NEETPhysicsSemiconductors
An unregulated DC power supply of 12 V is connected across a series combination of a resistor of 1 k and a Zener diode. The Zener diode has a breakdown voltage of 5 V and is connected in reverse bias. Assuming no load resistor is connected across the Zener diode, what is the current flowing through the series resistor?
Options
- A12 mA
- B17 mA
- C7 mA
- D0 mA
Correct answer
C. 7 mA
Step-by-step solution
The Zener diode is connected in reverse bias and the input voltage ( 12 V ) is greater than the Zener breakdown voltage ( 5 V ). Therefore, the Zener diode operates in the breakdown region and maintains a constant voltage of 5 V across itself. The voltage drop across the series resistor is the difference between the input voltage and the Zener voltage: V_R = V_ in - V_Z = 12 V - 5 V = 7 V Using Ohm's law, the current flowing through the series resistor is: I = V_R R = 7 V 1 k = 7 mA Answer: 7 mA