The following diagram shows a Zener diode as | vedclass

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(D) Given: Zener voltage $V_Z = 5 	ext{ V}$, load current $I_L = 5 	ext{ mA}$.Maximum Zener current $I_Z = 4 	imes I_L = 4 	imes 5 	ext{ mA} = 20

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$800$

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(D) Given: Zener voltage $V_Z = 5 	ext{ V}$, load current $I_L = 5 	ext{ mA}$.Maximum Zener current $I_Z = 4 	imes I_L = 4 	imes 5 	ext{ mA} = 20 	ext{ mA}$.Total current through the series resistor $R_S$ is $I = I_L + I_Z = 5 	ext{ mA} + 20 	ext{ mA} = 25 	ext{ mA} = 25 	imes 10^{-3} 	ext{ A}$.Maximum input voltage $V_{	ext{in}} = 25 	ext{ V}$.The voltage drop across the series resistor $R_S$ is $V_{R_S} = V_{	ext{in}} - V_Z = 25 	ext{ V} - 5 	ext{ V} = 20 	ext{ V}$.Using Ohm's law, $R_S = \frac{V_{R_S}}{I} = \frac{20 	ext{ V}}{25 	imes 10^{-3} 	ext{ A}} = 800  \Omega$.

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