### 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 $R _{s \text { max. }}=\frac{V _1}{I _{s \text { min. }}}=\frac{3}{5} k \Omega$

$\therefore \frac{3}{7} k \Omega<R _s<\frac{3}{5} k \Omega$