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Electronics / signal lab

Clipper circuit tutorial

Author: Dr Dangor

Practise the method used in the worked example. For each circuit, mark a KVL loop, write the loop equation, derive the diode ON and OFF conditions, and state Vo in each state. Sketch one output cycle and label the output maximum and minimum.

For all questions, Vi = Vp sin(ωt), Vb > 0, and the diode has a constant ON voltage VDon > 0. The lower rail is 0 V. Measure Vo between the marked + and − terminals. Assume no output load or diode reverse breakdown. At the switching boundary, diode current is zero.

Part A

Non-numerical questions

Leave your answers in terms of Vp, Vb and VDon. Take Vp > Vb + VDon so that switching occurs.

1

Output across the diode

Question 1 circuit An AC source feeds a resistor. The vertical branch has a battery with the positive terminal above, followed by a diode with the anode above. The output is measured from the battery-diode junction to ground. Vi+− R +−Vb VDon +Vo−
Figure 1. Use the output terminals shown.
2

Diode and battery reversed

Question 2 circuit An AC source feeds a resistor. The vertical branch has a diode with the cathode above, followed by a battery with the positive terminal below. The output is measured from the resistor-diode junction to ground. Vi+− R VDon −+Vb +Vo−
Figure 2. Use the output terminals shown.

Part B

Numerical questions

Use the same method. Check only your final results against the answer table below.

3

Lower clip at the diode

Question 3 circuit An AC source feeds a resistor. The vertical branch has a battery with the positive terminal above, followed by a diode with the cathode above. The output is measured from the battery-diode junction to ground. Vi+− R +−Vb VDon +Vo−
Figure 3. Vi = 6 sin(ωt) V; Vb = 2 V; VDon = 0.7 V.
4

Biased upper clip

Question 4 circuit An AC source feeds a resistor. The vertical branch has a diode with the anode above, followed by a battery with the positive terminal below. The output is measured from the resistor-diode junction to ground. Vi+− R VDon −+Vb +Vo−
Figure 4. Vi = 8 sin(ωt) V; Vb = 2 V; VDon = 0.7 V.

Check your work

Non-numerical final answers

Table 1. Final answers for Questions 1 and 2
QuestionDiode ON / OFFOutput when ONOutput when OFFOutput extrema
1ON: Vi > Vb + VDon
OFF: Vi < Vb + VDon		Vo = VDonVo = Vi − VbMinimum −Vp − Vb
Maximum VDon
2ON: Vi < −Vb − VDon
OFF: Vi > −Vb − VDon		Vo = −Vb − VDonVo = ViMinimum −Vb − VDon
Maximum Vp

Check your work

Numerical final answers

Table 2. Final answers for Questions 3 and 4
QuestionDiode ON / OFFOutput when ONOutput when OFFOutput extrema
3ON: Vi < 1.3 V
OFF: Vi > 1.3 V		Vo = −0.7 VVo = Vi − 2.0 VMinimum −0.7 V
Maximum +4.0 V
4ON: Vi > −1.3 V
OFF: Vi < −1.3 V		Vo = −1.3 VVo = ViMinimum −8.0 V
Maximum −1.3 V

For all four circuits, equality is the zero-current switching boundary; the ON and OFF output expressions agree there.