Physics - Semiconductors Question with Solution | TestHub

Given:
Input voltage, $V_{in}=24\text{ V}$
Zener voltage, $V_Z=15\text{ V}$
Series resistance, $R_S=300\Omega$
Load resistance, $R_L=1200\Omega$

First, check if the Zener diode is in the breakdown region. The voltage across the load without the Zener diode is:
$V=V_{in}\times \frac{R_L}{R_S+R_L}=24\times \frac{1200}{300+1200}=24\times \frac{1200}{1500}=19.2\text{ V}$

Since $19.2\text{ V}>15\text{ V}$, the Zener diode is in breakdown and regulates the voltage across the load to $V_Z=15\text{ V}$.

The voltage drop across the series resistor $R_S$ is:
$V_S=V_{in}-V_Z=24-15=9\text{ V}$

The current through the series resistor $R_S$ is:
$I_S=\frac{V_S}{R_S}=\frac{9}{300}\text{ A}=30\text{ mA}$

The current through the load resistor $R_L$ is:
$I_L=\frac{V_Z}{R_L}=\frac{15}{1200}\text{ A}=12.5\text{ mA}$

Applying Kirchhoff's current law, the current through the Zener diode $I_Z$ is:
$I_Z=I_S-I_L=30\text{ mA}-12.5\text{ mA}=17.5\text{ mA}$

Answer: 17.5 mA