2 =3.9kΩ vin(t) C2=0.01µF vout(t) RF =1.5kΩ C2=0.01µF RF =27kΩ RA =10kΩ RA =22kΩ Simulate in PSPICE to verify. Write the set of Matlab commands to obtain Magnitude Bode plot of filter. SEE ALSO HLAB2: FIG25-2 Problem Set #07 page 6 ANU ENGN 3227 Q3 Partial Solution (a) n = 2, Bessel filter, fc = 4 kHz. Assuming C1 = C2 = C and R1 = R2 = R, the Sallen-Key lowpass filter transfer function in standard form is H(s) = G 21 2 3−GRC s2 + CR s + R21C2 The n = 2 lowpass Bessel transfer function in standard form is K H(s) = s2 + 3ωc s + 3ω2c Comparing the constant term in the denominators, we have 1 3ω2c = R2C2 1 ωc = √ 3RC 1 √ fc = 2π 3RC Comparing the coefficient of s terms in the denominators, we have 3−G 3ωc = RC√ G = 3− 3 RF = 0.2679 RA The filter implementation is C1=0.01µF R1=3.9kΩ R2=3.9kΩ vout (t) vin(t) C2=0.01µF RF =1kΩ RA =3.9kΩ Compare with Butterworth implementation in Q1. (b) >> >> >> >> >> >> num=[826042246.16]; den=[1 44707.43 657462195.92]; [h,w] = freqs(num,den); f = w/(2*pi); phase = angle(h)*180/pi; semilogx(f,phase); See also Matlab: P07_Q03.m Problem Set #07 page 7 ANU ENGN 3227 Appendix n 1 2 3 4 Table 1: Butterworth Polynomials Factored Form Transfer Function H(s) K s + ωc K √ 2 s + 2ωc s + ω2c K 3 2 s + 2ωc s + 2ω2c s + ω3c K 4 3 s + 2.61313ωc s + 3.14142ω2c s2 + 2.61313ω3c s + ω4c n 1 2 3 4 K1 s + ωc K2 2 2 s + ωc s + ω c K1 2 s + 1.848ωc s + ω2c K2 2 s + 0.765ωc s + ω2c Table 2: Chebyshev Polynomials Transfer Function H(s) K s + 1.9652ωc K s2 + 1.0977ωc s + 1.1025ω2c K s3 + 0.9883ωc s2 + 1.2384ω2c s + 0.4913ω3c K 4 3 s + 0.9528ωc s + 1.4539ω2c s2 + 0.7426ω3c s + 0.2756ω4c n 1 2 3 4 Table 3: Bessel Polynomials Transfer Function H(s) K s + ωc K s2 + 3ωc s + 3ω2c K 3 2 s + 6ωc s + 15ω2c s + 15ω3c K 4 3 2 s + 10ωc s + 45ωc s2 + 105ω3c s + 105ω4c The above tables can be obtained using Matlab butter, cheby1 and besself commands. See also Matlab: L11_Example01.m Problem Set #07 page 8 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #08 555 Timer Q1 Consider the 555 astable circuit shown in Figure 1. Assume R1 = 10 kΩ, R2 = 5 kΩ, C = 300 pF. (a) Explain the working of the circuit. Sketch the voltage across capacitor C and output voltage. (b) Derive the equations for the ON time TH (time the output is HIGH), OFF time TL (time the output is LOW) and Duty cycle D. (c) Find the values of TH , TL and D. Figure 1: The circuit for Question 1. Q2 Consider the 555 monostable circuit shown in Figure 2. Assume R = 12 kΩ, C = 0.025 µF. (a) Explain the working of the circuit. Sketch the voltage across capacitor C and output voltage. (b) Derive the equations for the pulse duration T (time the output is HIGH). (c) Find the value of pulse duration T . Figure 2: The circuit for Question 2. Problem Set #08 page 1 ANU ENGN 3227 Q3 Design a square waveform generator using 555 Timer with duty cycle D = 60% and oscillation frequency f = 1 kHz. Q4 Design a square waveform generator using 555 Timer with duty cycle D = 25% and oscillation frequency f = 1 kHz. Problem Set #08 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #08 Solution Q1 Partial Solution The given circuit data is R1 = 10 kΩ, R2 = 5 kΩ, C = 300 pF. (a) See Lecture 12 for solution (b) = 0.693(R1 + R2 )C = 0.693R2C TH R1 + R2 × 100% = × 100% Duty cycle = TH + TL R1 + 2R2 TH TL See Lecture 12 for derivation steps (c) = 0.693(R1 + R2 )C = 0.693(15k)(300p) = 3.1185 µs = 0.693R2C = 0.693(5k)(300p) = 1.0395 