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Worked non-numerical solution

Author: Dr Dangor

A biased parallel clipper limits the output at a voltage set by the diode and battery. Follow the marked KVL loop, test each diode state, then draw the output waveform.

Practice question

Figure 1 gives the output across the diode-and-battery branch. Let Vi be the instantaneous input, Vb the battery voltage, and VDon the diode ON voltage. Derive the diode states and sketch one output cycle.

Assume Vb > VDon > 0, an input peak Vi(pk) greater than Vb − VDon, a constant diode ON voltage, no output load, and no reverse breakdown.

Biased parallel diode clipper with marked voltage polarities A source and resistor feed a diode-and-battery branch. The KVL loop is marked inside the circuit. The output is measured from the branch's upper terminal to ground. + − Vi + VR − + VDon − − Vb + + Vo − GND KVL
Figure 1. The marked KVL loop and voltage references set the signs in the solution.

Use Figure 1 to answer five questions.

  1. For which input voltages does the diode conduct?
  2. For which input voltages does the diode block?
  3. What is the output voltage while the diode conducts?
  4. What is the output voltage while the diode blocks?
  5. Sketch one complete output cycle and label the maximum and minimum output voltages.

Solution

Step 1: Follow the marked polarities

Assume the diode is ON. Follow the KVL loop marked in Figure 1:

Vi − VR − VDon + Vb = 0

Rearrange for the resistor voltage:

VR = Vi − VDon + Vb

Step 2: Diode ON

For the diode to be ON, the input must exceed the diode ON voltage after accounting for battery bias. At the switching boundary, VR = 0 and Vi + Vb = VDon. Above that boundary, Vi + Vb > VDon, or Vi > VDon − Vb. The conducting branch fixes the output at VDon − Vb.

ON: Vi > VDon − Vb   ⇒   Vo = VDon − Vb.

Step 3: Diode OFF

The diode blocks at or below the switching input. The ON-state diode drop no longer applies. No current flows, VR = 0, and the output follows the input.

OFF: Vi ≤ VDon − Vb   ⇒   Vo = Vi.

At Vi = VDon − Vb, both output expressions agree.

Step 4: Combine the states and draw the waveform

Combine the two diode states to obtain the output relationship:

Vo = VDon − Vb   when   Vi > VDon − VbVi   when   Vi ≤ VDon − Vb

Figure 2 shows one cycle. The output maximum is VDon − Vb; the output minimum is −Vi(pk).

Sinusoidal input and upper-clipped output over one cycle The sinusoidal input reaches positive and negative peaks. The output remains at V D on minus V b above the clipping level, then follows the input near the negative peak. +Vi(pk) 0 VDon − Vb −Vi(pk) one cycle Vi Vo
Figure 2. The output follows the negative trough and remains at the clipping level elsewhere. Both curves meet at Vi = VDon − Vb.

Final answers

Table 1 answers the five question parts.

Table 1. Answers to the practice question
PartResult
(a)Diode ON when Vi > VDon − Vb.
(b)Diode OFF when Vi ≤ VDon − Vb.
(c)While ON, Vo = VDon − Vb.
(d)While OFF, Vo = Vi.
(e)Figure 2; maximum VDon − Vb, minimum −Vi(pk).