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55:041 Electronic Circuits. The University of Iowa. Fall 2014. 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? (a) 12.73 V 2. (b) 25.5 V (c) 18 V (d) 12.73 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 18√2 = 25.5 V, so the answer is (b). The op-amp in the circuit is ideal, except for non-zero input bias currents. Further, 𝑅1 = 10K and 𝑅2 = 30K. What should 𝑅3 ′𝑠 value be? (a) (b) (c) (d) 10K 𝑅1 + 𝑅2 = 30K 0Ω 𝑅1 ||𝑅2 = 7.5K Answer: R 3 compensates for input bias currents and a rule of thumb is to choose R 3 = R1 ||R 2, 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? (a) 0.02K 4. (b) −0.02K (c) −0.2% (d) (e) 0.2% Answer: Percentage error is ((𝟗. 𝟗𝟖 − 𝟏𝟎)⁄𝟏𝟎) × 𝟏𝟎𝟎 = −𝟎. 𝟐%, so the answer is (d) The op-amp in the circuit is ideal, except that the open-loop gain is finite, namely 𝐴𝑑 = 100,000. Further, 𝑅2 = 100K and 𝑅1 = 10K. If, for some input signals, the output is 𝑣𝑂 = −5 V, then (a) (b) (c) (d) 𝑣− = 𝑣+ = 0V (virtual ground concept) 𝑣− = +50 𝜇V 𝑣− = −50 𝜇V Need additional information: 𝑅3 , 𝑣𝑠 Answer: The input bias current is zero and input offset voltage is zero, so v+ = 0 V. However, the gain is finite, so v− ≠ v+ . For a −5 V output, v− must be v− = vO ⁄Ad = 5⁄100,000 = 50 µV, so the answer is (b). 1 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 5. In the circuit below, the op-amp is ideal, except for an input bias current |𝐼𝑏 | = 1 nA. Further, 𝑅𝐹 = 10K, 𝑅1 = 100 Ω and 𝐶 = 1 𝜇F. The switch is opened at 𝑡 = 0. What is the output voltage after 5 seconds? (3 points) (a) (b) (c) (d) 6. ≈ +500 mV ≈ −500 mV ≈ ±500 mV Need additional information Answer: For 𝑡 ≥ 0, the voltage across the capacitor is 𝑣𝐶 = (±𝐼𝑏 Δ𝑡)⁄𝐶 which is �(±1 × 10−9 ) (5)�⁄(1 × 10−6 ) = ±5 mV for 𝑡 = 5 s. The gain of the amplifier is 1 + 𝑅𝐹 ⁄𝑅1 = 101, so that the output voltage is ± 505 mV. Thus (c) is the answer. What frequency is 2.2 decades higher than 500 Hz? (a) 1.01 kHz 7. (c) 522 Hz (d) 79.24 kHz Answer: 2.2 = log(𝑓𝑥 ⁄500), so that 𝑓𝑥 = 79.24 Hz, so (d) is the answer. What is frequency is 3 decades down from 220 Hz? (a) 22 mHz 8. (b) 644 Hz (b) 220 mHz (c) 6.4 mHz (d) 190 Hz Answer: 3 = log(220⁄𝑓𝑥 ), so that 𝑓𝑥 = 220 mHz, so (b) is the answer. A signal with amplitude 𝑣 = 4 V at 4 kHz decreases as frequency increases at −2 dB/octave. What is the amplitude in V at 13 kHz? (3 points) Answer: There are log 2 (13⁄4) = 1.7 octaves between 4 kHz and 13 kHz. Thus, the amplitude decreases by 1.7 × 2 = 3.4 dB. The new amplitude is 20 log 4 − 3.4 = 8.6 dB. This is equivalent to 2.7 V. 9. A 10 mW signal is equivalent to a power of (a) −20 dB (b) −40 dB (c) 20 dB Answer: 𝑃𝑑𝐵 = 10 log(10 × 10−3 ) = −20 dB, so the answer is (a) 2 (d) 0dB 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 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? (a) 12.73 V (b) 25.5 V (c) 18 V (d) 12.73 V Answer: The ac source’s effective (or rms) value should also be 9 V. This means the peak value should be 9√2 V, so the peak-to-peak value should be 18√2 = 25.5 V, so the answer is (b). 11. In the circuit 𝑉𝑅𝐸𝐹 = 2.5 V, and 𝑉𝐼𝑁 = 1.5 V. The output voltage is (a) (b) (c) (d) (e) 1.5 V 2.5V Close to the positive supply rail Close to the negative supply rail Need additional information Answer: |𝑉+ | > |𝑉− |, so that the op-amp will pull 𝑉𝑂 > 0, and turn the diode on. The cathode (𝑉𝑂𝑈𝑇 ) will follow until 𝑉− reaches 𝑉+ . Any further increase of will reduce 𝑉𝑂 and 𝑉𝑂𝑈𝑇 . Thus, the output stabilizes at 𝑉𝑂𝑈𝑇 = 2.5 V. The T.A. was instructed not to grade this question. 12. The voltage gain of the amplifier shown is (a) (b) (c) (d) 13. ≈ −5.7 ≈ 5.7 ≈ 6.77 ≈ 13.4 Answer: 𝐴𝑣 = −𝑅𝑓 ⁄𝑅1 = 68K⁄12K = 5.7, so the answer is (a) 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? (a) 7.82 V (b) Saturation (c) Cutoff (d) 9.52 V Answer: 𝐴𝑣 = �1 + 𝑅𝑓 ⁄𝑅1 � = (1 + 4.6) = 5.6. The output voltage is then 𝑉𝑂 = 1.7 × 5.6 = 9.52 V. Thus, the answer is (d) 3 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 14. Consider the amplifier below. 𝑉𝑖𝑛 = 1.5 V, what is 𝑉𝑜𝑢𝑡 ? 15. Answer: The gain of the first amplifier is 𝐴𝑓 = − (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 𝑉𝑜𝑢𝑡 = 1.5 × 3 = 4.5 V. In the circuit shown, the output voltage is (a) (b) (c) (d) (e) 5(1 + 8⁄2) = 25 V 5(8⁄2) = 20 V ≈ 15 V ≈ −15 V (8⁄2) = 120 V Answer: This is a non-inverting amplifier with gain (1 + 8⁄2) = 5, 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 (c). 16. Consider the circuit shown. Assume ideal op-amp behavior. (a) (b) (c) (d) V− = 𝑉+ = 5 V (op-amp operation) 𝑉− = 10 × 2⁄(2 + 8) = 2 V (voltage division) V− = 0 (op-amp input current = 0) Need additional information Answer: These is no feedback in the circuit to create a virtual short (𝑉− = 𝑉+ ). No current flows into the input terminals so that 𝑉− follows from voltage division, so the answer is (b). 4 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 17. An engineer measures the (step response) rise time of an amplifier as 𝑡𝑟 = 0.7 𝜇s. Estimate the 3 dB bandwidth of the amplifier. Answer: 0.35 𝑡𝑟 0.35 = 0.7 × 10−6 = 500 kHz 𝐵𝑊 ≅ 18. Define the CMRR for a differential amplifier. What is the ideal value? 19. Answer: CMMR = 20log10 |𝐴𝑑 ⁄𝐴𝑐𝑚 | ). Ideally, CMMR → ∞ An amplifier has a differential gain of -50,000 and a common-mode gain of 2. What is the common-mode rejection ratio? (a) –87.96 dB (b) 44 dB (c) -44 dB (d) 87.96 dB Answer: CMMR = 20 log10 |𝐴𝑑 ⁄𝐴𝑐 | = 20 log10 |50 × 103 ⁄2| = 87.96 dB, so the answer is (d). 20. A differential amplifier has a common-mode gain of 0.2 and a common-mode rejection ratio of 3250. What would the output voltage be if the single-ended input voltage was 7 mV rms? (a) 1.4 mV rms 21. (b) 650 mV rms (c) 4.55 mV rms (d) 0.455 V rms Answer: 𝐶𝑀𝑅𝑅 = |𝐴𝑑 ⁄𝐴𝑐𝑚 | so that 3250 = 𝐴𝑑 ⁄0.2 and 𝐴𝑑 = 650. The output voltage is 650 × 7 = 4.55 mV rms, so the answer is (c). What is 𝑣𝑜 in the following circuit if 𝑣𝑅𝐸𝐹 = 1.2 V, 𝑅1 = 680 Ω, and 𝑅2 = 200 Ω? Answer: The current through 𝑅2 is 1.2/200 = 6 mA, which also flows through 𝑅1 . Thus, the output voltage is 1.2 + 0.006× 680 = 5.28 V 5 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 22. Consider the voltage regulator below, implemented with a reference voltage 𝑉𝑅𝐸𝐹 = 1.25 V, an ideal op-amp and BJT with very large gain (𝛽 → ∞). Determine the output voltage to 4 significant figures. (5 points) Answer: The op-amp and feedback loop maintains a voltage 𝑉𝑅𝐸𝐹 across 𝑅1 so the current through both resistors is 𝐼 = 𝑉𝑅𝐸𝐹 ⁄𝑅1. The output voltage is then 𝑉𝑂 = 𝐼(𝑅1 + 𝑅2 ) = 23. 𝑉𝑅𝐸𝐹 (𝑅1 + 𝑅2 ) = 4.959 V 𝑅1 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 𝑅1 = 10K, 𝑅2 = 15K, and 𝑅3 compensates for the op-amp’s input bias current. What should it’s value be to be effective? (a) (b) (c) (d) (e) 10K 15K 6K 25K Need 𝐼𝑂𝑆 Answer: Choose 𝑅3 = 𝑅1 ‖𝑅2 = 6K, so (c) is the answer. 6 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 25. Which of the following depicts the correct current direction? Circle one. (1 point) 26. This is a current-to-voltage converter with 𝑣𝑂 = −𝑖𝑆 𝑅𝐹 = (10 × 10−6 )(1 × 106 ) = −10 V 27. 𝑖𝑆 = 10 µA, 𝑅𝐹 = 1 MΩ V, 𝑣𝑜 =? This is a follower where 𝑣𝑂 = 𝑣+ . Thus 28. 𝑣𝑂 = 𝑣+ = 𝑣𝐼 = 6 V, 𝑣𝑜 =? 20 6 =2V 20 + 40 This is a noninverting amplifier where 𝑣+ = Thus 𝑣𝐼1 𝑣𝐼2 + =1+2=3 2 2 𝑣𝑂 = �1 + 𝑣𝐼1 = 2 V, 𝑣𝐼2 = 4 𝑉, 𝑣𝑜 =? 7 50 � 𝑣 = 2𝑣+ = 6 V 50 + 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 29. A current source supplies a nominal current 𝐼𝑅𝐸𝐹 = 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 × 103 )(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 30. A bench power supply is set to an output voltage of 5 V. When it is connected to a circuit that draws 𝐼𝑂 = 2.5 A, the output voltage drops to 4.95 V. What is the output resistance 𝑅𝑂 of the power supply? (a) ≈ 20 mΩ (b) ≈ 1.98 Ω (c) Need additional information 31. Answer: 𝑅𝑂 = Δ𝑉⁄Δ𝐼 = 0.05⁄2.5 = 20 mΩ, so (a) 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 𝐼𝐿 = 1.595⁄100 = 15.95 mA. The internal resistance is 𝑅𝑂 = Δ𝑉⁄Δ𝐼 = (1.605 − 1.595)⁄(15.95 × 10−3 ) = 0.627 Ω. Thus, (a) is the answer. 