In a zener regulated power supply, a zener diode with $$V_z = 6V$$ is used for regulation. The load current should be $$4.0 mA$$ and zener current is five times the load current for unregulated input of $$10 V$$. Find the series resistor $$R_S$$ used in the power supply.
Given that $$I_z = 5I_L$$
$$I_2 = 4.0mA$$
$$V_s = 10V$$
$$V_z = 6V$$
Apply KVL, in Loop $$ABCDA$$
$$V_s - R_sI - V_z = 0$$
$$R_3 = \dfrac{V_s - V_z}{I} = \dfrac{10-6}{I_z + I_2} = \dfrac{4V}{5I_2 + I_2} = \dfrac{4V}{6\times 4 \times 10^{-3}A}$$
$$R_s = 166.67\Omega$$