Question

For the following circuit, choose the correct waveform corresponding to the output signal (V_out). Given V_in = 5 sin(200πt) V, forward bias voltage of the diodes (D and Z) = 0.7 V and reverse Zener voltage = 3 V.

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is A

To find the correct waveform for the output signal (V_out), we need to analyze the given circuit, which includes a diode (D) and a Zener diode (Z). The input voltage is given as V_in = 5 sin(200πt) V, with a forward-bias voltage of 0.7 V for both diodes and a reverse Zener voltage of 3 V.

1. The input voltage V_in is an AC sine wave with a peak voltage of 5 V. 
2. The diode D will conduct when V_in exceeds 0.7 V (forward-bias condition). Therefore, during the positive half-cycle of the input signal, the diode conducts and regulates the current.
3. When D conducts, V_out = V_in - 0.7 V, as there is a 0.7 V drop across the diode.
4. For the negative half-cycle, the diode D is reverse-biased and does not conduct.
5. The Zener diode Z will conduct only when the reverse voltage across it exceeds 3 V. If V_in reaches beyond -3 V (considering diode drop), the Zener clamps V_out at -3 V due to its breakdown characteristic.
6. The resulting V_out will have positive halves clamped at (5 - 0.7) V and negative halves limited at -3 V.

Thus, the waveform for V_out should show both a clipping at approximately 4.3 V on the positive side and -3 V on the negative side. The correct waveform matches Graph 1.