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Question

(C) The circuit diagram of a Zener diode as a voltage regulator is shown in the figure.Given: Supply voltage $V_s = 16 \, V$, Zener voltage $V_Z = 10

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$200 \, \Omega$

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(C) The circuit diagram of a Zener diode as a voltage regulator is shown in the figure.Given: Supply voltage $V_s = 16 \, V$, Zener voltage $V_Z = 10 \, V$, Zener current $I_Z = 5 I_L$, and load resistance $R_L = 2 \, k\Omega = 2000 \, \Omega$.The current through the load resistance is:$I_L = \frac{V_Z}{R_L} = \frac{10 \, V}{2000 \, \Omega} = 5 	imes 10^{-3} \, A = 5 \, mA$.The total current through the series resistance is:$I = I_Z + I_L = 5 I_L + I_L = 6 I_L$.$I = 6 	imes (5 \, mA) = 30 \, mA = 3 	imes 10^{-2} \, A$.The series resistance $R_S$ is given by:$R_S = \frac{V_S - V_Z}{I} = \frac{16 \, V - 10 \, V}{3 	imes 10^{-2} \, A} = \frac{6 \, V}{0.03 \, A} = 200 \, \Omega$.Thus, the series resistance for the Zener diode is $200 \, \Omega$.

When a Zener diode is used as a regulator with a Zener voltage of $10 \, V$, nearly five times the load current passes through the Zener diode. What should be the series resistance for the Zener diode if the load resistance is $2 \, k\Omega$ and the unregulated voltage supplied is $16 \, V$?

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