A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400 resistor connected in series, the zener current is found to be 4 times load current. The load current (IL) and load resistance (RL) are :
- IL = 0.02 mA; RL = 250
- IL = 10 A; RL = 0.5
- IL = 10 mA; RL = 500
- IL = 20 mA; RL = 250
Answer
(3) IL = 10 mA; RL = 500
Solution
Current through series resistor: I = 25 - 5400 = 0.05 A = 50 mA
I = IZ + IL = 4IL + IL = 5IL
IL = 10 mA
Load is in parallel with the zener VL = 5 V
RL = 510 10-3 = 500
Answer: (3)