Final Exam. 1. An engineer measures the (step response) rise time of an amplifier as t r = 0.1 μs. Estimate the 3 db bandwidth of the amplifier.

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1 Final Exam Name: Score /100 Question 1 Short Takes 1 point each unless noted otherwise. 1. An engineer measures the (step response) rise time of an amplifier as t r = 0.1 μs. Estimate the 3 db bandwidth of the amplifier. Answer BW = = 3.5 MHz t r Below as possible ordering for V γ for several LEDs. Circle the correct one. (a) V γ(red) < V γ(green) < V γ(blue) < V γ(white) (b) V γ(white) < V γ(green) < V γ(blue) < V γ(yellow) (c) V γ(blue) < V γ(red) < V γ(yellow) < V γ(white) (d) V γ(red) < V γ(green) < V γ(white) < V γ(blue) Answer: (a) 3. Which of the following depicts the correct current direction? Circle one. Answer: (a) 4. A forward-biased diode has a current I F and a diffusion capacitance C D. The diode current is halved doubled. The resulting diffusion capacitance is then (circle one): (a) C D (b) C D (c) C D (d) C D (e) Unchanged Answer: (a) 5. Give one phrase/sentence that describes the primary advantage of an active load. Answer: Large effective resistance large voltage gain 6. A MOSFET in a circuit is replaced with another that has a W L that is 50% larger. Assuming everything else stays the same, the drain current will 1 (a) Decrease by 50% (b) Increases by 50% (c) Quadruple (d) Stay the same Answer: (b)

2 7. Assume that your SPICE simulation software (such as Micro-Cap SPICE) do not have a photodiode part. Explain in 1 sentences how you can nevertheless simulate a photodiode. Answer: One can model a photodiode with a current source. 8. Classify the filter circuit below. (a) Low-pass filter (b) High-pass filter (c) Band-pass filter (d) Notch filter Answer: This is a two-pole high-pass filter, so (b) is the answer. 9. Consider the following drive circuit for an IR remote control. The IR diode is replaced with another IR diode that has a turn-on voltage that is 0% lower. The new peak current through the IR diode will be a) Unchanged b) Increased by 0% c) Decreased by 0% d) Decreased much than 0%, since I D = I S e V D/V T Answer: (a) 10. The output voltage of a three-terminal voltage regulator is ma load, and A load. What is the regulator s output resistance? (a) 33 mω (b) 660 Ω (c). Ω Answer: R = ΔV ΔI = = 33 mω, so (a) is the answer. 11. 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)

3 1. In the circuit below, R 1 = 10K, R = 15K, and R 3 compensates for the op-amp s input bias current. What should it s value be to be effective? (a) 10K (b) 15K (c) 6K (d) 5K (e) Need additional information (I OS ) Answer: Choose R 3 = R 1 R = 6K, so (c) is the answer. 13. In the circuit below, what is the maximum current that can flow through R L? Make reasonable assumptions. ( points) Answer: Assume that for Q, V BE(ON) = 0.7 V. Thus, Q will turn on and starve Q 1 from additional base current when the current through R 1 (which is also the current through R L ) is I = = 0.47 A. 14. 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 ( points) (a) 6.8 (b) 3.4 (c) g m R C = 0.04R C = 7 (d) g m R C = 0.04(R C R L ) = 136 Answer: (d) 15. A power MOSFET has rated power of 150 W at an ambient temperature T A = 5 o C and a maximum specified junction temperature of 175 o C. What is the thermal resistance between the junction and device case? ( points) Answer: θ dev case = (175 5) 150 = 0.1 C/W 3

4 16. What is the voltage gain A v = v o v s of the amplifier below if g m = 0.04 S and r o = 100K? ( points) (a) 400 (b) 400 (c) Need additional information (i.e., r π ) (d) 364 (e) 364 Answer: A v = g m (r o 10K) = 0.04(100K 10K) = , so (e). 17. An engineer designs a MOSFET-based class-ab amplifier to deliver 6.5 W (sinusoidal) signal power to a 4 Ω resistive load. What is the required peak-to-peak voltage swing across the load? ( points) (a) 9.77 V (b) V (c) 10 V (d) V (e) 7.07 V Answer: P = V rms R, so that V rms = 5 V, so that V pp = V, so (d). 18. 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. ( points) Answer: The gain-bandwidth product is Hz. The bandwidth of the closed-loop amplifier is then is /50 = 8 khz. 19. 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? ( points) 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. 0. An amplifier has gain of 800. After adding negative feedback, the gain is measured as 5. Find the loop gain. ( points) Answer: A f = A OL (1 + βa OL ) so that 5 = 800 ( β). Solving for T = 800β yields the loop gain T = 31. 4

