Calculating DC Output Current From Unregulated AC Transformer

I usually give someone the benefit of doubt and try to come up with various ways of looking at a problem and proposing a reasonable solution, or pointing out the need for more information. The OP has chosen not to comment on those posts, but only those where he feels compelled to argue and assume a self-righteous attitude. So I second your motion. All in favor?

Paul

The usual limiting factor for the current output of a transformer is heat dissipation.

The factors that need to be considered are:

Heat is produced in a transformer in several different ways. There are electrical losses in the windings due to the currents in the windings. There are electrical losses in the core materials dues to induced currents in the core There are probably other factors related to frequency and materials.

The easiest one of these to handle quantitatively is the power (heat) lost in the windings' resistance. As the current increases, so does the power that is lost in the resistance of the windings. For more information (and equations) about the power loss in a resistance see:

Short of giving you the equations for RMS current we cannot answer your question since we do not know that the winding currents are for your application. (You mentioned DC currents but that is pretty much meaningless for a transformer.) Please note that the RMS current is not the same as the average current. There can be a big difference in RMS current (and thus power lost) between a current waveform that consists of pulses with a low duty cycle and high peak current versus a constant DC with the same average current. If you are building power supplies, the RMS current will depend upon (among other things) the output current, the filter capacitance, and the internal resistance of the transformer. You have given us no information about any of the details of what you are trying to accomplish and we really do not want to repeat the contents of textbooks on basic electrical engineering.

How the heat is removed from where it is produced depends upon the specific details of the design of the transformer and the environment in which is being used. You have given us no details on your application and we really do not want to repeat the contents of the textbooks on heat flow.

The maximum operating temperature for a transformer depends upon the materials and upon operating considerations. As the operating temperature increases the rates of break down of the insulation increase. In general, the higher the operating temperature of a transformer, the shorter its life will be. There can also be other operating considerations. For instance, the materials that are used to build a transformer, may allow for a long life while operating at 100 C but I would not like to have that transformer in close proximity to my skin at that temperature.

Well, not completely, no. Consider a 1A (RMS) rated transformer, with a diode/capacitor output. When attached to a 1A (DC) load, the transformer actually forward-biases the diode for a brief time at the peak voltage, and otherwise current from the transformer is nil. If the conduction period is one-tenth of the full cycle period, that means 10A current from the transformer during the active time.

Here's where it gets mathematical: the power dissipated in a 1A transformer with (for instance) 1 ohm winding resistance is 1 watt, when the load is taking a simple

Heat =3D 1A **2 * 1 ohm =3D 1 watt

When the same transformer feeds the DC load as described above, the power is

Heat =3D (0.9 * 0) + (0.1 * 10A**2 * 1 ohm) =3D 10 watt

So a perfectly good transformer can burn up feeding a DC rectifier and load, when a similar AC load wouldn't bother it. That isn't always covered by a 'factor of two' or any other rule of thumb. I've seen manufacturers offer tables of the permissible DC output ratings for their transformers, but you can't count on that. The AC rating looks better, so salesmen will feed you that info first.

"whit3rd" wrote in message news: snipped-for-privacy@k30g2000hse.googlegroups.com...

rated transformer, with a diode/capacitor output. When attached to a 1A (DC) load, the transformer actually forward-biases the diode for a brief time at the peak voltage, and otherwise current from the transformer is nil. If the conduction period is one-tenth of the full cycle period, that means 10A current from the transformer during the active time.

in a 1A transformer with (for instance) 1 ohm winding resistance is 1 watt, when the load is taking a simple

power is

DC rectifier and load, when a similar AC load wouldn't bother it. That isn't always covered by a 'factor of two' or any other rule of thumb. I've seen manufacturers offer tables of the permissible DC output ratings for their transformers, but you can't count on that. The AC rating looks better, so salesmen will feed you that info first.