µs TH 3.1185 Duty cycle = × 100% = × 100% = 75% TH + TL 3.1185 + 1.0395 TH TL Problem Set #08 page 3 ANU ENGN 3227 Q2 Partial Solution The given circuit data is R = 12 kΩ, C = 0.025 µF. (a) See Lecture 12 for solution (b) T = 1.1RC See Lecture 12 for derivation steps (c) T = 1.1RC = 1.1(12k)(0.025µ) = 330 µs Problem Set #08 page 4 ANU ENGN 3227 Q3 Complete Solution The design data is: square waveform generator using 555 Timer Duty cycle D = 60% oscillation frequency f = 1 kHz. As D > 50%, the 555 circuit is From the time period, we have T TH + TL 1 f = 1ms = From the duty cycle, we have TH Duty cycle = × 100% T TH 0.6 = T TH = 0.6T (1) (2) Solving (1) and (2), we have TH TL = 0.6ms = 0.4ms We know for above 555 Timer, TH TL = 0.693(R1 + R2 )C = 0.693R2C Design Let C = 0.1 µF. Then R2 R1 = 5.772kΩ = 2.886 kΩ See also Matlab: P08_Q03.m Problem Set #08 page 5 ANU ENGN 3227 Q4 Partial Solution The design data is: square waveform generator using 555 Timer Duty cycle D = 25% oscillation frequency f = 1 kHz. As D < 50%, the 555 circuit is From the time period, we have T TH + TL 1 f = 1ms = (3) From the duty cycle, we have TH × 100% TH + TL TH 0.25 = T TH = 0.25T Duty cycle = (4) Solving (1) and (2), we have TH TL = 0.25ms = 0.75ms We know for above 555 Timer, TH TL = 0.693R1C = 0.693R2C Design Let C = 0.1 µF. Then R1 R2 = 3.607kΩ = 10.822 kΩ See also Matlab: P08_Q03.m Problem Set #08 page 6 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #09 Oscillators Q1 Explain briefly, in words, the operation of the following: (a) Wien-bridge oscillator. (b) Phase-shift oscillator. (c) Square-wave relaxation oscillator. (d) Triangle-wave relaxation oscillator. Q2 Consider the circuit shown in Figure 1. Assume R = 10000 2π Ω, C = 0.01 µF. (a) Derive an expression for the transfer function H( f ) of the given circuit. (b) Find the frequency fr for which H( f ) is real. (c) Find output voltage vout (t) if vin (t) = sin(2π10000t). (d) Find output voltage vout (t) if vin (t) = sin(2π1000t). C R vout(t) vin(t) R C Figure 1: The circuit for Question 2. Q3 √ Ω, C = 0.01 µF. Consider the circuit shown in Figure 2. Assume R = 10000 2π 6 (a) Derive an expression for the transfer function H( f ) of the given circuit. (b) Find the frequency fr for which H( f ) is real. (c) Find output voltage vout (t) if vin (t) = 29 sin(2π10000t). C C C vout(t) vin(t) R R R Figure 2: The circuit for Question 3. Problem Set #09 page 1 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #09 Solution Q1 Solution See Lecture 13: Oscillators and Textbook: Chapter 10. Q2 Partial Solution The given circuit data is R = 10000 2π Ω, C = 0.01 µF. R C vout(t) vin(t) R C (a) Using phasors, → − V out − 0 1 jωC + → − → − → − V out − 0 V out − V in + 1 R R + jωC = 0 Solving, → − V out H( f ) = → − V in = ωRC 3ωRC − j(1 − ω2 R2C2 ) (b) For real H( f ), the imaginary part must be 0. Hence 1 − ω2r R2C2 = 0 1 ωr = RC 1 fr = 2πRC 1 H( fr ) = 3 Problem Set #09 page 2 ANU ENGN 3227 (c) vin (t) = sin(2π10000t) f = 10 kHz Using phasors, → − V in = 1∠0◦ = 1 1 H(10000) = 3 1 → − → − V out = H(10000) V in = 3 Converting back, vout (t) = 1 sin(2π10000t) 3 (d) vin (t) = sin(2π1000t) f = 1 kHz Using phasors, → − V in = 1∠0◦ = 1 0.1 H(1000) = 0.3 − j(1 − 0.01) 0.1 = 0.3 − j0.99 = 0.028 + j0.0925 = 0.0967∠73.16◦ → − → − V out = H(1000) V in = 0.0967∠73.16◦ Converting back, vout (t) = 0.0967 sin(2π1000t + 73.16◦ ) See PSPICE: P09_Q02.sch In