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Full Wave Rectifier

The break dow...

Question

The break down voltage of zener diode shown in figure below is 4 volts.

The current $I_{z}$ through zener diode is_____mA.
1.7

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Solution

The correct option is A 1.7
Assuming diode is open circuited

$I = \frac{12}{(2.2 + 2.3) k} = 2.667 m A$
$V_{1} = 2.3 \times 2.667 = 6.1341 V > V_{z}$
Zener diode is working in breakdown region

Applying KCL at node-1, we get
$\frac{V_{1} - 12}{2.2 k} + \frac{V_{1} - 4}{0.1 k} + \frac{V_{1}}{2.3 k} = 0$
$V_{1} = 4.17 V$
Current. $I_{Z} = \frac{4.17 - 4}{0.1 k} = 1.7 m A$

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