I have a few small-ish unmarked power transformers. I would like to know what their ratings would be if I had the manufacturer's specifications. So far, I have determined that I can safely connect them to the mains without tripping a breaker.

For the voltage, I can use a DMM. However, an unloaded transformer produces a different reading than one with a load. Does the spec for a known transformer refer to the loaded or the unloaded reading?

I have heard about methods for approximating the transformer current capacity by measuring weighing them, but:

  • the transformers are well under a pound, making a weight less accurate
  • one type of transformer I have, for which I have 3 units, has two secondaries, and I would like to know the current capacity for each; the overall weight of the transformer will give only an approximate total power

I am about to experiment with an empirical approach in which I place power resistors of known values across the secondary in series with a DMM set for current reading, and then use an IR temperature gauge to watch how much the transformer heats up. The only problem here is I don't know what a typical maximum temperature would be.

Is this a viable approach? Any suggestions on making it accurate, such as an appropriate maximum temperature for the transformer under load?

I welcome answers with other techniques.

  • 1
    \$\begingroup\$ Unless you know the wire insulation is high temperature capable, I'd stay less than 70C on load. To measure wire temperature, a 50C rise in temperature is roughly +20% resistance change for copper (obviously measure unpowered). If you've already checked they power up off load, and you're keeping an eye on temperature, it sounds like your empirical approach is well founded. Excess current doesn't hurt transformers, heat does, so you can rate them higher if duty cycle is low. \$\endgroup\$ – Neil_UK Feb 23 '17 at 18:43
  • \$\begingroup\$ I would just throw them all away; they connect to an AC supply and could become a fire hazard and burn your house down or electrocute you. Do you know their history? Do you value your time? Would you expect a 2nd hand car (with unknown history) to be as reliable as a new one? Have you seen fatal attraction the film? Did you see what she turned into? It's your risk. \$\endgroup\$ – Andy aka Feb 23 '17 at 18:52
  • \$\begingroup\$ Regarding Andy's comment, the three dual-secondary transformers have very handy voltages, and I would like to use them. They appear to never have been used. One was definitely taken out of some used equipment. But a fire hazard? I have been working on electronics in some form or another for >50 years and never heard of a power transformer being a fire hazard, assuming it is in a circuit that does not overload it. \$\endgroup\$ – Mark Colan Feb 23 '17 at 20:03

About 50 years ago there was a rule of thumb that 1 sq.cm of main magnetic core can handle 10W of power. This was for 50 Hz AC. For 60 Hz it should be 20% better. The current capacity of a winding obviously depends of wire gauge, there could be low-current sections of the transformer, but it should be a good indication of carrying capacity (since no manufacturer would waste copper unnecessary). You should use transformer design rules when estimating current capability of a winding.

  • \$\begingroup\$ main magnetic cores are three-dimensional. Shouldn't the rule of thumb be expressed in volume? \$\endgroup\$ – Mark Colan Feb 27 '17 at 0:45
  • \$\begingroup\$ @MarkColan, this rule assumes that the magnetic core has proper proportions to keep flux density even. So one measure is good for all. I assume that, based on the same assumption, a rule expressed in terms of core volume should be equally ok, but much harder to measure quickly. \$\endgroup\$ – Ale..chenski Feb 27 '17 at 1:44
  • \$\begingroup\$ I will experiment with this estimation, with known transformers. \$\endgroup\$ – Mark Colan Feb 28 '17 at 18:50
  • \$\begingroup\$ @MarkColan, what criteria will you use for such parameters as core/winding temperature and level of saturation of magnetic core? What is your goal for this validation effort? The 1cm2=10W rule was for ordinary "transformer steel" of the past. Modern magnetic materials might differ. \$\endgroup\$ – Ale..chenski Feb 28 '17 at 21:10
  • \$\begingroup\$ One goal is simply to safely use transformers I already have in a safe manner, i.e. not drawing too much current and avoiding over-heating. If I were to put a fuse in the power supply output, what rating would it be. The other goal is to satisfy my curiosity about a good way to do this kind of thing, i.e. learn something new. \$\endgroup\$ – Mark Colan Mar 1 '17 at 22:11

To determine the current capacity, you can take as a reference the diameter of the wire used in the winding. From that diameter, you get the cross section of the wire.

$$S = \dfrac{\pi\cdot D^2}{4}$$

The current criterion for line frequency transformers is approximately 3 A per square millimeter of section.

Is this a viable approach? Any suggestions on making it accurate, such as an appropriate maximum temperature for the transformer under load?

Yes, but I think it is quite imprecise. The maximum operating temperature is set by the type of transformer insulation, which corresponds to class F transformers, class H, etc. If you do not have corroboration about the materials used for its manufacture, you will not be able to define the type of operation and therefore, you can not define the maximum operating temperature.

As for the power, as it was answered by another participant, you can use the rule of 10 W per square inch of magnetic section.

  • \$\begingroup\$ The problem with this formula is knowing the wire diameter. Seems like you have to take the transformer apart to know how much power it handled before you did that. \$\endgroup\$ – Mark Colan May 8 '19 at 18:34
  • \$\begingroup\$ @MarkColan, of course. Sometimes, the wire is soldered on external pins, and you can measure the diameter with a Vernier. This measurement must be just as exact to give you an estimation \$\endgroup\$ – Martin Petrei May 12 '19 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.