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: [figure]
- 0 V
- 3 V
- 2 V
- 1 V
Answer
(3) 2 V
Solution
When reverse biased beyond breakdown, the voltage drop across the ideal Zener diode remains fixed at its breakdown voltage:
VZ = 3 V
Applying Kirchhoff's voltage law across the closed circuit:
|Vi| = VZ + VBA
5 V = 3 V + VBA VBA = 5 V - 3 V = 2 V
Answer: (3)