Transformer secondary max currents - techniques for determining

[pjd3]

Hello.

This is really just a "topic for discussion" post but is somewhat relevant to a situation I have.

I approached the principle systems engineer at my day work with a question and that was, "Is there any method to determine the max current limit in PT secondary coils in a transformer that has no available spec sheets? And this is in fact the case with the PT in a Carvin donor amp that a friend gave to me. I'm turning it into a sort of JCM800 2204 with JEL mods plus a few other things (HV Fet inp boost, relay switching, DC heater pcb and HV serial FX loop pcb). One of the secondary's of the Carvin PT is approx 20vac which I thought/hoped/wished that would provide enough current for relay coils, some FX loop fets. While the wire gauge of that 20vac secondary seems to be around 20-22awg (which on its own would carry far more current than I need for the application), my many many hours of scoping the internet on the PT came up with nothing - and I was informed that a secondary coil's gauge is not necessarily a final call out on safe or effective max current).

While I was not expecting to get an answer from the principle systems engineer on a method of determining max PT coil currents, he did infact surprise me with a fairly definitive approach to finding this.

Two different approaches. (I'll try to recall as accurately as possible but,)

1. With proper loads on the other PT secondaries, gradually load down the coil of interest until it reaches 90% of its initial voltage. Take that voltage divided by the resistance needed to reach the 90% value, and that should be about the max current that should be drawn from that secondary.

2. Short out all of the secondaries and raise the input voltage to the primary to 10% of its rated operating voltage. Use some form of current meter to read the current in the secondary coil of interest at that point. That should be about the max current expectations of that particular secondary coil.

The engineer here use to design transformers in a former electronics company so while I held a fair amount of confidence in the viability of his explanation, I did admit that the idea of "shorting all of the secondary coils" seemed a bit a bit scary, he claimed that this test/determination is something that works for properly designed transformers, those that are designed to industry specs. That seemed reasonable. So, if I'm really on the fence about the current handling capabilities of the 20vac mystery secondary coil on the PT, I have something I can do in attempt to have some indication of current capacity of the coil.

Well, I found this kind of interesting so, thought I'd pass this on should somebody land a mystery PT that could be good for a project.

Thanks for coming by!
Best,
PJD3

[johnnyreece]

I've heard of the first test before. I'm with you about the second test; that does sound scary!

[R.G.]

A third test is the right way. Sadly, it's hugely difficult to do, so most people who are not transformer specialists, even experienced engineers, use some form of alternate testing.

The real limit on a transformer's power rating is temperature. The iron and copper will keep transforming correctly right up to the Curie point of the iron, on the order of 770C. The iron and copper are glowing a dull red at this temperature. So the metals are not the issue. What is the real limit is the insulation film on the wire and the interlayer insulation. When those break down, windings short, and local shorts really can and do go above the Curie point and/or melting point of copper. Then the transformer is really dead. Below the insulation breakdown temperature, the transformer will not die, even if the voltages have fallen a lot. There's no clear percentage drop on a a winding that tells you that the winding is overloaded.

There are two temperature limits, actually. One is the local hot spot temperature, the other is the whole transformer. A transformer is designed (f done properly, that is) so that the predictable hot spot at the center of the windings, or nearest the interior of the iron and copper, does not go over the wire/layer insulation's rated temperature. The other is that any winding does not let a section of its wire get over the insulation temperature rating. This second one is what you're asking. The insulation rating (such as Class 105, Class 130, Class 180 and so on) will generally not be known for an existing transformer if you can't find the manufacturer's technical data.

Interior temperatures are easy to measure in theory, difficult in practice because you need a good four-wire milliohm meter to do it. Copper's resistance goes up slightly with temperature, so you measure the wire resistance at, say, 25C, then load the transformer and let it heat up. Temporarily disconnecting the transformer and measuring the hot copper wire resistance tells you the temperature of the wire you're measuring. This is usually done on the whole interior winding, but you can do the same for one winding, like the 20Vac tap.

Temp measurements are slightly more difficult to do because the windings are copper, so they share heat/temperature with one another, and because the whole transformer has a thermal resistance to the outside air or mounting surface. A normal transformer may take an hour or up to several hours for its internal temperature to stabilize. So an internal hot spot, like a single winding, may take a long time to get to final temperature. By taking temps at regular intervals and assuming that the temperature rise follows a curve like a rising R-C exponential (and it does...), you can take a few temperature points and calculate a good estimate of the final temperature. As I said, the real way is a pain and takes a long time, so even experienced and knowledgeable people abbreviate it or use another way to estimate.