PSPICE, add trace 0.0967*SIN(2*3.14*1000*TIME+73.16*3.14/180) to compare simulation and predicted result. Problem Set #09 page 3 ANU ENGN 3227 Q3 Partial Solution The given circuit data is R = 10000 √ 2π 6 Ω, C = 0.01 µF. C C C vout(t) vin(t) R R R (a) Using phasors and writing node equation, → − → − → − → − → − V y − V in V y − 0 V y − V x + + = 0 1 1 R jωC jωC → − → − → − → − → − V x − V y V x − 0 V x − V out + + = 0 1 1 R jωC jωC → − → − → − V out − V x V out − 0 + = 0 1 R jωC Solving, using Cramer’s rule, we have → − 1 V out H( f ) = → − = 5 6 V in 1 − ω2 R2C2 − j ωRC − ω3 R13C3 See also Lecture 13: Oscillators (b) For real H( f ), the imaginary part must be 0. Hence 6 1 − ωr RC ω3r R3C3 = 0 1 √ 6RC 1 √ fr = 2π 6RC 1 H( fr ) = − 29 ωr Problem Set #09 = page 4 ANU ENGN 3227 (c) vin (t) = 29 sin(2π10000t) f = 10 kHz Using phasors, → − V in = 29∠0◦ = 1 1 H(10000) = − 29 → − → − V out = H(10000) V in = −1 = 1∠180◦ Converting back, vout (t) = sin(2π10000t + 180◦ ) See PSPICE: P09_Q03.sch In PSPICE, add trace SIN(2*3.14*10000*TIME + 180*3.14/180) to compare simulation and predicted result. Problem Set #09 page 5 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #10 Instrumentation Amplifiers Q1 Consider the circuit shown in Figure 1. (a) Derive an exact expression for the output voltage vout (t) in terms of input voltages v1 (t) and v2 (t) and op-amp open loop gain Aol . (b) Find an expression for the output voltage vout (t) when Aol → ∞. (c) Find an expression for the output voltage vout (t) when Aol → ∞ and RR34 = RR21 . (d) Find an expression for the output voltage vout (t) when Aol → ∞ and R4 = R3 = R2 = R1 = R. R4 v1 R3 v2 R1 vout RL R2 Figure 1: The circuit for Question 1. Q2 Consider the two op-amp instrumentation amplifier circuit shown in Figure 2. (a) Derive an exact expression for the output voltage vout (t) in terms of input voltages v1 (t) and v2 (t) and op-amp open loop gain Aol . (b) Find an expression for the output voltage vout (t) when Aol → ∞. (c) Find an expression for the output voltage vout (t) when Aol → ∞ and RR34 = RR12 . (d) Find an expression for the output voltage vout (t) when Aol → ∞ and R4 = R3 = R2 = R1 = R. R4 R3 vx v1 vout vout R1 RL R2 v2 Figure 2: The circuit for Question 2. Problem Set #10 page 1 ANU ENGN 3227 Q3 Consider the three op-amp instrumentation amplifier circuit shown in Figure 3. (a) Derive an expression for the output voltage vout1 (t). (b) Derive an expression for the output voltage vout2 (t). (c) Assuming R4 = R3 = R2 = R1 = R, find the expression for output voltage vout (t) in terms of input voltages v1 (t) and v2 (t). vout1 R4 v1 R3 RG v2 R R vout R1 R2 RL vout2 Figure 3: The circuit for Question 3. Problem Set #10 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #10 Solution Q1 R4 Partial Solution v1 R3 v2 R1 vout RL R2 (a) Writing KCL equations, v+ v− vin R2 v2 R1 + R2 R4 R3 = v1 + vout R3 + R4 R3 + R4 = (v+ ) − (v− ) = The output voltage is vout vout vout = Aol vin R2 Aol R4 Aol R3 Aol = v2 − v1 − vout R1 + R2 R3 + R4 R3 + R4 (R3 + R4 )Aol R2 R4 Aol = v2 − v1 R3 + R4 + R3 Aol R1 + R2 R3 + R4 + R3 Aol (b) When Aol → ∞, vout = R3 + R4 R3 +R4 Aol + R3 = R4 R3 + R4 R2 v2 − v1 R3 R1 + R2 R3 R2 v2 − R1 + R2 R4 R3 +R4 Aol + R3 v1 (c) Assuming Aol → ∞ and vout = R4 R3 = R2 R1 , R2 (v2 − v1 ) R1 (d) Assuming Aol → ∞ and R4 = R3 = R2 = R1 = R, vout = v2 − v1 Problem Set #10 page 3 ANU ENGN 3227 Q2 Partial Solution