# calculation capacitor size

Discussion in 'Mathematics and Physics' started by watosh, Jan 25, 2009.

1. ### watoshNew Member

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new to this forum; old..old navy et ..but have forgotten lots..i am looking for help in calculating the size of capacitors to use as filters in ac to dc powersupply.. have 4 to1 step down xfrmr (1500watt). bridge (full wave rect)diode rectifier 80 amp 1000v.. want to smooth out the pulsating dc some. would like someone to refresh me on the math to calculate the cap type and size for optimal use..thks art..

2. ### PhillDubyaNew Member

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Here is a thread where I got help with a similar question, this may help:

3. ### Hero999Banned

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It depends on the load, a constant current load will require a larger capacitor than a resistive load and a constant power load (i.e. a switching regulator) will require an even larger capacitor.

For a constant current load I use the following formula which I have programmed into a spreadsheet.
[Error: String\ is\ too\ long\ (338, limit 300)]

This complicated formula can be reduced to this:

This can be simplified again to:

This is only valid for 50/60Hz but it has the advantage of being easy to remember and uses easy units like mA and µF

I haven't come up with a formula or resistive load or constant power because I normally deal with linear regulators run off a rectifier.

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5. ### watoshNew Member

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calculating capacitor size

thank you sir.. ill roll this around and be back.. im sure-- with some more questions.. thanks agin!!

6. ### MrAlWell-Known MemberMost Helpful Member

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Hello,

Hero:
Interesting formula (the one with the arcsin in it).
What is that formula based on?

The reason i ask is because i compared it to my formula:

Vpp=6000*I/C

with C in uf and I in amps and Vpp in voltage peak to peak.

and it comes out similar.

7. ### Hero999Banned

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Hi MrAl,

The formula is several formulae joined together.

First we need to know how long the capacitor needs to supply.

The arcsin term gives the angle (in radians) from the zero crossing point when the voltage on the capacitor will be less than the voltage on the AC side, i.e. when Vin - Vr = Vc. Divide the result by 2pi×f and you have the time from the zero crossing to when the capacitor stops suppling current.

The 1/(4pi×f) part assumes that the capacitor starts discharging at 90 degrees. It gives the time from 90 degrees to the zero crossing point, one quarter of the AC cycle. This of course isn't completely accurate it'll be a bit less than that.

The I/Vr term calculates the capacitor size required assuming a constant current is drawn by the regulator given Vr is the voltage drop with the time given by the remainting terms.

If a resistive load were used the I/Vr term would be replaced by the Vc = Vs×e^(-t×RC) with Vr being Vs - Vc and rearranged to make C the subject.

The above formula also doesn't take into account of the diode's forward voltage.

For interest's sake I've added this to the formula below.

[Error: String\ is\ too\ long\ (382, limit 300)]

My latest spreadsheet includes this more accurate formula. I'll post it when I feel ready to.

8. ### MrAlWell-Known MemberMost Helpful Member

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Hi again,

Oh ok makes some sense.

What i was going to do next with the formula i found was
to include any series resistance. I had found that even
small resistances like 1 ohm could greatly affect the ripple
voltage level for obvious reasons.
Perhaps you can include some series resistance in your new
formula once you get to it.