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 the diode can handle?

Question

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 the diode can handle? (A) 40 mA (B) 60 mA (C) 50 mA (D) 45 mA

Tardigrade Admin · Accepted Answer

The maximum permissible current is IZmax = (P/VZ) = (364× 10-3/9.1) = 40 mA

Q. 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 the diode can handle?

$40 m A$

$60 m A$

$50 m A$

$45 m A$

The maximum permissible current is
$I_{Z_{ma x}} = \frac{P}{V _{Z}}$
$= \frac{364 \times 1 0 ^{- 3}}{9.1}$
$= 40 m A$

Q. A Zener diode is specified as having a breakdown voltage of 9.1V, with a maximum power dissipation of 364mW. What is the maximum current the diode can handle?

40mA

60mA

50mA

45mA