Answer:
The maximum power dissipation of the zener diode 112mV.
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 = [tex]\frac{15 - 5.6}{20 mA}[/tex] = 470 Ω.
Maximum power dissipated in the diode occours when, IL=0 is
Pmax = 20 * [tex]10^{3}[/tex] * 5.6 = 112mV.
