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Question: Answered & Verified by Expert
A zener diode of voltage $\mathrm{V}_{\mathrm{Z}}(=6 \mathrm{~V})$ is used to maintain a constant voltage across a load resistance $\mathrm{R}_{\mathrm{L}}(=1000 \Omega)$ by using a series resistance $R_{\mathrm{s}}(=100 \Omega)$. If the e.m.f. of source is $\mathrm{E}(=9 \mathrm{~V})$, what is the power being dissipated in Zener diode?
PhysicsSemiconductorsBITSATBITSAT 2021
Options:
  • A $0.144$ watt
  • B $0.324$ watt
  • C $0.244$ watt
  • D $0.544$ watt
Solution:
2099 Upvotes Verified Answer
The correct answer is: $0.144$ watt
Here, $E=9 \mathrm{~V} ; \mathrm{V}_{\mathrm{z}}=6 ; \mathrm{R}_{\mathrm{L}}=1000 \Omega$ and $\mathrm{R}_{\mathrm{s}}=100 \Omega$,

Potential drop across series resistor $\mathrm{V}=\mathrm{E}-\mathrm{V}_{\mathrm{Z}}=9-6=3 \mathrm{~V}$

Current through series resistance $R_{S}$ is $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{3}{100}=0.03 \mathrm{~A}$

Current through load resistance $R_{L}$ is $\mathrm{I}_{\mathrm{L}}=\frac{\mathrm{V}_{\mathrm{Z}}}{\mathrm{R}_{\mathrm{L}}}=\frac{6}{1000}=0.006 \mathrm{~A}$

Current through Zener diode is $\mathrm{I}_{\mathrm{Z}}=\mathrm{I}-\mathrm{I}_{\mathrm{L}}=0.03-0.006=0.024 \mathrm{amp} .$ Power dissipated in Zener diode is $\mathrm{P}_{\mathrm{Z}}=\mathrm{V}_{\mathrm{Z}} \mathrm{I}_{\mathrm{Z}}=6 \times 0.024=0.144 \mathrm{Watt}$

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