--- title: "Clamper circuit output waveform" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 4 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:clamper-output-waveform" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/clamper-output-waveform site: BCT Engineering --- # Clamper circuit output waveform Fundamental of Electrical and Electronics Engineering · Chapter 4: Diodes - Asked in: 2081 Chaitra (new course), 1+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_1/fundamental_of_electrical_and_electronics_engineering/clamper-output-waveform - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-4 ## Question What is a clamper circuit? Draw the output waveform of the circuit by applying the given input waveform. [Figure: input vin is a square wave switching between +6 V and −6 V. Circuit: source vin in series with resistor R to the output node; from the output node to ground a Si diode (anode at the output node, cathode downward) in series with a 3 V battery (positive terminal toward the diode cathode); vo is taken across the diode–battery branch.] ## Short answer The circuit with the series resistor acts as a biased clipper: during positive half-cycle the diode conducts and vo is clamped to 3.7 V (3 V battery + 0.7 V diode drop); during negative half-cycle diode is off, vo = −6 V. ## Answer #### Clamper A **clamper** (DC restorer) is a circuit of a capacitor, a diode and a resistor that shifts the whole input waveform up or down by a DC level **without changing its shape or peak-to-peak value**. A battery in series with the diode sets the level at which the waveform is clamped. The time constant RC must be much larger than the half-period of the input. #### Output waveform **Assumption:** a clamper needs a series capacitor, so the series element is taken as a capacitor C (large RC). Si diode: V_D = 0.7 V. The diode with the 3 V battery conducts when vo ≥ 3 + 0.7 = 3.7 V. **Positive half (vin = +6 V):** the diode is forward biased. The output is held at ```text vo = 3 + 0.7 = 3.7 V C charges to: Vc = vin − vo = 6 − 3.7 = 2.3 V (left plate +) ``` **Negative half (vin = −6 V):** the diode is reverse biased (open). The capacitor keeps its 2.3 V, so ```text vo = vin − Vc = −6 − 2.3 = −8.3 V ``` Check: peak-to-peak output = 3.7 − (−8.3) = 12 V = input peak-to-peak ✓. ```text vin vo +6 |‾‾‾| |‾‾‾| +3.7 |‾‾‾| |‾‾‾| 0 +---+---+---+-- 0 +---+---+---+-- -6 | |___| |___ | | | | -8.3 | |___| |___ ``` **Answer:** the output is a square wave switching between **+3.7 V and −8.3 V** (shifted down by 2.3 V), same shape and 12 V peak-to-peak. **Note:** if the series element is really a resistor (as drawn), the circuit acts as a biased clipper, not a clamper: vo = +3.7 V during the positive half (diode on) and vo = −6 V during the negative half (diode off). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What components does a clamper require? A. Diode and resistor only B. Capacitor, diode and resistor with RC much larger than the half-period C. Inductor and diode D. Capacitor only Answer: B. The capacitor holds a DC level; a large RC keeps it from discharging. (Key point: Clamper components) 2. In the positive half (vin = +6 V), with a 3 V battery and Si diode, vo is held at: A. 3 V B. 6 V C. 3.7 V D. 2.3 V Answer: C. Diode conducts and vo = 3 + 0.7 = 3.7 V. 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