Clipping circuit output, 5 V bias

Clipping circuit: a circuit that removes (clips) the part of the input signal above or below a chosen level without changing the rest of the waveform. It is made from diodes, resistors and dc batteries.

Output waveform of the given circuit

Assumptions: Vin = 15 sin ωt, Rs = 2 kΩ, RL = 1 kΩ, silicon diode with VD = 0.7 V in series with a 5 V battery, in a shunt branch. The diode anode is at the output side, so it conducts when Vo ≥ 5 + 0.7 = 5.7 V.

 Vin o--[2k]--+----------+--o Vo
              |          |
              D         RL 1k
              |          |
             5 V         |
 GND o--------+----------+

Diode OFF: the circuit is a divider: Vo = Vin × 1k/(2k + 1k) = Vin/3.

Condition for conduction: Vo = 5.7 V requires Vin = 3 × 5.7 = 17.1 V.

The peak input is only 15 V, so the diode never turns ON; the output peaks are

Vo(max) = 15/3 = +5 V, Vo(min) = −15/3 = −5 V

Result: with these values the output is an unclipped sine wave of amplitude 5 V (the 5 V battery acts as a limiter at +5.7 V but the divider keeps Vo below this level). The clipping level of this circuit is +5.7 V.

 Vo  +5.7 V -------------------- (clip level, not reached)
     +5 V     /\        /\
   0 --------/--\------/--\-----
                  \    /
                  -5 V

If RL were large (open circuit), the positive peaks would be clipped flat at +5.7 V, and the negative half would pass unchanged.