Homework Assignment 01 In this homework set students review some basic circuit analysis techniques, as well as review how to analyze ideal op-amp circuits. Numerical answers must be supplied using engineering notation. For example, I o = 19 ma, and not I o = 0.019 A,or I o = 1.9 10 2 A. Question 1 (2 points each unless noted otherwise) 1. What is the 3-dB bandwidth of the amplifier shown below if r π = 2.5K, r o = 100K, g m = 40 ms, and C L = 1 nf? (a) 65.25 khz (b) 10 khz (c) 1.59 khz (d) 10.4 khz Answer: The capacitor sees an equivalent resistance r o = 100K. (If one turns off V I, g m v π = 0, and the current source is effectively removed from the circuit.) The time-constant is τ = RC = 100 μs. The bandwidth is 1 (2πτ) = 1.59 khz, so the answer is (c). 2. What is the 3-dB bandwidth of the circuit below? (a) 63.6 Hz (b) 31.8 Hz (c) 5.07 Hz (d) 10.1 Hz Answer: The capacitor sees an equivalent resistance R S R L = 5K. The time-constant is τ = RC = 5 ms, so that the bandwidth is 1 (2πτ) = 31.82 Hz, and (b) is the answer. 3. Consider the amplifier below. V in = 1.5 V, what is V out? Answer: The gain of the first amplifier is A f = (30) 10 = 3 and the gain of the second amplifier is 1, gving an overall gain of ( 3)( 1) = 3. The output voltage is thus V out = 1.5 3 = 4.5 V. 1
4. 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 Answer: A = R f R i so one should increase R i to reduce the voltage gain. 5. 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 Answer: For an ideal op-amp, no current flows into the + input terminal, so the the input resistance is. This, the answer is (b). 6. 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. Answer: BW 0.35 t r 0.35 = 0.7 10 6 = 500 khz 7. A current source supplies a nominal current I REF = 1 ma. When connected to a 5K load, only 0.95 ma flows through the load. What is the internal resistance of the current source? Answer: The voltage across the load is (5 10 3 )(0.95 10 3 ) = 4.750 V. A current 0.05 ma flows through the current source s internal resistance, which has value 4.75 (0.05 10 3 ) = 95K 8. A bench power supply is set to an output voltage of 5 V. When it is connected to a circuit that draws I O = 2.5 A, the output voltage drops to 4.95 V. What is the output resistance R O of the power supply? (a) 20 mω (b) 1.98 Ω (c) Need additional information Answer: R O = ΔV ΔI = 0.05 2.5 = 20 mω, so (a) 2
9. An AAA cell has a no-load voltage of 1.605 V. When a 100 Ω resistor is connected across its terminals, the voltage drops to 1.595 V. What is the cell s internal resistance? a) 620 mω b) 10 mω c) Need additional information Answer: The current flowing through the load resistance is I L = 1.595 100 = 15.95 ma. The internal resistance is R O = ΔV ΔI = (1.605 1.595) (15.95 10 3 ) = 0.627 Ω. Thus, (a) is the answer. 10. What is the impedance of a 0.1 μf capacitor at f = 1 khz? (a) j1.6 10 3 Ω (b) j10 10 3 Ω (c) +j1.6 10 3 Ω (d) 1.6 10 3 Ω (e) 10K Answer: Z C = j (2πfC) = j (2π 1 10 3 0.1 10 6 ) = j1.592k. Thus, (a) is the answer. 11. A I REF = 1 ma current source has an output resistance R o = 100 kω and drives a 1 kω load. What current flows through the load? Answer: I load = I REF [100 (100 + 1) ] = 0.99 ma. 12. What is the impedance of a 10 mh inductor f = 100 khz? (a) 1 10 3 Ω (b) j6.28k (c) j(6.2.8 10 3 Ω) (d) 6.28 10 3 Ω (e) 1K Answer: Z C = j2πfl = j(2π)(10 10 3 )(100 10 3 ) = 6.282K Thus, (b) is the answer. 13. Four resistor in ascending order are (a) 22R, 270K, 2K2 1M (b) 4K7, 10K, 47R, 330K (c) 3R3, 4R7, 22R, 5K6 (d) 100R, 10K, 1M, 3K3 Answer: Option (c). 3
14. A 100-mV source with internal resistance R s = 1K drives an amplifier with gain A v = v o v i = 10 (see figure). The output voltage is 750 mv. What is the amplifier s input resistance R i? (a) (b) 1K (c) 3K (d) Need additional information (e) 0 Ω Answer: The source s and amplifier s internal resistances form a voltage divider and the output voltage is v O = A v v s (R i R i + R s ). Substituting for v O, v s, A v, and R s and solving for R i yields R i = 3K. 15. A schematic shows a capacitor s value as 100n. This is equivalent to a capacitor with value (a) 0.1 μf (b) 1,000 pf (c) 0.0001 μf (d) 0.01 μf Answer: Option (a). 16. A silicon diode measures a low value of resistance with the meter leads in both positions. The trouble, if any, is that (a) The diode is broken and internally open (b) The diode is broken and internally shorted (c) The diode is working but shorted to ground (d) The diode is working correctly Answer: Option (b). 17. An engineer tests a silicon diode with a multimeter using the Ohm-meter function. The meter measures a low value of resistance with the meter leads in both positions. The trouble, if any, is that (a) The diode is broken and internally open (b) The diode is broken and internally shorted (c) The diode is working but shorted to ground (d) The diode is working correctly Answer: Option (b). 18. A diode conducts when it is forward-biased, and the anode is connected to the through a limiting resistor. Answer: Option (b). (a) Anode (b) Positive supply (c) Negative supply (d) Cathode 4
Question 2 For the circuit shown, R 1 = 20 Ω, R 2 = 10 Ω, and C = 10 μf. Determine the equivalent resistance the capacitor sees. In other words, determine the Thevenin resistance of the network to the left of the capacitor. (8 points) Solution To determine the Thevenin equivalent resistance, inject a current I x and determine the voltage V x, see below. Then, R TH = V x I x KCL at A, using the convention that currents flowing away from the node is positive, gives Further, Ohm s law gives I 1 = V x 30, so that I 1 1.5I 1 I x = 0 I x = 0.5I 1 I x = 0.5V x 30 R TH = V x I x = 60 Ω 5
Question 3 (Op-Amps) 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 6