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14.Semiconductor Electronics
hard
In a given circuit diagram, a $5 \,{V}$ zener diode along with a series resistance is connected across a $50 \,V$ power supply. The minimum value of the resistance required, if the maximum zener current is $90\, {mA}$ will be $.....\,\Omega .$
A
$100$
B
$1000$
C
$500$
D
$50$
(JEE MAIN-2021)
Solution
Voltage across ${R}_{{L}}=5\, {V}$
$\Rightarrow {i}_{2}=\frac{5}{{R}_{{L}}}$
Also voltage across ${R}=50-5=45$ $volt$
By $v=i R \Rightarrow R=\frac{v}{i}=\frac{45}{i_{1}+i_{2}}$
${R}=\frac{45}{90 {m} {A}+\frac{5}{{R}_{{L}}}}$
Current in zener diode is maximum when ${R}_{{L}} \rightarrow \infty$
$\left(i_{2} \rightarrow 0\right.$ and $\left.i_{i}=i\right)$
So ${R}=\frac{45}{90 {mA}}=500 \,\Omega$
Standard 12
Physics
