Homework Assignment 08

Homework Assignment 08 Question 1 (Short Takes) Two points each unless otherwise indicated. 1. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance large voltage gain 2. In the circuit below I C = 1 ma and all the capacitors are large enough to be considered shorts. Estimate the midband gain A v = v o v i (3 points) (a) 6.8 (b) 3.4 (c) 272 (d) 136 (e) 12.1 Answer: (d) 3. In the circuit below I C = 1 ma and all the capacitors are large enough to be considered shorts. Estimate the midband gain A v = v o v i (3 points) (a) 12.1 (b) 8 (c) = 272 (d) 136 (e) 6.1 Answer: A v (R L R C ) 560 = 3.4 0.56 = 6.07, so (e). 1

4. In the circuit below I C = 1 ma and all the capacitors are large enough to be considered shorts. Estimate the midband gain A v = v o v i. (a) 6.8 (b) 3.4 (c) g m R C 0.04R C = 272 (d) Need additional information Answer: (b) 5. Sketch an npn BJT Darlington pair. 6. Sketch a two-transistor configuration using npn and pnp BJTs that is equivalent to a single pnp BJT. 7. True or false: the β of a transistor decreases with decreasing temperature. Answer: True, it decreases with decreasing temperature. 2

8. True or false: the β of a transistor is a function of temperature, but essentially independent of collector current. Answer: False 9. True or false: consider a BJT in the CE configuration, biased at I C = 1 ma. The smallsignal input resistance r π is in the order of 500K Answer: False. r π = β g m = β (40I C ). With β = 100, r π = 2.5K 10. True or false: consider a BJT in the CE configuration, biased at I C = 1 ma. The smallsignal input resistance r π is in the order of 500K. Answer: False. r π = β g m = β (40I C ). With β = 100, r π = 2.5K 11. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance large voltage gain 12. Estimate the voltage gain of the amplifier below if I CQ = 6.3 ma, β = 200, and C C. (a) 10 (b) 10 (c) 252 (d) 252 Answer: A v R C R E = 10, so (a). 3

13. Estimate the input resistance R i of the amplifier below if I CQ = 6.3 ma, β = 200, r π 800Ω, and C C. (3 point) (a) 17.35K (b) 18.15K (c) 37.35K (d) 9.46K Answer: Using BJT scaling, R i = 22K 82K (r π + (1 + β)r E ) = 22K 82K (r π + (1 + β)r E ) = 9.46K, so (d) 14. Consider the Bode plot of a 1 st order RC network. What is the attenuation of the network at f = 60 Hz? Provide your answer in db. (3 points) Answer: 60 Hz is log(60 2.5) = 1.38 decades higher than the 2.5 Hz corner frequency. The attenuation increases by 20 db per decade, so that at 60 Hz v o v i (in db) is 3.5 20 1.38 = 31.1 db. The attenuation is 31.1 db. An alternate calculation is 3.5 + 20 log 1 + (60 2.5) 2 = 3.11 db. 4

15. Assume the input voltage is a 1-V step function u(t). What is the long-term value of v o (t)? That is, what is v o (t) for t? Answer: In the steady state (t ), the capacitor has no effect on the circuit. The resistors form a voltage divider and v o ( ) = v s (R L (R L + R S )) = 0.5 V 16. Consider the current mirror below, and neglect base currents. What is I copy? Answer: I copy = 0.25 ma 3 = 83 μa 17. In the current mirrors below, neglect base currents and take I REF = 30 μa, What is I copy3? (a) 30 μa (b) 30 μa 3 = 10 μa (c) 30 μa 4 = 7.5 μa Answer: (a) 18. In the current mirrors below, neglect base currents and I REF = 10 μa, What is I copy? Answer: 30 μa 5

19. In the current mirrors below, neglect the base currents. What is I REF? Answer: 0.25 ma 20. True or false: the β of a transistor decreases with decreasing temperature. Answer: True, it decreases with decreasing temperature. 21. True or false: the β of a transistor is a function of temperature, but essentially independent of collector current. Answer: False 22. True or false: consider a BJT in the CE configuration, biased at I C = 1 ma. The smallsignal input resistance r π is in the order of 500K (2 points) Answer: False. r π = β g m = β (40I C ). With β = 100, r π = 2.5K 23. True or false: The input resistance of a BJT amplifier in the CB configuration, biased at 1 ma is about 25 Ω Answer: True 24. A single-pole op-amp has an open-loop low-frequency gain of A = 10 5 and an open loop, 3-dB frequency of 4 Hz. If an inverting amplifier with closed-loop low-frequency gain of A f = 50 uses this op-amp, determine the closed-loop bandwidth. (2 points) Answer: The gain-bandwidth product is 4 10 5 Hz. The bandwidth of the closed-loop amplifier is then is 4 10 5 /50 = 8 khz. 6

