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The Figure represents a voltage regulator circuit using a Zener diode the break down voltage of Zener diode is 6V and the load resistance is , $R_{L}=4K\Omega$ The series resistance of the circut is $R_{i}=1K\Omega$ . If the battery voltage $V_{B}$ varies from $8V\, \, to\, \, 16V$ ,What are the minimum and maximum values of the current through Zener diode?

Option 1)
$0.5mA;6mA$
Option 2)
$1mA;8.5mA$
Option 3)
$0.5mA;8.5mA$
Option 4)
$1.5mA;8.5mA$

Answers (1)

best_answer

$Given V_{B}=8V$

$i_{L}=\frac{6\times 10^{-3}}{4}=1.5\times 10^{-3}A$

$i_{R}=\frac{8-6\times 10^{-3}}{1}=2\times 10^{-3}A$

$\therefore i\, \, Zener diode=i_{R}-i_{L}= 0.5\times 10^{-3}A$

$When\, \, V_{B}=16V$

$i_{L}=1.5\times 10^{-3}A$

$i_{R}=\frac{\left ( 16.6 \right )\times 10^{-3}}{1}=10\times 10^{-3}A$

$\therefore i\, \, Zener\, diode=i_{R}-i_{L}$

$=8.5\times 10^{-3}A$

Option 1)

$0.5mA;6mA$

Option 2)

$1mA;8.5mA$

Option 3)
$0.5mA;8.5mA$
Option 4)

$1.5mA;8.5mA$

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