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The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is $6 \mathrm{~V}$ and the load resistance is $\mathrm{R}_{\mathrm{L}}=4 \mathrm{k} \Omega$ . The series resistance of the circuit is $R_i=1 \mathrm{k} \Omega$ . If the battery voltage $V_B$ varies from $9 \mathrm{~V}$ to $17 \mathrm{~V}$ , what are the minimum and maximum values of the current through Zener diode?
Option: 1
$\quad 0.5 \mathrm{~mA} ; 7 \mathrm{~mA}$

Option: 2
$\quad 0.5 \mathrm{~mA} ; 9.5 \mathrm{~mA}$

Option: 3
$\quad 1.5 \mathrm{~mA} ; 9.5 \mathrm{~mA}$

Option: 4
$\quad 1 \mathrm{~mA} ; 8.5 \mathrm{~mA}$

Answers (1)

best_answer

$\begin{aligned} & \text { At } \mathrm{V}_{\mathrm{B}}=9 \mathrm{~V} \\ & \mathrm{i}_{\mathrm{L}}=\frac{6 \times 10^{-3}}{4}=1.5 \times 10^{-3} \mathrm{~A} \\ & \mathrm{i}_{\mathrm{R}}=\frac{(9-6) \times 10^{-3}}{1}=3 \times 10^{-3} \mathrm{~A} \end{aligned}$

$\begin{aligned} & \therefore \mathrm{i}_{\text {zener diode }}=\mathrm{i}_{\mathrm{R}}-\mathrm{i}_{\text {load }} \\ & =1.5 \times 10^{-3} \mathrm{~A} \end{aligned}$

$\begin{aligned} & \text { At } \mathrm{V}_{\mathrm{B}}=17 \mathrm{~V} \\ & \mathrm{i}_{\mathrm{L}}=1.5 \times 10^{-3} \mathrm{~A} \\ & \mathrm{i}_{\mathrm{R}}=\frac{(17-6) \times 10^{-3}}{1}=11 \times 10^{-3} \mathrm{~A} \end{aligned}$

$\begin{aligned} & \therefore \mathrm{i}_{\text {zener diode }}=\mathrm{i}_{\mathrm{R}}-\mathrm{i}_{\mathrm{L}} \\ & =9.5 \times 10^{-3} \mathrm{~A} \end{aligned}$

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