Question

A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be 1 mA. The zener is to be used in the design of a shunt regulator fed from a 15-V supply. The load current varies over the range of 0 mA to 15 mA. Find a suitable value for the resistor R. What is the maximum power dissipation of the zener diode?

Answer 1

The maximum power dissipation of the zener diode 112mV.

Step-by-step explanation:

The minimum zener current should be:

5 * Iza= 5 * 1= 5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,

R =
`(15 - 5.6)/(20 mA)` = 470 Ω.

Maximum power dissipated in the diode occours when, IL=0 is

Pmax = 20 *
`10^(3)` * 5.6 = 112mV.