Question 44: Semiconductors and Logic Gates

Question

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 :

Accepted Answer

Pd across $R_s$
$$
\mathrm{V}_1=8-5=3 \mathrm{~V}
$$
Current through the load resistor
$$
\mathrm{I}=\frac{5}{1 	imes 10^3}=5 \mathrm{~mA}
$$
Maximum current through Zener diode
$$
\mathrm{I}_{\mathrm{z} \max .}=\frac{10}{5}=2 \mathrm{~mA}
$$
And minimum current through Zener diode
$$
\begin{gathered}
\mathrm{I}_{\mathrm{z} 	ext { min. }}=0 \\
	herefore \mathrm{I}_{\mathrm{s} 	ext { max. }}=5+2=7 \mathrm{~mA}
\end{gathered}
$$
And $\mathrm{R}_{\mathrm{s} 	ext { min }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} 	ext { max }}}=\frac{3}{7} \mathrm{k} \Omega$
Similarly
$$
\mathrm{I}_{\mathrm{s} \min .}=5 \mathrm{~mA}
$$
And $\mathrm{R}_{\mathrm{s} 	ext { max. }}=\frac{\mathrm{V}_1}{\mathrm{I}_{\mathrm{s} 	ext { min. }}}=\frac{3}{5} \mathrm{k} \Omega$
$$
	herefore \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega
$$

Answer

$10 \mathrm{k} \Omega$

Answer

$10 \Omega$

Answer

$1 \mathrm{k} \Omega$

Answer

$$ \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega$$

Semiconductors and Logic Gates

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 :

Choose the correct answer: