Homework Assignment 04 - University of...

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 (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).

2. 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) 10K (b) 𝑅1 + 𝑅2 = 30K (c) 0 Ω (d) 𝑅1||𝑅2 = 7.5K

Answer: R3 compensates for input bias currents and a rule of thumb is to choose R3 = R1||R2, 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 (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)

4. 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) 𝑣− = 𝑣+ = 0V (virtual ground concept) (b) 𝑣− = +50 𝜇V (c) 𝑣− = −50 𝜇V (d) 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).

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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) ≈ +500 mV (b) ≈ −500 mV (c) ≈ ±500 mV (d) Need additional information

Answer: For 𝑡 ≥ 0, the voltage across the capacitor is 𝑣𝐶 = (±𝐼𝑏Δ𝑡) 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.

6. What frequency is 2.2 decades higher than 500 Hz?

(a) 1.01 kHz (b) 644 Hz (c) 522 Hz (d) 79.24 kHz

Answer: 2.2 = log(𝑓𝑥 500⁄ ), so that 𝑓𝑥 = 79.24 Hz, so (d) is the answer.

7. What is frequency is 3 decades down from 220 Hz?

(a) 22 mHz (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.

8. 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 log2(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 (d) 0dB Answer: 𝑃𝑑𝐵 = 10 log(10 × 10−3) = −20 dB, so the answer is (a)

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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) 1.5 V (b) 2.5V (c) Close to the positive supply rail (d) Close to the negative supply rail (e) 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) ≈ −5.7 (b) ≈ 5.7 (c) ≈ 6.77 (d) ≈ 13.4

Answer: 𝐴𝑣 = −𝑅𝑓 𝑅1⁄ = 68K 12K⁄ = 5.7, so the answer is (a)

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?

(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)

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14. Consider the amplifier below. 𝑉𝑖𝑛 = 1.5 V, what is 𝑉𝑜𝑢𝑡?

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.

15. In the circuit shown, the output voltage is

(a) 5(1 + 8 2⁄ ) = 25 V (b) 5(8 2⁄ ) = 20 V (c) ≈ 15 V (d) ≈ −15 V (e) (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) V− = 𝑉+ = 5 V (op-amp operation) (b) 𝑉− = 10 × 2 (2 + 8) = 2⁄ V (voltage division) (c) V− = 0 (op-amp input current = 0) (d) 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).

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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:

𝐵𝑊 ≅0.35𝑡𝑟

=0.35

0.7 × 10−6

= 500 kHz

18. Define the CMRR for a differential amplifier. What is the ideal value? Answer: CMMR = 20log10|𝐴𝑑 𝐴𝑐𝑚⁄ | ). 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?

(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 (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).

21. 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

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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) =𝑉𝑅𝐸𝐹𝑅1

(𝑅1 + 𝑅2) = 4.959 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 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) 10K (b) 15K (c) 6K (d) 25K (e) Need 𝐼𝑂𝑆

Answer: Choose 𝑅3 = 𝑅1‖𝑅2 = 6K, so (c) is the answer.

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25. Which of the following depicts the correct current direction? Circle one. (1 point)

26.

𝑖𝑆 = 10 µA,𝑅𝐹 = 1 MΩ V, 𝑣𝑜 =?

This is a current-to-voltage converter with

𝑣𝑂 = −𝑖𝑆𝑅𝐹 = (10 × 10−6)(1 × 106) = −10 V

27.

𝑣𝐼 = 6 V, 𝑣𝑜 =?

This is a follower where 𝑣𝑂 = 𝑣+. Thus

𝑣𝑂 = 𝑣+ =20

20 + 406 = 2 V

28.

𝑣𝐼1 = 2 V, 𝑣𝐼2 = 4 𝑉, 𝑣𝑜 =?

This is a noninverting amplifier where

𝑣+ =𝑣𝐼12

+𝑣𝐼22

= 1 + 2 = 3 Thus

𝑣𝑂 = �1 +5050� 𝑣+ = 2𝑣+ = 6 V

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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

Answer: 𝑅𝑂 = Δ𝑉 Δ𝐼⁄ = 0.05 2.5 = 20 mΩ⁄ , so (a)

31. 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⁄ .

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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 = 𝐴𝑣𝑣𝑠(𝑅𝑖 𝑅𝑖 + 𝑅𝑠⁄ ). Substituting for 𝑣𝑂 , 𝑣𝑠,𝐴𝑣, and 𝑅𝑠 and solving for 𝑅𝑖 yields 𝑅𝑖 = 3K.

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Question 2 The op-amp in the circuit is ideal except for non-zero 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

𝑉𝑁1K

+𝑉𝑁 − 𝑉𝑂

10K+ 𝐼𝐵 = 0

Substitution of 𝑉𝑁 = −1 𝜇V yields

−1 × 10−6

1K+−1 × 10−6 − 𝑉𝑂

10K+ 10 × 10−9 = 0

⇒ 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 op-amp 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 = (±𝐼𝑏Δ𝑡) 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.

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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 time-constant 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

CMMR = �𝐴𝑑𝐴𝑐𝑚

� = �10.0430.083

� = 120.85

Expressed in dB

CMMRdB = 20 log10 120.85 = 41.65 dB

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Question 6 Consider the circuits below, and then complete the table (9 points).

Voltage follower Circuit (a) Simple difference amplifier (c) Current-to-voltage converter (b) Voltage to current converter (d) Logarithmic amplifier (e) Exponential amplifier (h) Differentiator (f) Instrumentation amplifier (j) Integrator (g) Summing amplifier (i)

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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)

2= 7.5 V

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

7.50.375 × 10−3

= 20 × 103

This is equivalent to 86 dB.

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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

The output of the integrator is 𝑣𝑂 = − 1𝑅𝐶 ∫ 𝑣𝐼(𝑡)𝑑𝑡

𝑡0 . During first pulse the output voltage

decreases linearly and at the end of the first pulse the output voltage is

𝑣𝑂 = −10 𝜇𝑠𝑅𝐶

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 = −𝑛10 𝜇𝑠𝑅𝐶

We have to design the circuit so that this voltage is -5 V when 𝑛 = 10. Thus

−5 = −1010 𝜇𝑠𝑅𝐶

⇒ 𝑅𝐶 = 20 𝜇s

Thus, the time constant is 20 𝜇s. If we pick 𝐶 = 0.01 𝜇F, then 𝑅 = 2 kΩ

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Homework Assignment 04 - University of...

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