Full Wave Rectifier

Full Wave Rectifier (FWR) contains two diodes $D_{1}$ and $D_{2}$.

FWR converts a.c voltage into pulsating DC in two-half cycles of the applied input signal.

Here  we use a  Transformer, whose secondary winding has been split equally into two half waves with a common center tapped connection ‘c’.

This configuration results in each diode conducting in turn when it’s anode terminal is positive with respect to Center point ‘c’ of  the Transformer.

Working of Full Wave Rectifier:-

During positive half cycle of applied i/p signal

• point ‘P’ is more positive w.r.to ‘c’.
• point ‘Q’ is more negative w.r.to ‘c’.

i.e, Diode $D_{1}$ is Forward Biased and $D_{2}$ is Reverse Biased , under this condition the equivalent circuit is as shown below

$\therefore&space;V_{o}&space;\approx&space;V_{i}&space;=i_{L}R_{L}$, when there is no diode resistance.

Similarly the conditions of diodes will be reversed for the negative half cycle of i/p signal.

• point ‘P’ is negative w.r.to ‘c’.
• point ‘Q’ is positive w.r.to ‘c’.

i.e, Diode $D_{1}$ is Reverse Biased and $D_{2}$ is Forward Biased , under this condition the equivalent circuit is  and output voltage is $V_{o}&space;\approx&space;i_{L}R_{L}$.

the i/p and o/p wave forms are as shown below

FWR is advantageous compared to HWR in terms of its efficiency and ripple factor.

Ripple Factor ($\Gamma$):-

$\Gamma&space;=&space;\frac{V_({ac})rms}{V_{dc}}&space;=&space;\sqrt{\frac{(V_{rms})^2}{V_{dc}}-1}$

to find out $V_{rms}$ and $V_{dc}$ of output signal

$\therefore&space;V_{rms}&space;=&space;\sqrt{\frac{1}{\pi&space;}\int_{0}^{\pi&space;}V_{m}^{2}&space;(\sin&space;^{2}\omega&space;t&space;)&space;d\omega&space;t}$                     ,     $\therefore&space;V_{dc}&space;=&space;\frac{1}{\pi&space;}&space;\int_{0}^{\pi&space;}V_{m}&space;(\sin&space;\omega&space;t&space;)&space;d\omega&space;t$

$V_{rms}&space;=&space;\sqrt{\frac{1}{\pi&space;}\int_{0}^{\pi&space;}V_{m}^{2}&space;(\frac{1-\cos&space;2\omega&space;t}{2}&space;)&space;d\omega&space;t}$          ,               $V_{dc}&space;=&space;\frac{-V_{m}}{\pi&space;}&space;[-2&space;]$

$V_{rms}&space;=&space;\sqrt{\frac{V_{m}^{2}}{\pi&space;}\int_{0}^{\pi&space;}&space;}(\frac{1}{2}\pi&space;)$                                           ,              $V_{dc}&space;=&space;\frac{2V_{m}}{\pi&space;}$

$V_{rms}&space;=&space;\frac{V_{m}}{\sqrt{2}}$.

$I_{rms}&space;=&space;\frac{V_{rms}}{R_{L}}=\frac{V_{m}}{\sqrt{2}R_{L}}&space;=&space;\frac{I_{m}}{\sqrt{2}}$      and  $I_{dc}&space;=&space;\frac{2I_{m}}{\pi&space;}$.

now the ripple factor results to be  $\Gamma&space;=&space;\sqrt{\frac{(\frac{V_{m}}{\sqrt{2}})^{2}}{(\frac{2V_{m}}{\pi&space;})^{2}}-1}$

$\Gamma&space;=&space;\sqrt{\frac{V_{m}^{2}\pi&space;^{2}}{8V_{m}^{2}}-1}$

$\Gamma&space;=&space;\sqrt{\frac{\pi&space;^{2}}{8}-1}$

$\Gamma&space;=&space;0.482$

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.