Take the 2-minute tour ×
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It's 100% free, no registration required.

I am curious how far can I go if I'd like to rework a small 12V/0.1A DC power supply to give higher current without changing the transformer inside it.

I understand that I have to upgrade the electronic part (a different linear voltage regulator, or maybe get rid of it completely to get maximum possible output). What I don't understand is what is the limit of the transformer? How can I find that limit? Should I expect that as a soft source under heavier load there will be a significat voltage drop and nothing bad happens? Or will the transformer under heavier load dangerously overheat? Is appropriate to put there a thermal fuse (e.g. 115°C) to protect if from that overheating? You know I don't want to buy a new transformer, I just want to try to use the existing one (without datasheet).

This question is probably more theoretical than practical, I just want to make things clear for me and not actually desperately need the new power supply. For example if I tried to push the same transformer to give .2 A instead of .1 A (two times more than originally designed), what result can I expect? Since it is made of a thin wire, I expect either it will give lower voltage and nothing else happen, or it will dangerously overheat.

I know that this question is similar to this: Finding maximum current experimentally But I think they ain't the same, just similar. My question is aimed at transformers.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

AFAIK transformers are disigned to withstand a 180% current overload for short periods. (They can be overloaded this way for something like 15 minutes every 24 hours or so.) But in the long run overloading it with 200% rated current will most likely cause serious overheating and eventually damage. Maybe a good (meaning in your case probably water - or oil - cooling, which can be way more expensive than a new transformer) can provide the necessary cooling to enable the trafo to carry that kind of current.

Additional info: If your supply provides 0.1 A DC as you say, the AC current (RMS value) that comes out of the trafo is not the same as the magnitude of the DC current! In general, assuming the output current of the trafo is not distorted (meaning it keeps its beautiful sine shape - not necessarily true) you have for full-wave diode rectifiers (from: Mohan - Power Electronics):


meaning that the AC current will be actually 0.22 A if you want 0.2 A DC. But, since the trafo is seriously overloaded, I'm sure the current will be strongly distorted due to the saturation of the trafo iron core. At the end of the day, this will lead to extra heat generation and some other odd behavior.

All in all I don't really think it's a very good idea.

The above calculation applies only if you have a line frequency transformer (50 or 60 Hz) followed by a rectifier. If the circuit includes a high frequency transformer, the situation can be entirely different. So, if this is the case, please edit your question, so we can adjust the answer to that.

share|improve this answer
Even if your equation is correct, it will give that I need only 200%, not 220%, because the original 0.1 A value is on power supply's DC output, not the transformer itself. Also, since there is a linear voltage regulator, there must be some spare power to let it do its job, so the actual overload will surely be far less than 200% anyway. –  Al Kepp Jan 25 '12 at 19:34
@AlKepp: You're correct, the efficiency of the different stages was not taken into account. But, if you think a bit about it, you'll be aware that the current at the DC output might have branched off earlier, e.g. to provide 'spare power' to some other component, hence you have to add the losses as you move upstream, which actually makes the situation worse. Even if you were right, the saturation would unavoidably take its toll. Btw: without any extra info about the circuit, I think this an educated guess you can make. –  Count Zero Jan 25 '12 at 20:16
But still, my situation is that I own the transformer, in fact several ones, I know that they were used in 12V/0.1A or similar supplies, but don't know their actual limits. I want to determine the actual safe limits, because I hope that tese transformers are able to provide more than just .1 Amp. It is sad to hear it is not possible or not safe to do. –  Al Kepp Jan 29 '12 at 15:20
The previous answer is incorrect in some respects. Some transformers, particularly some isolation transformers in the power range the OP inquires about, are designed to withstand continuous short circuit without anything bad happening to them. You can see the IEC labels over here. Old/salvaged material probably won't be so labelled though. –  Respawned Fluff Oct 15 at 12:02

Use a beefy enough (both in watts) variable resistor (pot/rheostat) on the output of the transformer and gradually lower the resistance until the winding temperature reaches 55C above ambient (to be on the safe side) or 65C above ambient (if the stuff is of somewhat newer vintage). That's how the VA (volt-ampere) rating of a trafo is established [source]. The max secondary current is simply that rating divided by secondary voltage; you could also measure the max current directly while doing this experiment. Alas pots that can handle > 100mA (which you need here) are pretty expensive; if don't need to establish a precise VA/current rating, you could probably get by with a network of ceramic resistors. It may also be difficult to insert a temperature probe in the winding to get an accurate reading as well. The IEC 61558-1:1997 standard gets around this by measuring the temperature rise indirectly from the change in the DC resistance of the winding using the following formula: \$ Δt = \frac{R_2 - R_1}{R_1} (x+t_1) - (t_2-t_1)\$, where \$x \$ is 234.5 for copper wire (or 225 for aluminum), \$R_1\$ is the winding resistance at the beginning of the test and \$R_2\$ at the end, once the temperature under load stabilized; \$t_1\$ and \$t_2\$ are beginning and end test ambient temperatures (ideally the same), but the formula is said to be valid (accurate enough) for \$t_2-t_1\$ not exceeding 10C.

There's also a rule of thumb formula for estimating the VA rating based on the size of the magnetic circuit cross section as 31A2, where A is the area of the (E-I) core in square inches [source], although I have yet to find this formula in a textbook, so it might be wrong. (Note that the width of the center leg is twice that of the outer frame in a E-I core.) A more precise design formula involves calculating a rather involved 6th-order polynomial; see over here for details. You actually want to do the inverse of that calculation, i.e. solve for the root(s) of the polynomial. There's also a table with common values at that link. There is generally no universally accepted method for sizing transformer cores, but (based on its dimensional aspect) the first method/equation I listed appears to approximate the slightly more elaborate design method called Core Area Product, which multiplies the cross-section area of the magnetic circuit with the area of the window for the copper/windings. The latter is the approach given in several textbooks: [1][2][3]. Bear in mind however that transformer cores are usually made to somewhat standardized sizes, so the electric circuit (windings) added on top of that may overheat before maxing out the core's magnetic circuit. To know this for sure, you'd need to also measure the wire gauge and number of turns, which may be impossible to do without taking the thing apart. So even after you get a rough idea of its power rating by measuring the magnetic circuit, which is much easier to do non-destructively, you should still perform the experiment that I suggested in the first paragraph.

Also, based on the subsequent comments to the first answer, where you say that you have several of these transformers: you could also put them in parallel to get the same output voltage and increased current rating, but be careful to match the polarities when doing so or you will blow them up. More details about putting transformers in parallel can be found in most textbooks e.g. here or even on the web. Given that a class II transformer with the characteristics you want (SEC: 12V, 0.2A) can be bought for $3 (in quantities of 1) from fairly reputable manufacturers, all this doesn't seem worth the hassle to actually put in practice, unless you want to learn/experiment.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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