25. A single-pole op-amp has an open-loop gain of 100 db and a unity-gain bandwidth frequency of 2 MHz. What is the open-loop bandwidth of the op-amp? (2 points) Answer. A gain of 100 db corresponds to 10 5 and the gain-bandwidth product is 2 MHz. Thus, the open-loop bandwidth is (2 MHz) 10 5 = 20 Hz 26. A single-pole op-amp has an open-loop gain of 100 db and a unity-gain bandwidth frequency 5 MHz. What is the open-loop bandwidth of the amplifier? The amplifier is used as a voltage follower. What is the bandwidth of the follower? Answer: A gain of 100 db corresponds to 10 5 and the gain-bandwidth product is 5 MHz. Thus, the open-loop bandwidth is (5 MHz) 10 5 = 50 Hz. A unity follower will have a bandwidth of 5 MHz. 27. Consider a first-order RC low-pass filter with 3-dB frequency f = 60 Hz. By how much does it delay a 50 Hz sine wave? Express you answer in ms. (3 points) Answer: The phase shift at 60 Hz is 45 and increases at 45 / decade. 50 Hz is log(50 60) = 0.08 decades higher than 60 Hz. (The negative sign implies 50 Hz is 0.08 decades before 60 Hz.) Thus, the phase shift is 45 0.08 45 = 41.4. The period of a 50 Hz sine wave is 20 ms, so the delay is (20)(41.4 360) = 2.3 ms An alternate and more accurate calculation for the phase is tan 1 (50 60) = 39.8 and delay of 2.2 ms. 7

Question 2 Consider the circuit below. The duty cycle and frequency of the 555 astable is 60% and 10 khz respectively. (a) Specify a value for R limit to ensure that the average current through the IR diode does not exceed 30 ma (4 points) (b) Explain (2 sentences maximum) the purpose of the decoupling capacitor (1 point) (c) Give a reasonable value for the decoupling capacitor (1 point) (d) If the frequency-setting capacitor is 0.1 μf, what should R A and R B be? Specify E24 series standard resistors. (4 points) (e) What are the frequency and duty cycle with the standard resistors? (2 points) Part (a) The peak current must be I peak = 30 (0.6) = 50 ma. This value will give an average of 30 ma with a 60% on time. Assuming the V BE(ON) = 0.7 V for the BJT, then R limit = 0.7 0.05 = 14 Ω. Choose the closest standard value of 15 Ω. Part (b) When the FET switches, large current spikes may appear on the supply rail, which can propagate into the IC and disturb its operation. The decoupling capacitor provides a local reservoir of energy, and ensures a clean power supply rail. Part (c) A good first try would be 0.1 μf. Part (d) Calculated (using an online applet) values are R A = 288 Ω, R B = 576 Ω with closest standard values 300 Ω and 560 Ω respectively. Part (e) With standard values, f = 10.1 khz, and D = 60.6 %. 8

Question 3 The datasheet for a 5 V, three-terminal regulator indicate that the output voltage typically changes by 3 mv when the input voltage is varied from 7 V to 25 V, and by 5 mv when the load current is varied from 0.25 A to 0.75 A. Further, the ripple rejection ratio is 78 db at 120 Hz. (a) Estimate the typical line- and load regulation for the regulator. (4 points) (b) What is the output resistance of the regulator? (2 points) (c) Estimate the output ripple amplitude for every volt of input ripple at 120 Hz. (3 points) Part(a) Line Regulation = ΔV O 100% = 3 10 3 100% = 0.017% ΔV I (25 7) Load Regulation = V O(NL) V O(FL) V O(NL) 100 = 5 10 3 5 100 = 0.1% Part (b) Output Resistance = ΔV O = 5 10 3 = 10 mω ΔI O (0.75 0.25) Part (c) Ripple Rejection (db) = 20 log 10 V RI V RO = 78 V RO = (0.126 10 3 )V RI Thus, a 1-V, 120-Hz ripple at the input will result in an output ripple of 0.126 mv 9

