The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is $\mathrm{R}_{\mathrm{L}}=4 \mathrm{k} \Omega$. The series resistance of the circuit is $\mathrm{R}_{\mathrm{i}}=1 \mathrm{k} \Omega$. If the battery voltage $\mathrm{V}_{B}$ varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?

(D) $0.5 \mathrm{~mA} ; 8.5 \mathrm{~mA}$

(A) $0.5 \mathrm{~mA} ; 0.6 \mathrm{~mA}$

(B) $1 \mathrm{~mA} ; 8.5 \mathrm{~mA}$

(C) $1.5 \mathrm{~mA} ; 8.5 \mathrm{~mA}$

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Current Electricity

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is

R_{L} = 4 k Ω

. The series resistance of the circuit is

R_{i} = 1 k Ω

. If the battery voltage

V_{B}

varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?

Answer: (D)

The Zener holds the load at

6 V

, so the load current is fixed at

i_{L} = \frac{6}{4000} = 1.5 mA

. The series resistor carries whatever the battery pushes through: when

V_{B} = 8 V

i = \frac{8 - 6}{1000} = 2 mA

, leaving the diode only

2 - 1.5 = 0.5 mA

; when

V_{B} = 16 V

i = \frac{16 - 6}{1000} = 10 mA

, leaving the diode

10 - 1.5 = 8.5 mA

. The Zener current therefore ranges from

0.5 mA

up to

8.5 mA

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option