JEE MainPhysicsSemiconductors
A Zener diode with a breakdown voltage of 12 V is connected in series with a resistance of 40 to form a voltage regulator circuit. The circuit is powered by an unregulated DC supply that fluctuates between 15 V and 22 V . To ensure the diode does not burn out under any condition, its minimum power rating must be
Options
- A3.0 W
- B0.9 W
- C2.5 W
- D5.5 W
Correct answer
A. 3.0 W
Step-by-step solution
The maximum power dissipation in the Zener diode occurs when the current through it is maximum. This happens when the unregulated input voltage is at its highest value. The maximum input voltage is V_ in ( max ) = 22 V . The voltage drop across the series resistor under this condition is: V_ s = V_ in ( max ) - V_ z = 22 V - 12 V = 10 V The maximum current flowing through the circuit (and entirely through the Zener diode, as there is no load) is: I_ s = V_ s R_ s = 10 V 40 = 0.25 A The maximum power dissipated by t