Physics - Semiconductors Question with Solution | TestHub

Given, input voltage $V_{in}=10\text{ V}$
Zener breakdown voltage $V_Z=8\text{ V}$
Power of Zener diode $P_Z=1.6\text{ W}$

The current through the Zener diode is given by:
$I_Z=\frac{P_Z}{V_Z}=\frac{1.6}{8}=0.2\text{ A}$

Since there is no load resistor, the entire current flows through the series resistor $R$.
Therefore, current through the resistor $I_R=I_Z=0.2\text{ A}$

The voltage drop across the resistor $R$ is:
$V_R=V_{in}-V_Z=10\text{ V}-8\text{ V}=2\text{ V}$

Using Ohm's law, the value of resistance $R$ is:
$R=\frac{V_R}{I_R}=\frac{2}{0.2}=10\Omega$

Answer: $10\Omega$