Question:
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, \(R_L=4\,\mathrm{k}\Omega\). The series resistance of the circuit is \(R_i=1\,\mathrm{k}\Omega\). 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?
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, \(R_L=4\,\mathrm{k}\Omega\). The series resistance of the circuit is \(R_i=1\,\mathrm{k}\Omega\). 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?
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The Zener holds the load at 6 V, so load current is fixed; Zener current = (current through R_i) - (load current).
Updated On: Nov 5, 2026
- 0.5 mA; 8.5 mA
- 1.5 mA; 8.5 mA
- 1 mA; 8.5 mA
- 0.5 mA; 6 mA
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The Correct Option is A
Solution and Explanation
Across the Zener (and load) the voltage stays at 6 V. Load current: \(I_L=\frac{6\ \mathrm{V}}{4\ \mathrm{k\Omega}}=1.5\) mA (constant).
Current through \(R_i\): \(I=\frac{V_B-6}{1\ \mathrm{k\Omega}}\).
Minimum, at \(V_B=8\) V: \(I=2\) mA, so \(I_Z=2-1.5=0.5\) mA.
Maximum, at \(V_B=16\) V: \(I=10\) mA, so \(I_Z=10-1.5=8.5\) mA.
Answer: (1) 0.5 mA; 8.5 mA
Current through \(R_i\): \(I=\frac{V_B-6}{1\ \mathrm{k\Omega}}\).
Minimum, at \(V_B=8\) V: \(I=2\) mA, so \(I_Z=2-1.5=0.5\) mA.
Maximum, at \(V_B=16\) V: \(I=10\) mA, so \(I_Z=10-1.5=8.5\) mA.
Answer: (1) 0.5 mA; 8.5 mA
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