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 mA
60 mA
50 mA
45 mA
A.
40 mA
Current increases with increase in voltage.
Power = voltage × current
When voltage is maximum then current will be maximum.
Therefore maximum ower dissipation occurs at maximum voltage
Pmax = Vmax IZmax
Pmax = 364 mW
= 364 × 10-3 W
Vmax = 9.1 V
The maximum permissible current is
IZmax =
=
IZmax = 40 mA