2. The. op-amp in and 10K. (a) 0 Ω. (c) 0.2% (d) (a) 0.02K. (b) 4. The. 5 V, then. 0V (virtual. (a) (c) Fall V. (d) V.

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1 Homework Assignment 04 Question 1 (2 points each unless noted otherwise) 1. A 9-V dc power supply generates 10 W in a resistor. What peak-to-peak amplitude should an ac source have to generate the same power in the resistor? V (b) 25.5 V 18 V (d) V Answer: The ac source s effective (or rms) value should also be 9 V. This measn the peak value should be 9 2 V, so the peak-to-peathe circuit is ideal, except for non-zero input bias currents. Further, 10K value should be V, so the answer is (b). 2. The op-amp in and 30K. What should value be? 10K (b) 30K 0 Ω (d) 7.5K Answer: compensates for input bias currents andd a rule of thumb is to choose, so the answer is (d). 3. A resistor has a nominal value of 10K, but its actual value is 9..98K. What is the percentage error? 0.02K (b) 0.02K 0.2% (d) (e) 0.2% Answer: Percentage error is.. % %, so the answer is (d) 4. The op-amp in the circuit is ideal, except that the open-loop gain is finite, namely 100,000. Further, 100K and 10K. If, for some input signals, the output is 5 V, then 0V (virtual ground concept) (b) 50 V 50 V (d) Need additional information:, Answer: The input bias current is zero and input offset voltage is zero, so. However, the gain is finite, so. For a output, must be,, so the answer is (b). 1

2 5. In the circuit below, the op-amp is ideal, except forr an input bias current 1 na. Further, 10K, 100 Ω and 1 F. The switch is opened at 0. What is the output voltage after 5 seconds? (3 points) 500 mv (b) 500 mv 500 mv (d) Need additional information Answer: For 0, the voltage acrosss the capacitor is Δ which is mV for 5s. The gain of the amplifier is 1 101, so that the output voltage is 501 mv. Thus ( c) is the answer. 6. What frequency is 2.2 decades higher than 500 Hz?? 1.01 khz (b) 644 Hz 5222 Hz (d) khz Answer: 2.2 log 500, so that Hz, so (d) is the answer. 7. What is frequency is 3 decades down from 220 Hz?? 22 mhz (b) 220 mhz 6.44 mhz (d) 190 Hz Answer: 3 log 220, so that 220 mhz, so (b) is the answer. 8. A signal with amplitude 4Vat 4 khz decreases as frequency increasess at 2 db/octave. What is the amplitude in V at 13 khz? (3 points) Answer: Theree are log octaves between 4 khz and 13 khz. Thus, the amplitude decreases by dB. The new amplitude is 20 log db. This is equivalent to 2.7 V. 9. A 10 mw signal is equivalent to a power of 20 db (b) 40 db 20 db (d) 0dB Answer: 10 log db, so the answer is 2

3 10. A 9-V dc power supply generates 10 W in a resistor. What peak-to-peak amplitude should an ac source have to generate the same power in the resistor? V (b) 25.5 V 18 V (d) V Answer: The ac source s effective (or rms) value should also be 9 V. This measn the peak value should be 9 2 V, so the peak-to-peak value should be V, so the answer is (b). 11. In the circuit 2.5 V, and 1.5 V. The output voltage is 1.5 V (b) 2.5V Close to the positive supply rail (d) Close to the negative supply rail (e) Need additional information Answer:, so that the op-amp will pull 0, and turn the diode on. The cathode ( will follow until reaches. Any further increase of will reduce and. Thus, the output stabilizes at 2.5V. 12. The voltage gain of the amplifier shown is 5.7 (b) (d) 13.4 Answer: 68K 12K 5.7, so the answer is ) 13. If the feedback/ /input resistor ratio of An op-amp feedback amplifier is 4.6 with 1.7 V applied to the noninverting input, what is the output voltage value? 7.82 V (b) Saturation Cutoff (d) 9.52 V Answer: The output voltage is then V. Thus, the answer is (d) 3

4 14. Consider the amplifier below. 1.5V, what is? Answer: The gain of the first amplifier is and the gain of the second amplifier is 1, gving an overall gain of The output voltage is thus V. 15. In the circuit shown, the output voltage is V (b) V 15 V (d) 15 V (e) V Answer: This is a non-inverting amplifier with gain , 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. 16. Consider the circuit shown. Assume ideal op-amp behavior. V 5 V (op-amp operation) (b) V (voltage division) V 0 (op-amp input current = 0) (d) Need additional information Answer: Thesee is no feedback in the circuit to create a virtual short ( ). No current flows into the input terminals so that follows from voltage division, so the answer is (b). 4

5 17. An the engineer measures the (step response) rise time of an amplifier as 0.7 s. Estimate 3 db bandwidth of the amplifier. Answer: khz 18. Define the CMRR for a differential amplifier. What is the ideal value? Answer: CMMR 20log. Ideally, CMMR 19. An amplifier has a differential gain of -50,000 and a common-mode gain of 2. What is the common-mode rejection ratio? db (b) 444 db ) -44 db (d) db Answer: CMMR 20log is ( d). 20log db, so the answer 20. A differential amplifier has a common-mode gain of 0.2 and a common-mode rejection ratio of What would the outputt voltage be if the single-ended inputt voltage was 7 mv rms? 1.4 mv rms (b) 650 mv rms 4.55 V rms (d) V rms Answer: mv rms. so that and 650. The output voltage is 21. What is in the following circuit if = 1.2 V, 680 Ω, and 200 Ω? Answer: The current through is 1.2/200 = 6 ma, which also flows through. Thus, the output voltage is = 5.28 V 5