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I beg to differ with your analysis, if you are talking about an ordinary rectifier and capacitor circuit. As an example, I simulated a FWB with a 12 VAC nominal output transformer with 1 ohm series resistance, and a load of

If you can show me a circuit where you will get these 10 ampere peaks at

Paul

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I have a transformer with a nominal 24 volt secondary, rated at 8 amps. It has a measured series resistance (secondary plus reflected primary) of about .125 ohms. I connected a bridge rectifier consisting of 4 80 amp Schottky diodes, and a real 100,000 uF capacitor.

If you simulate this, a DC load which gives 8 amps RMS in the secondary may give a DC current of less than 4 amps. The ratio of secondary RMS current to DC load current will exceed 2 to 1 if the transformer is much larger than this, with a series resistance less than this transformer has.

I posted a partial analysis over on ABSE in which I indicate that the grid waveform has a large effect on the RMS to DC current ratio in these rectifier circuits.

"The Phantom"

A 192VA rated transformer does not have 4% regulation - correctly rated it has 8%.

Look up makers data if you doubt this.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full load and for a correctly rated transformer.

BTW:

Percent voltage regulation and percent power loss with resistive load are virtually the same numbers (ignoring I mag loss).

..... Phil

This one does.

With 121.7 VAC applied to the primary, the secondary voltage measures 26.9 VAC, unloaded.

With 121.7 VAC applied to the primary, and rated load of 8 amps drawn from the secondary with a pure resistive load, the secondary voltage measures

This is a change of 1.05 volts for a load of 8 amps; 1.05/8 = .13125 ohms.

This transformer has no maker's identification, so I can't look it up.

However, the primary (it's a 60 Hz transformer) says 120 VAC and it measures .961 ohms, cold.

The secondary (24 volts, 8 amps nominal) measures .080 ohms.

With 121.7 VAC applied to the primary and no load, the secondary voltage measures 26.9 VAC, for an approximate turns ratio of 4.524:1. The .961 ohm primary, divided by the turns ratio squared, gives .04695 ohms reflected to the secondary. Adding the measured .080 ohm secondary gives a total .12695 ohms, which is *about* .125 ohms.

This is what I would expect to measure at the secondary if the primary were shorted. What I actually measure is .131 ohms; the transformer is still a little warm from having been under load a few hours ago. See the picture of the impedance meter's display over on ABSE.

"The Phantom" "Phil Allison"

The *correctly* rated secondary load is not 8 amps - but more like 15.

The oft quoted ratio of 1.6 for AC amps to DC amps applies ONLY at full load and for a correctly rated transformer.

..... Phil

The manufacturer rated it, and printed the rating on the outer paper.

I am taking the manufacturer's word for it. Printed on the transformer is the designation 24 volts, 8 amps.

That would depend on the insulation system the manufacturer used, and the resultant allowable temperature rise, wouldn't it?

Maybe it's only class O. Maybe they wanted to be very conservative.

The *correct* rating depends on the allowable temperature rise, which, presumably, the manufacturer knew.

Wouldn't this ratio depend somewhat on the regulation of the transformer?

A transformer with poorer regulation would have a wider conduction angle, and a smaller ratio of AC amps to DC amps.

For example, I have a cheap Radio Shack transformer rated at about 15 VA, and its measured regulation is about 15%. The AC amps to DC amps ratio at rated secondary RMS current is about 1.53.

The 192 VA transformer discussed above, with its better regulation has a ratio of AC amps to DC amps of about 1.7, at the manufacturer's rated secondary current.

That's not much variation, and the 1.6 figure would be a good rule of thumb, although the actual number varies with the regulation of the transformer. Perhaps slightly different ratios could be specified for different values of regulation.

I varied the load current with the 192 VA transformer discussed above, and measured slightly varying ratios at the various load levels:

DC current ratio of AC amps to DC amps

If all transformers were manufactured to a single regulation and temperature rise standard, the formula sought by the O.P. would be a lot closer to practical, but your two examples span only a part of the considerable range of such specifications. This range is what makes a universal formula either inaccurate if simple, or so detailed that it is impractical to fill in all the variables if complete.

