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JEE Main 2025 · Physics · Semiconductors and Communication · easy

A Zener diode with breakdown voltage 10 V is used in a voltage regulator. A 20 V source feeds the circuit through a series resistor of 200 Ω, and a…

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A Zener diode with breakdown voltage 10 V10\ \mathrm{V} is used in a voltage regulator. A 20 V20\ \mathrm{V} source feeds the circuit through a series resistor of 200 Ω200\ \Omega, and a load resistor of 1 kΩ1\ \mathrm{k\Omega} is connected across the Zener diode. The current through the Zener diode (in breakdown) is
  1. A. 10 mA10\ \mathrm{mA}
  2. B. 40 mA40\ \mathrm{mA}
  3. C. 50 mA50\ \mathrm{mA}
  4. D. 60 mA60\ \mathrm{mA}
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Answer: (B) 40 mA40\ \mathrm{mA}

In breakdown the voltage across the Zener and the load is held at 10 V10\ \mathrm{V}. The drop across the series resistor is 20−10=10 V20-10=10\ \mathrm{V}, so the total current is I=10/200=50 mAI = 10/200 = 50\ \mathrm{mA}. The load takes IL=10 V/1 kΩ=10 mAI_L = 10\ \mathrm{V}/1\ \mathrm{k\Omega} = 10\ \mathrm{mA}. By Kirchhoff's current law the Zener carries IZ=50−10=40 mAI_Z = 50-10 = 40\ \mathrm{mA}.
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