5 1. An amplifier with gain of 00 has a 10% variation in gain over a certain frequency range. Using negative feedback, what value of β should one use to reduce the gain variation to 1%? ( points) Answer: The improvement factor we want from the negative feedback is ΔA OL ) ΔA f = 10% 1% = 10. Therefore, (1 + βa OL ) = 10 (1 + 00β) = 10 β = 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 ( points) (a) 1.1 (b) 5.7 (c) g m R C = 0.04R C = 7 (d) g m R C = 0.04(R C R L ) = 136 Answer: A v (R L R C ) 560 = = An engineer wants to reduce the output amplitude of the Wien bridge oscillator below by adjusting R 5. Should she increase or decrease the resistor s value? Briefly explain you answer. ( points) Answer: She should decrease R 5 s value, as this will cause the diodes to turn on at a lower output voltage. With the diodes turned on, the negative feedback resistance consisting of R 5, R 5, and the diodes are reduced, which reduces the overall gain. 4. An amplifier has gain of 100,000, and a 0% variation in gain over a certain temperature range. Negative feedback is used to reduce the gain to 10. What is the variation in gain with temperature of the feedback amplifier? ( points) Answer. The gain is reduced by 1 + βa OL = 100, = 10,000. The temperature variations are reduced by the same factor, so the feedback amplifier s gain varies by 0% 10 4 = 0.00% 5

6 Question The transistor in the amplifier shown has β = 350 and V BE(ON) = 0.65 V. Further, C c = 1 μf. (a) Show that I CQ 1 ma ( points) (b) Show that R i 13.7K (3 points) (c) Estimate the lower 3-dB frequency (3 points) Part (a) Since β is large, ignore I BQ so that V B = (9)(7 ( ) ) = 1.9 V. Since V BE(ON) = 0.65 V, then V RE = = 1.5 V. Consequently, I CQ I E = K = 0.96 ma 1 ma. Part (b) r π = β g m = I CQ = 8.75K. Using BJT scaling, Part (c) f 3dB = R i = 65K 18K r π + (1 + β)(1.3k) = 13.68K 1 1 = πr i C c (π)(13.68k)( = Hz ) For comparison, SPICE gives r π = 8.15K, R i = 13.59K, f 3dB = 11.8 Hz 6

7 Question 3 For the non-inverting op-amp circuit, the parameters are A OL = 10 5, A vf = 0, R i = 100K, and R o = 100 Ω. Determine R if and R of respectively. The op-amp has a single open-loop pole at 10 Hz. Determine the closed-loop bandwidth. (8 points) A f = A OL 1 + βa OL 0 = βa OL 1 + βa OL = 5,000 R if = (1 + βa OL )(100K) = 500.1M R of = βa OL = 0 mω BW = (1 + βa OL )(10) = 50 khz Question 4 A power MOSFET has thermal characteristics given below and dissipates 5 W. Design, by specifying the thermal resistance, a heat sink that will ensure the MOSFET does not overheat. The ambient temperature is 5. Assume one can keep the thermal resistance between the MOSFET case and heat sink (θ case sink or θ CS ) below 1 W. (4 points) θ juntion case = θ JC = 1.75 /W, θ case ambient = θ CA = 50 /W T j,max = 150 A thermal model that captures the information is shown below: T j = T A + P D θ JC + θ CS + θ SA 150 = 5 + 5( θ SA ) θ A =.5 /W The heat sink s thermal resistance must be less than this value. Note: θ SA is not part of the picture, since we will replace it with a much lower θ SA 7

8 Question 5 Below is a small-signal model of a BJT amplifier. Determine the so-called Miller capacitance C M, and draw an equivalent small-signal circuit that incorporates C M. Next, determine the circuit time constant, 3-dB frequency, and the midband gain. Finally, does this amplifier have a high-pass or low-pass response? (8 points) R L = K g m = 0.04 A V r π = 5 K R S = 5 K C π = 10 pf C μ = pf The gain that works across C μ is g m R L = 80. The Miller capacitance is C M = (1 A V )C μ = (81)() = 16 pf. A small-signal model that incorporates C M is shown below. The circuit time constant is τ = C π C μ (r π R s ) = (16 pf + 10 pf)(.5k) = 430 ns. The 3-dB frequency is f 3dB = 1 (πτ) = 370 khz. The midband gain is A v(mid) = The amplifier has a low-pass response. r π R s + r π ( g m R L ) = 40 8

