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11th January 2005, 03:40 AM  #1 
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Full wave vs full wave bridge rectifier
Do these two rectifier circuits have any real difference between them regarding the charging characteristics of the following capacitor?
The standard formulae for calculating their output voltages and currents shows the FWB to be around 20% more efficient in terms of the transformer VA rating required to produce a fixed output VA. How does this manifest itself in the way the first filter capacitor is charged? 
11th January 2005, 05:22 AM  #2 
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They are electrically identical.
FWB has two diode drops at any given time, so produces somewhat greater loss. Nowhere near 20% though (unless you are using tubes for the bridge ). Tim
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11th January 2005, 05:35 AM  #3  
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Quote:
Here is what I meant by efficiency difference of 20%: Assume you require 24VDC at 1A. Without taking diode drops into account, the textbook formulae say that you need the following transformer ratings: FWB: Vac = 24 Vdc / 1.41 = 17 Vac Iac = 1 Adc / 0.62 = 1.6 Aac VA rating = 27VA FW: Vac = 24Vdc / .71 = 34 Vac Iac = 1 Adc / 1 = 1 Aac VA rating = 34VA So what gives? 

11th January 2005, 06:10 AM  #4 
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The bridge uses all of the secondary all of the time
Full wave uses half of the secondary all of the time Bridge gives you higher voltage out Full wave gives you lower voltage out AFAIK, The extra kick from "all" of the secondary is where you get the lower VA rating.
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11th January 2005, 06:18 AM  #5  
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Quote:
I'd always assumed (without really thinking about it) the derating applied to the current capability of the transformer in the FWB arrangement had something to do with capacitor charging currents. I'm not sure I ever read this; I inferred it from the formulae. 

11th January 2005, 08:25 AM  #6 
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Hi,
surely the formulae are not comparable? 34Vac rectified & smoothed will give about 50Vdc less 1 diode drop. Did you mean 2 times 17Vac? then you have 1 diode drop off 25V but in the bridge two diode drops off, so effectively about 2.8% less efficient. The heating effect in the transformer secondary and to an extent the primary as well is based on the accumulation of Isqrd*R losses over the full charging cycle (integral but I don't know the maths). Nonsinusoidal pulses with a high peak to average ratio will have a much higher heating effect averaged over the full cycle. This leads to the derating shown in your first formula for but for an unknown reason is missing from the second. All rectifier / capacitor charging systems will suffer from this derating, not just the bridge. regards Andrew T. 
11th January 2005, 09:07 AM  #7  
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Quote:
The VA rating is determined by the inner temperature of the core and the windings, don't forget that. May I ask why you ask? Are you thinking of using a center tapped transformer? I see that you have cross posted this question to "the other forum" . Noboby there knows? Fred or Jocko maybe?
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11th January 2005, 09:09 AM  #8  
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Quote:
In a full wave rectifier (not a bridge) one must have double the AC voltage (centertapped, of course) to make the same DC voltage. That is clear. I've been using full wave rectifier circuits with valves for years. I'm curious why the calculated current rating of the transformer is different from what is expected. The formulae I used are published in practically every transformer manufacturer's literature as an aid to selection. 

11th January 2005, 09:27 AM  #9  
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Quote:
I was afraid that no one would ever understand what I was getting at. This thought was inspired by the discussion of capacitor charge currents in the "Snubberized" thread on the chip amp forum. There I suggested choke input power supply filters and full wave rectifiers as a means of reducing peak charge currents. Later I was reflecting on my suggestion regarding the full wave rectifier and was wondering why I thought this would reduce the peak charge current. I realized it was because of the mathematical relationship within these formulae and not because of any mechanism of which I'm aware. So now, of course, I "have to know" if you understand what I mean. I suppose it could be due to resistive and core losses, though 20% seems like a lot to me. Anyone else have any info? 

11th January 2005, 09:50 AM  #10 
diyAudio Member

If the transformer is big enough this isn't a big issue really because you won't drive the transformer hotter than h**l.
One thing you might consider is not to have too much capacitance, "lagom"(*) as we say in sweden. My gut feeling says that a dual secondary transformer with two recitifier bridges is the best but I'll guess someone can confirm that. *) not too little, not too much, just about enough. Is there a good single word for that in english?
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