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Clipper output for 5 sin wt, 2 V

Short answer

The circuit is a biased clamper, not a clipper. Assuming the diode conducts when output goes below +2 V, the capacitor charges to 7 V. Output is 5 sin ωt + 7 V, ranging from +2 V to +12 V, with negative peaks clamped at +2 V.

What is a clipper circuit? Find the output waveform for the following circuit. [Figure: Vin = 5 sin ωt, series capacitor C, diode in parallel with RL, 2 V battery in series with the diode; output V0 across RL.]

A clipper is a circuit that removes (clips) the part of an input signal above or below a chosen level, leaving the rest of the waveform unchanged. It does not need a capacitor.

Note on the given circuit: because it has a capacitor in series with the signal and the diode with battery across the output, it actually behaves as a biased clamper, not a clipper. It shifts the waveform instead of cutting it.

Output for Vin = 5 sin ωt, diode with 2 V battery

Assumption: the diode conducts when the output goes below +2 V (anode to the positive terminal of the 2 V battery, cathode to the output); ideal diode; RC ≫ period.

  • In the first negative half cycle the diode is ON and C charges to Vc = 5 + 2 = 7 V.
  • The diode is then OFF and the capacitor keeps its 7 V.
  • Vout = 5 sin ωt + 7
 Vout
 12 V |     /\       /\
      |    /  \     /  \
  7 V |---/----\---/----\---
  2 V |__/      \_/      \_  (negative peaks at 2 V)

Result: the output is the sine wave shifted up, with a minimum of +2 V and a maximum of +12 V (dc level 7 V; the peak-to-peak value of 10 V is unchanged).

If both the diode and the battery are reversed, the positive peaks are clamped at −2 V instead, and Vout = 5 sin ωt − 7, ranging from −12 V to −2 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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