Homework Assignment 13

Transcription

1 Question 1 Short Takes 2 points each. Homework Assignment Classify the type of feedback uses in the circuit below (i.e., shunt-shunt, series-shunt, ) Answer: Series-shunt. 2. True or false: an engineer uses series-shunt negative feedback to extend the bandwidth of a voltage amplifier this will also increase the input resistance. Answer: True 3. What type of negative feedback (series-shunt, series-series, ) is used in the following amplifier? Answer: Shunt-shunt 4. True or false: voltage regulators use negative feedback to stabilize/regulate their output voltages: a side effect is that their output resistances are high. Answer: False 5. An amplifier has gain of 800. After adding negative feedback, the gain is measured as 25. Find the loop gain. Answer. A f = A OL (1 + βa OL ) so that 25 = 800 ( β). Solving for T = 800β yields the loop gain T = 31. 1

2 6. An amplifier has gain of 800. After adding negative feedback, the gain is measured as 25. Find the feedback factor. Answer A f = A OL (1 + βa OL ) so that 25 = 800 ( β). Solving for β yields β = An amplifier with gain of 200 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%? Answer. The improvement factor we want from the negative feedback is ΔA OL ) ΔA f = 10% 1% = 10. Therefore, (1 + βa OL ) = 10 ( β) = 10 β = An amplifier has gain of 100,000, and a 20% 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? 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 20% 10 4 = 0.002% 9. An op-amp has an open-loop gain of 120 db and an input resistance of 50 MΩ. An engineer wants to use negative feedback to obtain an amplifier with input resistance of 5 GΩ. What is the gain (in db) of the feedback amplifier? (2 points) Answer. Negative feedback increases the resistance by ( ) ( ) = 100 (or 40 db) and reduces the gain by the same factor, so the feedback amplifier s gain is 80 db. 10. 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. Answer. The gain-bandwidth product is Hz. The bandwidth of the closed-loop amplifier is then is /50 = 8 khz. 11. 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. 2

3 Question 2 A certain audio power amplifier with a signal gain of 10 V/V is found to produce a 2-V peak-to-peak 60-Hz hum. We wish to reduce the output hum to less than 1 mv peak-topeak without changing the signal gain. To this end, we precede the power stage with a preamplifier stage with gain a 1 and then apply negative feedback around the composite amplifier. What are the required values of a 1 and β? Provide β to four significant digits. (6 points) Original amplifier with 60-Hz hum problem. Preamplifier and negative feedback to fix hum problem. We need to reduce the hum by a factor (2 V) (1 mv) = 2,000. Thus, the improvement factor is 1 + βa = 2,000 (1) where A is the open-loop gain of the composite amplifier the original 10 amplifier and the preamplifier, namely A = 10a 1. The closed-loop gain of the composite amplifier must be 10, so A f = A 10 = A 1 + β Aol 1 + βa (2) We have two unknowns β and A, and two equations. Solving yields β = , and A = 20,000. Thus, a 1 = A 10 = 2,000. 3

4 Question 3 The amplifier below has an open-loop gain A OL = 80 db. What is β, the loop gain T, and the closed-loop gain A f? (6 points) R 1 = 1K R 2 = 47K R L = 4.7K R 1, R 2 form a voltage divider and feeds back a fraction β = 1 (1 + 47) = of the output voltage. The loop gain is T = βa OL = ( )(10 4 ) = The closed-loop gain is A OL (1 + βa OL ) = 10 4 ( ) = Question 4 For the non-inverting op-amp circuit below, the parameters are A = 10 5, A vf = 20, R i = 100K, and R o = 100 Ω. Determine R if and R of respectively (6 points) A f = A OL 1 + βa OL 20 = βa OL 1 + βa OL = 5,000 R if = (1 + βa OL )(100K) = 500.1M R of = βa OL = 20 mω 4

5 Question 5 The parameters of the ideal shunt-series amplifier below are I i = 20 μa, I fb = 19 μa, R i = 500 Ω, R o = 20K, and β i = A/A.. The open-loop gain is A i = 2,000 A A. Determine the values and units for I ε, I o, A f, R if, and R of. (8 points) I ε = 20 μa 19 μa = 1 μa I fb = β i I o I o = I fb = β i = 2 ma For the overall (feedback) amplifier A f = A i 2,000 = = 100 A/A 1 + β i A i 1 + (0.0095)(2,000) Alternatively, use the approximation A f 1 β i = = A/A. One could also determine I O as follows R if = I O = A f I i = ( )(100) = 2 ma R i 500 = 1 + β i A i 1 + (0.0095)(2,000) = 25 Ω R of = (1 + β i A i )R o = 1 + (0.0095)(2,000) (20K) = 400K 5

