The Zener diode has a $V_{Z} = 30 V$ . The current passing through the diode for the following circuit is ______ $mA$ .

Open in App

Solution

Zener diode is in reverse biased, so potential drop of $30 V$ will appear across it.

So,
$I = \frac{60}{4 \times 10^{3}} = 0.015 A$

Also,
$I_{1} = \frac{30}{5 \times 10^{3}} = 0.006 A$

Therefore, current through the diode,
$I_{2} = I - I_{1} = 0.015 - 0.006 = 0.009 A = 9 mA$

Suggest Corrections

Similar questions

V_{Z} = 4 V)

(V_{Z} = 10 volts for the Zener diode)

mA

5.6 V

I_{Z}

A zener diode, having breakdown voltage equal to $15 V$ , is used in a voltage regulator circuit shown in figure.

The current through the diode( in $m A$ ) is

Join BYJU'S Learning Program