The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is, R$$_L$$ = 4 kΩ. The series resistance of the circuit is R$$_i$$ = 1 kΩ. If the battery voltage V$$_B$$ varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?

$$\text{The load current is constant because the voltage across it is fixed at } V_Z:$$

$$I_L = \frac{V_Z}{R_L} = \frac{6\text{ V}}{4\text{ k}\Omega} = 1.5\text{ mA}$$

$$\text{For minimum Zener current (at } V_B = 8\text{ V}):$$ $$I_{i,\text{min}} = \frac{V_{B,\text{min}} - V_Z}{R_i} = \frac{8 - 6}{1\text{ k}\Omega} = 2\text{ mA}$$

$$I_{Z,\text{min}} = I_{i,\text{min}} - I_L = 2\text{ mA} - 1.5\text{ mA} = 0.5\text{ mA}$$

$$\text{For maximum Zener current (at } V_B = 16\text{ V}):$$

$$I_{i,\text{max}} = \frac{V_{B,\text{max}} - V_Z}{R_i} = \frac{16 - 6}{1\text{ k}\Omega} = 10\text{ mA}$$

$$I_{Z,\text{max}} = I_{i,\text{max}} - I_L = 10\text{ mA} - 1.5\text{ mA} = 8.5\text{ mA}$$