Transformers in parallel but different VA ratings.

[davelectronic]

I have two toroidal transformers, both are 120 volt dual primary, and dual 15 volt secondary windings. I have put them in parallel and all seems well. one is 300VA the other is 225VA  rating. so can both be added together for 525VA rating ? both transformers are from the same manufacturer, the only difference is there power ratings. I have never connected dissimilar transformers before. Any help appreciated, and thanks for reading. 

[bob91343]

It depends.  The winding currents probably won't share as you'd like, but to find out, connect it and measure them.

In addition, the turns ratio may not be the same, which causes circulating currents which serve only to heat the transformer.  Even if you are lucky to have the same turns ratio, the winding resistances will be different and again, cause circulating currents under load.

[Whales]

It depends.  The winding currents probably won't share as you'd like, but to find out, connect it and measure them.

+1 to this.

We can guess based off the theory (they "probably" won't match), but the best thing to do is measure.  I'd also suggest individually fusing the outputs of each toroid, so if one goes offline for some reason (or gets reverse wired) you don't overheat the other.

[KD0CAC John]

What is the intended use ?
I would say in series adding voltage - do not exceed the rating of the lowest rating .
" In parallel " adding current - do not exceed - double - the rating of the lowest rating .
Open to more thinking / opinions

[bob91343]

John those are good rules of thumb.  Like all rules of thumb, they are approximations and only correct most of the time.

[davelectronic]

Thanks for your replies. Double the lowest power rating would be 450VA, I can measure current with a clamp meter for both primary and secondary windings. Both secondary windings go through its own bridge rectifier befor they go in parallel. It's another large power supply. I'm getting 22.00 Volts DC from the secondary parallel combination. And I have 20000uf of filter capacitance. I haven't loaded it up yet, but can do this to measure current in both primary and secondary windings.

[Brumby]

Both secondary windings go through its own bridge rectifier befor they go in parallel.

That's a useful piece of information you left out in your original scenario.  Without such a detail being shared, we will all (initially, at least) respond on the topology of joined windings - which is the most common situation which we are asked about.

As I understand it, in this situation, you are going to have less of an issue and, in fact, could probably run up near your 525VA rating.*

* I wouldn't design a circuit to run to such a limit.  I'd plan a typical max load of around 450VA which would give a bit of headroom.

[Siwastaja]

Current sharing is still dependent on the transformers having the same output voltage, and same ESR-to-capability ratio, so that the output voltages track with load variations, and if one of the transformers is taking too much load, its voltage drops more, making it negative feedback.

Diode negative Vf temperature coefficient makes current sharing even worse.

I think we can say nothing more but the available output power is at least the power of the smaller of the two transformers. Everything beyond that is uncertain. In other words, don't parallel voltage sources.

Resistance helps but it needs to be of similar value. Explicit series resistors would be best.

I'd say, just get a larger transformer.

[David Hess]

Under ideal conditions with identical turns ratios and proportional series resistance they should share current proportionally.

Nothing prevents verifying how well they share current.  Measure the current and voltage drop under load of each separately; this is easier and more accurate than measuring the primary and secondary resistances.  If the voltages match, then they will share current proportionally.  In practice the power ratings are so close that I do not think you will have a problem.

[Siwastaja]

In practice the power ratings are so close that I do not think you will have a problem.

Quite a stretch, I would say. I'm not worried about power ratings being dissimilar; but output voltages in transformers are notoriously all over the place. How many turns will be wound when a company designs a "15V transformer"? No one can said, because 15V is nominal rating only. Is it 15V under full load? 10% load? Or is it even close to 15V at any load, really?

If you have two identical transformers, then sure, they are very likely wound with same number of turns.

But two different products, they are unlikely designed using exact same formulas. Other might output +/- 16V and other +/-  15V under the same relative load. In such case, no current sharing happens at all, the one with higher voltage supplies way over its ratings and it's already beyond ratings when the voltage drops so much that the lower voltage transformer starts to participate even a little bit.

Copper's positive tempco in resistance helps a bit, diodes negative tempco in Vf hinders.

The best simple test you can make is to maybe load both of them with say 75% of the rated current separately and see if the voltages are close at all. They'd need to be within some hundred mV in order to self-regulating resistances have any chance.