6 22. Consider the voltage regulator below, implementedd with a reference voltage V, an ideal op-ampp and BJT with very large gain ( ). Determine the output voltage to 4 significant figures. (5 points) Answer: The op-amp and feedback loop maintains a voltage acrosss so the current through both resistors is 1. The outpu voltage iss then V 23. When researching part numbers for three-terminal regulators, an engineer encounters the term LDO. What does LDO stand for? Answer: Low Drop Out 24. In the circuit below 10K, 15K, and compensatess for the op-amp s input bias current. What should it s value be to be effective? 10K (b) 15K 6K (d) 25K (e) Need Answer: Choose 6K, so is the answer. 6

7 25. Which of the following depicts the correct current direction? Circle one. (1 point) 26. This is a current-to-voltage converter with V 10 μa, 1MΩ V,? 27. This is a follower where. Thus V 6 V,? 28. This is a noninverting amplifier wheree 2 V, 4,? Thus V 7

8 29. A current source supplies a nominal current 1mA. 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 K 30. A bench power supply is set to an outpu voltage off 5 V. When it is connected to a circuit thatt draws 2.5 A, the output voltage drops to 4.95 V. What is the output resistance of the power supply? 20 mω (b) 1.98 Ω Need additional information Answer: Δ Δ mω, soo 31. 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 iss the cell s internal resistance? a) b) c) 620 mω 10 mω Need additional information Answer: The current flowing through the load resistance is ma. The internal resistance is Δ Δ Ω. Thus, is the answer. 32. In the circuit 10 V, 10, and 5. What current flows through? Answer: By op-amp action the voltage across is and the current through and is 10 10K 1 ma. 8

9 33. A 100-mV source with internal resistance 1K drives an amplifier with gain 10(see figure). The output voltage is 7500 mv. What is the amplifier s input resistance? (b) 1K 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. Substituting for,,, and and solving for yields 3K. 9

10 Question 2 the op-amp in the circuit is ideal except for non- zero input bias current 10 na. In the circuit, 10K, 1K, and 100 Ω. Determine the maximum and minimum output voltage resulting from. Remember that could be positive or negative. (6 points) Solution The op-amp is ideal with respect to gain so that and we assume the input bias current is the same for both the non-inverting and the inverting inputs. For the case where flows into the op-amp V. KCL at the inverting input, assuming current flows awayy from the node gives 1K 10K 0 Substitution of 1 V yields K 10K 89 V For the case where flows out of the op-amp, 89 V. Thus, the maximum output voltage is 89 V and the minimum output voltage is 89 V. Question 3 (Own, non-ideal op-amp) In the circuit thee op- Further, 10K, 100 Ω and 6.8 F. The switch is opened at 0. amp is ideal, except for an input bias current 10nA. What is the output voltage after 10 seconds? (3 points) Solution For 0, the voltage across the capacitor is Δ which is mv for 10 s. The gain of the amplifier is1 101, so that thee output voltage is V. 10

11 Question 4 (Varactor) For the circuit shown, 5 V and 10K. The varactor characteristics are shown in the graph. What is the bandwidth of the circuit? Solution From the graph, the varactor has a capacitance of 100 pff with a 5-V reverse voltage. The time constant of the circuit is s. The bandwidth is then khz. Question 5 With inputs 50 mv, and 50 mv, a difference amplifier has output V. With inputs 5 V, the output is V. Determine the CMRR, expressed in db. Solution The differential input voltage is 100 mv, and the differential-mode gain is With 5 V the common-mode voltage gain iss The common-mode rejection ratio is Expressed in db CMMR CMMR 20log db 11

12 Question 6 Consider the circuits below, and then complete the table (9 points). Voltage follower Simple difference amplifier Current-to-voltage converter Voltage to current converter Logarithmic amplifier Exponential amplifier Differentiator Instrumentation amplifier Integrator Summing amplifier Circuit 12

13 Question 7 We would like to measure the voltage in the circuit below with a voltmeter. What is the value of, and what is the common-mode voltage associated with? What CMMR is required of the voltmeter if we are too measure to within 0.01%? Express you answer in db. Solution The current through the resistance is 15 40K0.375 ma. The voltage across the 10K resistor is therefore 3.75 V. Further, V, and The common-mode voltage is then 7.5 V 2 The errorr 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

14 Question 8 (N TYU 9.13) An integrator is driven by a series of pulse shown below. At the end of the 10 th pulse, the output voltage is to be 5 V. Assume 0 at 0. Determine the time constant and values for and that will meet thesee specifications. Solution The output of the integrator is. During first pulse the output voltage decreasess linearly and at the end of the first pulse the output voltage is 10 The circuit holds this voltage until the next pulse, duringg which it again increases linearly. At the end of n pulses, the voltage is We have to design the circuit so that this voltage is -5 V when 10. Thus s Thus, the time constant is 20 s. If we pick 0.01 F, then 2 kω 14