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A zener diode in the circuit shown below, has a knee current of 5 mA, and a maximum allowed power dissipation of 300 mW. What are the minimum and maximum load current that can be drawn safely from the circuit, keeping the output voltage V0 at 6 V?
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- 0 mA, 180 mA
- 5 mA, 110 mA
- 10 mA, 55 mA
- 60 mA, 180 mA
Correct Option: C
Given, IZ (min) = 5 mA
V0 = 6 V = Vz
(according to given condition).
Maximum power dissipation means maximum zener current
i.e. Iz (max) = Pmax V = 300 mW 6 V = 50 mA
from figure
I = Vin – Vz 50 = 9 – 6 50 = 60 mA
and I = IZ + IL
IL (min) = I – Iz (max) = 60 – 50 = 10 mA
IL (max) = I – Iz (min) = 60 – 5 = 55 mA
Hence alternative (C) is the correct choice.