[Zero999]

The secondary voltage of a transformer is normally specified, under full load. A transformer with a 240V primary and 12V secondary, will not have a turns ratio of 240:12 or 20:1. It might have an unloaded voltage of just over 13V, thus have a turns ratio of 18:1.

Higher power transformers generally have better regulation, than lower transformers. The regulation factor is how much the voltage drops under load. A transformer with a regulation factor of 10%, will have a 10% higher secondary voltage when unloaded, than the nominal specified value, so the secondary will have 10% more turns, than the the ratio of primary:secondary voltage ratings would suggest.

In this case, the two transformer have different power ratings, thus different turns ratios, even if the primary and secondary voltage ratings are identical. If they've just connected in parallel, the lower power transformer would have excessive losses, when unpowered, as it will be pushing current into the secondary of the higher power transformer. This is unlikely to cause overheating, just excessive losses, when unloaded. When loaded, the higher power transformer will provide most of the current. I'd say this is probably acceptable, if the transformers are always used near the maximum ratings and never left unloaded.

Now in the case of the original poster's circuit. The transformers are connected in parallel, via bridge rectifiers. This will eliminate the issue of excessive secondary currents, whilst unloaded. Under light loads, the lower power transformer will pass most of the current, as it has the higher secondary voltage. If the load is increased, the secondary voltage of the smaller transformer will start to fall towards that of the larger transformer, which will start to pass more current. I say don't worry. It'll be fine.

[Siwastaja]

Good description how current is NOT shared evenly, and then the conclusion: don't worry, it'll be fine.

So what will be the total power output without exceeding the individual ratings. Is it going to be (300+225) VA? Or something else?

Of course, the solution is derating, but you don't mention it at all. How much to derate? I know why you don't mention it: because you don't know. Me neither.

I have the right answer: total power output will be anything between 225 .. 525 VA. Tough luck, in worst case it works worse than the bigger transformer alone.

The idea that diodes magically help is funny without proof because it's very well known that diodes do not share current when paralleled, and just thermally run away to even poorer current sharing. Here the winding resistance of course is in our favor, but does it save the day of diodes?

There is one solution which likely works, and I believe the "don't worry" guys also simply count on this: exceed the ratings. It will be fine! Don't worry! If you don't worry, you also don't need to measure; just try if it works. On a lab table, for 5 minutes is enough. Try it, this is the most likely outcome: it works. The irony here is: you don't need to parallel at all. Just use the bigger transformer as is, alone, with one simple diode bridge, and exceed its ratings. Try it, you might be surprised to see it works. Don't worry!

Seriously though, do not forget extra derating needed because of the bridge rectifier + capacitive load. Don't remember the fudge factor offhand but do google it.

[Kleinstein]

The transformers are not that much different in power rating. So the current sharing should be reasonable, though not perfect. To some degree the diodes work against current sharing, but at least the avoid current flow between the secondaries with slightly different windings ratio.
It depends on the voltage how bad this will be. With low voltage, like 6 V this would likely be a problem, with some 50 V I would not worry that much.

How much derating is needed is hard to tell and would likely depend on the power factor / how large the current peaks are.
The tendency is that under light load the smaller transformer will take over most of the load as it should have a slightly high no load voltage. With higher current the larger transformer should take over more of the load, as it has less series resistance relative to the power rating.

With a rectifier much of the current would flow in relatively high current peaks, often higher than the RMS rating. So the larger transformer will tend to get a higher load share than the smaller one.

The actual resistance ration depends on the exact way the transformers are made calculated. Even the same manufacturer can have different sereis and the wire only comes in discrete sizes. A rought rule of thumb / general tendency rule  is to scale the power rating with the 4 th power of the linear dimensions and the relative resistanc going down with the 1 over the linear dimensions.  So the smaller transformer is expected to have about 43% higher resistance, though only a factor 1.33 smaller power.
So I would consider something like a 8 % derating for the smaller transformer a 1st guess.  So maybe expect some 500 VA total rating.

Anyway the rating of transformers is not a hard bickwall limit - it is the power when the norminal votlage is reached and the temperature reaches the preset limit for the temperature class of the transformer. Higher load would increase the temperature and lower the transformer life.

[Zero999]

Good description how current is NOT shared evenly, and then the conclusion: don't worry, it'll be fine.

You're completely mistaken, if you believe current sharing is a good thing.

