Semiconductors Question 1
Question 1 - 2024 (01 Feb Shift 1)
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:
[We changed options. In official NTA paper no option was correct.]
(1) $5 \mathrm{k} \Omega$
(2) $10 \Omega$
(3) $1 \mathrm{k} \Omega$
(4) None of these
Show Answer
Answer: (4)
Solution:
$P d$ across $R_{s}$
$\mathrm{V}_{1}=8-5=3 \mathrm{~V}$
Current through the load resistor
$\mathrm{I}=\frac{5}{1 \times 10^{3}}=5 \mathrm{~mA}$
Maximum current through Zener diode
$\mathrm{I}_{\mathrm{z} \text { max. }}=\frac{10}{5}=2 \mathrm{~mA}$
And minimum current through Zener diode
$\mathrm{I}{\mathrm{z} \text { min. }}=0$ $\therefore \mathrm{I}{\mathrm{s} \text { max. }}=5+2=7 \mathrm{~mA}$
And $\mathrm{R}{\mathrm{s} \text { min }}=\frac{\mathrm{V}{1}}{\mathrm{I}_{\mathrm{smax}}}=\frac{3}{7} \mathrm{k} \Omega$
Similarly
$\mathrm{I}_{\mathrm{s} \text { min. }}=5 \mathrm{~mA}$
And $\mathrm{R}{\mathrm{s} \text { max. }}=\frac{\mathrm{V}{1}}{\mathrm{I}_{\mathrm{s} \text { min. }}}=\frac{3}{5} \mathrm{k} \Omega$
$\therefore \frac{3}{7} \mathrm{k} \Omega<\mathrm{R}_{\mathrm{s}}<\frac{3}{5} \mathrm{k} \Omega$
