Filter calculation for rectifier..

onlyvinod56

Joined Oct 14, 2008
369
Hello,

I am doing electronic projects past few years.
Almost all circuits require a DC power supply rectified from AC (230V, 50Hz).
Usually a 1000uF or 2200uF caps are using as filters. How can i calculate the exact value required? I have seen some inductor based filters too. What is the difference between those two filters?

Thankyou.

MrChips

Joined Oct 2, 2009
30,806
There is no EXACT value. The solution is a compromise. The larger the capacitor, the lower the output ripple. But this puts increasing demands on the rectifiers, i.e. the peak current goes way up. The solution is usually based on how much ripple you can tolerate.

#12

Joined Nov 30, 2010
18,224
radical 2 C Er F = I

The square root of 2 times capacitance times peak to peak voltage sag times frequency equals current (in amps).

Hi-Z

Joined Jul 31, 2011
158
If we assume a full-wave rectifier, the raw output from this would be a waveform like that shown here:

http://www.kpsec.freeuk.com/powersup.htm

As you can see, there will be a ripple equal to the peak output voltage. The article describes the addition of a capacitor to reduce this ripple, and calls it a "smoothing" capacitor. I prefer to use the term "reservoir" capacitor, because that's the function it's providing: it stores charge for use by the load when the rectifier isn't supplying current.

Whatever; you need to be able to calculate the value of reservoir capacitor, and you can do this using the following formula:

C = t*I/V

where V is the "droop" in voltage between the rectifier "topping-up" the reservoir, t is the time (seconds) between top-ups, I is the load current (Amps) and C is capacitance (Farads).

If you assume a large C, then the time between top-ups will be approximately the half-period of the input waveform, so 1/100 of a second if your mains is at 50Hz.

Once you've got a decent-sized reservoir capacitor, further filtering could possibly involve a choke (inductor), followed by a further smoothing capacitor (and this was popular in the days of valves/tubes), but nowadays it's much more common to find a solid-state voltage regulator.

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MrChips

Joined Oct 2, 2009
30,806
But think all that charge that is removed during the 10ms period between top-ups has to be replaced by the rectifier in a much shorter time. The diodes only conduct when the peaks of the sine wave exceed the output droop by the forward diode drop. The current spikes are huge.

wayneh

Joined Sep 9, 2010
17,498
The OP might also like to review this thread, a lengthy debate on this same topic.

Hi-Z

Joined Jul 31, 2011
158
But think all that charge that is removed during the 10ms period between top-ups has to be replaced by the rectifier in a much shorter time. The diodes only conduct when the peaks of the sine wave exceed the output droop by the forward diode drop. The current spikes are huge.
They are indeed - probably limited mainly by transformer winding resistance. It's certainly a mistake to think of a "linear" power supply as non-noisy - as you say, they're very spiky.