JEE MainPhysicsSemiconductors
A protection circuit consists of a Zener diode having a breakdown voltage of 5 V and a red light emitting diode (LED) with a turn-on voltage of 2 V connected in series with a current-limiting resistor of 300 . This entire combination is placed across a variable DC power supply. If the maximum power rating of the Zener diode is 50 mW , the maximum safe voltage of the DC power supply is _____ V .
Correct answer
10
Step-by-step solution
The maximum safe current through the Zener diode is determined by its maximum power rating: I_ max = P_ max V_ Z = 50 10⁻³ 5 = 10 mA Since the components are connected in series, this is the maximum allowable current for the entire circuit. The maximum safe voltage of the DC supply is the sum of the voltage drops across the Zener diode, the LED, and the resistor at this maximum current: V_ total = V_ Z + V_ LED + I_ max R V_ total = 5 + 2 + (10 10⁻³ 300) V_ total = 7 + 3 = 10 V . Answer: 10