---
title: "Slew-rate check for 40 dB amplifier at 20 kHz"
subject: "Advanced Electronics"
chapter: 2
key: "semester_4/advanced_electronics:slew-rate-40db-20khz"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/slew-rate-40db-20khz
site: BCT Engineering
---

# Slew-rate check for 40 dB amplifier at 20 kHz

Advanced Electronics · Chapter 2: Operational Amplifier Characterization

- Asked in: 2074 Asoj, 7 marks
- Video: [Op-Amp Slew Rate Explained (with Examples)](https://www.youtube.com/watch?v=2DFIr6t1hbc) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/slew-rate-40db-20khz
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2

## Question

An amplifier using an op-amp with a slew rate SR = 1V/μs has a gain of 40 dB. Peak value of 100 mV, 20 KHz sinusoidal wave is applied to the input of amplifier. Determine whether the output will be distorted due to slew rate limiting of op-amp. If so, find remedies.

## Short answer

Gain of 40 dB is 100, giving 10 V peak output. Required slew rate 2π×20 kHz×10 V = 1.257 V/µs exceeds the 1 V/µs SR, causing distortion. Remedy: use an op-amp with SR ≥1.26 V/µs or reduce gain to ≤79.6 (output 7.96 V peak).

## Answer

Given: SR = 1 V/µs, gain = 40 dB, input = 100 mV peak, f = 20 kHz.

```text
Gain: 20 log10 Av = 40 dB  →  Av = 10^(40/20) = 100
Output peak: Vm = 100 × 0.1 = 10 V
Required slope of output (sine wave):
  (dVo/dt)max = 2πf·Vm
              = 2π × 20 000 × 10
              = 1.2566 × 10⁶ V/s = 1.257 V/µs
```

Since 1.257 V/µs > SR = 1 V/µs, **the output will be distorted** by slew-rate limiting (the peaks are rounded into a triangular shape and the amplitude falls).

Check with full-power bandwidth: fmax = SR/(2πVm) = 10⁶/(2π × 10) = 15.92 kHz < 20 kHz, confirming distortion.

#### Remedies
1. **Use a faster op-amp:** SR ≥ 1.257 V/µs (for margin, 2 V/µs or more, e.g. LF351 with 13 V/µs).
2. **Reduce the gain:** the largest undistorted output at 20 kHz is Vm = SR/(2πf) = 10⁶/(2π × 20 000) = 7.96 V, so gain ≤ 7.96/0.1 = 79.6 (about 38 dB). For example, with R1 = 1 kΩ use Rf ≤ 79.6 kΩ (standard 75 kΩ).
3. **Reduce the input amplitude:** at gain 100, Vin(peak) ≤ 7.96/100 = 79.6 mV.
4. Operate below 15.92 kHz if the signal frequency can be lowered.

**Answer: output is distorted (needs 1.257 V/µs > 1 V/µs); use SR ≥ 1.26 V/µs or gain ≤ 79.6.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A gain of 40 dB with a 100 mV peak input gives what output peak?
   A. 10 V (gain 100)
   B. 1 V (gain 10)
   C. 4 V (gain 40)
   D. 100 V (gain 1000)
   Answer: A. 20 log Av = 40 gives Av = 10^2 = 100, so Vm = 100 x 0.1 V = 10 V. (Key point: dB to gain and output peak)

2. What slope does a 10 V peak, 20 kHz sine output need?
   A. f*Vm = 0.2 V/us
   B. 2*pi*f*Vm = 1.257 V/us
   C. Vm/(2*pi*f) = 79.6 uV/us
   D. 2*pi*Vm/f = 3.14 V/us
   Answer: B. (dVo/dt)max = 2*pi*20000*10 = 1.2566x10^6 V/s = 1.257 V/us. (Key point: Required slope)

3. Why is the output distorted for SR = 1 V/us?
   A. The gain of 100 is above the open-loop gain
   B. The input frequency exceeds the unity-gain frequency
   C. The required slope 1.257 V/us exceeds the SR of 1 V/us
   D. The required slope is below SR so the op-amp clips
   Answer: C. When required slope is greater than SR the sine peaks become rounded/triangular with reduced amplitude. (Key point: Distortion decision)

4. Which is a valid remedy for the distortion?
   A. Increase the gain above 100
   B. Use an op-amp with a lower slew rate
   C. Raise the input frequency above 20 kHz
   D. Reduce the gain to about 79.6 or less (Rf <= 79.6 kohm for R1 = 1 kohm)
   Answer: D. Max undistorted output is SR/(2*pi*f) = 7.96 V, so gain <= 7.96/0.1 = 79.6; or use a faster op-amp. (Key point: Remedy)

## Related in Advanced Electronics

- [Slew-rate check for four input signals](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aslew-rate-multiple-signals) — Advanced Electronics, chapter 2
- [Voltage follower square-wave output with slew rate](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Avoltage-follower-square-wave) — Advanced Electronics, chapter 2
- [Slew-rate distortion check, 330 kΩ/10 kΩ inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aslew-rate-330k-10k) — Advanced Electronics, chapter 2
- [Design three op-amp IA with gain 20](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adesign-3opamp-ia-gain-20) — Advanced Electronics, chapter 4, asked 3×
- [Slew-rate distortion check, 470 kΩ/10 kΩ inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aslew-rate-470k-10k) — Advanced Electronics, chapter 2, asked 3×
- [Maximum peak input for undistorted output](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aslew-rate-max-peak-input) — Advanced Electronics, chapter 2

## Related in other subjects

- [Ideal op-amp properties; non-inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Ideal op-amp properties; inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-inverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Voltage gain of non-inverting op-amp](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 2×
- [Op-amp properties and summing amplifier](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-summer-v1-2v2-3v3) — Basic Electronics Engineering, chapter 4