In your case, what I would do if I were really determined to find this out is that I would (1) get/borrow/rent a good milliohm meter (2) set up test loads for each winding equal to the maximum currents you think it would have in normal/max operation (3) do the time/temperature heat-up test, noting the data and computing an expected final temperature for each section from the ohmmeter readings and finally (4) start overloading the 20vac winding and seeing how much it and the other windings heat at various loads, as well as watching the output voltage value drop. If nothing goes over 105C or maybe 130C if you're feeling lucky, you're good up to that amount of overload.

[bepone]

too much speculating.. secondary wire cannot be found without dismantling the transformer..

if you know about trafo winding, specially double secondaries with tube rectifier, wire on secondary is for purpose made smaller, to increase resistance (for the series resistance which tube rectifier must see for normal operation).

also some secondaries are made with bigger wire to avoid voltage drops (heater winding)

some trafos are made on 1T , some on 1.25 , some even more!

how you can know what was the design idea/goals in production? which test will give this data to you?

so it is not possible to extract those data. some improvisation is possible to make, knowing the resistance of the winding and comparing with commercial or other trafos for the same purpose, also some core alowance can be found from the core cross area

[R.G.]

too much speculating.. secondary wire cannot be found without dismantling the transformer..

No, but its temperature rise can be found.

if you know about trafo winding, specially double secondaries with tube rectifier, wire on secondary is for purpose made smaller, to increase resistance (for the series resistance which tube rectifier must see for normal operation).

Transformer designers can do a lot of special purpose things for a custom design. This is one of those tricks. But I guarantee that no good designer ever designs for temperature rise in a winding over the temperature rating of the wire insulation.

also some secondaries are made with bigger wire to avoid voltage drops (heater winding)

Transformer designers can do a lot of special purpose things for a custom design. This is one of those tricks. But I guarantee that no good designer ever designs for temperature rise in a winding over the temperature rating of the wire insulation.

some trafos are made on 1T , some on 1.25 , some even more!

???? What is "1T"?

so it is not possible to extract those data. some improvisation is possible to make, knowing the resistance of the winding and comparing with commercial or other trafos for the same purpose, also some core alowance can be found from the core cross area

As I said, the real way to figure out whether an additional load will overload the transformer is to load it up to its rated maximum, measure the wire's temperature rise, then see if an overload pushes it over the temperature maximum of its insulation. Where you don't know the insulation temperature rating, it is safe to guess at Class 105, (105C) ; no commercial transformers are wound without at least this rating. Most of them today are wound with Class 130 (130C) rated insulation.

[bepone]

???? What is "1T"?

i dont believe that you are asking this question, this is MAIN VARIABLE in transformer designing.. magnetic flux density, B= 1 TESLA, 1T.

do you have more ???????' s?

[R.G.]

A simple "it stands for flux density in Teslas" would have sufficed. No need to raise your voice.

I don't know when you learned transformer design, but when I learned transformer design back in the Dark Ages ( ~1970-1975) here in the USA, the unit used was Gauss. Ten thousand Gauss equals one Tesla, and yeah, there's been a concerted effort to convert everyone over to the newly-correct international units. I know that academically, but I design in Gauss.

So I could equivalently say:
i dont believe that you are surprised by this, the MAIN VARIABLE in transformer designing is Gauss.. magnetic flux density, B= 10k GAUSS.
But I would probably just say "Gauss is the unit of magnetic flux density" instead of being amazed, given how polite generally I try to be.

Do you have any more pithy comments?


As an aside, given my age, I should say down with metric; don't give an inch.

[pjd3]

I do recall some talk about temperature being the ultimate limiter of what a max current could be but, I figured that the guy tutoring me in this figured that may be beyond my scope of doing. (although I do have daily access to Meter/thermistors, heat scanning guns and any sorts of the common equipment found in a good electronics R&D lab).

What I do know is that this 'Mystery secondary coil" was probably for the digital delay and reverb processing board plus all the relay channel switching that was going on in the amp. It had 3 full channels and a fair sized board with what looked like medium scale processors and chip set on it for the effects. My inkling is that the coil may provide enough current for a hand full of relay coils and about 3 of those high voltage FETS - LND150 high voltage FETS. I recall that these relays use about 36mA of current, times 4 = 144mA plus what ever the FETs use which I don't suppose is very much. Hard to say for me, I have no idea what level of energy those digital reverb/delay boards use.

