Question 18.14: For the input waveform to the clipping circuit in Fig. 18.40...
For the input waveform to the clipping circuit in Fig. 18.40, find the output voltage waveform.
The battery of 5 V reverse biases the diode. The point *A must go positive to 5 V before the diode turns on. For all voltages at point A equal to or greater than 5 V, the diode conducts and the output voltage stays at 5 V. For all negative voltages at A and positive voltages less than 5 V, the diode is reverse biased. When reverse biased, the diode acts like an open circuit and V_{\text {out }}=V_{\text {in }}. Thus circuit in Fig. 18.40 is an adjustable positive peak clipper that clips all positive peaks greater than battery voltage (i.e. 5V).
* Assuming the diode to be ideal. Actually point A must go positive to 5.6 V before the diode turns on. Here 0.6 V accounts for potential barrier.