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JEE Mains · Physics · STD 12 - Semicondutor electronics
  • JEE Mains 2023
  • Medium
  • Numerical

A zener diode of power rating \(1.6\,W\) is to be used as voltage regulator. If the zener diode has a breakdown of \(8\,V\) and it has to regulate voltage fluctuating between \(3\,V\) and \(10 \,V\). The value of resistance \(R_5\) for safe operation of diode will be \(.........\Omega\)

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Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

\(V_b=8 \text { volt }\)

\(V_A-V_B=8 \text { volt }\)

Current through zener diode,

\(i =\frac{ P }{ V }=\frac{1.6 W }{8 V }=0.2\,A\)

\(V _{ C }- V _{ A }=10-8 \text { volt }\)

\(\therefore R =\frac{ V _{ C }- V _{ A }}{ i }=\frac{2 V }{0.2 A }=10\,\Omega\)

[Note : A zener diode can regulate only if input voltage is \(\geq\) zener breakdown voltage the range of input voltage should be \(8\) to \(10\,V\) so that output voltage remains constant \(=8\,V\) ]

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