Perhaps the engineer at work was just giving me a "playing it safe" arena of testing that would give ballpark results. The damn guy is the main over-see'r for some of the most popular defibrillators and ventilators that we see everywhere, that issues he contends with electronically are far beyond anything I can throw his way! But its up to me to know or figure out if my mystery transformer is up for the job!
Thanks everyone, very interesting outcome to this thread! I do appreciate your engagment.
Best,

Phil D

[Phil_S]

I'm not playing in the same pond as the others who responded. Those are great responses. I've been wondering about this question for a long time. It's good to hear some answers based on real engineering.

However, given the nature of the transformer you've got and the likelihood that you don't have access to that very expensive meter, I suggest going the most practical route. You don't really need to know the maximum current. You only need to know if the winding will support the circuit you want to build. These are, most likely, two different current amounts.

For grins, I guess I should mention one more way to determine the answer. Load up the winding until you let the smoke out. I'd say this gives a definitive answer if you've been meticulous in taking readings and ramping up slowly. Of course, that certainty has unwelcome consequences.

So, I'm only half joking about this. If I were in your shoes, I'd go for what your work colleague suggested. Load it up until you have a 10% drop. (Actually, if you know the load you want to support, you might be able to quit at a drop of less than 10%, reducing risk of damage.) That ought to get you close enough to at least decide if the winding offers some promise that it will work for your intended use. It's quick. It's inexpensive. I'm doubtful it will let the smoke out. The idea of shorting the windings and running the primary at 10% (12V in the US) is probably not going to damage the transformer, but the idea of shorting the windings is unsettling to me.

Finally, at some point, years ago, this is a compendium on information I collected on the topic. There are several suggestions for obtaining reasonable approximation of current limits. I make no claim to how reliable any of these methods are.

From the Primary Winding
For a 120 volt AC supply the VA rating and primary resistance is as
follows.
30 VA = 30 to 40 ohms
50 VA = 13 to 16 ohms
80 VA = 7 to 9 ohms
120 VA = 5 to 6 ohms
160 VA = 2.5 to 3.5 ohms
225 VA = 1.8 to 2.2 ohms
300 VA = 1.0 to 1.3 ohms
500 VA = 0.45 to 0.55 ohms
Simply multiply all ohmage values by four (4) for a 230 / 240 volt supply.
Derate to 65%, which is probably reasonable and to allow 15VA for the filament windings.

From the Secondary Winding
One very general way, based on a copper loss of say 4%. The ht winding rating is probably the most important. Find the secondary voltage, e.g. 300Vac. Take 4%, giving 12V. Measure the winding resistance; one half if 300-0-300V (i.e. a 300V winding), if bridge then the whole winding. The current will now be that which causes a 12V drop across the winding d.c. resistance. Thus dividing 300V by the resistance would give a ball-park figure for current (Ohms Law). Heaters more difficult; same method, but low voltage winding resistance is usually impossible to measure accurately. The heater current could be expected to be in line for the output stage that would require the previously calculated anode current.

One degree further would entail loading of the transformer. Sometimes this is possible with mains globes. Such a load across the ht winding for >1 hour should cause the transformer to get only slightly warm, as you still have no heater load, unless you simulate that too. Very generally heater and h.t. load can be assumed to have similar power figures.

Temperature Method
The operating temperature should not exceed the boiling point of water, 212F. Just load it up until you get close to that on a sustained basis. Remember to consider the VA rating applies to the whole transformer and allow something for the filament windings.

Core size
A good rule of thumb for the current limit of a transformer can be found by measuring the cross-sectional area of its core in inches-squared, mutliplying that value by 5.58 and squaring the result. (Refer to the Radiotron Designers Handbook 4th ed. page 235.) This formula gives a reasonable approximation of the total VA rating of the transformer. You may have to remove the transformer bell-ends to obtain these measurements.