It's good that the transformer with the lower power rating, will pass less current, than the higher power transformer, which is exactly what's going to happen because it will have a higher effective series resistance. The diodes are good, because they prevent the current circulating between the secondary windings.

[davelectronic]

I should have mentioned the two bridge rectifier in my opening post. I did it this way as I understand it, the transformers won't see each other once there in parallel. I'd read that power supply could be put in parallel if they go through a bridge rectifier first. What I didn't look at was there current sharing ratio. I have two of those uni-T small clamp meters, they can do AC or DC current. So I should be able to see what's going on with both transformers at the same time. I never realised it was as complex as your posts read above. There is a lot going when they are idle, and under load conditions. I would like 30 Amps from the finished power supply, but I'd be happy with anything up to 20 - 25 Amps, but 20 Amps would be usable for powering my intended load, a HF linear amplifier. Thanks for all the insight, it's very interesting.

[Siwastaja]

You're completely mistaken, if you believe current sharing is a good thing.

RELATIVE current sharing, relative to their ratings. At full load, both should deliver 100% of their ratings. Because if not, then another of them is delivering >100%.

And you exactly described this situation: self-regulation by resistive voltage drop under load. But this is consequence of not having the relative current sharing. It corrects the situation in the right direction, but not with infinite gain. It all depends on how close the voltages are, and how much resistance you have. If voltages are equal, no resistance is needed, but this is theoretical, because voltages can't be equal. If voltages are not the same, the relative current sharing can never be satisfied. Now the amount of voltage mismatch (less is better) and amount of resistance (more is better) defines how close to relative equal sharing we can get.

Without derating, exceeding ratings is guaranteed. Now the only problem is to figure out the amount of derating needed.

[Brumby]

As I understand it, in this situation, you are going to have less of an issue

... but there are still considerations.

and, in fact, could probably run up near your 525VA rating.

Perhaps a little optimistic - but not impossible.  The key factor is the voltage on each secondary at maximum load.  If these voltages are the same for both transformers, then you could easily run the full VA.

Relatively unequal current contribution may not be a problem at sub maximal load.  If the voltage of one transformer is higher than the other at lower load, then that transformer will be providing more current to the load than its partner - it may even be providing ALL the current to the load.  This is fine, so long as that transformer is still operating within its ratings.  As the load increases, this transformer's output will sag and, in doing so, the other transformer's contribution will become increasingly significant.

If the first transformer would continue to sag as load increases further, the second transformer will contribute even more current.  Worst case being the second transformer providing most of the current, with the first only contributing what it can at the voltage present.  Such inequalities are the precise reason why it is crucial to ensure the operation of EACH transformer is within its specifications.

The ideal is to have the same voltage on each secondary at maximum load.  If not, a lower limit will exist.

[Zero999]

You're completely mistaken, if you believe current sharing is a good thing.

RELATIVE current sharing, relative to their ratings. At full load, both should deliver 100% of their ratings. Because if not, then another of them is delivering >100%.

And you exactly described this situation: self-regulation by resistive voltage drop under load. But this is consequence of not having the relative current sharing. It corrects the situation in the right direction, but not with infinite gain. It all depends on how close the voltages are, and how much resistance you have. If voltages are equal, no resistance is needed, but this is theoretical, because voltages can't be equal. If voltages are not the same, the relative current sharing can never be satisfied. Now the amount of voltage mismatch (less is better) and amount of resistance (more is better) defines how close to relative equal sharing we can get.

Without derating, exceeding ratings is guaranteed. Now the only problem is to figure out the amount of derating needed.

You could be right, or wrong.

It should be easy to calculate/simulate.

The output from each transformer will look like a constant voltage source, in series with a resistor.

R_S = (V_UNLOADED-V_LOADED)/I_LOAD

V_LOADED is the nominal secondary voltage.

Calculate V_UNLOADED by multiplying the nominal secondary voltage, by the 1 + the regulation factor on the data sheet, or it can be measured, in which case the actual primary voltage should also be measured and the result scaled, to account for the mains being different to the nominal.

Now you have two voltage sources, in series with resistors and two diodes, connected together in parallel. I'd just plug it into LTSpice.

Assuming the original poster has a filter capacitor, on the output of the bridge rectifiers, it would need to be de-rated anyway, because of the fact that current is only draw during the peak of the AC cycle.