9 Question 6 Consider the amplifier shown. The maximum power the transistor may dissipate is P Q,max = 5 W, and C. (a) Determine a load resistance R L so that the amplifier can deliver maximum power to it. ( points) (b) What is the maximum signal power the amplifier can deliver to R L? Assume sinusoidal input signal. ( points) (c) For V p = 5 mv, determine the signal power dissipated in the load. (4 points) For the power calculations, neglect the base current Part (a) The transistor will dissipate the maximum power (5 W) and deliver maximum power to a load when V C = V CC = 1 V. One can derive this on-the-fly, but this is a result that we derived in class and used many times in this course. From this follows that I CQ = 5 1 =.0843 A, and R L = = 5.76 Ω. Part (b) The power supplied by the power supply is P Supply = V CC I CQ = 50%. This is a class A amplifier that a maximum efficiency of 5%, so the maximum power the amplifier can supply to R L = 1.5 W. Alternatively, the peak-to-peak voltage swing across the 5.67 Ω load resistor is 4 V witch corresponds to an V rms load voltage, and a power dissipation of R L = = 1.5 W V rms Part (c) The gain of the amplifier is A V = g m R L = 40I CQ R L = (40)(.083)(5.76) = 480, so that the amplitude of the signal output voltage is =.4 V. The signal power dissipated in the resistor is R L = V p (R L ) =.4 ( 5.76) = 500 mw = 0.5 W V rms 9

10 Question 7 The circuit shown is the input buffer of the IR receiver you built in the final lab, and R L is the input resistance of the next amplifier in the chain. Assume that your dc analysis show that I D = 0.5 ma, and that K n = 0.5 ma V, V TN = V, and λ = 0. Ignore the transistor s internal capacitances and assume that C c1 is large enough to be considered a short. (a) Show that R O = (1 g m ) R S 1 g m (6 points) (b) Determine C so that f 3dB = khz. (4 points) Part (a) A small-signal model to determine R o is shown: the input is shorted and a test source V x drives the FET s source, resulting in a current I x, and R o = V x I x. Further, R G = R 1 R R ph KCL at the source is g m v gs V x R s + I x = 0 However, from the circuit it follows that v gs = V x, so that V x g m V x R s + I x = 0 R o = V x I x = 1 g m + 1 = 1 R g s = 1K 10K 1K m R s We used the fact that g m = K n I D = 1 ma V so that 1 g m = 1K. 10

11 Page Intentionally Left Blank Part (b) The circuit time constant is τ = C C (R o + R L ). For a -khz corner frequency, f 3dB = 1 πτ 103 = Solving yields C C = 8.8 nf. Use standard values of 8. nf. 1 π(1k + 8K)C C 11

12 Question 8 Below is a screen dump of a SPICE simulation of a Wien bridge oscillator that uses an incandescent lamp for gain control. The graph depicts the lamp resistance as a function of the voltage across the lamp. (a) Determine the frequency of oscillation in Hz ( points) (b) Determine the voltage at which the output stabilizes (8 points) (a) (b) Part (a) R 1 = R = R and C 1 = C = C and f = 1 (πrc) = Hz Part (b) At startup R lamp 9 Ω, and the loop gain T = R 4 R lamp = 3.5, enough for a Wien bridge oscillator to start oscillating. As the output amplitude grows, V lamp grows, R lamp increases. However when R lamp increases, the loop gain T = R 4 R lamp decreases. The loop stabilizes when T = R 4 R lamp = 3 R lamp = 1 Ω From the plot R lamp = 1 Ω when the V lamp 0.38 V. The current flowing through the lamp is = ma. The voltage at the output is (0.018)( ) = 3.4 V. 1

13 Question 9 The open-loop voltage amplification of an amplifier is A v = jf With β v = 0.005, find the phase- and gain margins. Determine if the amplifier is stable. Give the gain margin in db. (10 points) For the phase margin, we check the phase of the loop gain T = βa v where T = 1. The frequency where the gain equals 1, is where T = βa V = β jf = f 10 5 Solving yields f = Hz. The phase at this frequency is φ = 3 tan 1 f = 163 Thus, the phase margin is = = 1 For the gain margin, we determine the magnitude of the loop gain at the frequency where the phase shift is 180. The frequency where the phase shift is 180, is where φ = 3 tan 1 f = 180 Solving yields f = Hz. The magnitude at this frequency is = 0.64 = 4 db The loop gain is less that 1 ( 0 db), so the system is stable. The gain margin is 4 db. 13

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