The given circuit is R4 R3 vx v1 vout vout RL R2 R1 v2 (a) Writing KCL equations for non-inverting op-amp v+ = v1 v− = vin vx vx R1 vx R1 + R2 = (v+ ) − (v− ) = Aol vin (R1 + R2 )Aol v1 = R1 + R2 + R1 Aol Writing KCL equations for inverting op-amp v+ = v2 v− = vin vout vout = = = vout = R4 R3 vx + vout R3 + R4 R3 + R4 (v+ ) − (v− ) Aol vin Aol v2 − Aol v− (R3 + R4 )Aol R4 Aol (R1 + R2 )Aol v2 − v1 R3 + R4 + R3 Aol R3 + R4 + R3 Aol R1 + R2 + R1 Aol (b) When Aol → ∞, vout = R4 1+ R3 R4 v2 − R3 R2 1+ R1 v1 (c) Assuming Aol → ∞ and RR43 = RR21 , R4 vout = 1+ (v2 − v1 ) R3 (d) Assuming Aol → ∞ and R4 = R3 = R2 = R1 = R, vout = 2(v2 − v1 ) Problem Set #10 page 4 ANU ENGN 3227 Q3 Partial Solution The given circuit is vout1 R4 v1 R3 R R RG v2 vout R1 R2 RL vout2 (a) v1 − vout1 v1 − v2 + R RG vout1 = 0 = R1 + RG R v1 − v2 RG RG (b) v2 − vout2 v2 − v1 + R RG vout2 = 0 = R1 + RG R v2 − v1 RG RG (c) The third op-amp is a unity gain difference amplifier [using result of Q1 part (d)]. vout = vout2 − vout1 2R = 1+ (v2 − v1 ) RG See PSPICE: L15_Example02.sch See PSPICE: L15_Example02_CMRR.sch Problem Set #10 page 5 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #11 Integrator and Differentiator Q1 Consider the circuit shown in Figure 1. (a) Derive an expression for the output voltage vout (t) in terms of input voltage vin (t). (b) Find an expression for the transfer function H( f ) of the circuit. (c) Find an expression for the output voltage vout (t) when vin (t) = Vin(p) sin(ωt). (d) Find an expression for peak output voltage Vo(p) when input voltage is a square wave with peak Vin(p) and frequency f . CI RI vin vout Figure 1: The circuit for Question 1. Q2 Consider the circuit shown in Figure 2. Assume Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs. (a) Design the circuit for an operating frequency of 2 kHz and a maximum gain of 20. Assume CI = 100 nF. (b) Find and sketch the output voltage vout (t) when input voltage vin (t) = 0.1 sin(2π2000t). (c) Find and sketch the output voltage vout (t) when input voltage is a square wave with ±2.5V peak and frequency 2 kHz. (d) Find the dc component in the output when there is a +50 mV dc input. Rf CI RI vin vout R2 Figure 2: The circuit for Question 2. Problem Set #11 page 1 ANU ENGN 3227 Q3 Consider the circuit shown in Figure 3. (a) Derive an expression for the output voltage vout (t) in terms of input voltage vin (t). (b) Find an expression for the transfer function H( f ) of the circuit. (c) Find an expression for the output voltage vout (t) when vin (t) = Vin(p) sin(ωt). (d) Find an expression for peak output voltage Vo(p) when input voltage is a square wave with peak Vin(p) and frequency f . RD CD vin vout Figure 3: The circuit for Question 3. Q4 Consider the circuit shown in Figure 4. Assume Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs. (a) Design the circuit for an operating frequency of 100 Hz. Assume CD = 0.1 µ F. (b) Find and sketch the output voltage vout (t) when input voltage vin (t) = 0.1 sin(2π100t). (c) Find and sketch the output voltage vout (t) when input voltage is a triangular wave with ±0.5V peak and frequency 100 Hz. CI RD CD RI vin vout R2 Figure 4: The circuit for Question 4. Problem Set #11 page 2 ANU ENGN 3227 AUSTRALIAN NATIONAL UNIVERSITY Department of Engineering ENGN 3227 Analogue Electronics Problem Set #11 Solution Q1 Solution The given circuit is CI RI vin vout (a) 0 − vin dvC +CI RI dt = 0 vout (t) = − 1 RI CI Z t t0 vin (t)dt + vC (t0 ) (b) → − V out H( f ) = → − V in = − 1 jωRI CI (c) vin (t) = Vin(p) sin(ωt) Vin(p) vout (t) = cos(ωt) ωRI CI Vin(p) Vo(p) = ωRI CI (d) Vo(p) = Vin(p) 4 f RI CI For detailed derivations, SEE: Lecture 15. Problem Set #11 page 3 ANU ENGN 3227 Q2 Partial Solution The given circuit is Rf CI RI vin vout R2 The op-amp data is Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs. (a) Given that fo = 2 kHz |Av(max) | = 20 CI = 100 nF Design: fo = 200 Hz 10 1 = 7.958 kΩ 2π fLCI Rf = 397.9 Ω |Av(max) | fL = Rf = RI = R2 = RI ||R f = 378.95 Ω Check: T τI = 500 µs = 39.79 µs We see that τI T . Hence condition of accurate integration is satisfied. (b) vin (t) = 0.1 sin(2π2000t) vout (t) = 0.5 cos(2π2000t) Check: SR = 2π foVo(p) × 10−6 = 0.00628 V/µs < 0.5 V/µs Problem Set #11 page 4 ANU ENGN 3227 The input and output voltages are shown below: 0.5 input output 0.4 0.3 0.2 Voltage 0.1 0 −0.1 −0.2 −0.3 −0.4 −0.5 0 250 500 Time (µs) 750 1000 See also PSPICE: P11_Q02_sine.sch (c) = 15.7 V = 7.85 V Vo(pp) Vo(p) Check: SR = Vo(pp) = 0.0628 V/µs < 0.5 V/µs T /2 The input and output voltages are shown below: 10 input output 8 X: 0.0005 Y: 7.854 6 X: 0.000161 Y: 2.5 4 Voltage 2 0 −2 −4 −6 −8 −10 0 250 500 Time (µs) 750 1000 See also PSPICE: P11_Q02_square.sch (d) Vin Vout = 50 mV Rf = − Vin = 1 V RI Problem Set #11 page 5 ANU ENGN 3227 Q3 Solution The given circuit is RD CD vin vout (a) CD dvC 0 − vout + dt RD = 0 vout (t) = −RDCD d vin (t) dt (b) → − V out H( f ) = → − V in = − jωRDCD (c) vin (t) = Vin(p) sin(ωt) vout (t) = −ωRDCDVin(p) cos(ωt) Vo(p) = −ωRDCDVin(p) (d) Vo(p) = 4 f RDCDVin(p) For detailed derivations, SEE: Lecture 15. Problem Set #11 page 6 ANU ENGN 3227 Q4 Partial Solution The given circuit is CI RD CD RI vin vout R2 The op-amp data is Aol = 2 × 105 , Rout = 75 Ω, Rin = 2 MΩ, fT = 1 MHz, SR = 0.5 V/µs. (a) Given that fo CD = 100 Hz = 0.1 µF Design: fD = fID = RD = RI = CI = fI = fD fD ≤ ≤ fo = 10 Hz 10 10 fo = 1 kHz 1 = 159.15 kΩ 2π fDCD 1 = 1.59 kΩ 2π fIDCD 1 = 1 nF 2π fID RD 1 = 100 kHz 2πCI RI Check: fID ≤ fI fo ≤ fID Problem Set #11 page 7 ANU ENGN 3227 (b) vin (t) = 0.1 sin(2π100t) vout (t) = −1 cos(2π100t) The input and output voltages are shown below: 1.5 input output 1 Voltage 0.5 0 −0.5 −1 −1.5 0 5 10 Time (ms) 15 20 See also PSPICE: P11_Q04_sine.sch (c) Vo(p) = 3.183 V The input and output voltages are shown below: 5 input output 4 3 X: 0.0022 Y: 3.183 2 X: 0 Y: 0.5 Voltage 1 0 −1 −2 −3 −4 −5 0 5 10 Time (ms) 15 20 See also PSPICE: P11_Q04_triang.sch Problem Set #11 page 8 ENGN3227 Analogue Electronics Matlab Scripts Dr. Salman Durrani November 2006 %% L 0 3 s i g n a l w a v e f o r m s %% ====================== %% %% S c r i p t t o g e n e r a t e d i f f e r e n t s i g n a l waveforms . %% See ENGN3227 : L e c t u r e 0 3 S l i d e s 15−20 %% %% C r e a t e d : 01 Aug 2006 %% M o d i f i e d : 01 Aug 2006 %% %% MODIFICATION HISTORY %% −−−−−−−−−−−−−−−−−−−−− %% 01 Aug 0 6 : F i l e c r e a t e d %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% C r e a t e t i m e v e c t o r ( 2 0 ms ) t s t a r t = 0; s t e p = 1e −6; t s t o p = 20 e −3; t = [ t s t a r t : step : tstop ] ; %% A 100 Hz s i n e wave f = 100; Vpeak = 1 ; p h a s e = 0∗ p i / 1 8 0 ; V s i g n a l 1 = Vpeak ∗ s i n (2∗ p i ∗ f ∗ t + p h a s e ) ; %% A 100 Hz s q u a r e wave f = 100; Vpeak = 5 ; V s i g n a l 2 = Vpeak ∗ s