Question 4 An n-channel MOSFET with V TN = 1 V, K n = 0.8 ma/v 2 is biased to operate in its saturation region with I D = 1 ma. Determine the transconductance g m. (3 points) If v GS changes with 1 mv, by how does the drain current change? (3 points) g m = 2 K n I D = 2 (0.8 10 3 )(1 10 3 ) = 1.78 10 3 A/V = 1.78 ma/v i d = g m v gs = (1.78 10 3 )(1 10 3 ) = 1.78 μa Question 5 For the circuit shown, V PS = 5 V and = 10K. The varactor characteristics are shown in the graph. What is the bandwidth of the circuit? (3 points) From the graph, the varactor has a capacitance of 100 pf with a 5-V reverse voltage. The timeconstant of the circuit is τ = RC = (10 10 3 )(100 10 12 ) = 1 μs. The bandwidth is then B = 1 (2πτ) = 159 khz. 10

Question 6 K n = 93 ma V 2 V TN = 2 V λ = 0 C S = 100 μf C C = 1 μf (a) Find I DQ for the circuit (6 points) (b) Determine g m (2 points) (c) Determine the voltage gain of the circuit (2 points) Part (a) The gate current is zero, so V G = Further V S = I D R S = K n (V GS V TN ) 2 Thus 56 15 = 5.39 V 100 + 56 V GS = V G V S = 5.39 K n (V GS V TN ) 2 (R S ) V GS = 5.39 0.093(V GS 2) 2 (100) Solving using trial-and-error gives 2.55 V. The drain current is I D = K n (V GS V TN ) 2 = (0.093)(2.55 2) 2 = 28.13 ma Part (b) g m = 2 K n I D = 2 0.093 0.0284 = 0.102 A/V. Part (c) The small-signal model is given below where R G = 100K 56K. From the model the voltage gain is A V = g m (220 2.2K) = 20.5 11

Question 7 A MOEFET amplifier along with the FET and circuit parameters are shown below. C C1, C C2 are coupling capacitors. Determine the quiescent values for I D and V DS (8 points) K n = 1 ma V 2 V TN = 2 V λ = 0 V DD = 12 V R S = 2K R D = 3K R 1 = 300K R 2 = 200K R Si = 2K R L = 3K The gate current is zero, so V G = 200 12 = 4.8 V 200 + 300 Further V S = I D R S = K n (V GS V TN ) 2 R S Thus Solving yields V GS = 2.96 V. Thus Further, V GS = V G V S = 4.8 K n (V GS V TN ) 2 (R S ) V GS = 4.8 0.001(V GS 2) 2 (2K) I D = K n (V GS V TN ) 2 = 0.001(2.96 2) 2 = 0.922 ma V DS = V D V S = 12 I D R D I D R S = 12 4.61 = 7.4 V 12

Question 8 For the circuit below, show that V GS = 3.55 V. State any assumptions that you make. (8 points) K n = 0.5 ma/v 2, V TN = 2 V, λ = 0, C gd = 0.1 pf, C gs = 1 pf Assume the FET operates in saturation mode. No current flows into gate, so that 166 V G = 10 166 + 234 = 4.15V I D = K n (V GS V TN ) 2 Further, V GS = V G I D R S = 4.15 I D R S, so that I D = 0.5 (4.15 I D R S ) 2 2 ma If V GS = 3.55 V, then I D = K n (V GS V TN ) 2 gives I D = 1.2 ma. Substituting I D = 1.2 ma shows that the right hand side and the left hand is equal so that V GS is indeed 3.55 V 13

Question 9 For the circuit below, determine the dc load line equation for the BJT, incorporating β. That is, do not assume that I E I C. (8 points) R C = 5K R E = 5K R L = 5K β = 100 V CC = +12 V V DD = 5 V The load line equation is an equation that expresses I C as a function of V CE. If R L were absent, one would start with, for example: V CC = I C R C + V CE + I E R E + V DD and note that I E = ((1 + β) β)i C, substitute that into the equation above an express I C as a function of V CE. With R L present, one has to take a different approach. The simplest is to replace V CC, R C, and R L with a Thevenin equivalent circuit as shown below. R TH = R L R C = 2.5K V TH = R L V R L + R CC = 6 V C Now V TH = I C R TH + V CE + I E R E + V DD I E = 1 + β β I C = 1.01I C Combining these equations, substituting values, and rearranging, gives 1 I C = 7.55 10 3 (11 V CE) 14

