---
title: "Clamping circuit output, Vin ±25 V"
subject: "Basic Electronics Engineering"
chapter: 2
key: "semester_2/basic_electronics_engineering:clamping-25v-10v"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clamping-25v-10v
site: BCT Engineering
---

# Clamping circuit output, Vin ±25 V

Basic Electronics Engineering · Chapter 2: Diodes

- Asked in: 2069 Bhadra, 1+3 marks
- Video: [Clamper Circuit Explained](https://www.youtube.com/watch?v=7O3Hbkkt624) — ALL ABOUT ELECTRONICS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clamping-25v-10v
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2

## Question

What is clamping circuit? Find the output waveform of the given circuit. [Figure: Vin is a sine wave of amplitude ±25 V; series capacitor C, then a diode in parallel with RL, with a 10 V battery in series with the diode; output Vout across RL.]

## Short answer

Assuming the diode conducts when output tries to go below +10 V, the capacitor charges to 35 V. The output is Vin+35 V, swinging between +10 V and +60 V, with negative peaks clamped at +10 V. The waveform shape is unchanged.

## Answer

A **clamping circuit** (clamper, dc restorer) adds a dc level to an ac signal so that the waveform is shifted up or down without change of shape. It uses a capacitor, a diode, a resistor and, for a biased clamper, a battery.

#### Output for Vin = 25 sin ωt, C in series, diode with a 10 V battery in parallel with RL

**Assumption** (the diode and battery polarity is not clear in the scan): the diode conducts when the output tries to go below +10 V (diode anode connected to the positive terminal of the 10 V battery, cathode to the output), ideal diode, and RC much larger than the signal period.

**Working**
- During the first negative quarter cycle, the diode is ON, and the capacitor charges until Vc = Vm + VB = 25 + 10 = 35 V.
- After this the diode is OFF and the capacitor holds 35 V, adding to the input.
- Vout = Vin + 35 V
- Negative peak: −25 + 35 = **+10 V** (clamped at the battery voltage)
- Positive peak: 25 + 35 = **+60 V**

```text
 Vout
 +60 |      /\      /\
     |     /  \    /  \
 +10 |----/----\--/----\---  (negative peaks at +10 V)
   0 +-------------------- t
```

**Result:** the output is the same sine wave shifted upward, swinging between +10 V and +60 V (peak-to-peak 50 V, dc level +35 V).

If the diode and battery are connected the opposite way (clamping the positive peak at −10 V), then Vout = Vin − 35 V and the output swings between −60 V and −10 V.

## Check yourself

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

1. With a 10 V bias and Vin = 25 sin ωt, to what voltage does the capacitor charge?
   A. 25 V
   B. 10 V
   C. 15 V
   D. 35 V
   Answer: D. Vc = Vm + VB = 25 + 10 = 35 V. (Key point: Capacitor voltage)

2. What is Vout?
   A. Vin + 35 V
   B. Vin - 35 V
   C. Vin + 10 V
   D. Vin + 25 V
   Answer: A. The capacitor adds 35 V to the input. (Key point: Output expression)

3. What is the output range?
   A. -25 V to +25 V
   B. +10 V to +60 V
   C. 0 V to +50 V
   D. +10 V to +35 V
   Answer: B. -25 + 35 = 10 V and 25 + 35 = 60 V. (Key point: Output swing)

4. What happens to the peak-to-peak value?
   A. It stays 50 V
   B. It doubles
   C. It halves
   D. It becomes 35 V
   Answer: A. A clamper only shifts the waveform, keeping its shape. (Key point: Waveform shape)

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