q u a r e (2∗ p i ∗ f ∗ t ) ; %% A 1000 Hz s i n e wave f = 1000; Vpeak = 1 ; p h a s e = 0∗ p i / 1 8 0 ; V s i g n a l 3 = Vpeak ∗ s i n (2∗ p i ∗ f ∗ t + p h a s e ) ; %% A 1000 Hz s q u a r e wave f = 1000; Vpeak = 5 ; V s i g n a l 4 = Vpeak ∗ s q u a r e (2∗ p i ∗ f ∗ t ) ; 2 %% P l o t v o l t a g e s subplot ( 2 , 2 , 1 ) p l o t ( t , V s i g n a l 1 , ’ r− ’ ) y l a b e l ( ’ V o l t a g e (V) ’ ) x l a b e l ( ’ Time ( s ) ’ ) a x i s ( [ 0 20 e−3 −1 1 ] ) g r i d on subplot ( 2 , 2 , 2 ) p l o t ( t , V s i g n a l 2 , ’ r− ’ ) y l a b e l ( ’ V o l t a g e (V) ’ ) x l a b e l ( ’ Time ( s ) ’ ) a x i s ( [ 0 20 e−3 −10 1 0 ] ) g r i d on subplot ( 2 , 2 , 3 ) p l o t ( t , V s i g n a l 3 , ’ r− ’ ) y l a b e l ( ’ V o l t a g e (V) ’ ) x l a b e l ( ’ Time ( s ) ’ ) a x i s ( [ 0 20 e−3 −1 1 ] ) g r i d on subplot ( 2 , 2 , 4 ) p l o t ( t , V s i g n a l 4 , ’ r− ’ ) y l a b e l ( ’ V o l t a g e (V) ’ ) x l a b e l ( ’ Time ( s ) ’ ) a x i s ( [ 0 20 e−3 −10 1 0 ] ) g r i d on 3 %% L 0 6 S c h m i t t T r i g g e r %% ====================== %% %% S c r i p t t o g e n e r a t e s i m u l a t e S c h m i t t T r i g g e r i n Matlab . %% See ENGN3227 : L e c t u r e 0 6 Example 03 %% %% C r e a t e d : 08 Aug 2006 %% M o d i f i e d : 08 Aug 2006 %% %% MODIFICATION HISTORY %% −−−−−−−−−−−−−−−−−−−−− %% 08 Aug 0 6 : F i l e c r e a t e d %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% S c h m i t t T r i g g e r Data V outmax = 1 0 ; R 1 = 50 e3 ; R 2 = 50 e3 ; %% C r e a t e t i m e v e c t o r ( 2 ms ) t s t a r t = 0; s t e p = 1e −6; t s t o p = 2e −3; t = [ t s t a r t : step : tstop ] ; %% Vin i s 1kHz Hz s i n e wave f = 1 e3 ; Vpeak = 5 ; p h a s e = 0∗ p i / 1 8 0 ; Vin = Vpeak ∗ s i n (2∗ p i ∗ f ∗ t + p h a s e ) ; %% S c h m i t t T r i g g e r R e f e r e n c e V o l t a g e s % Vutp = ( ( R 1 ) / ( R 1+R 2 ) ) ∗ V outmax % V l t p = ( ( R 1 ) / ( R 1+R 2 ))∗( − V outmax ) %% d e f i n e a r b i t r a r y Vutp= 4 Vltp = 0; %% Pre−a l l o c a t e S c h m i t t T r i g g e r o u t p u t Vout = −V outmax . ∗ o n e s ( 1 , length ( Vin ) ) ; %% c a l c u l a t e s l o p e o f i n p u t s i n e wave 4 g =g r a d i e n t ( Vin ) ; %% mimic S c h m i t t T r i g g e r o p e r a t i o n f o r k = 1 : 1 : length ( Vin ) i f ( ( Vin ( k ) <Vutp ) && ( Vin ( k ) > 0 ) && g ( k ) >0 ) Vout ( k ) = V outmax ; e l s e i f ( Vin ( k ) < V l t p && g ( k ) < 0 ) Vout ( k ) = V outmax ; e l s e i f ( Vin ( k ) < Vutp && g ( k ) >= 0 ) Vout ( k ) = V outmax ; end end %% P l o t v o l t a g e s eps openfig p l o t ( t , Vin , ’ k−− ’ , ’ L i n e w i d t h ’ , 1 ) hold on p l o t ( t , Vout , ’ k− ’ , ’ L i n e w i d t h ’ , 1 ) hold on p l o t ( [ 0 max( t ) ] , [ Vutp Vutp ] , ’ k : ’ , ’ L i n e w i d t h ’ , 1 ) p l o t ( [ 0 max( t ) ] , [ V l t p V l t p ] , ’ k : ’ , ’ L i n e w i d t h ’ , 1 ) y l a b e l ( ’ V o l t a g e (V) ’ ) x l a b e l ( ’ Time ( s ) ’ ) a x i s ( [ 0 2e−3 −11 1 1 ] ) legend ( ’ I n p u t ’ , ’ Output ’ ) s e t ( gca , ’ XTick ’ , [ 0 : 0 . 5 e −3:2 e −3]) % s e t ( gca , ’ YTick ’ , [ −10 −6.55 −5 −1.15 0 1 . 1 5 5 6 . 5 5 1 0 ] ) 5 %% L09 Example02 %% ============== %% S c r i p t t o p l o t m a g n i t u d e and p h a s e f o r op−amp open l o o p t r a n s f e r f u n c t i o n %% %% C r e a t e d : 23 Aug 2006 %% M o d i f i e d : 23 Aug 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% D e f i n e op−amp d a t a fT =10ˆ6; Aol = 2 ∗ 1 0 ˆ 5 ; %% D e f i n e t h e f r e q u e n c y v e c t o r f 1=l i n s p a c e ( 0 . 