Homework Assignment True or false. For both the inverting and noninverting op-amp configurations, V OS results in

Transcription

1 Question 1 (Short Takes), 2 points each. Homework Assignment An op-amp has input bias current I B = 1 μa. Make an estimate for the input offset current I OS. Answer. I OS is normally an order of magnitude smaller than I B. Thus, I OS = 100 na is a reasonable estimate. 2. True or false. For both the inverting and noninverting op-amp configurations, V OS results in an output offset voltage V OS (1 + R 2 R 1 ). Answer. True 3. An inverting op-amp amplifier is designed with R 1 = 1K and R F = 39K. What value of resistance should be placed in series with the noninverting input terminal for bias current compensation. Answer R c = R 1 R F = 975Ω 4. 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 (e) 10K Answer: Z C = j (2πfC) = j (2π ) = j1.592k. Thus, (a) is the answer. 1

2 5. What is the impedance of a 10 mh inductor f = 100 khz? (a) Ω (b) j6.28k (c) j( Ω) (d) Ω (e) 1K Answer: Z C = j2πfl = j(2π)( )( ) = 6.282K Thus, (b) is the answer. 6. 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. 7. 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). 2

3 8. A schematic shows a capacitor s value as 100n. This is equivalent to a capacitor with value Answer: Option (a). (a) 0.1 μf (b) 1,000 pf (c) μf (d) 0.01 μf 9. 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). 10. 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). 11. For a forward-biased diode, the barrier potential as temperature increases. Answer: Option (a). (a) decreases (b) increases (c) stays the same 12. In a 20 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 V (b) 55 V (c) 50 V (d) 45 V Answer: The applied voltage is V IN = V R + jv C, so that V IN = V R 2 + V C 2 = 2, V. Thus (d) is the answer. 13. 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 ( ) = Thus, (c) is the answer. 3

4 14. What is the voltage across a capacitor after being charged from a 100 V source for a period of one time constant? The initial voltage across the capacitor is 0 V. (a) 37.8 V (b) 38 V (c) 63.2 V (d) 90 V Answer: The voltage across the capacitor is v c (t) t τ = e. Thus, v c (τ) = 100(1 1 e 1 ) = 63.3 V. Thus, (c) is the answer. Note, it is a standard result that one should know by heart a capacitor charges to 63% of its final value after one time constant. 15. What is the charging time constant of the following circuit? (a) 294 ps (b) 13.5 ms (c) 21.5 ms (d) 2.16 Gs Answer: τ = RC = ( )( ) = , so the answer is (c). 16. Determine absolute value of the peak current through the load resistor? Assume V γ = 0.7 V for the diodes. (3 points) (a) ma (b) 4.65 ma (c) 0 ma (d) 1.25 ma Answer: When v i = 10 V, D 1 is reverse-biased and an open ciruit. D 2 is forward biased and has a 0.7 V voltage drop across it. It is in series with R L and the left 2K resistor, so the current that flows is I = (10 0.7) (2K + R L ) = 9.3 (4K) = ma. When V i = 10 V, then D 2 is reverse-biased but D 1 is forward biased. The current that flows is again ma, but now it flows in the opposite direction. Regardless, (a) is the answer. 4

5 17. 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 = 46.8 Ω. Thus, (a) is the answer. 18. Which of the circuit is a current-to-voltage converter? Answer: Circuit (a) 19. Which circuit is a voltage-to-current converter? Answer: Circuit (b) 5

6 20. 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. 21. What is the current through the ideal diode? (a) (b) (c) (d) 1 ma ma ma ma Answer: For an ideal diode there is no forward voltage drop, so I = 12 12K = 1 ma, so option (a) is the answer. 22. What is the current through the Zener diode? (a) 7 ma (b) 6 ma (c) 12.3 ma (d) 13 ma Answer: I z = (13 6) 1K = 7 ma, so option (b) is the answer. 6

7 23. With a 12-V supply, a silicon diode, and a 370 Ω resistor in series, what voltage will be dropped across the diode? Answer: Option (b) (a) 0.3 V (b) 0.7 V (c) 0.9 V (d) (a) 1.4 V 24. The Thevenin voltage V TH for the circuit external to R L is (3 points) (a) V (b) V (c) V (d) V 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 V, so the answer is (b). 7

