Physics - Semiconductors Question with Solution | TestHub

First, we check if the Zener diode is in the breakdown region. The voltage across $R_2$ without the Zener diode is:

$V^{\prime }=\frac{R_2}{R_1+R_2}\times V=\frac{1800}{800+1800}\times 18=\frac{1800}{2600}\times 18\approx 12.46\text{ V}$

Since $V^{\prime }>V_Z$ ($12.46\text{ V}>9\text{ V}$), the Zener diode is in the breakdown region and the voltage across it is fixed at $V_Z=9\text{ V}$.

The voltage across resistor $R_1$ is:

$V_{R1}=V-V_Z=18\text{ V}-9\text{ V}=9\text{ V}$

The total current $I$ flowing through $R_1$ is:

$I=\frac{V_{R1}}{R_1}=\frac{9}{800}\text{ A}=11.25\text{ mA}$

The current flowing through resistor $R_2$ is:

$I_{R2}=\frac{V_Z}{R_2}=\frac{9}{1800}\text{ A}=5\text{ mA}$

By Kirchhoff's current law, the current through the Zener diode $I_Z$ is:

$I_Z=I-I_{R2}=11.25\text{ mA}-5\text{ mA}=6.25\text{ mA}$

Answer: $6.25\text{ mA}$