0 1 , 1 0 ˆ 2 , 1 0 ˆ 3 ) ; f 2=l i n s p a c e ( 1 0 ˆ 2 , 1 0 ˆ 4 , 1 0 ˆ 3 ) ; f 3=l i n s p a c e ( 1 0 ˆ 4 , fT , 1 0 ˆ 3 ) ; f= [ f 1 f 2 f 3 ] ; %% C a l c u l a t e opn l o o p cut−o f f f r e q u e n c y f c o l = fT / Aol ; %% C a l c u l a t e open l o o p t r a n s f e r f u n c t i o n A o l f = Aol ./(1+ i . ∗ ( f . / f c o l ) ) ; Magnitude = 20∗ log10 ( abs ( A o l f ) ) ; Phase = angle ( A o l f )∗180/ p i ; %% Magnitude r e s p o n s e figure semilogx ( f , Magnitude ) a x i s ([10ˆ −0 10ˆ6 0 1 1 0 ] ) y l a b e l ( ’ Magnitude ( dB ) ’ ) x l a b e l ( ’ F r e q u e n c y ( Hz ) ’ ) g r i d on %% Phase r e s p o n s e figure semilogx ( f , Phase ) a x i s ([10ˆ −2 10ˆ6 −90 0 ] ) y l a b e l ( ’ Phase ( deg ) ’ ) x l a b e l ( ’ F r e q u e n c y ( Hz ) ’ ) g r i d on %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 6 %% L10 Example02 %% ============== %% S c r i p t t o p l o t bode p l o t f o r a c t i v e low−p a s s f i l t e r . %% See L e c t u r e 1 0 : Example 02 %% %% C r e a t e d : 30 Aug 2006 %% M o d i f i e d : 30 Aug 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% D e f i n e c i r c u i t d a t a C=1e −6; R=1e3 ; R2 = 9 e3 ; R1 =1e3 ; %% D e f i n e t h e t r a n s f e r f u n c t i o n G = 1+(R2/R1 ) ; num = [ G/ (R∗C ) ] den = [ 1 1 / (R∗C ) ] ; H=t f (num , den ) %% C a l c u l a t e bode p l o t bode (H) 7 %% L11 Example02 %% ============== %% S c r i p t t o a n a l y s e B u t t e r w o r t h f i l t e r d e s i g n . %% See L e c t u r e 1 1 : B u t t e r w o r t h F i l t e r D e s i g n Example %% %% C r e a t e d : 18 Sep 2006 %% M o d i f i e d : 18 Sep 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% C a l c u l a t e t h e o r e t i c a l B u t t e r w o r t h t r a n s f e r f u n c t i o n n =2; f c= 2 e3 ; omegac = 2∗ p i ∗ f c ; [ num , den ] = b u t t e r ( n , omegac , ’ low ’ , ’ s ’ ) ; H th = t f ( num , den ) %% C a l c u l a t e s i m u l a t e d S a l l e n Key C i r c u i t t r a n s f e r f u n c t i o n R = 8200; C = 0 . 0 1 e −6; RA = 3 9 0 0 ; RF = 2 2 0 0 ; G = 1+ RF/RA ; num sk =[G/ ( ( R∗C ) ˆ 2 ) ] ; d e n s k = [ 1 (3−G) / ( R∗C) ( 1 / (R∗C ) ) ˆ 2 ] ; H sk = t f ( num sk , d e n s k ) %% P l o t s f o r s i m u l a t e d c i r c u i t [ h , w ] = f r e q s ( num sk , d e n s k ) ; f = w/(2∗ p i ) ; mag = 20∗ log10 ( abs ( h ) ) ; semilogx ( f , mag ) ; 8 %% P05 Q02 %% ======== %% S c r i p t t o c a l c u l a t e op−amp c l o s e d l o o p t r a n s f e r f u n c t i o n f o r %% Problem S e t 0 5 : Q u e s t i o n 02 %% %% C r e a t e d : 23 Aug 2006 %% M o d i f i e d : 23 Aug 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% D e f i n e op−amp d a t a fT =10ˆ6; Aol = 2 ∗ 1 0 ˆ 5 ; B = 1/23; %% D e f i n e t h e f r e q u e n c y v e c t o r f =[0 10 100 1 e3 30 e3 4 3 . 