8 Question 2 (Principles) For the circuit shown, the peak of the input voltage is 10 V. Determine the peak voltage across R 1. (10 points) Solution The impedance of the series elements R 1 and X C1 is Z 1 = 180 j240. The impedance of the parallel elements R 2 and X C2 is Z 2 = ( jx C2 R 2 ) (R 2 jx C2 ). Substituting values gives Z 2 = ( j ) (333 j250) = 120 j160 Ω. The source supplies a current I s = 10 (Z 1 + Z 2 ) = 10 (180 j j160) = j0.016 A. The voltage across R 1 is R 1 I s = 180( j0.016 A) = j2.88 = V. The peak voltage across R 1 is therefore 3.6 V. Below is a Matlab script that performs these calculations. Z1 = j*240; % Impedance of series elements Z2 = ((333)*(-j*250))/(333 - j*250) % Impedance of elements Is = 10/(Z1 + Z2); % Current flowing out of Vs VR1 = 180*Is; % Voltage across R1 mag = abs(vr1); phase = angle(vr1)*180/pi; s = sprintf('magnitude = %5.2f V, phase = %5.2f degrees\n',mag,phase); display(s) % Check. V1 = Is*Z1; V2 = Is*Z2; V1+V2 % Voltage across series branch % Voltage across branch % Adds up to 10 V 8

9 Question 3 (Principles) For the circuit shown, determine the current through R 3. (5 points). Solution Use node (a) as a reference (i.e., ground) and apply KCL at node (b) using the convention that current flows away from the node. V b V 1 R 1 + V b + V b V 1 = 0 V b ( 15) + V b 0 R 3 R V b 7 = 0 5 Solving yields V b = 13 7 V so that I 3 = V b R 3 = = V Question 4 Find the output voltage for the following circuit. Assume an ideal op-amp. (6 points) Solution KCL at v N, assuming currents flow away from the node: v N v o 4K + ( 1 ma) = 0 v N = 4 V + v 0 Voltage at v P = 1 V because 1 ma flows through the 1K resistor. v P = v N = 1 V 1 = 4 V + v o v o = 5 V 9

10 Question 5 An op-amp Miller integrator with 1 ms time constant, has input and output voltage that are initially zero. The input signal shown below is applied. Sketch and label the output waveform. (8 points) Solution For the Miller integrator: v O (t) = 1 t τ v I(t)dt = dt dt 0 10

11 Question 6 The circuit below uses an op-amp having V OS = 4 mv. What is its output offset voltage? (8 points) Solution Ground the input and call the output v o. Write a KCL equation at the inverting input: V OS 1M + V OS V X 1M = 0 This does not touch the output node, so write a KCL equation at x V X 1K + V X V OS 1M + V X v o 1M = 0 From the first equation, V X = 2V OS. Substituting this into the second equation gives Simplifying leads to 2V OS 1K + 2V OS V OS + 2V OS v o = 0 1M 1M v o = 2003V OS = (2003)(4 mv) = V 11

12 Question 7 Suppose the amplifier below has a differential-mode gain of 2,500 and a CMMR of 80 db. What is the output voltage if v 1 = V, and v 2 = V? What is the error introduced by the finite CMMR, expressed as a % of the differential input voltage? (6 points) Solution v id = v 1 v 2 = 2 mv, v icm = v 1 + v 2 2 = 5 V The differential-mode gain is 2,500 and the common-mode gain is 80 db less, which corresponds to a factor 10 = Thus, the common-mode gain is 2,500 ( ) = The output voltage is The error is v o = A d v id + A cm v icm = (2,500)(2 mv) + (0.25)(5) = = 6.25 V %error = = 25% 12

13 Question 8 With inputs v I1 = 50 mv, and v I2 = +50 mv, a difference amplifier has output v O = V. With inputs v I1 = v I2 = 5 V, the output is v O = V. Determine the CMRR, expressed in db. (6 points) Solution The differential input voltage is v I2 v I1 = 100 mv, and the differential-mode gain is = With v I1 = v I2 = 5 V the common-mode voltage gain is A cm = = The common-mode rejection ratio is Expressed in db CMMR = A d = = A cm CMMR db = 20 log = db 13

14 Question 9 We would like to measure the voltage V = V 1 V 2 in the circuit below with a voltmeter. What is the value of V, and what is the common-mode voltage V cm associated with V? What CMMR is required of the voltmeter if we are to measure V to within 0.01%? Express you answer in db. Solution The current through the resistance is I = 15 (40K) = ma. The voltage across the 10K resistor is therefore 3.75 V. Further, V 1 = 15 ( ) = 9.37 V, and V 2 = = The common-mode voltage is then V cm = (V 1 + V 2 ) 2 = 7.5 V The error must be less than 0.01% or 0.01% or 3.75 V, which is mv. Thus, the multimeter must suppress the 7.5 V common-mode voltage to less than mv. In other words, the CMMR must be at least This is equivalent to 86 db. 7.5 =