The zener diode has a $$V_z = 30$$ V. The current passing through the diode for the following circuit is ______ mA.

Potential across load without Zener diode:

$$V = V_{\text{in}} \left( \frac{R_L}{R_s + R_L} \right) = 90 \left( \frac{5}{4 + 5} \right) = 50\ \text{V}$$

Since $$V > V_z$$ ($$50\ \text{V} > 30\ \text{V}$$), the Zener diode operates in the breakdown region: $$V_L = V_z = 30\ \text{V}$$

$$I_s = \frac{V_{\text{in}} - V_z}{R_s} = \frac{90 - 30}{4 \times 10^3} = 15 \times 10^{-3}\ \text{A} = 15\ \text{mA}$$

$$I_L = \frac{V_z}{R_L} = \frac{30}{5 \times 10^3} = 6 \times 10^{-3}\ \text{A} = 6\ \text{mA}$$

$$I_z = I_s - I_L = 15 - 6 = 9\ \text{mA}$$