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Can someone explain about calculation of filtering capacitor for power supply in detail? Answered

plz explain about Vp-p, freq, Vrms, in DC ! because in power supplys, when the current pass through the bridge rectifier, it is rectified into DC current by the rectifier! I'm really confused about it ! Wanna Know in Detail!!!!!!!!!

Tags:Thomas

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Jack A Lopez
Jack A Lopez

5 years ago

A bridge rectifier does not do a perfect job of converting AC into DC, and this statement is best explained using pictures. The AC waveform is a sine wave. The waveform that emerges from the bridge rectifier is this funny shape that has all the positive peaks of the AC waveform, plus all the negative ones flipped upside down.

This image,
https://en.wikipedia.org/wiki/File:Gratz.rectifier...

from the Wikipedia article titled "Rectifier"
https://en.wikipedia.org/wiki/Rectifier#Full-wave_...

shows what rectified AC looks like.

But this rectified AC waveform is far from a perfect DC waveform. A perfect DC waveform looks like a flat line, voltage is constant with time.

By the way, the word "ripple" is used to describe, and quantify, a waveform that is not perfect DC. Assuming you start with a waveform that is always positive, ripple magnitude is just the voltage difference between the maximum voltage seen, and the minimum voltage seen, in this waveform.

The usual trick for smoothing out rectified AC, i.e. making it flatter, i.e. making the ripple smaller, is to put a big capacitor on the output of the rectifier stage.

It turns out the size of the capacitor needed depends on the amount of current (in amperes) this power supply must, uh, supply.


The usual trick for approximating ripple voltage is just to assume the capacitor is being drained by a constant current I, the load current, for a time equal to the period of the rectified AC, which is T/2, half the period of the original AC waveform.

The charge on a capacitor is Q = C*V, and

I = dQ/dt = C*(dV/dt)

That is to say, a constant current I will cause constant (dV/dt) =I/C.

So the change in capacitor voltage, after (T/2) seconds, is
deltaV = (dV/dt)*(T/2) = (I/C)*(T/2)

deltaV = (I*T)/(2*C) = I/(2*f*C)

And this is the same formula, seen in this Wiki article about "Ripple (electrical)"
https://en.wikipedia.org/wiki/Ripple_%28electrical...

and that page is probably worth looking at, especially if you think their explanation of ripple voltage is more clear than mine.

Anyway, the important points to take away from this formula for ripple voltage:

Ripple magnitude is directly proportional to load current. If your power supply does not need to supply any current, then any size capacitor will work.

It is much easier to remove the ripple from high frequency rectified AC. That's why for smoothing the output of a switching power supply, with typical frequency of several 10s of KHz, the capacitors can be teensy, but for smoothing out rectified mains AC (50 or 60 Hz) the capacitors tend to be huge.

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Downunder35m
Downunder35m

5 years ago

We should request a dedicated homework channel for the forum ;)

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verence
verence

5 years ago

As a student of Electrical and Electronics Engineering that should be part of your classes. So this is either homework or you should just ask your teachers.

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steveastrouk
steveastrouk

5 years ago

It DOESN'T get rectified into a Constant level after the rectifier does it ?