Physics - Semiconductors Question with Solution | TestHub

Given:
Input voltage, $V=20\text{ V}$
Zener voltage, $V_Z=8\text{ V}$
Series resistance, $R_1=400\Omega$
Load resistance, $R_2=1600\Omega$

First, check if the Zener diode is in the breakdown region. The voltage across $R_2$ without the Zener diode is:
$V^{\prime }=\frac{R_2}{R_1+R_2}\times V=\frac{1600}{400+1600}\times 20=16\text{ V}$

Since $V^{\prime }>V_Z$ ($16\text{ V}>8\text{ V}$), the Zener diode is in the breakdown region and the voltage across it will be maintained at $8\text{ V}$.

The voltage across $R_1$ is:
$V_{R_1}=V-V_Z=20-8=12\text{ V}$

The current through $R_1$ is:
$I_{R_1}=\frac{V_{R_1}}{R_1}=\frac{12}{400}=0.03\text{ A}=30\text{ mA}$

The current through $R_2$ is:
$I_{R_2}=\frac{V_Z}{R_2}=\frac{8}{1600}=0.005\text{ A}=5\text{ mA}$

Applying Kirchhoff's current law, the current through the Zener diode is:
$I_Z=I_{R_1}-I_{R_2}=30\text{ mA}-5\text{ mA}=25\text{ mA}$

Answer: 25 mA