Homework Assignment 01

Transcription

1 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 = A,or I o = A. Question 1 (2 points each unless noted otherwise) 1. 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) = so (e) is the correct answer. 2. An engineer designs an amplifier to deliver 6.25 W (sinusoidal) signal power to a 4 Ω resistive load. What is the required peak-to-peak voltage swing across the load? 2 Answer: P = V rms R, so that V rms = 5 V, so that V pp = V 3. The output voltage of a three-terminal voltage regulator is 5 5 ma load, and A load. What is the regulator s output resistance? (a) 27 mω (b) 1K (c) 3.3 Ω Answer: R = ΔV ΔI = = 27 mω, so (a) 4. 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 ( )( ) = V. A current 0.05 ma flows through the current source s internal resistance, which has value 4.75 ( ) = 95K 1

2 5. 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 = = 20 mω, so (a) 6. An AAA cell has a no-load voltage of V. When a 100 Ω resistor is connected across its terminals, the voltage drops to 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 = = ma. The internal resistance is R O = ΔV ΔI = ( ) ( ) = Ω. Thus, (a) is the answer. 7. What is the impedance of a 0.1 μf capacitor at f = 1 khz? a) j Ω b) j Ω c) +j Ω d) Ω a) 10K Answer: Z C = j (2πfC) = j (2π ) = 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 ( ) ] = 0.99 ma. 2

3 Question 2 (Principles) 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 3

4 Question 3 For the following circuit, determine I D and V o. Make reasonable assumptions. (6 points) Solution Assume the diodes internal resistance is negligible and that V γ = 0.7 V. Assume that both diodes are forward-biased. Replace the diodes with linear models as shown below. This is now a linear circuit that one can solve using nodal analysis, KCL, KVL, superposition, Thevenin or Norton equivalent circuits, etc. A KCL equation for the output node is V O 2K + V O (10 0.7) + V O (10 0.7) = 0 2K 2K Solving yields V O = 6.2 V. The sign of the voltage is consistent with our assumption: the diodes are forward biased. The current through the output resistor is V O 2K = 3.1 ma. By symmetry, half of this current flows through each diode, so that I D = 1.55 ma 4

5 Question 4 (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 + = = 2 V (b) Same answer as (a) (c) This is a noninverting amplifier where v O = ( )v + = 2v +. Thus 6 v O = 2v + = 2 6 = V

6 Question 5 (Op-Amps) In the circuit below, the offset voltage for each op-amp is V OS = 10 mv. Find the worst case output voltage v 02 for v I = 0. (4 points) Solution The equivalent circuit with v I = 0 and the offset voltages indicated, is shown below. With respect to its offset voltage, the first amplifier is a noninverting amplifier with gain 11, so that the worst-case v 01 is v 01 = 110 mv. This is then amplified by the second amplifier by a factor 5. With respect to its offset voltage, the gain of the second amplifier is 6, so that the worst-case v 02 (using superposition) is v o2 = = 610 mv 6