Question

A zener diode having breakdown voltage equal to 15 V, is used in a voltage regulator. Circuit as shown in the figure, the current through the diode is 

(a) 5 mA 

(b) 10 mA 

(c) 15 mA 

(d) 20 mA

Answer 1

Answer is (a) 5 mA

The voltage drop across 1 kΩ 

= V₂ = 15 V

The current through 1 kΩ  is

I' = \(\frac{15V}{1\times10^{-3}}\) = 15 x 10^-3A = 15 mA

The voltage drop across 250 Ω

= 20 V - 15 V = 5 V 

The current through 250 Ω is

I = \(\frac{5V}{250Ω}\) = 0.02 A = 20mA

The current through Zener diode is

I₂ = I - I, = (20 - 15)mA - 5mA