Homemade Power Supply Smoothing Capacitor Help

[kenw232]

I took a variac, stepped the output down with a microwave oven transformer, rectified it and then needed to smooth the rectified AC out with a smoothing capacitor.  This power supply needs to provide a lot of amps (about 25A) so the smoothing cap bank needed to be large.
Image: http://sites.extremehosting.ca/trash/ps.jpg

So I go to a web site like below and entered DC amps of about 25A, allowable voltage drop of something small like .5V, and time of 8.3ms because its rectified mains AC so 60Hz * 2 or 1000ms/120Hz = 8.3ms.  This says I needed a large cap bank of 415,000uF. 
http://digitalskywave.com/filter_capacitor_size.html

So I bought 18 of these in parallel which is 33,000uF * 18 = 594,000uF.
http://www.digikey.ca/product-detail/en/0/338-2260-ND

I then power it up, it charges the cap bank to 14V DC because of wherever my variac is dialed too,  and I then turn on my circuit and try to draw some current.  Here's a video ( ).  The question is, why does it drop 4 Volts from 14V to 10V?  Especially when I'm only drawing like 4A as shown on the LCD display?  I would expect it to drop to about 13.5V or even less, hence the entire point of calculating the cap bank size...

Can anyone enlighten me as to why I cannot maintain a very minor voltage drop when I turn my circuit on?  I'm trying to get something as close to this commercial DC power supply which drops no voltage at all as shown here ( ) with the exact same circuit.

Thanks for any help.

[Andy Watson]

stepped the output down with a microwave oven transformer

How exactly? All transformers have regulation factor - have you measured the resistance of its windings?

[DanielS]

More caps will reduce ripple voltage. They do nothing to reduce I*R voltage drop in the transformer windings, other wiring and the rectifiers' Vf. If you want a stable DC output voltage independent of load, you will need some form of regulator.

[ManCave]

More caps will reduce ripple voltage. They do nothing to reduce I*R voltage drop in the transformer windings, other wiring and the rectifiers' Vf. If you want a stable DC output voltage independent of load, you will need some form of regulator.

...after you do that, you will also need to beef up the wires going from your secondary to the capacitor bank. Those wimpy wires will definitely heat up a lot (if not burn) at 25A and will give you noticeable voltage drop even at 4 amps.

[kenw232]

How exactly? All transformers have regulation factor - have you measured the resistance of its windings?

The primary had 120 winds, so I wrapped the secondary with about 24 windings.  I have not measured the resistance of the windings yet.

[kenw232]

EDIT
Forget what I said in my first paragraph I thought you wanted 35VDC@ 25A but I guess you are after 14 or 18VDC?

I won't need 35V.  The voltage is not a specific value right now.

[kenw232]

So what is the point of the "Allowable Voltage Drop (V)" value in the formula? 
http://digitalskywave.com/filter_capacitor_size.html

I have no way to stop this 4V drop then? 

[BradC]

So you charge up the bank unloaded to ~14v, and when you load it you get 10V?

10V RMS ~= 14vVPeak. So you are charging up to the peak voltage of the transformer output and when loaded you are dropping to the RMS voltage. Sounds about right for a rectified and un-regulated transformer. Circuit is working perfectly.

[Psi]

So what is the point of the "Allowable Voltage Drop (V)" value in the formula? 
http://digitalskywave.com/filter_capacitor_size.html

I have no way to stop this 4V drop then?

And after rectification

[Dave]

Having a massive capacitor bank is incredibly stupid, as it causes massive current spikes in the transformer, causing it to overheat. 100mF should be more than enough for a 25A power supply. I'd also add a nice fat inductor after the rectifier to further reduce the current spikes.

The regulation factor of a transformer is not only influenced by resistive losses, but also by spurious magnetic fields.

[kenw232]

I'd also add a nice fat inductor after the rectifier to further reduce the current spikes.

What value inductor would you recommend for this?  Would this slow down the inrush current, like a poor mans soft start?

[mikerj]

So what is the point of the "Allowable Voltage Drop (V)" value in the formula? 
http://digitalskywave.com/filter_capacitor_size.html

I have no way to stop this 4V drop then?

You have an unregulated supply, so naturally the output voltage will vary according to the amount of current drawn.  Your commercial supply is regulated, the circuit inside maintains the output voltage between tight limits.  A 25 amp regulator is going to be a fairly involved design however, especially a switching regulator which you will need to minimise losses.

[The Electrician]

Can anyone enlighten me as to why I cannot maintain a very minor voltage drop when I turn my circuit on?

Thanks for any help.

Did you knock out the magnetic shunts before you wound the high current secondary?

[Kevin.D]

I dont think the whole 4v drop is due to transformer regulation (implies a  poor 30% tran regulation) unless using VERY inadequate guage wire on your secondary. Some of that drop is due to rectifiers ,when no load current through them Vf drop is small and caps will get almost upto Vpeak, any load then you get VPeak- 2*Vf  ( => 2V) .

[Kleinstein]

At high power a inductor is more efficient than large capacitors in smoothing ripple. However this will also increase the output resistance, so this is not really helping if you are after a stable voltage.
Having a variac and transformer already results in a rather high output impedance. Otherwise the high peak current due to the large capacitor might overload the transformer:  The power factor gets rather poor. Though you only want 25 A DC current, the transformer might see an RMS current of about 50 A, and possibly even more if the transformer is stiff enough.

Inrush current is due to 2 effects: charging the capacitors and possible magnetic saturation of the transformer core, especially if it's rather large or toroid form.

The easy way would be something like a SMPS, possibly even an old PC supply. They offer regulation and thus resonably stable voltage without excessive capacitance.

[kenw232]

Did you knock out the magnetic shunts before you wound the high current secondary?

No, I put them back in.  You mean the smaller bars this guy is trying to chisel out at 1:08 in the video?


I thought they would help so I left them in when I rewired.   

[kenw232]

yes. I'm beginning to see how it works.  I don't necessarily need to stop the voltage drop, its just what I thought was the inrush current due to the difference at startup.  The cap bank is at 14V (Vpeak) to start, I turn it on, thats more power for a second or so as it drops to 10V and which slows my circuit down.  I need to stop that spread somehow and make it more consistent, slow or fast.  I guess an inductor after the smoothing cap bank can't help...