Question

In the circuit shown in Fig. the zener diode draws a current of 6 mA at breakdown voltage of 12 V. The zener diode is rated an 12 V, 0.36 W. The load resistance R_L has a range of

(1) 2 KΩ < R_L ≤ ∞

(2) 1 KΩ ≤ R_L < 100 KΩ

(3) 0.5 KΩ ≤ R_L ≤ ∞

(4) 1.5 KΩ ≤ R_L ≤ ∞

Answer 1

(3) 0.5 KΩ ≤ R_L ≤ ∞

In loop ABEF; p.d across zener diode = 12 V. Therefore p.d across resistance R = 15–12 = 3 V.

The current, I; as shown in Fig.; is

At branch point B; current I divides, I₁ = current in zener diode = 6 mA (given); therefore

Since zener diode is rated as 12 V, 0.36 W, the maximum current it can with stand = 0.36/12 A = 30 mA

when RL increases, IL decreases and I, increases. At R_L = ∞; the zener diode gap burnt because across it is 15 V which is more than its rated value.