---
title: "Design clamper for given waveforms"
subject: "Basic Electronics Engineering"
chapter: 2
key: "semester_2/basic_electronics_engineering:clamper-20v-to-30v"
source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clamper-20v-to-30v
site: BCT Engineering
---

# Design clamper for given waveforms

Basic Electronics Engineering · Chapter 2: Diodes

- Asked in: 2070 Magh, 2+4 marks
- Video: [Clamper Circuits (Basics, Working & Waveforms) Explained | Positive Clamper & Negative Clamper](https://www.youtube.com/watch?v=o_3shPSI6Ro) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/clamper-20v-to-30v
- Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-2

## Question

What is a clamper circuit? Design a clamper circuit to perform the function indicated in the figure. [Figure: input Vin is a square wave swinging between +20 V and -20 V; the output Vout of the clamper is a square wave swinging between +30 V and -10 V.]

## Short answer

The output is shifted up by 10 V. Use a positive clamper with a 10 V battery (negative terminal to diode anode, positive to ground). The capacitor charges to 10 V when Vin=-20 V, then Vout=Vin+10, giving -10 V to +30 V. Choose large RC, e.g., 100 kΩ, 1 µF.

## Answer

A **clamper circuit** is a circuit that shifts the dc level of an input waveform up or down without changing its shape or peak-to-peak value. It uses a capacitor, a diode, a resistor and, if needed, a dc battery.

#### Design for the given waveforms
- Input: square wave from −20 V to +20 V (peak-to-peak 40 V).
- Output: square wave from −10 V to +30 V (peak-to-peak 40 V, the same shape).
- So the waveform is shifted **up by 10 V**: Vout = Vin + 10 V. A positive clamper is needed whose negative peak is clamped at −10 V.

**Design (ideal diode)**
- The diode must turn ON when Vout goes below −10 V, so the diode anode is connected to a −10 V point (a 10 V battery with its negative end to the anode and its positive end to ground) and the cathode to the output.
- Capacitor voltage: when Vin = −20 V, Vout = −10 V, so Vc = Vin − Vout = −20 − (−10) = −10 V. The capacitor is charged to 10 V.
- Then Vout = Vin + 10: at Vin = +20 V, Vout = +30 V.

```text
              C
 Vin o-------| |-------+--------+---o Vout
                       |        |
                     -----      |
                      / \   D   R
                     /___\      |
                       |        |
                     - |        |
                      ---       |
                     -----  10 V|
                     + |        |
 GND o-----------------+--------+---o
```

- The anode of D joins the negative terminal of the 10 V battery; its positive terminal is at ground. The cathode of D is at the output.
- R is large (RC much greater than the period of the square wave), for example R = 100 kΩ, C = 1 µF.

For a real silicon diode, use a 9.3 V battery so that the clamp level stays at −10 V.

## Check yourself

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

1. Input is -20 V to +20 V and output is -10 V to +30 V. What shift is needed?
   A. Up by 30 V
   B. Up by 10 V
   C. Down by 10 V
   D. Up by 20 V
   Answer: B. -20 V becomes -10 V, so Vout = Vin + 10 V. (Key point: Required shift)

2. At what level must the negative peak be clamped?
   A. -10 V
   B. 0 V
   C. -20 V
   D. +10 V
   Answer: A. The negative peak of the output is -10 V. (Key point: Clamp level)

3. What voltage does the capacitor hold?
   A. 30 V
   B. 10 V
   C. 20 V
   D. 40 V
   Answer: B. Vc = Vin - Vout = -20 - (-10) = -10 V, i.e. 10 V magnitude. (Key point: Capacitor voltage)

4. With a silicon diode, how is the battery changed to keep the clamp at -10 V?
   A. Use 10.7 V
   B. Use 0.7 V
   C. Use 9.3 V
   D. No change
   Answer: C. The 0.7 V diode drop is compensated by lowering the battery to 9.3 V. (Key point: Real diode)

## Related in Basic Electronics Engineering

- [Clamping circuit output, Vin ±25 V](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aclamping-25v-10v) — Basic Electronics Engineering, chapter 2
- [Output waveforms of two diode circuits](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Adiode-circuits-output-waveforms) — Basic Electronics Engineering, chapter 2
- [Diode clamper circuit operation](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Adiode-clamper-operation) — Basic Electronics Engineering, chapter 2
- [Short note: Diode clamper circuit](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Ashort-note-diode-clamper) — Basic Electronics Engineering, chapter 2
- [Pulse shaping circuit and output waveform](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Apulse-shaping-circuit) — Basic Electronics Engineering, chapter 2
- [Clipping circuit output, 5 V bias](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aclipping-circuit-5v-bias) — Basic Electronics Engineering, chapter 2

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