A zener diode is specified as having a breakdown voltage of 9.1 V , with a maximum power dissipation

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14.Semiconductor Electronics

easy

A zener diode is specified as having a breakdown voltage of $9.1 \;V$, with a maximum power dissipation of $364 \;mW$. What is the maximum current ($mA$) the diode can handle?

A

$28$

B

$20 $

C

$35 $

D

$40$

(AIIMS-2016)

Solution

The maximum permissible current is

$I_{Z_{\max }}=\frac{P}{V_Z}$

$=\frac{364 \times 10^{-3}}{9.1}$

$=40 mA$

Standard 12

Physics

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