Physics - Semiconductors Question with Solution | TestHub

Potential drop across $R_s$, 

$V_1=8-5=3 \mathrm{V}$

Current through the load resistor

$I=\frac{5}{1\times {10}^3}=5 \mathrm{mA}$

Maximum current through Zener diode

$I_{Z, \max }=\frac{10}{5}=2 \mathrm{mA}$

And minimum current through Zener diode

$I_{Z, \min }=0$

$\therefore I_{S, \max }=5+2=7mA$

And $R_{S, \min }=\frac{V_1}{I_{S, \max }}=\frac{3}{7}k\Omega$

Similarly,

$I_{S, \min }=5 \mathrm{mA}$

And $R_{S, \max }=\frac{V_1}{I_{S, \min }}=\frac{3}{5} k\Omega$

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