An ideal Zener diode with breakdown voltage of -3 V is reverse biased with a negative input voltage Vi = -5 V. The magnitude of voltage difference between point B and A is: (a) 0 V (b) 3 V (c) 2 V (d) 1 V

Answer: (c). The Zener is reverse biased with |Vi| = 5 V greater than 3 V, so it breaks down and holds 3 V across itself (B to C). The rest of the supply drops across R, which lies between B and A. |VB - VA| = 5 - 3 = 2 V So the answer is (c) 2 V.

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Special Purpose Diodes

An ideal Zener diode with breakdown voltage of -3 V is reverse biased with a negative input voltage $V_i = -5\text{ V}$ . The magnitude of voltage difference between point $B$ and $A$ is:

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Answer: (c)

The Zener is reverse biased with $|V_i| = 5$ V greater than $3$ V, so it breaks down and holds $3$ V across itself (B to C).
The rest of the supply drops across $R$ , which lies between B and A.
$|V_B - V_A| = 5 - 3 = 2$ V
So the answer is (c) $2$ V.

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