Physics - Semiconductors Question with Solution | TestHub

Given:
Input voltage, $V_{in}=20\text{ V}$
Zener voltage, $V_Z=15\text{ V}$
Series resistance, $R_s=250\Omega$
Load resistance, $R_L=1\text{ k}\Omega =1000\Omega$

Since the Zener diode is in parallel with the load resistor, the voltage across the load is regulated at $V_L=V_Z=15\text{ V}$.

The voltage drop across the series resistor is:
$V_s=V_{in}-V_Z=20\text{ V}-15\text{ V}=5\text{ V}$

The total current from the source (through $R_s$) is:
$I_s=\frac{V_s}{R_s}=\frac{5}{250}\text{ A}=20\text{ mA}$

The current through the load resistor is:
$I_L=\frac{V_L}{R_L}=\frac{15}{1000}\text{ A}=15\text{ mA}$

Applying Kirchhoff's current law, the current through the Zener diode is:
$I_Z=I_s-I_L=20\text{ mA}-15\text{ mA}=5\text{ mA}$

Answer: 5 mA