32. In the circuit 𝑉𝐼𝑁 = 10 V, 𝑅1 = 10𝐾, and 𝑅𝐿 = 5𝐾. What current flows through 𝑅𝐿 ? Answer: By op-amp action the voltage across 𝑅1 is 𝑉𝑖𝑛 and the current through 𝑅1 and 𝑅𝐿 is 10⁄10K = 1 mA. 8 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 33. A 100-mV source with internal resistance 𝑅𝑠 = 1K drives an amplifier with gain 𝐴𝑣 = 𝑣𝑜 ⁄𝑣𝑖 = 10 (see figure). The output voltage is 750 mV. What is the amplifier’s input resistance 𝑅𝑖 ? (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 𝑣𝑂 = 𝐴𝑣 𝑣𝑠 (𝑅𝑖 ⁄𝑅𝑖 + 𝑅𝑠 ). Substituting for 𝑣𝑂 , 𝑣𝑠 , 𝐴𝑣 , and 𝑅𝑠 and solving for 𝑅𝑖 yields 𝑅𝑖 = 3K. 9 55:041 Electronic Circuits. The University of Iowa. Fall 2014. Question 2 The op-amp in the circuit is ideal except for nonzero input bias current 𝐼𝐵 = 10 nA. In the circuit, 𝑅𝐹 = 10K, 𝑅1 = 1K, and 𝑅3 = 100 Ω. Determine the maximum and minimum output voltage 𝑉𝑂 resulting from 𝐼𝐵 . Remember that 𝐼𝐵 could be positive or negative. (6 points) Solution The op-amp is ideal with respect to gain so that 𝑉𝑃 = 𝑉𝑁 and we assume the input bias current is the same for both the non-inverting and the inverting inputs. For the case where 𝐼𝐵 flows into the op-amp 𝑉𝑁 = 𝑉𝑃 = −𝑅3 (𝐼𝐵 ) = −(100)(10 × 10−9 ) = −1 𝜇V. KCL at the inverting input, assuming current flows away from the node gives Substitution of 𝑉𝑁 = −1 𝜇V yields 𝑉𝑁 𝑉𝑁 − 𝑉𝑂 + + 𝐼𝐵 = 0 1K 10K −1 × 10−6 −1 × 10−6 − 𝑉𝑂 + + 10 × 10−9 = 0 1K 10K ⇒ 𝑉𝑂 = 89 𝜇V For the case where 𝐼𝐵 flows out of the op-amp, 𝑉𝑂 = −89 𝜇V. Thus, the maximum output voltage is −89 𝜇V and the minimum output voltage is +89 𝜇V. I gave some students incomplete guidance on this question. The T.A. was instructed to make this an extra-credit question by lowering the total score by 6 points. Thus, even if a student did not do this question, he/she could still get 100% on the homework assignment. Question 3 (Own, non-ideal op-amp) In the circuit the opamp is ideal, except for an input bias current |𝐼𝑏 | = 10 nA. Further, 𝑅𝐹 = 10K, 𝑅1 = 100 Ω and 𝐶 = 6.8 𝜇F. The switch is opened at 𝑡 = 0. What is the output voltage after 10 seconds? (3 points) Solution For 𝑡 ≥ 0, the voltage across the capacitor is 𝑣𝐶 = (±𝐼𝑏 Δ𝑡)⁄𝐶 which is �(±10 × 10−9 ) (10)�⁄(6.8 × 10−6 ) = ±14.71 mV for 𝑡 = 10 s. The gain of the amplifier is1 + 𝑅𝐹 ⁄𝑅1 = 101, so that the output voltage is ± 1.485 V. 10 55:041 Electronic Circuits. The University of Iowa. Fall 2014. Question 4 (Varactor) For the circuit shown, 𝑉𝑃𝑆 = 5 V and = 10K . The varactor characteristics are shown in the graph. What is the bandwidth