6 Question 6 The open-loop gain and input resistance of the op-amp below is 10 6 and 1 MΩ respectively. Further, R 1 = 99K, R 2 = 1K. What is the closed-loop gain and input resistance? (5 points) This is series-shunt (voltage-voltage) feedback, with β = R 2 (R 1 + R 2 ) = Further, 1 + βa OL = 1 + (0.01)(10 6 ) = Thus A f = A OL 1 + βa OL = = 100 R if = R i (1 + βa OL ) = (10 6 )(10 4 ) = 10 4 MΩ Question 7 An op-amp having a single-pole at 100 Hz, and a low-frequency gain of 10 5 is operated is a feedback loop with β = (a) What is the factor which feedback shifts the pole? (2 points) (b) To what frequency? (2 points) (c) If β is changed to a value that results in a closed loop gain of +1, to what frequency does the pole shift? (2 points) Part (a). Feedback shifts the pole by a factor (1 + βa) = =1001 Part (b). The pole is shifted to = khz Part (c) A f = 1 = A OL 1 + βa OL = β10 5 Thus (1 + βa OL ) = Feedback scales the bandwidth/shifts the pole by this factor to = 10 MHz. Alternate solution. The gain-bandwidth product is MHz, so that when the gain is 1, the bandwidth is 10 MHz. 6

7 Question 8 Part (a) of the figure below shows a non-feedback amplifier with gain A that delivers 5 W into a 5 Ω speaker when the amplifier input is 50 mv rms. The nonlinear distortion in the amplifier output is 1% of the total signal. In part (b) negative feedback is employed to reduce the nonlinear distortion. A preamplifier is used to compensate for changes in gain the feedback introduces (a) Find the numerical value of the voltage gain A. (3 points). (b) Find the value of β required to reduce the distortion to 0.1% with the same output signal amplitude. Find the value of the preamplifier voltage gain A PR. You can assume the preamplifier nonlinear distortion is negligible. (5 points) Part (a) v L(rms) = PR L = 5 5 = 5 V A = 5/( ) = 100 (open loop) Part (b) We need to reduce the nonlinear distortion by a factor 10. Thus The closed-loop gain is 1 + βa = 10 β = 0.09 A f = βa = 10 To provide the same output voltage, the pre-amplifier must have a voltage gain of 10. 7

8 Question 9 Consider an op-amp having a single-pole open-loop response with A o = 10 5 and an open-loop 3-dB bandwidth of 10 Hz. The amplifier is ideal otherwise. The amplifier is connected in the non-inverting configuration with a nominal low-frequency closed-loop gain of 100. (a) Find the feedback factor β. (2 points) (b) Make neat Bode plots showing the open-loop gain and the phase of the open-loop amplifier. (6 points) (c) Add a plot of the loop gain T to you figure and find the frequency at which T = 1. (3 points) (e) Find the phase margin of the closed loop amplifier. (3 points) (f) Is the amplifier stable? (1 point) A o A f = 1 + βa o 100 = β10 5 β 0.01 A plot of βa o is identical to that of A o, except that the magnitude is scaled by β. The phase is the same. From the plot, T = βa o = 1 at f = The phase at f = 10 4 is 90, leaving another is 90 before the phase shift becomes is 180 and making the amplifier unstable. Thus, the phase margin is 90. The feedback amplifier is stable. 8

9 Question 10 The open-loop voltage gain of an amplifier is given by A v = j f f j 10 5 An engineer used the amplifier to design a feedback amplifier with closed-loop gain A fv = 100. Will the amplifier be stable? If so, what is the phase margin? (15 points) Determine β for a closed-loop gain of 100: A v A vf = 1 + βa v = 1 + β(10 5 ) β = Find the frequency where the magnitude of the loop gain function is 1: T(f) = (10 5 ) 1 + f f Program this value into a programmable calculator and try different values for f to find f Hz. The phase of the loop gain function T(f) = βa v at this frequency is φ = tan 1 f f tan The phase margin is then = 18.2 o = = tan 10 3 tan 10 5 = o 9

10 Question 11 Consider a feedback amplifier with loop gain transfer function β(100) T(s) = s Determine the stability with β = 0.2 (12 points) Substitute s = jω = j2πf to find the loop gain and β = 0.2 β(100) T(f) = 3 2πf = 20 3 tan 1 f 1 + j πf 2 3/ The frequency where the phase become 180 is 3 tan 1 f = 180 f 180 = 173 khz The magnitude of loop gain at this frequency is T(f) = 20 = πf 2 3/ Thus, the amplifier is unstable. 10