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Old 17th July 2007, 04:13 AM   #1
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Default FWCT Rectification

Ok. While I have mentioned it in another post, I find the lack of an explination available to others a disadvatage when it comes to understanding power supplys.


Q 1:How exactly do you calculate output from a FWCT rectifier?

While on non CT transformers, when the two ends are ran using a bridge rectifier you get ~1.414*v, and .707*I.

This is all fine and dandy if you are using a FWB with an x-frmr designed for non CT use. However, I would like an explination on how to calculate voltage from a FWCT circuit, as I have been unable to come up with one so far.



Lets use a common example of 195-0-195v. Theoreticly, you would just add the two together to come up with 380v, but in practice this is not the case. When using a FWCT , solid state produces about 280v , while a tube rectifier like the EZ81 produces around 278v. (normal tube rect. voltage drop.)

I basically need to calculate it for virtual ( no diode or tube recto losses) output.

There has got to be a mathematical way to produce an answer to this problem, however it has sofar eluded me, and if anyone would care to shed light on this matter it would be greatly appreciated.

The best thing I have come up with so far is take one half of the winding and multiply by 1.414 to get the output, however sometimes the calculation is either too high or too low.

Q2: Is there an advantage to using FWB on the windings of a CT transformer? For instance: since only one secondary winding is used at a time, the power rating of the transformer has to be about 30% greater than
for a full wave bridge transformer. This would mean , theoreticly, that if FWB rectified that the transformer could produce 30% more current.

Lets say, that instead of using the full winding ( which would give us twice the voltage), we use FWB rectifiers on each winding, and then paralell the two outputs. What would this instance yelid as far as voltage/current?
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Old 17th July 2007, 08:01 AM   #2
Colt45 is offline Colt45  Serbia
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1.414 * voltage in each leg - diode drop


so for 195-0-195,

SS - 275
tube ~275, dropping down with current rise.. as tube diode drop is not fixed .7v like silicon. At a decent load, AC RMS input ~ DC output, for most normal vacuum (not merc, xenon) rectifier.



So.. 1.414, then read tube rect whitepaper.. see curves.
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Old 17th July 2007, 11:16 AM   #3
Tweeker is offline Tweeker  United States
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How close to *1.414 you get depends on diode drop, current, and the size of the first filter capacitor. SS gets much closer as not only does it drop much less, you can use a larger capacitor. There is a price to pay for this, diode conduction narrows and peak currents rise as you push the capacitor size. This makes for more diode switching noise and possibly leads to saturation and/or overheating in the power transformer.
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Old 17th July 2007, 12:46 PM   #4
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There are various ways to approximate this - it isn't a simple calculation. The voltage drop due to the transformer is more than you might think, since the current waveform is highly peaked, nowhere near a sine wave. And a tube rectifier drop is not linear with current, so it can't be modeled with a resistor.

I use SPICE or the free PSU Designer software from duncanamps.com. There's a nice table of transformer circuits and voltages at http://www.hammondmfg.com/pdf/5c007.pdf that can give you a starting point - but it's an approximation.
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Old 19th July 2007, 08:57 PM   #5
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Thanks for the replies, this is just what I was looking for, at least to get approximations of what to expect.

Thanks!

TSD88~
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Old 20th July 2007, 09:22 PM   #6
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Default CT power supply

The 1.41 figure is good for a capacitor input filter.

For a choke input filter, the figure should be 0.9
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Old 24th July 2007, 04:05 AM   #7
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Is that Capacitor Choke or just choke and no input cap? I know a lot of circuits have a cap right after the rectifier, and then a resistor for a RC filter, but some use a choke in place ofthe resistor.
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Old 24th July 2007, 05:14 AM   #8
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CLC or CRC or just C = cap input

LC = choke input
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