of the circuit in Hz? (5 points) Solution From the graph, the varactor has a capacitance of 100 pF with a 5-V reverse voltage. The timeconstant of the circuit is 𝜏 = 𝑅𝐶 = (10 × 103 )(100 × 10−12 ) = 1 𝜇s. The bandwidth is then 𝐵 = 1⁄(2𝜋𝜏) = 159 kHz. Question 5 With inputs 𝑣𝐼1 = −50 mV, and 𝑣𝐼2 = +50 mV, a difference amplifier has output 𝑣𝑂 = 1.0043 V. With inputs 𝑣𝐼1 = 𝑣𝐼2 = 5 V, the output is 𝑣𝑂 = 0.4153 V. Determine the CMRR, expressed in dB. (4 points) Solution The differential input voltage is 𝑣𝐼2 − 𝑣𝐼1 = 100 mV, and the differential-mode gain is 1.0043⁄0.1 = 10.043 With 𝑣𝐼1 = 𝑣𝐼2 = 5 V the common-mode voltage gain is 𝐴𝑐𝑚 = 0.4152⁄5 = 0.083 The common-mode rejection ratio is Expressed in dB CMMR = � 𝐴𝑑 10.043 �=� � = 120.85 𝐴𝑐𝑚 0.083 CMMR dB = 20 log10 120.85 = 41.65 dB 11 55:041 Electronic Circuits. The University of Iowa. Fall 2014. 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 (a) (c) (b) (d) (e) (h) (f) (j) (g) (i) 12 55:041 Electronic Circuits. The University of Iowa. Fall 2014. Question 7 We would like to measure the voltage 𝑉 = 𝑉1 − 𝑉2 in the circuit below with a voltmeter. What is the value of 𝑉, and what is the common-mode voltage 𝑉𝑐𝑚 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. (8 points) Solution The current through the resistance is 𝐼 = 15⁄(40K) = 0.375 mA. The voltage across the 10K resistor is therefore 3.75 V. Further, 𝑉1 = 15 − (0.375 × 15) = 9.37 V, and 𝑉2 = 0.375 × 15 = 5.625. The common-mode voltage is then 𝑉𝑐𝑚 = (𝑉1 + 𝑉2 ) = 7.5 V 2 The error must be less than 0.01% or 0.01% of 3.75 V, which is 0.375 mV. Thus, the multimeter must suppress the 7.5 V common-mode voltage to less than 0.375 mV. In other words, the CMMR must be at least This is equivalent to 86 dB. 7.5 = 20 × 103 0.375 × 10−3 13 55:041 Electronic Circuits. The University of Iowa. Fall 2014. Question 8 (N TYU 9.13) An integrator is driven by a series of pulse shown below. At the end of the 10th pulse, the output voltage is to be 𝑣𝑂 = −5 V. Assume 𝑉𝐶 = 0 at 𝑡 = 0. Determine the time constant and values for 𝑅 and 𝐶 that will meet these specifications. (6 points) Solution 1 𝑡 The output of the integrator is 𝑣𝑂 = − 𝑅𝐶 ∫0 𝑣𝐼 (𝑡)𝑑𝑡. During first pulse the output voltage decreases linearly and at the end of the first pulse the output voltage is 10 𝜇𝑠 𝑣𝑂 = − 𝑅𝐶 The circuit holds this voltage until the next pulse, during which it again increases linearly. At the end of n pulses, the voltage is 𝑣𝑂 = −𝑛 10 𝜇𝑠 𝑅𝐶 We have to design the circuit so that this voltage is -5 V when 𝑛 = 10. Thus 10 𝜇𝑠 𝑅𝐶 ⇒ 𝑅𝐶 = 20 𝜇s −5 = −10 Thus, the time constant is 20 𝜇s. If we pick 𝐶 = 0.01 𝜇F, then 𝑅 = 2 kΩ 14