4 8 e3 1 e6 ] ’ ; %% C a l c u l a t e open l o o p cut−o f f f r e q u e n c y f c o l = fT / Aol ; %% C a l c u l a t e c l o s e d l o o p t r a n s f e r f u n c t i o n A c l f = Aol ./(1+ ( Aol ) . ∗ ( B)+ i . ∗ ( f . / f c o l ) ) Magnitude =(abs ( A c l f ) ) Phase = angle ( A c l f )∗180/ p i 9 %% P05 Q02 %% ======== %% S c r i p t t o c a l c u l a t e op−amp c l o s e d l o o p t r a n s f e r f u n c t i o n f o r %% Problem S e t 0 5 : Q u e s t i o n 02 %% %% C r e a t e d : 23 Aug 2006 %% M o d i f i e d : 23 Aug 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% D e f i n e op−amp d a t a fT =10ˆ6; Aol = 2 ∗ 1 0 ˆ 5 ; B = 1/100; %% D e f i n e t h e f r e q u e n c y v e c t o r f =[0 10 100 1 e3 10005 1 e6 ] ’ ; %% C a l c u l a t e open l o o p cut−o f f f r e q u e n c y f c o l = fT / Aol ; %% C a l c u l a t e c l o s e d l o o p t r a n s f e r f u n c t i o n A c l f = Aol ./(1+ ( Aol ) . ∗ ( B)+ i . ∗ ( f . / f c o l ) ) Magnitude =−(abs ( A c l f ) ) Phase = angle ( A c l f )∗180/ p i 10 %% P07 Q01 %% ============== %% S c r i p t t o a n a l y s e B u t t e r w o r t h f i l t e r d e s i g n . %% See L e c t u r e 1 1 : F i l t e r D e s i g n %% %% C r e a t e d : 18 Sep 2006 %% M o d i f i e d : 18 Sep 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% C a l c u l a t e t h e o r e t i c a l B u t t e r w o r t h t r a n s f e r f u n c t i o n n =2; f c= 4 e3 ; omegac = 2∗ p i ∗ f c ; [ num , den ] = b u t t e r ( n , omegac , ’ low ’ , ’ s ’ ) ; H th = t f ( num , den ) %% C a l c u l a t e s i m u l a t e d S a l l e n Key C i r c u i t t r a n s f e r f u n c t i o n R = 3900; C = 0 . 0 1 e −6; RA = 3 9 0 0 ; RF = 2 2 0 0 ; G = 1+ RF/RA ; num sk =[G/ ( ( R∗C ) ˆ 2 ) ] ; d e n s k = [ 1 (3−G) / ( R∗C) ( 1 / (R∗C ) ) ˆ 2 ] ; H sk = t f ( num sk , d e n s k ) %% P l o t s f o r s i m u l a t e d c i r c u i t f = logspace ( 1 , 6 , 1 0 0 0 ) ; [ h , w ] = f r e q s ( num sk , d e n s k , f ) ; f = w/(2∗ p i ) ; mag = 20∗ log10 ( abs ( h ) ) ; p h a s e = angle ( h )∗180/ p i ; figure semilogx ( f , mag ) ; figure semilogx ( f , p h a s e ) ; 11 %% P07 Q03 %% ============== %% S c r i p t t o a n a l y s e B e s s e l f i l t e r d e s i g n . %% See L e c t u r e 1 1 : F i l t e r D e s i g n %% %% C r e a t e d : 18 Sep 2006 %% M o d i f i e d : 18 Sep 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% C a l c u l a t e t h e o r e t i c a l B e s s e l t r a n s f e r f u n c t i o n n =2; f c= 4 e3 ; omegac = 2∗ p i ∗ f c ; [ num , den ] = b e s s e l f ( n , omegac ) ; H th = t f ( num , den ) %% C a l c u l a t e s i m u l a t e d S a l l e n Key C i r c u i t t r a n s f e r f u n c t i o n R = 3900; C = 0 . 0 1 e −6; RA = 3 9 0 0 ; RF = 1 0 0 0 ; G = 1+ RF/RA ; num sk =[G/ ( ( R∗C ) ˆ 2 ) ] ; d e n s k = [ 1 (3−G) / ( R∗C) ( 1 / (R∗C ) ) ˆ 2 ] ; H sk = t f ( num sk , d e n s k ) %% P l o t s f o r s i m u l a t e d c i r c u i t f = logspace ( 1 , 6 , 1 0 0 0 ) ; [ h , w ] = f r e q s ( num sk , d e n s k , f ) ; f = w/(2∗ p i ) ; mag = 20∗ log10 ( abs ( h ) ) ; p h a s e = angle ( h )∗180/ p i ; figure semilogx ( f , mag ) ; figure semilogx ( f , p h a s e ) ; 12 %% P08 Q03 %% ============== %% S c r i p t t o d e s i g n 555 a s t a b l e c i r c u i t s . %% See L e c t u r e 1 2 : 555 Timer C i r c u i t s %% %% C r e a t e d : 26 Sep 2006 %% M o d i f i e d : 26 Sep 2006 %% %% C o p y r i g h t ( c ) 2006 Salman D u r r a n i . %% −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− clc clear a l l close a l l %% e n g i n e e r i n g f o r m a t format s h o r t eng %% D e s i g n d a t a f =1e3 ; D= 0 . 6 %% C a l c u l a t e ON and OFF t i m e T = 1/ f ; TH = D∗T ; TL = (1−D)∗T ; %% F i n d R1 , R2 and C C = 0 . 1 e−6 i f (D R2 R1 else R2 R1 end > 0.5) = TL / ( 0 . 6 9 3 ∗ C) = (TH/ ( 0 . 6 9 3 ∗ C))−R2 = TL / ( 0 . 6 9 3 ∗ C) = (TH/ ( 0 . 6 9 3 ∗ C ) ) %% d e f a u l t f o r m a t format 13