Question 10 For the following circuit, assume that the BJT is operating in the forward-active region. V BE(on) = 0.7 V V A = β = 50 (a) Determine I C (5 points) (b) Determine the power dissipated by the BJT (3 points) Part (a) The voltage at the base is 9 0.7 = 8.3 V. KCL at the collector (using the convention that currents into the node is positive) gives I C V C 1K + I B = 0 Using I B = (8.3 V C ) 18K, and I C = 50I B, this equation becomes Solving yields V C = 6.14 V, so that Part (b) 50 8.3 V C 18K V C 1K + 8.3 V C 18K = 0 I B = (8.3 V C ) 18K = 0.12 ma I C = 50I B = 6 ma V C = (I B + I C )(1K) = 6.12 V V CE = 9 V C = 2.88 V P = V CE I C = 17.28 mw 15

Question 11 Consider the amplifier below, which amplifies the signal from a sensor with an internal resistance of 1K. Ignore BJT s output resistance, and assume C 1 = C 2 = C 3. β = 100 I C = 0.245 ma (a) Determine g m, r π (4 points) (b) Using BJT scaling, determine R i see figure (4 points) (c) Using the ratio of the collector and emitter resistors, estimate the overall voltage gain A v = v o v s (4 points) Part (a) Part (b) g m = 40 I C = 9.8 ms, r π = β g m = 10.2 K R i = R 1 R 2 [r π + (1 + β)r E ] = 300K 160K [10.2K + 101 3K] = 78.3K Part (c) The effective collector resistance is R C = 22K 100K = 18K and the sensor s internal resistance and R i form a voltage divider. Thus A v = v o R i R C = 78.3 v s R S + R i R E 1 + 78.3 18K 3K = 5.93 16

Question 12 (Principles, Own) An engineer measures the bandwidth of the circuit below by driving it with a sinusoidal signal and measuring the attenuation at various frequencies. She uses a scope with an input impedance of 1 MΩ with a 1 probe, and then a 10 probe. Complete the following table (6 points) True BW in Hz Measured BW in Hz ( 1) probe Measured BW in Hz ( 10) probe This is a simple 1 st order system with bandwidth B = 1 (2πRC). Here R is the equivalent resistance that the capacitor C sees. True bandwidth calculation: the Thevenin equivalent resistance that C sees is simply R 1, so B True = 1 2πRC = 1 2π(910 10 3 )(1 10 9 = 175 Hz ) 1 Probe bandwidth calculation: the probe+scope has a 1M resistance that is effectively in parallel with R 1, so that the Thevenin equivalent resistance that C sees is 1M R 1, so that B 1 probe = 1 2π(R R scope )C = 1 2π(910 10 3 ) (1 10 6 )(1 10 9 = 334 Hz ) 10 Probe bandwidth calculation: 10 probes increase the 1 MΩ probe+scope resistance to 10 MΩ, so that the Thevenin equivalent resistance that C sees is 10M R 1, so that B 1 probe = 1 2π(R R scope )C = 1 2π(910 10 3 ) (10 10 6 )(1 10 9 = 191 Hz ) True BW in Hz Measured BW in Hz ( 1) probe Measured B in Hz ( 10) probe 175 Hz 334 Hz 1901 Hz 17

Question 13 For the circuit below β = 120, V BE(ON) = 0.7 V, and V A = R 1 = 110 kω R 2 = 82 kω β = 120 V A = V BE(ON ) = 0.7 V (a) Determine I CQ, g m, and r π (8 points) (b) Determine the output resistance R o using BJT scaling (4 points) Part (a) The Thevenin equivalent circuit for the base-emitter bias circuit is R TH = R 1 R 2 = 47K R 2 V TH = 12 = 5.1 V R 1 + R 2 V BE = 0.7 V β = 120 V TH 0.7 I B = = 8.3 μa R TH + (1 + β)r E I C = βi B = 1 ma Part (b) g m = 40 I C = 40 ms, and r π = β g m = 3K Part (c) Using BJT impedance scaling R o = R E r π 1 + β = 4K 3K 25 Ω 1 + 120 18

The table below compares the design values with a SPICE simulation that started with the 2N2222 BJT with the Early voltage and β replaced with VAF = 1G and BF = 120. Parameter Calculated/Design SPICE I CQ 1 ma 1.02 ma V BE 0.7 V 0.642 V I BQ 8.33 μa 12.5 µa V RE 4 V 3.89 V 19