Transformers in parallel?

[G6Tanuki]

I guess the "double section" is for isolation between the primary and secondary [to help with 'double insulation' approvals - it also helps reduce the risk of RFI 'hop-over' between primary and secondary in the absence of an interwinding screen] rather than being used to ensure there are relatively-equal-numbers of turns on the windings.

[Ed_Dinning]

Double section bobbin will also make the regulation worse as it reduces the coupling between the windings

Ed

[G6Tanuki]

Looking again, it seems that the two transformers are essentially identical except that one has two windings 0-115V as primary, the other only a single 230V winding.

So I'm now really much much less concerned about any differences in loading between them when paralleled!

As I said, my tests gave positive results and no signs of different temperatures. Will wire them parallel and use a single bridge-rectifier even though I have dozens of nice stud-mounted diodes on ex-MoD heatsinks...

[ms660]

Max. load is 45 Watts (15V*3A) and the parallel connected transformer is rated at 20VA.

Lawrence.

[mhennessy]

Originally Posted by G6Tanuki

What would you do?

I know what I wouldn't do: ask a question on a public forum and then dismiss almost every bit of advice I receive

As has been said by many people already, if the requirement is for a continuous 3A at 12V then you simply don't have enough VAs for this to work reliably. It might work for days, weeks or even months, but core temperature is not the same as winding temperature, and eventual failure is inevitable.

With this sort of stuff, there's no point trying to be too precise, but in broad terms, you'll have around 20V after the rectifier, depending on load current. That's 60VA. From a 40VA source.

I'm less concerned about the parallel connection - assuming the correct phasing, of course. Small transformers like these have enough DCR to "ballast" the load between windings. But while that DCR helps this aspect, it's the thing that will eventually kill one of these transformers. I wonder how long the other will survive, given that it'll be back-feeding a transformer with a shorted turn? That alone is reason to consider separate rectifiers, with LEDs on the AC side of each, showing that each transformer is working.

On the other hand, if the 3A requirement is an occasional peak, well above the continuous average load current, then you'll probably get away with it.

As I mostly deal with audio, I'm well used to seeing transformers that are far too small, according to those who don't understand the nature of audio signals. Probably the most extreme example being the Musical Fidelity B200, which has a 120VA mains transformer and produces around 100W per channel. Failures are very common indeed, and as there's no room in the box for anything larger, an "outboard" PSU containing a 200VA transformer is the only reliable way forward.

[ms660]

Another way to figure out what VA rating is needed is by using Schades curves.

Lawrence.

[kalee20]

It would, and that would be the way to go if a transformer is being designed.

But here, the OP already has his transformers!

Given the number of replies received...

Originally Posted by mhennessy

I know what I wouldn't do: ask a question on a public forum and then dismiss almost every bit of advice I receive...

... I honestly think the speediest way forward is to lash it up and give it a whirl! Add a realistic dummy load, switch it on, and wait.

Monitor temperature after 5 minutes, 15 minutes, an hour, 3 hours (easiest way is to measure winding resistance cold; switch off and measure it again quickly, then switch back on. Use the tempco of copper to work out the winding temperature) and see if it gets alarming.

Modern enamel will be good to at least 125deg C as will bobbin plastics.

The environment makes a big difference to temperature rise, such as if it's in a box with not much airflow or totally open on the bench, so you need to set it up in a representative manner.

And then - share results with us all!

[trsomian]

Since mains transformers fed from a low impedance source (the mains) will have quite a low output impedance surely the normal rules apply:-
Low impedance devices can be put in series but not in parallel
High impedance devices can be put in parallel, but not in series.

Obviously this is a generalisation, and there is a grey area between the high and low impedances, but I believe the above is valid in this instance.

[mhennessy]

Originally Posted by kalee20

... I honestly think the speediest way forward is to lash it up and give it a whirl! Add a realistic dummy load, switch it on, and wait.

Well, in post #15 he did that, but he only measured the core. Well, not measured, but assessed. IR thermometers often pop up in Lidl/Aldi for between £10 and £20 - they're not exactly high end, but better than nothing.

The tempco of copper is about 0.39%/C, so a rise of 100 degrees is 39%. The primary is easier to measure with a regular DMM. Don't forget this gives an average; there will be a temperature gradient across the windings, so the hottest parts of the winding will be hotter than the measurement implies.

But as I say, it's a question of "when" rather than "if". Been there, got the t-shirt.

Perhaps I'm missing a joke somewhere? The mention of a "few tens-of-thousand-microfarads of smoothers" makes me wonder if experienced PSU designers are being trolled. Either way, best of luck



PS: I have a second-hand 50VA toroidal transformer sitting around doing nothing - 2 times 12V at 2.1A each. That'll be 4.2A going into the bridge, which is still pushing your luck for 3A DC but is much better than what you've got. 60VA is the least I'd use for continuous operation, preferably 80VA to reduce the temperature rise (there's relatively little cost difference for a one-off project). Yours for £10 plus P&P.

[kalee20]

Originally Posted by mhennessy

Well, in post #15 he did that, but he only measured the core.

