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. Consider the amplifier below. V in = 0.5 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 = 0.5 3 = 1.5 V. 2. Increasing 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 f to increase the voltage gain. 3. 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). 1
4. A current source supplies a nominal current I REF = 2 ma. When connected to a 5K load, only 1.95 ma flows through the load. What is the internal resistance of the current source? Answer: The voltage across the load is (5 10 3 )(1.95 10 3 ) = 9.75 V. A current 0.05 ma flows through the current source s internal resistance, which has value 9.75 (0.05 10 3 ) = 195K. 5. A bench power supply is set to an output voltage of 9 V. When it is connected to a circuit that draws I O = 2.5 A, the output voltage drops to 8.95 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) 6. An AAA cell has a no-load voltage of 1.605 V and internal resistance of 0.62 Ω. When a 100 Ω resistor is connected across the battery terminals, the voltage drops to (pick best option): a) 1.595 V b) 1.6 V c) The voltage does not drop d) Need additional information Answer: The current flowing through the load resistance is I L = 1.605 (100 + 0.62) = 15.95 ma. The voltage across the resistor (and battery terminals) is V = IR = (15.95 10 3 )(100) = 1.595 V, Thus, (a) is the answer. 7. 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. 8. 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. 2
9. An certain design calls for an inductive reactance with magnitude 6.28 kω at f = 100 khz. What value inductor will meet this requirement? (a) Need additional information (b) 10 mh (c) 10 H (d) 62.8 mh Answer: Z L = j2πfl so that L = Z 2πf = (6.28 10 3 ) (2π100 10 3 ) = 9.996 mh 10 mh, so (b) is the correct answer. 10. 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). 11. 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. 12. 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). 3
13. 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). 14. 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). 15. A diode conducts when it is forward-biased, and the anode is connected to the through a limiting resistor. (a) Anode (b) Positive supply (c) Negative supply (d) Cathode Answer: Option (b). 16. The op-amp in the circuit is ideal, and R 1 = 10K, R 2 = 100K, and R 3 = 10K. The input resistance that the source sees is (a) R 1 = 10K (b) R 1 + R 3 = 20K (virtual short between + and ) (c) (Ideal op-amp has R i = ) (d) R 1 R 2 R 3 = 4.72K (KCL at terminal) Answer: R 1 = 10K, so (a) is the answer. 4
Question 2 Determine the Thevenin voltage V TH for the circuit external to R L. Use phasor notation. For example, V TH = 122 0 V. (4 points) 5
Question 3 The circuit shown uses an ideal op-amp, except that the input offset voltage is V OS = 4 mv. What is its output offset voltage? Provide your answer to four significant digits. (8 points) 6