First slide

Diodes

Question

From the circuit shown below, the maximum and minimum values of Zener diode current are

Moderate

A

$6 mA, 5 mA$
B

$14 mA, 5mA$
C

$9 m A, 1 m A$
D

$3 m A, 2 m A$

Solution

$\begin{array}{l} R = 5 k Ω = 5000 Ω \\ V_{i n (\min)} = 80 V \end{array}$
$\begin{array}{l} V_{i n (\max)} = 120 V \\ V_{Z} = 50 V \end{array}$
$R_{L} = 10 k Ω = 10000 Ω$
$\begin{array}{l} I_{Z (\min)} = ? \\ I_{Z (\max)} = ? \end{array}$
Let $V_{in} =80V$
$V_{i n} {=IR+V}_{Z}$
$\begin{array}{l} \Rightarrow 80 = I 5000 + 50 \\ \Rightarrow 30 = I 5000 \end{array}$

$\Rightarrow I = \frac{30}{5000} = \frac{30}{5 \times 1000} = 6 \times 10^{- 3} A = 6 m A$
$\begin{matrix} W e k n o w, I = I_{Z} + I_{L} \\ \Rightarrow 6 m A = I_{Z} + \frac{V_{o}}{R_{o}} = I_{Z} + \frac{50}{10000} = I_{Z} + 5 \times 10^{- 3} \end{matrix}$
$\Rightarrow {6mA=I}_{Z} +5mA \Rightarrow I_{Z} = 1 mA$
${Let V}_{i n} =120V$
$V_{i n} {=IR+V}_{Z}$
$\Rightarrow 120=I5000+50$
$\Rightarrow 70 = I 5000$
$\Rightarrow I = \frac{70}{5000} = 14 m A$
$W e k n o w, I = I_{Z} + I_{L}$
$\Rightarrow 14 m A = I_{Z} + 5 \times 10^{- 3} = I_{Z} + 5 m A$
$\Rightarrow 9 m A = I_{Z (\max)}$

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