A Zener diode with a breakdown voltage of $V_z=10\text{V}$ is used in a voltage regulator circuit. If the input voltage $V_i$ varies from $12\text{V}$ to $16\text{V}$ and the load resistance is $R_L=1\text{k}\Omega$, the maximum current through the Zener diode is (assuming series resistance $R_s=200\Omega$)

Solution

1. The load current $I_L$ is constant because the Zener diode maintains a fixed voltage $V_z=10\text{V}$ across the load $R_L=1\text{k}\Omega$: $$\begin{array}{rl}{I}_L& =\frac{V_z}{R_L}\\ & =\frac{10\text{V}}{1000\Omega }\\ & =10\text{mA}\end{array}$$.
2. The total current $I_s$ flowing through the series resistor $R_s=200\Omega$ is given by $I_s=\frac{V_i-V_z}{R_s}$.
3. To find the maximum Zener current $I_{z,\text{max}}$, we use the maximum input voltage $V_i=16\text{V}$: $$\begin{array}{rl}{I}_{s,\text{max}}& =\frac{16\text{V}-10\text{V}}{200\Omega }\\ & =\frac{6\text{V}}{200\Omega }\\ & =30\text{mA}\end{array}$$.
4. The Zener current is $I_z=I_s-I_L$, so $$\begin{array}{rl}{I}_{z,\text{max}}& =30\text{mA}-10\text{mA}\\ & =20\text{mA}\end{array}$$.
5. Hence the answer is (C).