Question

A half wave rectifier is using a diode with a forward resistance of $50\Omega $. If input is given as $200{{V}_{rms}}$ and the load resistance of $1K\Omega $. Determine,
(a) ${{I}_{m}},{{I}_{dc}}\text{ and }{{I}_{rms}}$
(b) $PIV$
(c) DC output power

Answer 1

Hint: First of all the maximum current, dc current and the rms current is to be found by using appropriate equations. Peak inverse voltage of a half wave rectifier is given as the product of square root of two and the rms voltage. And the DC output power is determined by taking the product of the square of the dc current and the resistance of the load. These all may help you to solve this question.

Complete step by step answer:

It is already mentioned in the question that,
The rms voltage is given as,
${{V}_{rms}}=200V$
The forward resistance can be written as,
${{R}_{f}}=50\Omega $
And the load resistance is mentioned as,
${{R}_{L}}=1000\Omega $
(a) The maximum current can be found out by using the equation,
${{I}_{m}}=\dfrac{{{V}_{m}}}{{{R}_{f}}+{{R}_{L}}}$
Where ${{V}_{m}}$be the maximum voltage in the diode. This can be given by the equation,
${{V}_{m}}=\sqrt{2}{{V}_{rms}}$
Substituting the values in it will give,
${{V}_{m}}=\sqrt{2}\times 200=282.843V$
Therefore the maximum current can be written by substituting all the value in it,
${{I}_{m}}=\dfrac{282.843}{50+1000}=0.269A=269.37mA$
Now the DC current is to be found which is the maximum current divided by pi,
That is,
${{I}_{DC}}=\dfrac{{{I}_{m}}}{\pi }$
Substituting the values in it,
\[{{I}_{DC}}=\dfrac{269.37}{\pi }=85.75mA\]
Now let us find the rms current. This can be found by taking the half of maximum current.
That is,
\[{{I}_{rms}}=\dfrac{{{I}_{m}}}{2}=\dfrac{269.37}{2}=134.685mA\]
(b) Peak inverse voltage is given as the product of square root of two and the rms voltage.
That is,
\[PIV=\sqrt{2}{{V}_{rms}}=\sqrt{2}\times 200=282.84V\]
(c) Finally we have to find the DC output power of the circuit,
This can be found by taking the product of the square of the dc current and the resistance of the load.
That is,
\[P={{I}_{DC}}^{2}\times {{R}_{L}}\]
Substituting the values in it,
\[P={{\left( 85.75 \right)}^{2}}\times {{10}^{-6}}\times {{10}^{3}}\]
Simplifying this will give,
\[P=7.35watt\]
Therefore the answer has been obtained.

Note: The rms value is found by taking the square root of the mean of the square of the values. Peak Inverse Voltage is referred to as the maximum voltage inside the diode if it is in reverse bias during the negative half cycle. It is usually abbreviated as PIV.