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 V_{o} \approx V_{i} =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} \approx 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 = \frac{V_({ac})rms}{V_{dc}} = \sqrt{\frac{(V_{rms})^2}{V_{dc}}-1}

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

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

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

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

V_{rms} = \frac{V_{m}}{\sqrt{2}}.

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

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

                                                                        \Gamma = \sqrt{\frac{V_{m}^{2}\pi ^{2}}{8V_{m}^{2}}-1}

                                                                        \Gamma = \sqrt{\frac{\pi ^{2}}{8}-1}

                                                                        \Gamma = 0.482


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Author: vikramarka

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