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, RL = 4k Ω. The series resistance of the circuit is Ri = 1 k Ω. If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode? (a) 0.5 mA; 8.5 mA (b) 1.5 mA; 8.5 mA (c) 1 mA; 8.5 mA (d) 0.5 mA; 6 mA

Answer: (a). Load current is fixed: IL = (6)/(4 kΩ) = 1.5 mA At VB = 8 V: I = (8 - 6)/(1 kΩ) = 2 mA, so IZ = 2 - 1.5 = 0.5 mA At VB = 16 V: I = (16 - 6)/(1 kΩ) = 10 mA, so IZ = 10 - 1.5 = 8.5 mA So the answer is (a) 0.5 mA; 8.5 mA.

The figure represents a voltage regulator circuit using a Zener… | JEE Main 2019 (10 Apr, Shift 2)

Q81JEE Main 2019 · 10 APR · SHIFT 2Medium

Special Purpose Diodes

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, RL = 4k $\Omega$ . The series resistance of the circuit is Ri = 1 k $\Omega$ . If the battery voltage VB varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?

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Answer: (a)

Load current is fixed: $I_L = \frac{6}{4\text{ k}\Omega} = 1.5$ mA
At $V_B = 8$ V: $I = \frac{8 - 6}{1\text{ k}\Omega} = 2$ mA, so $I_Z = 2 - 1.5 = 0.5$ mA
At $V_B = 16$ V: $I = \frac{16 - 6}{1\text{ k}\Omega} = 10$ mA, so $I_Z = 10 - 1.5 = 8.5$ mA
So the answer is (a) $0.5$ mA; $8.5$ mA.

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