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Operational amplifiers: board exam questions with solutions

On the ECE exam under Electronics Engineering. Try each question before you open its solution, then practise with fresh numbers below: 3 computed question types on this topic.

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Build it in the lab

Inverting amplifier · Non-inverting amplifier · Summing amplifier · Difference amplifier · Instrumentation amplifier: the same circuits on a live bench, with meters you can move.

Questions

1. An ideal inverting amplifier uses Rin = 10 kΩ and Rf = 100 kΩ. What is its voltage gain?

  1. +10
  2. +11
  3. −11
  4. −10
Show the answer and solution

Answer: D. −10

Av = −Rf/Rin = −100k/10k = −10.

2. A non-inverting amplifier has Rf = 47 kΩ and R1 = 4.7 kΩ (to ground). What is its voltage gain?

  1. 10
  2. −10
  3. 11
  4. 9
Show the answer and solution

Answer: C. 11

Av = 1 + Rf/R1 = 1 + 47/4.7 = 11.

3. An inverting summer has Rf = 10 kΩ. Input V1 = 0.5 V is applied through 10 kΩ and V2 = 0.3 V through 5 kΩ. What is the output?

  1. −0.8 V
  2. −1.1 V
  3. +1.1 V
  4. −0.65 V
Show the answer and solution

Answer: B. −1.1 V

Vo = −Rf(V1/R1 + V2/R2) = −(10/10 × 0.5 + 10/5 × 0.3) = −(0.5 + 0.6) = −1.1 V.

4. An op-amp has a slew rate of 0.5 V/µs. What is the highest frequency at which it can deliver an undistorted 10 V peak sine wave?

  1. 7.96 kHz
  2. 50 kHz
  3. 3.98 kHz
  4. 7.96 MHz
Show the answer and solution

Answer: A. 7.96 kHz

fmax = SR/(2πVp) = (0.5 × 10⁶)/(2π × 10) = 7958 Hz ≈ 7.96 kHz.

5. An op-amp with a 1 MHz gain-bandwidth product is used as a non-inverting amplifier with a closed-loop gain of 20. What is the closed-loop bandwidth?

  1. 20 MHz
  2. 1 MHz
  3. 5 kHz
  4. 50 kHz
Show the answer and solution

Answer: D. 50 kHz

For a single-pole op-amp, BW = GBW/Acl = 1 MHz/20 = 50 kHz.

6. An op-amp has a differential gain of 100 000 and a common-mode gain of 1. What is its CMRR?

  1. 50 dB
  2. 1000 dB
  3. 100 dB
  4. 10 dB
Show the answer and solution

Answer: C. 100 dB

CMRR = 20 log(Ad/Acm) = 20 log(10⁵) = 100 dB.

7. A difference amplifier uses matched resistors R1 = 10 kΩ (input) and R2 = 50 kΩ (feedback and ground leg). V1 = 1.2 V is applied to the inverting input network and V2 = 1.5 V to the non-inverting network. What is Vo?

  1. −1.5 V
  2. 1.5 V
  3. 13.5 V
  4. 0.3 V
Show the answer and solution

Answer: B. 1.5 V

Vo = (R2/R1)(V2 − V1) = 5 × (1.5 − 1.2) = 1.5 V.

8. An ideal op-amp integrator has R = 10 kΩ and C = 0.1 µF, with the output initially at 0 V. A constant +1 V is applied to the input for 2 ms. What is the output at the end of that interval?

  1. −2.0 V
  2. +2.0 V
  3. −0.2 V
  4. −20 V
Show the answer and solution

Answer: A. −2.0 V

Vo = −(1/RC)∫Vin dt = −(1/(10⁴ × 10⁻⁷)) × 1 × 0.002 = −1000 × 0.002 = −2.0 V.

Worked examples with computed numbers

These numbers are fixed on this page; in practice they change every time.

9. An inverting amplifier has Rin = 82 kΩ and Rf = 680 kΩ on ±15 V rails. What is the output for a 0.1 V DC input?

  1. -829 mV
  2. 829 mV
  3. -12.1 mV
  4. -929 mV
Show the answer and solution

Answer: A. -829 mV

Inverting: Vo = −(Rf/Rin)·Vin = −(680 kΩ/82 kΩ) × 0.1 = -829 mV.

10. A non-inverting amplifier uses Rf = 12 kΩ and Rg = 12 kΩ to ground. What is its voltage gain?

  1. 1
  2. 0.5
  3. 3
  4. 2
Show the answer and solution

Answer: D. 2

Non-inverting: G = 1 + Rf/Rg = 1 + 12 kΩ/12 kΩ = 2.

11. An op-amp with a 3 MHz gain-bandwidth product is used at a closed-loop gain of 100. What is the approximate −3 dB bandwidth?

  1. 15 kHz
  2. 30 kHz
  3. 3 MHz
  4. 300 MHz
Show the answer and solution

Answer: B. 30 kHz

Closed-loop bandwidth ≈ GBW/G = 3 MHz/100 = 30 kHz.

Other topics in the same subjects

DC circuits · AC circuits · Transients (RC, RL, RLC) · Semiconductors and diodes · BJT and FET circuits · Oscillators and feedback · Power electronics · Digital electronics · Microprocessors and microcontrollers · Control systems · Instrumentation and measurement

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