For example, for a transformer that uses E-I laminations to form the core, you would measure the width of the middle leg of the E part across the face of the laminations. If you were looking at the E as if it were a letter, this would the height of the middle leg. Next measure the total thickness of the stack. Multiply those two values to obtain the cross-sectional area. Suppose you have a transformer that uses an EI-76.2 standard core, where the width of the middle leg of the E is 1 inch and suppose the stack is 1.25 inches thick (yes this is a Fender Vibrochamp transformer, part numbers 125P1B or 022772). This gives a cross-sectional area of 1.25 (1.00*1.25) and an approximate VA rating of 48.65 [(1.25*5.58)^2]. That rather handily corresponds to Hammond's rating for their replacement transformer of this type whose secondary windings are: 325-0-325 @ 81 mA, 6.3 @ 2 A and 5 @ 2 A:

(325*.081)+(6.3*2)+(5*2) = 48.92
This gives a decent approximation of maximum total current you can expect the core to support. Other factors such as the gauge of wire used in the windings, number of turns, number of windings, insulation thickness (which may drive up core size), etc. place additional limits on the maximum current a given transformer winding can deliver.

[pompeiisneaks]

???? What is "1T"?

i dont believe that you are asking this question, this is MAIN VARIABLE in transformer designing.. magnetic flux density, B= 1 TESLA, 1T.

do you have more ???????' s?

I think I've mentioned this several times before bepone, but nobody, literally nobody, needs this level of condescension here. IF you can't be polite about things, you should refrain from talking. I find it particularly odd that you're being rude and condescending to one of the most knowledgeable people here on the forums. Please, as my grandmother used to say "If you can't say something nice, say nothing at all".

~Phil

[R.G.]

Good advice, Phil, and very practical.

Yeah, a practical approach would be to just load it up with the expected full load and added load on the sub-section of winding, and see if it gets too hot. The only catch with that is that the time constant of the trannie means you may have to let it cook for a looooooong time to see where it stabilizes. A 400-500VA trannie might have a time constant of over an hour. If the trannie's exterior stays under 140F or so under long time loading, it's going to be fine for the whole-transformer rating.

More thinking on the topic: determining whether a winding can take an additional X milliamperes well enough is different from finding out how much the winding can take without overheating the transformer, as you correctly point out, Phil. This is largely because the OP has control of how big X is.

@pdj3: Signal switching relays can be very, very small current devices. One of my favorites is the low-power signal switching relay in a DIP styled package. You can get these in 12, 18, and 24V coils. The coil power can be as low as 150mW. The nominal 24V coil usually has a "must operate" voltage of 18vdc and a coil current of nominally 8.3ma.
20Vac rectified gives 28.28Vdc peak, but this will sag with loading depending on the coil resistance - which you can measure independent of the termperature to get an idea of the sag.
So assuming three of these relays, you'd only need about 25ma if they were all on. Presumably the JFETs are buffers and receives for loops; this could be 1-1ma per device. Call the extra load about 35ma. So

One of the secondary's of the Carvin PT is approx 20vac which I thought/hoped/wished that would provide enough current for relay coils, some FX loop fets.

your wish will probably come true if you're conservative about the load you add.

[Phil_S]

Good advice, Phil, and very practical.

So, I'm pretty sure you're talking to me, but we now have 3 people named Phil in this thread! Anyhow, I appreciate your thoughts on what I said.

[R.G.]

So, I'm pretty sure you're talking to me

Yep, that was at you. Good advice you gave.

[pjd3]

Oh no all you other Phils, he was certainly talking to me - I am the OP (the "good Phil") but, you probably knew that already.

Well, once again, I got boat loads more out of the thread than I anticipated but, that'*s a good thing. It will take me a bit more time to digest everything that has come this way but, I have been given multiple ways and methods for determining the status of my mystery secondary.

I was intending on using the relay boards from Headfirst amplification that I purchased and they are layed out to use those Omron G5V-2-DC12 kind of relays. They happen to be upwards of 40mA so 3 of those are looking at 120mA activated. Yeek, its a bit. There are probably some lower power relays to be had and I certainly will look around.

Thanks again everyone, this as been fun and quite a learning thread.

Best,

Phil D.

[R.G.]

On the PCBs: It is possible that you could use other relays, like the low power ones I mentioned, with a technique known among technicians as "dead-bugging".

In dead-bugging, you glue the component down on its back/top in a safe place on the PCB, then run wires from each of the legs/pins waving in the air to the correct hole in the PCB. This might work on the boards. I'd be happy to take a look and see if you can send schematic and other tech info.

Edit: I thought that Headwise might offer some tech detail on line, so I checked the web site. Nope, they're not an outfit that does info on line. However, some of the relay board photos show the bare PCBs, and some of the relay footprints match the footprint of the low power relays, so it's at least possible that the low power stuff could drop into the PCBs. Can't tell from what little I found on line.