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JEE Main 2018 · Physics · Semiconductors and Communication · easy

It is connected across a load of 600 Ω and fed from a 15 V supply through a series resistor of 300 Ω.

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A Zener diode of breakdown voltage 9 V9\ \mathrm{V} is used as a voltage regulator. It is connected across a load of 600 Ω600\ \Omega and fed from a 15 V15\ \mathrm{V} supply through a series resistor of 300 Ω300\ \Omega. The current through the Zener diode is
  1. A. 15 mA15\ \mathrm{mA}
  2. B. 20 mA20\ \mathrm{mA}
  3. C. 5 mA5\ \mathrm{mA}
  4. D. 35 mA35\ \mathrm{mA}
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Answer: (C) 5 mA5\ \mathrm{mA}

In regulation the Zener and the load share the Zener voltage of 9 V9\ \mathrm{V}. The drop across the series resistor is 15−9=6 V15-9=6\ \mathrm{V}, so the total current is I=6300=20 mAI=\dfrac{6}{300}=20\ \mathrm{mA}. The load current is IL=9600=15 mAI_L=\dfrac{9}{600}=15\ \mathrm{mA}. The Zener current is therefore IZ=I−IL=20−15=5 mAI_Z=I-I_L=20-15=5\ \mathrm{mA}, which is positive, confirming that the diode is indeed in breakdown.
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