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)

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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