Homework Assignment 03 Question 1 (Short Takes), 2 points each unless otherwise noted. 1. Two 0.68 μf capacitors are connected in series across a 10 khz sine wave signal source. The total capacitive reactance is: (a) 46.8 Ω (b) 4.68 Ω (c) 7.45 Ω (d) 21 Ω Answer: The total capacitance is 0.34 μf. The reactance at 10 khz is X c = 1 (2πfC) = 1/(2π 10 10 3 3.4 10 6 = 46.8 Ω. Thus, (a) is the answer. 2. In a VAC series RC circuit, if 20 VAC is measured across the resistor and 40 VAC is measured across the capacitor, the magnitude of the applied voltage is: (a) 60 VAC (b) 55 VAC (c) 50 VAC (d) 45 VAC Answer: The applied voltage is V IN = V R + jv C, so that V IN = V R 2 + V C 2 = 2,000 45 VAC. Thus (d) is the answer. 3. What is the magnitude of the current phase angle for a 5.6 μf capacitor and a 50-Ω resistor in series with a 1.1 khz, 5 VAC source? (a) 72.9 (b) 62.7 (c) 27.3 (d) 17.1 Answer: The impedance of the RC circuit is = R 1 j2πfc = 50 j25.84 Ω. The magnitude of the phase angle is tan 1 ( 25.84 50) = 27.3. Thus, (c) is the answer. 4. In the circuit, V 1 = 1.5 V, V 2 = 2.5 V, and all the resistors have value 20K. The output voltage is (a) 4.5 V (b) +4.5 V (c) 4 V (d) 0.5 V (e) +0.5 V Answer: This is a summing inverter with V out = V 1 (R 2 R 1 ) V 2 (R 2 R 1 ). Since all the resistors have same value, V out = V 1 V 2 = 4 V 1
5. An engineer measures the (step response) rise time of an amplifier as t r = 0.7 μs. Estimate the 3 db bandwidth of the amplifier. 6. What is the time constant of the circuit? Answer: BW 0.35 t r 0.35 = 0.7 10 6 = 500 khz Answer: The resistance the capacitor sees is R TH = 10K 10K = 5K, so the time constant is τ = R TH C L = (5 10 3 )(1 10 6 ) = 5 ms. 7. Consider the amplifier below. V in = 1.5 V, what is V out? Answer: Gain of 1 st amplifier is A f = (30) 10 = 3 and gain of the 2 nd amplifier is 1. Thus, overall gain is ( 3)( 1) = 3. The output voltage is thus V out = 1.5 3 = 4.5 V. 8. A differential amplifier has a common-mode gain of 0.2 and a common-mode rejection ratio of 3250. What would the output voltage be if the single-ended input voltage was 7 mv rms? (a) 1.4 mv rms (b) 650 mv rms (c) 4.55 V rms (d) 0.455 V rms Answer: CMMR = A d A c so that 3250 = A d 0.2 and A d = 650. The output voltage is 650 7 = 4.55 mv rms. 9. An amplifier has a differential gain of -50,000 and a common-mode gain of 2. What is the common-mode rejection ratio? (a) 87.96 db (b) 44 db (c) -44 db (d) 87.96 db Answer: CMMR = 20 log 10 A d A c = 20 log 10 50 10 3 2 = 87.96 db, so the answer is (d). 2
10. If the feedback/input resistor ratio of a feedback amplifier is 4.6 with 1.7 V applied to the noninverting input, what is the output voltage value? (a) 7.82 V (b) Saturation (c) Cutoff (d) 9.52 V Answer: A v = 1 + R f R 1 = (1 + 4.6) = 5.6. The output voltage is then V O = 1.7 5.6 = 9.52 V. Thus, the answer is (d) 11. The output of the circuit shown is (a) Sine wave with frequency ω rad/s (b) Square wave with frequency ω 2π Hz (c) Triangular wave with frequency ω rad/s (d) Need additional information Answer: With no feedback, the circuit is highly non-linear and operates as a comparator, comparing the input amplitude against 0 V. The output is a square wave for frequency ω 2π Hz, so the answer is (b). 12. Decreasing the magnitude of the gain in the given circuit could be achieved by (a) Reducing amplitude of the input voltage (b) Increasing R f (c) Removing R f (d) Increasing R i 13. Answer: A = R f R i so one should increase R i to reduce the voltage gain. 14. Assuming ideal op-amp behavior, the input resistance R i of the amplifier is (a) R x (b) Ω (c) 0 Ω (d) R 1 (e) R 1 + R x 15. Answer: For an ideal op-amp, no current flows into the + input terminal, so the the input resistance is. This, the answer is (b). 3
