Take the breakdown voltage of the zener diode used in the given circuit as 6 V. For the input voltage shown in figure here, the time variation of the output voltage is (graphs drawn are schematic and not to scale) [Figure: input is a sine wave of peak ±10 V. It is applied through a series resistor…

Question

Take the breakdown voltage of the zener diode used in the given circuit as 6 V. For the input voltage shown in figure here, the time variation of the output voltage is (graphs drawn are schematic and not to scale) [Figure: input is a sine wave of peak ±10 V. It is applied through a series resistor R₁ to two identical Zener diodes connected in series back to back across the output V₀. Options: (a) positive half clipped flat, negative half unclipped; (b) positive half unclipped, negative half clipped flat; (c) both halves clipped flat at the same level; (d) unclipped sine.]

Accepted Answer

(3) Both halves clipped at the same level. In each half-cycle one Zener is forward biased (about 0.7 V) and the other is in breakdown (6 V). So the pair conducts once |Vᵢₙ| exceeds about 6.7 V, in either direction. The output follows the input below that level and is clipped flat at about ±6.7 V: a symmetric clipped wave.

Step-by-step solution

Why the other options are wrong