Semiconductors

JEE Main level, numerical answer, medium

58 of 209

Question 58 of 209

A Zener of breakdown voltage VZ=8V{V}_{Z}=8V and maximum Zener current, IZM=10mA{I}_{ZM}=10mA is subjected to an input voltage Vi=10V{V}_{i}=10V with series resistance R=100ΩR=100\Omega. In the given circuit RL{R}_{L} represents the variable load resistance. The ratio of maximum and minimum value of RL{R}_{L} is _____ .

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Answer: 2

Minimum value of RL{R}_{L} for which the diode is shorted is
RLRL+100×10=8⇒RL=400Ω\dfrac{{R}_{L}}{{R}_{L}+100}\times 10=8\Rightarrow {R}_{L}=400\Omega
For maximum value of RL{R}_{L}, current through diode is 10mA10mA
So, iR=iRL+IZM{i}_{R}={i}_{{R}_{L}}+{I}_{ZM}
⇒2100=8RL+10×10−3⇒10×10−3=8RL⇒RL=800Ω\Rightarrow \dfrac{2}{100}=\dfrac{8}{{R}_{L}}+10\times {10}^{-3} \Rightarrow 10\times {10}^{-3}=\dfrac{8}{{R}_{L}} \Rightarrow {R}_{L}=800\Omega
So, RLmaxRLmin=2\dfrac{{R}_{Lmax}}{{R}_{Lmin}}=2

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