I don't think it's that difficult for the reasons you cite.

Knowing the rated secondary current means we don't have to know the allowable temperature rise. We need only see to it that the RMS secondary current when used in a bridge rectifier circuit is the same as the rated secondary current. Then we will get the maximum possible DC current with the nearly the same temperature rise as when a pure AC load drawing rated current is applied. The actual temperature rise may not be exactly the same because the higher crest factor pulses of current drawn by the bridge cause a little extra IR drop in the primary wire, and change the core loss slightly. I tried to measure this effect, but it is so small as to be completely negligible.

The regulation is easily measured and provides what would seem to be the remaining needed information, but I found that the grid voltage waveshape has a large effect on the result, and knowing and specifying that may be the most uncertain part of the process.

Did you look at my post on alt.binaries.schematics.electronic? I show a method of deriving a "formula", and discuss the results I got compared to actual measurements.

Only if you assume the design environment (sealed container, forced air velocity, ambient temperature range, etc. ). Many transformers are designed for very specific (and varied) environments.

the

Yes, that effect is more pronounced for cheap, low efficiency transformers than it is for high efficiency units that have plenty of iron and are rated to perform well with

And changes day to day and even time of day. These wave shape effects are almost unnoticeable with a resistive load., but can be major players for rectifiers with capacitor input filters.

I did a fly over and saved it, but do not wade through your math.

"The Phantom" wrote in message news: snipped-for-privacy@4ax.com...

I have not looked at the analysis, but I did find an error in my analysis as stated above, although it does not change the essential fact that the transformer will not be overloaded if you keep the DC current out to about

My error was that I used the voltage and current out of the transformer as a measure of the power it was delivering, and that is correct in a sense, but the internal resistance sees an RMS current of about 1.8 amps, for a power dissipation of 3.24 watts, and not 0.8. I found it easier to use an external resistance for the simulation. This model would be for a 12 VAC transformer rated at 2 amps (24 VA) with 2/12 = 16.7% regulation. Larger transformers will generally have better regulation, partly because they do not have as much surface area to volume, and cannot as easily get rid of internal heat by convection.

Simulating your circuit with a 3.3 ohm load, I get Pin = 142W, Pout = 127W, Iin = 8.14A, Iout=4.39A. The internal resistance of the tranny dissipates

Paul

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Version 4 SHEET 1 880 680 WIRE 144 144 128 144 WIRE 240 144 224 144 WIRE 288 144 240 144 WIRE 384 144 352 144 WIRE 416 144 384 144 WIRE 512 144 416 144 WIRE 528 144 512 144 WIRE 528 160 528 144 WIRE 128 192 128 144 WIRE 416 192 416 144 WIRE 240 256 240 144 WIRE 288 256 240 256 WIRE 416 256 352 256 WIRE 416 304 416 256 WIRE 528 304 528 240 WIRE 528 304 416 304 WIRE 640 304 528 304 WIRE 640 336 640 304 WIRE 128 368 128 272 WIRE 288 368 128 368 WIRE 384 368 384 144 WIRE 384 368 352 368 WIRE 128 480 128 368 WIRE 288 480 128 480 WIRE 416 480 416 304 WIRE 416 480 352 480 FLAG 640 336 0 FLAG 512 144 V+ SYMBOL schottky 288 160 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D2 SYMATTR Value MBR745 SYMATTR Description Diode SYMATTR Type diode SYMBOL polcap 400 192 R0 SYMATTR InstName C2 SYMATTR Value 100000µ SYMATTR Description Capacitor SYMATTR Type cap SYMATTR SpiceLine V=63 Irms=2.51 Rser=0.025 MTBF=5000 Lser=0 ppPkg=1 SYMBOL voltage 128 176 R0 WINDOW 3 -11 133 Left 0 WINDOW 123 0 0 Left 0 WINDOW 39 -90 104 Left 0 WINDOW 0 -73 31 Left 0 SYMATTR Value SINE(0 34 60 0 0 0 200) SYMATTR SpiceLine Rser=0 SYMATTR InstName V1 SYMBOL res 512 144 R0 SYMATTR InstName R1 SYMATTR Value 6.6 SYMBOL schottky 352 272 M270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D3 SYMATTR Value MBR745 SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 288 384 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D1 SYMATTR Value MBR745 SYMATTR Description Diode SYMATTR Type diode SYMBOL schottky 352 496 M270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D4 SYMATTR Value MBR745 SYMATTR Description Diode SYMATTR Type diode SYMBOL res 128 160 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName R2 SYMATTR Value .13125 TEXT 184 528 Left 0 !.tran 3