He did... but re-reading post #15 it still doesn't look as though it's set up as it would be - I got the idea that the combined secondary was loaded, not by a bridge rectifier and several thousand microfarads, but just by a resistive load.

And as several previous posts have pointed out, that is far more benign to the transformer than the same current, taken from a rectifier/capacitor.

I'm uncomfortable...

Originally Posted by trsomian

Since mains transformers fed from a low impedance source (the mains) will have quite a low output impedance surely the normal rules apply:-
Low impedance devices can be put in series but not in parallel
High impedance devices can be put in parallel, but not in series.

Obviously this is a generalisation, and there is a grey area between the high and low impedances, but I believe the above is valid in this instance.

It's a good rule of thumb! The transformers started out different (230V and 240V primaries) which would have given sufficiently different output voltages when commoned, for there to be a problem.

By post #23 this was revised, both are the same, so the secondary voltages are likely to be close enough for there not to be a problem with circulating currents.

The issue now, is whether the VA rating is high enough to cope with the intended load.

[ms660]

Originally Posted by ms660

Another way to figure out what VA rating is needed is by using Schades curves.

Originally Posted by kalee20

It would, and that would be the way to go if a transformer is being designed.

You don't have to design the transformer in order to figure out what VA rating is likely to be required (give or take a few percent) using Schades curves.

I've done a comparison for 15 VDC @ 3 Amps using a guide from a transformer manufacturer and Schades curves.

Manufactures guide:

VAC = 0.85*VDC + 2*Vdrop = 14.75 Volts (Vdrop being the DC RMS voltage across one diode)

IAC = 1.65*IDC = 4.95 Amps

VAC*IAC = 73 VA

Schades curves.....75.8 VA....Based on Rs/Rload of 6% and a ripple factor of 1%

Near enough, given the errors that can be incurred when reading Schades curves.

Lawrence.

[stainless]

Does the fact that when the rectifier (or bridge) is feeding a sizeable capacitor, the current flows only in spikes near the top of the sine wave throw a spanner in the works? Heating effects in the transformer being I squared times R averaged over time?
I have always been mystifed, and caught out by power supply calculations!

[ms660]

Originally Posted by stainless

Does the fact that when the rectifier (or bridge) is feeding a sizeable capacitor, the current flows only in spikes near the top of the sine wave throw a spanner in the works? Heating effects in the transformer being I squared times R averaged over time?

Yes, hence the various formulas and Schade's curves etc.

The classic Otto Schade Analysis of Rectifier Operation here:

https://www.hifisystemcomponents.com...les/schade.pdf

Lawrence.

[kalee20]

Originally Posted by ms660

You don't have to design the transformer in order to figure out what VA rating is likely to be required (give or take a few percent) using Schades curves.

Agree! Actually, I was trying to make the point that ten minutes work with a soldering iron, some wire, diodes, lots of microfarads and a load is probably quicker than 15 minutes with Otto Schade's techniques.

But your own simplified analysis has indicated even faster there's likely to be a problem... transformers are loaded together at 73VA, and the OP's first post informs us there's only 40VA rating to play with.

It does, of course, depend on why the manufacturer has rated the transformers at 10VA per secondary. If it's because of temperature rise, the OP is in trouble. If it's because that's the loading at which the voltage has dropped more than 5%, but temperatures are still entirely comfortable, then happy days! (Though I doubt whether the second scenario is the case!)

[ms660]

Originally Posted by kalee20

Agree! Actually, I was trying to make the point that ten minutes work with a soldering iron, some wire, diodes, lots of microfarads and a load is probably quicker than 15 minutes with Otto Schade's techniques.

But your own simplified analysis has indicated even faster there's likely to be a problem... transformers are loaded together at 73VA, and the OP's first post informs us there's only 40VA rating to play with.

It does, of course, depend on why the manufacturer has rated the transformers at 10VA per secondary. If it's because of temperature rise, the OP is in trouble. If it's because that's the loading at which the voltage has dropped more than 5%, but temperatures are still entirely comfortable, then happy days! (Though I doubt whether the second scenario is the case!)

I chose 15 Volts to allow for a regulator, using Schades curves only takes a few minutes:

Make an assumption that Rs/Rload is 6% and choose the ripple percentage of choice, with those two figures 2pi*f*C*Rload can be got from the ripple % curves (for full wave) Using that figure, determine the ratio for the DC output to the peak AC output from the transformer using the appropriate set of Schade's curves, use those two figures to determine the VAC output needed from the transformer.

For the RMS current through the diodes, half the Rs/Rload percentage and double the 2pi*f*C*Rload figure to obtain the diode RMS current to it's average current (DC) ratio from the appropriate set of Schade's curves, multiply the diode DC current by that figure then multiply the result by the square root of 2 (1.414) to determine IAC.

It's taken me at least 15 minutes to write this but 5 minutes for the calcs....

Lawrence.

[mhennessy]

Originally Posted by stainless

Does the fact that when the rectifier (or bridge) is feeding a sizeable capacitor, the current flows only in spikes near the top of the sine wave throw a spanner in the works? Heating effects in the transformer being I squared times R averaged over time?
I have always been mystifed, and caught out by power supply calculations!

