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Calculating voltage output of Capacitor in AC Circuit


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Hi there,

I'm trying to work out how to calculate the voltage output of a capacitor that I want to use as a filter after using a bridge rectifier.

The output from the secondary coil of the transformer is 12vAC max 4amps. The rectifier has a max of 400V and 8amps.

I want to understand how to work out what capacitor size I need to get a filtered 12V when I connect it in parellel with the rectifier outputs.

I used a 2200uf capacitor and my volt meter said about .3V. I then added a second 2200uf capacitor in parellel and I read 18V!

Can anyone point me in the right direction?

Cheers,
Hamish




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Hi Hamish,
Welcome to our forum.
The capacitor must provide the 3A to the load between charging pulses from the rectifier bridge so the voltage will have ripple while it discharges into the load then is quickly charged by the rectifier bridge.
The formula is approximately 1V p-p of ripple for each amp of current and each 2000uF. So your 3A load and 2200uF cap would have about 2.7V p-p of ripple.

Your 12VAC will have a peak voltage of about 15V. The rectifier bridge forward voltage drop will reduce it to about 13V. The ripple will reduce its average to about 11.6V. When you add a second 2200uF cap, the average voltage will rise to about 12.3V.

Your meter isn't reading the voltage correctly. Use a scope on it.

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Thank you for help with this! I have been having fun lifting weights by blowing up ballons under a bowl with my newly powered mini air compressor. :) One thing though, I didn't understand the forumla, just apprecited the outcome - could someone clarify what p-p stands for?

Hamish

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Hi Hamish,
Balloons lifting weights with an air compressor? I just use hot air or helium. ;D
Have you seen those radio-controlled indoor blimps? Helicopters lift stuff pretty well too.

1) The capacitor charges to the peak voltage of the transformer's output sine-wave, minus about 2V voltage drop of the rectifiers.
The formula for calculating the peak voltage of a sine-wave is to multiply its RMS voltage by 1.414 (the root of 2).
2) p-p is peak-to-peak. The capacitor isn't big enough to filter perfectly, so its voltage will have ripple, it will charge up quickly by the rectifier then discharge about 1V into the load before the next rectifier pulse charges it again. Therefore the sawtooth waveform is 1V p-p, measured from its lowest voltage to its highest.

You didn't catch my mistake with the formula. I subtracted the rectifier voltage drop twice. The average output voltage will be 13.6V with 2200uF and 14.3V with a second cap added. Drop the voltage down to about 12V by connecting 2 power rectifiers in series with its output.

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I went over capacitor input filters with my son around the mid-part of last year, and finally got my hands back on the books I gave him. I thought a couple of scans from the chapter on diode circuits might be of some help--either here, or in the future.
(It makes a decent-enough reference sheet for one's notebook on such things.)

post-6324-14279142056447_thumb.png

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Hi Enigma,
Your simple formula for ripple voltage doesn't allow for the source resistance of the transformer. It seems to give an excessive ripple calculation like it is assuming a high resistance transformer.

A high-current transformer usually has a pretty good voltage regulation because it has a very low resistance. Therefore it would produce a fairly low ripple voltage.
However, a cheap wall-wart AC-DC adapter has a very poor voltage regulation, some doubling the output voltage without a load. Their high resistance does not allow the filter cap to charge quickly to the peak of the AC.

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Among notable others, my Boylestad & Nashelsky text is missing at the moment (my son claims innocence), so I'm left with Malvino's "Devices" text--and the like--from which the 3 graphics were scanned.

Thus far, all other references that I have at my disposal concur with Malvino; either verbatim, or with a 'C =' or 'Iripple =' formula that derives directly to the one presented. I will dredge-up the B & N text--simply because I like that one better anyway--and compare. I doubt that it will, according to my recollection, differ substantively.

A high-current transformer usually has a pretty good voltage regulation because it has a very low resistance. Therefore it would produce a fairly low ripple voltage.

Until you draw current close to the design spec of the transformer; which increases the percentage ripple output--the value for 'C' remaining constant.

Personally speaking, I prefer transformers to putt-along at about 50% of their manufactured output current rating. It's all part of the balancing-act of staying within your (hopefully) generous engineering overheads.

For a single-phase power supply, the equation errs on the side that most engineers would prefer to err on: Providing a flatter output waveform across the designed range of load currents. One would then simply pay attention to the surge current for high values of C, adding a regulator, some shorting protection and a set of output caps.
I can live with the fact that attempting to refine a more exact solution--in the face of typical tolerances for e-caps--amounts to an exercise in false economy and wasted time.

I still want to get my hands on who took my textbook(s).
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Hey folks, thanks for input. It is all making much more sense. I need to get myself an o-scope though. Am also reading the AC circuits pdf on the site which is enlightening if not long and technical. :-)

One question that someone might be able to answer is does running a DC motor with voltage ripple damage the motor at all, or just affect its performance?

Hamish

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Well I can tell you that unfiltered AC will not damage a DC motor as I often run 12VDC motors of a bridge rectifier connected to a 12V transformer with no filter capacitors. I don't know about performance though it ran jus as fast as when it was connected to a 12V battery and the torque was also slightly higher.

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Have you 'scope'd the secondary of a power transformer that has a rectifier bridge, big filter cap and a fairly heavy load? The peaks of the sine-wave are flat-topped when the rectifier conducts.

As a matter of fact, I recall seeing something like that, although I can't recall if I was looking at the voltage across the secondary, or the current through it. I'd imagine that it could have been the former at some point, because the occasion I'm remembering was when I was comparing the overshoot of standard rectifiers with that of hot carrier rectifiers (FRRs).
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