Semiconductors

JEE Main level, numerical answer, medium

53 of 209

Question 53 of 209

A 8V8V Zener diode along with a series resistance RR is connected across a 20V20V supply (as shown in the figure). If the maximum Zener current is 25mA25mA, then the minimum value of RR will be _____ Ω\Omega.

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Answer: 480

Let I1{I}_{1} be the current through Zener diode and I2{I}_{2} through load resistor.
​​​​​​​
As maximum current flows through Zener diode, hence
I2=0{I}_{2}=0.
∴∴Maximum current, I=I1+I2=25mA+0=25mAI={I}_{1}+{I}_{2}=25mA+0=25mA
Here,
Vo{V}_{o} is voltage drop across Zener =8V=8V.
Since
II is maximum, hence minimum value of RR
R=Vi−Vo=20−825×10−3=480ΩR=\dfrac{{V}_{i}-}{{V}_{o}}=\dfrac{20-8}{25\times {10}^{-3}}=480\Omega

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