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Clamping circuit output, Vin ±25 V

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.

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.]

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
 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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