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JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2024
  • Hard
  • MCQ

In the given circuit if the power rating of Zener diode is \(10 \mathrm{~mW}\), the value of series resistance \(R_s\) to regulate the input unregulated supply is _______.

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Answer & Solution

Correct Answer

(A) \(5 \mathrm{k} \Omega\)

Step-by-step Solution

Detailed explanation

Pd across \(R_{\text {s }}\)

\(V_1=8-5=3 \mathrm{~V}\)

Current through the load resistor

\(\mathrm{I}=\frac{5}{1 \times 10^3}=5 \mathrm{~mA}\)

Maximum current through Zener diode

\(I_{z \max .}=\frac{10}{5}=2 \mathrm{~mA}\)

And minimum current through Zener diode

\(\mathrm{I}_{z_{\text {min. }}}=0\)

\(\therefore \mathrm{I}_{5 \text { max. }}=5+2=7 \mathrm{~mA}\)

\(\text { And } \mathrm{R}_{5 \text { min }}=\frac{\mathrm{V}_1}{\mathrm{I}_{5 \max } \cdot}=\frac{3}{7} \mathrm{k} \Omega\)

\(\text { Similarly }\)

\(\mathrm{I}_{5 \text { min. }}=5 \mathrm{~mA}\)

\(\text { And } \mathrm{R}_{5 \text { max. }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} \text { min. }}}=\frac{3}{5} \mathrm{k} \Omega\)

\(\therefore \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_5<\frac{3}{5} \mathrm{k} \Omega\)

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