Full wave vs full wave bridge rectifier

[jeff mai]

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?

 

[Sch3mat1c]

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

 

[jeff mai]

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 ).

That they are electrically identical makes sense to me in some respects, but read on.

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?

 

[Stocker]

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.

 

[jeff mai]

AFAIK, The extra kick from "all" of the secondary is where you get the lower VA rating.

But why not 50% less VA rating instead of only 20% less?

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.

 

[AndrewT]

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). Non-sinusoidal 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.

 

[peranders]

But why not 50% less VA rating instead of only 20% less?

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.

I'll guess your question is not so easy to answer. The main difference between a center tapped transformer and two diodes and a fullwave bridge is I*I*R looses in the transformer together with losses in the core together with chosen caps. I think, not knowing for sure, that 20% seems to be a good approximation. This is a gut feeling only.

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?

 

[jeff mai]

34Vac rectified & smoothed will give about 50Vdc less 1 diode drop. Did you mean 2 times 17Vac

You might read my post more carefully. If I had meant 17Vac to make 24Vdc instead of 34Vac to make 24Vdc, I would be comparing two identical setups which makes no sense.

In a full wave rectifier (not a bridge) one must have double the AC voltage (center-tapped, 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.

 

[jeff mai]

May I ask why you ask? Are you thinking of using a center tapped transformer?

Thank you P-A!!!

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?

 

[peranders]

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?

 

[jeff mai]

My gut feeling says that a dual secondary transformer with two recitifier bridges is the best but I'll guess someone can confirm that.

There are many things to consider here. What does the diode switching noise look like? Maybe having only 2 noise makers are better than 4? Or maybe the noise is reduced in severity when 4 are used and that is more important?

How does the transformer respond? The filter capacitors?

Hmmm...an english word for just the right amount...I'm stumped. English was never a strong point of mine; too bad it's the only language I know!

 

[peranders]

Over at "the other" forum, Fred Dieckmann has kindly answered your question and he is right but I'm not sure he has understood your question completely (not I either!).

What is your "end product" here? A dual supply or a single supply?

If it's a dual supply things aren't that bad as Fred want to claim.

The right amount, do you mean "capacitance"?

BTW2: Fred what is a "föremedlare"? Do you mean "förmedlare" and what do you mean by that beacuse I don't "get it".

 

[Sch3mat1c]

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.

It must be because you have two half wave rectified windings, half wave has a worse power factor than full wave. I personally still maintain they are electrically identical, but that goes against the experimental evidence the companies have developed, which you quoted.

Tim

 

[Sch3mat1c]

My gut feeling says that a dual secondary transformer with two recitifier bridges is the best but I'll guess someone can confirm that.

Step 1. Connect windings as CT'd.
Step 2. Connect FWB across ends. Ground CT.

Half the diodes, better performance (half the diode drop), electrically identical.

That, ridiculously large capacitors, and RESISTIVE FILTERS...wtf!? are my pet peeves.

There are many things to consider here. What does the diode switching noise look like? Maybe having only 2 noise makers are better than 4? Or maybe the noise is reduced in severity when 4 are used and that is more important?

How does the transformer respond? The filter capacitors?

Noise is BS. If you are getting diode noise, you screwed up your circuit or construction, and I mock you.

The transformer core, and primary, see the same FWCT or FWB. The difference is that you have an extra winding, equal and opposite, being used at a similarly low duty cycle (i.e., half wave).
Think of using the other pair of diodes (in the FWB) as turning the winding upside-down every half cycle.

Tim

 

[Jocko Homo]

REALLY??????? I never knew that......

Noise is BS. If you are getting diode noise, you screwed up your circuit or construction, and I mock you.

You said it, not me.......

What was someone saying about SNR?

Jocko

 

[jan.didden]

[snip]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?

There's something wrong here. What are those "17Vac" and "34Vac" values? I guess the FWB is a 24Vac secondary with a full bridge, and that will give some 34Vac minus 2 diode drops.

What is the FW config? 2 x 24Vac with centertap? Or 24Vac with center tap = 2 x 12Vac, with two diodes? That would give only 17Vac as shown. So you cannot compare the two directly.

Jan Didden

 

[jeff mai]

There's something wrong here.

Yes, there is. I chose a confusing way to write out the formulae. The information within them is correct.

24Vdc @ 1A is the target. 17Vac and 34 Vac are the windings required to achieve 24Vdc in FWB and FW rectifier circuits, respectively, according to the formulae.

 

[jeff mai]

I personally still maintain they are electrically identical, but that goes against the experimental evidence the companies have developed, which you quoted.

From the info gathered here and elsewhere, it seems the electrical output (a fully rectified waveform) is identical, but the transformer ratings required to achieve this output are different.

 

[AndrewT]

Hi,
thanks Janneman, I think you are confirmming some doubt re comparability of the formulae.
The full wave circuit is two 17Vac windings not 34Vac. Each 17Vac winding is working on alternate half cycles.
Each winding is effectively a 17VA or thereabout. This implies a thinner wire and necessarily higher resistance. The Isqrd *R loss for each winding will be higher than the FWB circuit and the total loss over the whole cycle should result in a lower efficiency for the FW vs FWB. I cannot do the integral maths as said before. But in my opinion I would be surprised if the loss is as bad as 20%, and I await being enlightened.
regards Andrew T.

 

[jeff mai]

You might read my post more carefully. If I had meant 17Vac to make 24Vdc instead of 34Vac to make 24Vdc

I might have read your post more carefully AndrewT! Sorry! I see now that we meant the same thing. I though you meant 2*17Vac in parallel. I see you probably meant 2*17Vac in series which is exactly what I meant (34Vac center-tapped.)