Current rating and full wave rectification

Thread Starter

bitrex

Joined Dec 13, 2009
79
I understand that if I have a transformer rated for a certain voltage at a certain amperage that it is usually the AC rating that's listed, and if the output is full wave rectified the available current will be reduced - I think the rule of thumb I heard was 80% of the AC current? I'd appreciate it if anyone could tell me the correct rule of thumb and perhaps a reference that gives a derivation of this rule, as I'm on a kick lately where I can't take things at face value without seeing a proof. :D
 

rjenkins

Joined Nov 6, 2005
1,013
The transformer voltage is given as the RMS value (at rated load).

The voltage after rectification will be roughly 1.4 times that, as the rectifiers conduct up to the peak of the AC waveform.

For an example, think of a transformer rated 10V at 10A. (100W or correctly 100VA).
The rectified voltage will be roughly 14V (not allowing for rectifier drop).

If you draw 10A at that 14V, you are pulling 140W from the transformer.

You need to reduce the current to about 0.7 x the transformer rating to stay within it's overall thermal power rating; 7V * 14A = 98W
 

eblc1388

Joined Nov 28, 2008
1,542
If you draw 10A at that 14V, you are pulling 140W from the transformer.
You simply can't do that. As long as current is drawn, the voltage will collapse back to 10V minus the diode drop.

While one can draw 10V@10A AC from the transformer, the point is one simply can't even draw DC 10A from the transformer after rectification as some current is lost as ripple current and total secondary winding current = ripple current + load current.

For a full-wave-center tap(FWCT) rectification application, one must allow 20% more in winding current. In the example above, if one wants 10A DC, then one has to use a 12A rated transformer. Full-wave-bridge(FWB) is even worst at 80% more so one needs a 18A rated secondary winding.

The interesting thing is, despite the 80% increase, overall transformer rating is *smaller* when using full-wave-bridge rectification because only a single winding is involved.

Let say the overhead diode drop is 1V at required current.

FWCT : 11V x2 @12A = 22*12 = 264VA (one diode drop)
FWB : 12V @ 18A = 216VA (two diode drop)

The above assume capacitor filters. There isn't any additional current requirement for FWB if an inductor is used instead. In fact, with a FWCT arrangement, only 70% of DC load current is suffice. Therefore a 11Vx2 @7A transformer can supply a 10V @10A load.
 
You can go thru a lot of calculations to figure transformer and rectifier sizes, but until transformer efficiencies are improved (unlikely), here are some general rules that always work. For single phase FWB circuits, figure 11% (1.11 x DC load KW) for AC transformer size. For 3 phase designs, figure 5% (1.05 x DC load KW) for AC transformer size.

ps: Don't overload your transformers, or the voltage will sag.

Cheers, DPW [ Everyone's knowledge is in-complete...Albert Einstein]
 
Last edited:

eblc1388

Joined Nov 28, 2008
1,542

Attachments

Source of information: Standard Handbook for Electrical Engineers, ISBN 007020974x, 1978 McGraw-Hill, pages 13-22, 13-23.

Many years designing power rectifiers show this data to be accurate.

Cheers, DPW [ Always remember that you are not going to live forever]
 

eblc1388

Joined Nov 28, 2008
1,542
Thanks.

I'll take a look at the handbook next time I'm visiting the library. I wonder why there are such big differences in the quoted figures.

Since the rating quoted above have "KW" mentioned as transformer rating, does the formula still hold true for small size transformers less than 1000VA?
 
It holds true for all transformers that I ever used, large and small. This data is available in many other references also and is published by the rectifier manufacturers too.

Cheers, DPW [ Spent years making heaters out of op-amps.]
 

k7elp60

Joined Nov 4, 2008
562
Gentlemen,
I find this thread very interesting. I have been building linear power supplies for many years. Almost all have been with brute force filtering, meaning capacitor input filters. A number of years ago I was concerned as one of the powersupplys I built the power transformer was overheating.
The transformer was made by Triad Transformer co. I personally talked to one of the engineers and he told me that with brute force filtering, one should derate the transformer by about 40%. At the same time I talked to a local transformer manufacture(Company no longer in business), and he also confirmed the same data. They both said to derate by 0.566.
Rebuilding the powersupply with their recommendations the power transformer didnot overheat.

Several years ago I did some current monitoring of both full-wave ct and bridge rectifier power supplys that used brute force filtering. I monitored both the charging current to the capacitors and the current on the load and the readings were a real suprize to me. I don't have my notes immediately available(I plan to see if I can find them) As I recall in some cases the charging current for the capacitor was over 2 times the load current. I also recall that the charging current for the full-wave ct and the bridge were different for an appropiate load current.
Ned Stevens
 

eblc1388

Joined Nov 28, 2008
1,542
Hi Duane P Wetick,

I have taken a quick look at the 15th edition of the above handbook but can't find the required information as the format of the handbook has changed in this new edition.

Which section would you suggestion I should look in depth to find such information?

 

Attachments

Section 22 of this edition, Power Electronics, Commutation, will provide the data you seek along with the voltage and current equations for 2-12 pulse rectifiers.

Cheers, DPW [ Everyone's knowledge is in-complete...Albert Einstein ]
 
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