16. In the circuit shown, the output voltage is (a) 5(1 + 8 2) = 25 V (b) 5(8 2) = 20 V (c) 15 V (d) 15 V (e) (8 2) = 120 V Answer: This is a non-inverting amplifier with gain (1 + 8 2) = 5, so with a 5-V input the output should be 25 V. However, the op-amp is powered by a +15-V power supply, so that the output will be clamped to a value close to +15 V, so the answer is (c). 17. Consider the circuit shown. Assume ideal op-amp behavior. (a) V = V + = 5 V (op-amp operation) (b) V = 10 2 (2 + 8) = 2 V (voltage division) (c) V = 0 (op-amp input current = 0) (d) Need additional information Answer: These is no feedback in the circuit to create a virtual short (V = V + ). No current flows into the input terminals so that V follows from voltage division, so the answer is (b). 18. In the circuit below R 1 = 10K, R 2 = 15K, and R 3 compensates for the op-amp s input bias current. What should its value be to be effective? (a) 10K (b) 15K (c) 6K (d) 25K (e) Need I OS Answer: Choose R 3 = R 1 R 2 = 6K, so (c) is the answer. 4
19. What is the purpose of R 3 in the circuit below, and what should the value be to be effective? Answer: This compensates for the op-amp s input bias current. The value should be R 1 R 2. 20. What are the units of slew-rate? Answer: Typically, slew rate is expressed in V/μs. 21. 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? (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) = 363.6 364 so (e) is the correct answer. 22. An engineer measures the gain of the circuit below and finds that with an input voltage v i = 3 V, the output voltage is v o = 18 V, so that the gain of the amplifier is 6. However, op-amp theory suggests the gain is 1 + 6.2 1 = 7.2. Give one phrase that could explain the difference. Answer: Slew Rate 5
23. Which of the circuit is a current-to-voltage converter? Answer: Circuit (a) 24. Which circuit is a voltage-to-current converter? Answer: Circuit (b) 6
25. In the circuit V IN = 10 V, R 1 = 10K, and R L = 5K. What current flows through R L? Answer: By op-amp action the voltage across R 1 is V in and the current through R 1 and R L is 10 10K = 1 ma. 26. The Thevenin voltage V TH for the circuit external to R L is (4 points) (a) 135 63.4 V (b) 13.4 63.4 V (c) 12.2 0 V (d) 122 0 V 27. Answer: V TH is the no-load voltage between terminals A and B. Using voltage division, V TH = (30 0 ) (j45 (90 + j45) ) = 6 + j12 V. This is equivalent to 13.4 63.4 V, so the answer is (b). Answer: I = 2 0.7 ( 8) (5K + 20K) = 0.37 ma V O = 8 + (20K)(I) + 0.7 = 0.14 V V O =?, I =? 7
28. Answer: This is a current-to-voltage converter with v O = i S R F = (10 10 6 )(1 10 6 ) = 10 V i S = 10 µa, R F = 1 MΩ V, v o =? 29. Answer: This is a follower where v O = v +. Thus v O = v + = 20 20 + 40 6 = 2 V v I = 6 V, v o =? 30. Answer: This is a noninverting amplifier where v + = v I1 2 + v I2 2 = 1 + 2 = 3 v I1 = 2 V, v I2 = 4 V, v o =? Thus v O = 1 + 50 50 v + = 2v + = 6 V 8
