Skip to main content

Clipper circuit with Vbias waveform

Short answer

Clippers cut off waveform parts above or below a reference level. In this circuit, the ideal diode conducts when Vin > Vbias, clamping the positive peaks to Vbias. Negative half cycles pass unchanged down to −Vm, so output peaks are +Vbias and −Vm.

What are clippers? Draw the sinusoidal waveform of the following circuit and indicate the output voltage. Assume diode is ideal. [Figure: sinusoidal source Vin, series resistor R, diode in series with battery Vbias in a shunt branch, output Vo.]

Clippers are circuits that cut off (clip) the portion of an input signal above or below a certain level, leaving the rest of the waveform unchanged. They are made of diodes and resistors, with a battery Vbias to set the clipping level.

Circuit and assumptions

There is a series resistor R and a shunt branch of an ideal diode in series with Vbias; the output Vo is taken across the shunt branch. Assume the diode anode is at the output side and the battery positive terminal is toward the diode cathode, so the diode conducts when Vo > Vbias. Input: vin = Vm sin ωt with Vm > Vbias.

 Vin o--[ R ]--+---o Vo
               |
               D (anode at Vo)
               |
              Vbias
 GND o---------+---o

Operation

  • Vin < Vbias: the diode is OFF (reverse biased) and no current flows in R, so Vo = Vin.
  • Vin > Vbias: the diode is ON (ideal, a short), so Vo = Vbias. The excess voltage Vin − Vbias drops across R.
  • Negative half cycle: the diode is OFF, so Vo = Vin, down to −Vm.
 Vo
 Vm      . . . . . . . . . .  (input peak, not reached)
 Vbias   ____          ____
        /    \        /    \
 0  ---/------\------/------\------
               \    /
                \  /
 -Vm             \/

Result: the peak positive output = Vbias and the peak negative output = −Vm. The positive peaks are clipped flat at Vbias. If both the diode and the battery are reversed, the negative peaks are clipped at −Vbias instead. For a real silicon diode, the clipping level is Vbias + 0.7 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 2 questions →