This is what I was getting at when I drew attention to the size of smoothing capacitors that the OP mentioned in the opening post.

As the current only flows at the peaks of the AC waveform, they are much larger than the average DC current. For a sensibly sized smoothing capacitor - say, 4700uF for a 3A supply - you can assume that the current peaks are about 4 times the average DC current. This is a rule of thumb based on observations over the decades - as you say, the calculations get complicated if you're trying to be precise which, given the many variables that are hard to measure and/or predict, is a fool's errand in my humble opinion.

This post discusses the relationship between charging currents and the average DC current:

https://www.vintage-radio.net/forum/...147#post940147

Trying to reduce the ripple by using large smoothing capacitors might seem like a good idea on the face of it, but the reduced ripple means that the conduction angle is reduced - therefore there is less time to "top up" the capacitors, so you have to draw more current from the transformer.

This image shows the relationship between ripple voltage and conduction angle:

https://www.markhennessy.co.uk/artic...tion_angle.gif

Taken from this basic article about PSUs:

https://www.markhennessy.co.uk/artic...r_supplies.htm

The trouble with charging currents is that they multiply with the DCR in the windings to cause voltage drops, meaning you end up with a lower DC voltage than you'd expect, and not necessarily with the reduction in ripple you were hoping for. Far better to allow for a couple of volts of ripple and use a voltage regulator (or design for good PSRR if it's an unregulated supply, such as an audio power amplifier).

It might be best to pick up an open-frame linear supply. I'll have a rummage in the attic later, as I have a few up there. 12V at 3A is a common enough output spec for them. They are usually based on an LM723 and a beefy power transistor...

[G6Tanuki]

The need for lots-of-upstream-overvoltage to feed regulators of the obsolete 78xx-types is long gone.

Now we have LDO types that need only something like 1.1V 'headroom' - which is great because it lets you use much more of the AC-into-the-rectifier waveform at peak load.

Programed output-voltage 12V; input-voltage [with plenty of ripple] 13.9V - and all is good!

[mhennessy]

I agree...

So why use 15V transformers?

"Because you have them" seems to be in direct contradiction to post #37 - if you stick to 15V AC, you don't need an LDO. But then, where has 13.9V come from - is that the DC voltage measured after the smoother from those transformers with a 3A DC load? If so, that's really scary! What's the AC voltage from the transformers?

Watch out of the quiescent current of some LDO types. The LM2940, for example, uses a PNP pass transistor, and the base current is sunk to the ground pin. As the drop-out voltage falls, it jumps up to nearly 140mA as it tries to saturate the pass transistor. Easy to miss that graph in the datasheet!

The LD1084 is a much better bet today - it uses a CFP as the pass device. There are many others.

Watch out for thermals though - for the LD1084 with a junction to case thermal resistances of 3C/W, these need a bigger heat sink than a power transistor would (which not only has a lower thermal resistance, but a much higher allowable junction temperature). With around 3V across the device, there will be a difference between junction and case of about 30 degrees. That's why so many open-frame linear PSUs have stuck with the beefy TO3 transistor(s) driven by an LM723 or similar. E.g. https://www.mouser.co.uk/ProductDeta...vdclau7w%3D%3D

[kalee20]

Originally Posted by mhennessy

Trying to reduce the ripple by using large smoothing capacitors might seem like a good idea on the face of it, but the reduced ripple means that the conduction angle is reduced - therefore there is less time to "top up" the capacitors, so you have to draw more current from the transformer.

This image shows the relationship between ripple voltage and conduction angle:

https://www.markhennessy.co.uk/artic...tion_angle.gif

It's true that upping the smoothing capacitor reduces the conduction angle and the peak current gets higher, but it's not without limit! The transformer (and rectifier) resistance ultimately limit the charging current, and the capacitor voltage never reaches the AC peak, as per your linked diagram.

Once you go round the 'knee' in Schade's curves, the peak current and charging-pulse duration hardly change with increase of C. The ripple of course does decrease without limit, but once it's a small fraction of the voltages under consideration, the transformer resistance dominates the situation and it makes no difference to the transformer, once steady-state conditions have established, whether the OP has his several tens of thousands of microfarads, or a hundred million farads. (This latter extreme would of course take a lot longer to reach steady-state, but once there, the transformer wouldn't notice the difference).

See attached sketch.

[mhennessy]

Yes, that's exactly what the penultimate paragraph in post #36 is getting at - ultimately transformer DCR limits the charging pulses. Which doesn't seem like good engineering to me, as this means you're using far more capacitance than you need, given that you're following up with a voltage regulator - and if you're not, then it's because ripple (and DC precision) doesn't matter in your application anyway.

Hard to put a number on it with all the other variables in play - easiest to measure with a current clamp 'scope probe.

Incidently, this is a much more serious problem with switched mode power supplies that don't have the 50Hz transformer to soften the charging pulses. Active PFC schemes based around boost converters solve this problem rather neatly (and add universal voltage operation and take care of inrush protection too). Active PFC is basically mandatory on anything bigger than 75 watts these days if you want to meet EN61000-3-2.