Question 2 Assume that the op-amp in the non-inverting buffer configuration below has infinite input resistance, zero output resistance, and an open-loop gain of A OL = 1,000. Determine the closed-loop gain A f = V o V s, and be sure to provide your answer to four decimal places. (6 points) Solution KVL around the loop shown below gives V s + V Rs + V i + V o = 0 Where V RS is the voltage across R s. However, no current flows into the op-amp, so V RS = 0, and the KVL equation becomes V s + V i + V o = 0 Now V o = 1,000V i = 1,000 V s, which means V i = V o 1,000, so the KVL equation becomes V s + V o 1,000 + V o = 0 Solving for A f = V o V s yields A f = 1,000 (1 + 1,000) = 0.9999 9
Question 3 An amplifier has an input resistance R i = 1K, and has a voltage gain of A v = 100 when driven from a signal with internal resistance R s 0. The amplifier is used to amplify a v s = 1 mv signal from a sensor that has an internal resistance of R s 20K. What is the output amplitude? (6 points) Solution The sensor s internal resistance and the amplifier s input resistance form a voltage divider so that the effective input voltage is v i = 1 1 mv = 47.6 μv 1 + 20 The output voltage is v o = A v v i = 100 47.6 10 6 = 4.76 mv 10
Question 4 The multimeter in the figure below has a common-mode rejection specification of 80 db. What possible range of output voltages can the meter indicate? (6 points) Solution V + = 5.01 V, V = 5 V The difference voltage is V d = V + V = 0.01 V = 10 mv, and the common-mode voltage is V cm = (V + + V ) 2 = 5.005 V A common-mode rejection specification of 80 db means that the multimeters will suppress V cm by a 80 db, which is equivalent to a factor 10 4. The contribution of the common-mode error is thus 5.005 10 4 = 0.5 mv. However, the sign is unknown. Thus, the multimeter could display anything in the range 9.5 mv V Display 10.5 mv 11
Question 5 The input voltage is v I for each ideal op-amp below. Determine each output voltage. Assume v I = 6 V. (2 points each) Solution (a) This is a follower where v O = v +. Thus v O = v + = 20 20 + 40 6 = 2 V (b) Same answer as (a) (c) This is a noninverting amplifier where v O = (1 + 10 10)v + = 2v +. Thus 6 v O = 2v + = 2 6 = 1.333 V 6 + 48 12
Question 6 The circuit below is driven by the series of pulses shown. Assume V C = 0 at t = 0. (a) Write an expression for v O. (2 points). (b) Determine the output voltage after n pulses. (4 points) (c) Use the results from (b) and design, by specifying values for R and C, so that the output voltage is -5 V after 5 pulses. (4 points) Solution Part (a) The output of the integrator is v O = 1 t v RC 0 I(t)dt. Part (b) v O decreases linearly with each pulse. A the end of the pulse the output voltage is 10 μs v O = RC The circuit holds this voltage until the next pulse, during which it again increases linearly. At the end of n pulses, the voltage is 10 μs v O = n RC Part (c) We have to design the circuit so that this voltage is -5 V when n = 5. Thus 10 μs 5 = 5 RC RC = 10 μs Pick C = 0.01 μf, then R = 1 kω. Other reasonable values will also work. 13
Question 7 Determine the voltage gain A v = v O v I for the ideal op-amp circuit shown. (10 points) (Hint: apply KCL at the junction of the T-network and the inverting input. Solution There is a virtual short between v + and v so that v + = v I. Call the voltage at the junction of the three resistors that form the T-network, v x. KCL at the T-network junction gives KCL at the inverting input gives v x v I 2R + v x R + v x v O = 0 2R v I R + v I v x 2R = 0 Solving for v x in the second equation gives v x = 3v I. Substituting this in the first KCL equation and some algebraic manipulation yields v O = 11v I A v = v O v I = 11. 14