---
title: "Slew-rate distortion check, 330 kΩ/10 kΩ inverter"
subject: "Advanced Electronics"
chapter: 2
key: "semester_4/advanced_electronics:slew-rate-330k-10k"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/slew-rate-330k-10k
site: BCT Engineering
---

# Slew-rate distortion check, 330 kΩ/10 kΩ inverter

Advanced Electronics · Chapter 2: Operational Amplifier Characterization

- Asked in: 2070 Chaitra, 2+5 marks
- Video: [#172: Basics of Op Amp Gain Bandwidth Product and Slew Rate Limit](https://www.youtube.com/watch?v=UooUGC7tNRg) — w2aew
- 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-330k-10k
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2

## Question

Define slew rate of an operational amplifier. The inverting op-amp with Rf = 330 KΩ and R1 = 10 KΩ has a slew rate of 0.5 V/μs with input signal equal to 0.1 sin(200000t). Determine whether the output will be distorted due to slew rate limitation, if so find remedy.

## Short answer

Output peak 3.3 V at 31.83 kHz requires 0.66 V/µs slew rate, exceeding the 0.5 V/µs limit, so the output distorts. Remedy: reduce Rf to ≤250 kΩ (gain ≤25) or use a faster op-amp.

## Answer

#### Slew rate
**Slew rate (SR)** is the maximum rate of change of output voltage per unit time that an op-amp can produce, SR = (dVo/dt)max, usually in V/µs (0.5 V/µs for the 741). It is caused by the limited current available to charge the internal compensation capacitor (SR = Imax/C).

#### Checking for distortion
Given: Rf = 330 kΩ, R1 = 10 kΩ, SR = 0.5 V/µs, vin = 0.1 sin(200000t) V.

```text
|Av| = Rf/R1 = 330/10 = 33
Vm   = 33 × 0.1 = 3.3 V   (output peak)
ω    = 2 × 10⁵ rad/s,  f = ω/2π = 31.83 kHz
Required slope = ω·Vm = 2 × 10⁵ × 3.3
               = 0.66 × 10⁶ V/s = 0.66 V/µs
```

Since 0.66 V/µs > 0.5 V/µs, **the output will be distorted** due to slew-rate limiting. (Full-power bandwidth for 3.3 V peak: fmax = SR/(2πVm) = 0.5 × 10⁶/(2π × 3.3) = 24.11 kHz < 31.83 kHz.)

#### Remedies
1. **Reduce the gain:** maximum undistorted output peak Vm = SR/ω = 0.5 × 10⁶/(2 × 10⁵) = 2.5 V, so gain ≤ 2.5/0.1 = 25, i.e. **Rf ≤ 25 × 10 kΩ = 250 kΩ**.
2. **Use an op-amp with higher slew rate:** SR ≥ 0.66 V/µs (e.g. LF351, 13 V/µs).
3. **Reduce the input amplitude:** Vin(peak) ≤ 2.5/33 = 75.8 mV.

**Answer: distorted (0.66 V/µs required > 0.5 V/µs); remedy Rf ≤ 250 kΩ or a faster op-amp.**

## Check yourself

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

1. Rf = 330 kohm, R1 = 10 kohm, input 0.1 V peak gives output peak:
   A. 3.3 V
   B. 33 V
   C. 0.33 V
   D. 0.1 V
   Answer: A. Gain 33 x 0.1 V. (Key point: Gain and peak)

2. The required slope for omega = 2x10^5 and Vm = 3.3 V is:
   A. 0.33 V/us
   B. 0.66 V/us
   C. 6.6 V/us
   D. 0.066 V/us
   Answer: B. omega Vm = 2e5 x 3.3 = 0.66e6 V/s. (Key point: Required slope)

3. Compared with SR = 0.5 V/us, the output is:
   A. Undistorted, since 0.66 > 0.5
   B. Undistorted, since gain is 33
   C. Distorted, since 0.66 > 0.5
   D. Distorted only above 1 MHz
   Answer: C. Required slope exceeds the slew rate. (Key point: Decision)

4. A suitable remedy is:
   A. Increase Rf to 500 kohm
   B. Increase the input amplitude
   C. Use an op-amp with SR below 0.5 V/us
   D. Rf <= 250 kohm (gain <= 25) or a faster op-amp
   Answer: D. Max undistorted Vm = SR/omega = 2.5 V, so gain <= 25. (Key point: Remedy)

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