MHT CET · Physics · Semiconductors
A \(5.0 \mathrm{~V}\) stabilized power supply is required to be designed using a \(12 \mathrm{~V} \mathrm{DC}\) power supply as input source. The maximum power rating of zener diode is \(2.0 \mathrm{~W}\). The minimum value of resistance \(\mathrm{R}_{\mathrm{s}}\) in \(\Omega\) connected in series with zener diode will be
- A \(16.5\)
- B \(17.5\)
- C \(18.5\)
- D \(15.5\)
Answer & Solution
Correct Answer
(B) \(17.5\)
Step-by-step Solution
Detailed explanation
Using the series resistance formula for a zener diode
\(\begin{aligned}
\mathrm{R}_{\mathrm{S}} & =\frac{\left(\mathrm{V}_{\mathrm{S}}-\mathrm{V}_{\mathrm{Z}}\right)}{\mathrm{I}_{\mathrm{Z}_{\max }}} \\
\mathrm{I}_{\mathrm{Z} \max } & =\frac{\mathrm{P}_{\mathrm{Z}}}{\mathrm{V}_{\mathrm{Z}}}=\frac{2}{5}=400 \mathrm{~mA} \\
\therefore \quad \mathrm{R}_{\mathrm{S}} & =\frac{(12-5)}{400 \times 10^{-3}}=\frac{7}{400} \times 10^3 \\
& =17.5 \Omega
\end{aligned}\)
\(\begin{aligned}
\mathrm{R}_{\mathrm{S}} & =\frac{\left(\mathrm{V}_{\mathrm{S}}-\mathrm{V}_{\mathrm{Z}}\right)}{\mathrm{I}_{\mathrm{Z}_{\max }}} \\
\mathrm{I}_{\mathrm{Z} \max } & =\frac{\mathrm{P}_{\mathrm{Z}}}{\mathrm{V}_{\mathrm{Z}}}=\frac{2}{5}=400 \mathrm{~mA} \\
\therefore \quad \mathrm{R}_{\mathrm{S}} & =\frac{(12-5)}{400 \times 10^{-3}}=\frac{7}{400} \times 10^3 \\
& =17.5 \Omega
\end{aligned}\)
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