I don't see what this has to do with it. Of course I am assuming that the manufacturer's rating was determined (by the manufacturer) knowing the design environment, and we don't need to do that ourselves, and I further assume that we know what the manufacturer's rating is, and that the transformer will be used in its design environment with the rectifier load.

I explain in the next paragraph that if the rectifier circuit causes the same secondary RMS current as a pure AC load, then the heating will be the same. Therefore, we don't need to know what the heating is; we already know it's safe. We only need to know that the transformer will be safe operating at its rated secondary current, whether that current is a result of a pure resistive AC load, or a rectifier load. If we don't know that (the rated secondary current in its intended environment), then we don't know enough to safely use the transformer even with a resistive load.

There's no point in trying to calculate the safe DC current available from a rectifier if we don't even know the safe AC secondary current; in that case we must resort to characterizing the transformer thermal behavior ourselves.

the

So is a negligible effect which is, in some cases, more pronounced, still negligible? :-)

I was unable to measure the effect using a cheap Radio Shack transformer; I think it's negligible in all cases.

As a side note, I measured the core loss of the cheap Radio Shack transformer with a wattmeter designed for accurate measurement of low power factor loads (unloaded transformers, in other words). The loss was 2.65 watts cold and 2.42 watts hot. After make a series of measurements with the transformer hot, I connected it, unloaded, to the wattmeter, and over the course of a couple of hours watched the core loss drift back up from 2.42 watts to 2.65 watts.

The saving grace is that the clipping of the grid waveform reduces the heating of the transformer rather than increasing it. Therefore, if a formula is used which assumes that the grid waveform is a good sinusoid, not flat-topped, the DC current calculated to give the rated secondary current will be lower than the true (measured) value, more so for transformers with good regulation.

For example, for the 24 volt, 8 amp transformer I've mentioned in this thread, a calculation assuming an undistorted grid waveform gives a value of 3.6 amps for the DC current to give an 8 amp RMS secondary current. But, the measured value is 4.6 amps; the calculation is very conservative. The calculated Irms/Idc is

If I change the calculation to use a grid waveform with a 2.1 millisecond flat spot at the top of the waveform, then I get a result of about 4.6 amps. The recommendation that John Fields made, to assume Irms/Idc = 2, which he says will always be safe, may not be safe if you are using a transformer with good regulation and if your grid waveform is a good sinusoid.

method

"The Phantom" "Phil Allison"

Your argument is entirely false.

Using higher temp grade will only increase the figure.

...... Phil

"John Popelish"

The oft quoted ratio of circa 1.6 applies to stock lines transformers.

..... Phil

"The Phantom"

Plucked out of his arse.

Which wipes you asinine case out.

Piss off.

...... Phil

A transformer with unusually good regulation was presumably rated that way by the manufacturer for some reason. Whatever the temperature rise with rated secondary current, it will be exceeded if the secondary current is greater than its rating. If the user wants to do that, it's his choice. He should be aware that under certain conditions, some transformers may give Irms/Idc greater than