The Zener diode has reverse breakdown voltage $V_{Z}$ . The current through the load resistance $R_{L}$ is $I_{L}$ . The current through Zener diode is
( $V_{0}$ = Input voltage , $R_{S}$ = Series resistance)

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TestHub · Accepted Answer

The voltage across the Zener diode in breakdown region is $V_{Z}$ .

The voltage drop across the series resistor $R_{S}$ is $V_{0} - V_{Z}$ .

The current through the series resistor is $I_{S} = \frac{V _{0} - V _{Z}}{R _{S}}$ .

Applying Kirchhoff's current law at the junction, the total current $I_{S}$ is the sum of the Zener current $I_{Z}$ and the load current $I_{L}$ .

$I_{S} = I_{Z} + I_{L}$

$I_{Z} = I_{S} - I_{L}$

$I_{Z} = (\frac{V _{0} - V _{Z}}{R _{S}}) - I_{L}$

Answer: $(\frac{V _{0} - V _{Z}}{R _{S}}) - I_{L}$

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