Question

The break down voltage of zener diode shown in figure below is 4 volts.

The current $I_z$ through zener diode is_____mA.

• A. 1.7

Answer 1

The correct option is A 1.7
Assuming diode is open circuited

$I=\frac{12}{(2.2+2.3)k}=2.667mA$
$V_1=2.3\times 2.667=6.1341V>V_z$
Zener diode is working in breakdown region

Applying KCL at node-1, we get
$\frac{V_1-12}{2.2k}+\frac{V_1-4}{0.1k}+\frac{V_1}{2.3k}=0$
$V_1=4.17V$
Current. $I_Z=\frac{4.17-4}{0.1k}=1.7mA$