Semiconductors Question 1
Question 1 - 2024 (01 Feb Shift 1)
In the given circuit if the power rating of Zener diode is $10 mW$, 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) $5 k \Omega$
(2) $10 \Omega$
(3) $1 k \Omega$
(4) None of these
Show Answer
Answer: (4)
Solution:
$P d$ across $R _s$
$V _1=8-5=3 V$
Current through the load resistor
$I=\frac{5}{1 \times 10^{3}}=5 mA$
Maximum current through Zener diode
$I _{z \text { max. }}=\frac{10}{5}=2 mA$
And minimum current through Zener diode
$I _{z \text { min. }}=0$ $\therefore I _{s \text { max. }}=5+2=7 mA$
And $R _{s \text { min }}=\frac{V _1}{I _{smax}}=\frac{3}{7} k \Omega$
Similarly
$I _{s \text { min. }}=5 mA$
And $R _{s \text { max. }}=\frac{V _1}{I _{s \text { min. }}}=\frac{3}{5} k \Omega$
$\therefore \frac{3}{7} k \Omega<R _s<\frac{3}{5} k \Omega$
