Question 45 of 209

In the given circuit if the power rating of Zener diode is $10mW$, the value of series resistance ${R}_{s}$ to regulate the input unregulated supply is:

A$\dfrac{3}{7}k\Omega$B$10\Omega$C$1k\Omega$D$10k\Omega$

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Answer: (A) $\dfrac{3}{7}k\Omega$

Potential drop across ${R}_{s}$,
${V}_{1}=8-5=3V$
Current through the load resistor
$I=\dfrac{5}{1\times {10}^{3}}=5mA$
Maximum current through Zener diode
${I}_{Z,\max }=\dfrac{10}{5}=2mA$
And minimum current through Zener diode
${I}_{Z,\min }=0$
$∴{I}_{S,\max }=5+2=7mA$
And ${R}_{S,\min }=\dfrac{{V}_{1}}{{I}_{S,\max }}=\dfrac{3}{7}k\Omega$
Similarly,
${I}_{S,\min }=5mA$
And ${R}_{S,\max }=\dfrac{{V}_{1}}{{I}_{S,\min }}=\dfrac{3}{5}k\Omega$
$∴\dfrac{3}{7}k\Omega \lt {R}_{s}\lt \dfrac{3}{5}k\Omega$

JEE Main Physics practice question on Semiconductors (Diodes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.