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I have obtained a small transformer with 7 windings, a 220V and 6 other individually isolated low voltages. How can I find out the current rating for the lower voltage windings? I was told these used to be made for "CD players". This is from the local market, so no datasheet. Transformer weighs about 500 grams and looks reasonably well made. If it was only one secondary 12-0-12V, I'd say it would be 1-1.5A. But, with 6 completely isolated secondaries, I'm not sure. The wires near the output seem probably slightly less than 0.5mm dia, but I'm not sure, it may be thinner. Output voltages were listed, and I measured the resistance with a DMM:

Volts Ohms
220  229
13.5 1.3
10   1.5
12   1.9
11.5 2.3
3.5  2.5
21.5 11.1

I got this to power op amps, and am currently building an electronic load (for a power supply that I will be building next). So, if an electronic load can be used to guess the output current rating (by plotting voltage drop etc...), that will also be a big help. Obviously op amps will only draw ~1mA, so I'm safe for now.... but circuits tend to get bigger...

This is my first question on stackexchange. Thanks in advance for all help and suggestions!

[edit] Thanks Neil and Tony!

Assuming 20VA for the transformer (as it's not a toroid), I have the following calculations:

wire    V       R       V2/R    VA      I
red     220     229             ~20
blue    13.5    1.3     140     7.2     0.53
purple  10      1.5     66      3.4     0.34
green   12      1.9     75      3.9     0.32
white   11.5    2.3     57      2.9     0.25
black   3.5     2.5     5       0.3     0.08
yellow  21.5    11.1    42      2.2     0.10

I have managed to make an electronic load on the breadboard, and although it's not terribly accurate because of the wire resistances, it seems to be working reasonably. I used it to load up the blue winding. Firstly, the transformer is becoming warm even with no load (all secondaries disconnected). With some back an forth loading over 30 min (while testing the electronic load, mostly below 200mA, maybe 2 minutes @ 400mA), it became quite warm (under ceiling fan, room temp 30C, rel humidity 75% here in Kolkata/Calcutta). I didn't measure temperature as it wouldn't be accurate under the ceiling fan. Forgot to measure the hot resistance as well. Also, the voltage drop was a lot. DC voltage (measured after a 1N4007 bridge and 1000uf cap) went down quickly (open circuit voltage close to 18.5V, went down to even 11.5V under load):

V      mA
18.22  0.32
17.0   33.5
15.8   150
15.0   180
14.88  245
13.8   350

All this for the blue winding. (Also I managed to fry an lm358 as I stupidly forgot that transformers put out AC, thankfully the IRF540 was spared). The open circuit AC voltage is 13.8V.

I'll see if I can load up all the windings simultaneously for the test you suggested, but I'm a bit scared considering how fast it's heating up. I'll probably have to measure the AC primary current to check what is happening. The other 9-0-9V transformer I have has 1K resistance on the primary winding, and it stays cool even with 175mA load.

Neil, thanks for the excellent answer, I'll accept it, and see if I can do the resistance test in future.

Added Feb 2018: After some more experience with transformers in the local markets in India... Transformers here are rated on the open circuit voltage (no load). i.e. a 12V 5A transformer will show 12V at no load. Super hyped up ratings? Customer beware? One needs to carry a multimeter and measure the secondary winding resistance to estimate voltage drop and heating before purchase.

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  • \$\begingroup\$ In small transformers they often allow 10% loss of the rated load. THis is called %Zo, and the inverse is the Isc/Irated. So if Zo%=10% then Isc=10x I rated for VA ratings. , so I might expect 10% of short cct current in your case is (Pri)~1A, (Sec)10A, 6.6A, 6.3A, 5A, 1.4A, 1.9A \$\endgroup\$ – Sunnyskyguy EE75 Jul 30 '17 at 1:51
  • \$\begingroup\$ I'm not sure I understand. Do you mean the I rated is (Pri)~1A, (Sec)10A, 6.6A, 6.3A, 5A, 1.4A, 1.9A ? Or do you mean I rated is (Pri)~100mA, (Sec)1A, .66A, .63A, .5A, .14A, .19A ? The latter does seem more like what a CD player transformer might be (from labelled transformer photos on the web).... with the voltages near 12V with high current, and the 3.5V and 21.5V with low current, presumably for op amps. I might try to run a short circuit test if there is no risk of frying the transformer. Will it help? \$\endgroup\$ – Indraneel Jul 30 '17 at 6:21
  • \$\begingroup\$ No it wont help, consider the 10% ROT (rule of thumb) \$\endgroup\$ – Sunnyskyguy EE75 Jul 30 '17 at 7:25
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My standard way to initially estimate the total VA rating of a transformer is to weigh it, and then compare the weight with published figures from a transformer catalogue. From for instance, RS conventional EI transformers weighing 500g are rated at 20VA, whereas toroidal transformers of the same weight are rated 30VA.

This is the total VA, what you would expect the primary to handle. You have to divide this amongst the secondaries.

You have the measured voltage \$V\$ and the measured resistance \$R\$ of the windings.

There are two limitations to the max current that can be drawn from secondaries, regulation (voltage drop) and temperature rise.

Regulation is easy to handle, as it can easily be estimated from the current we want to draw \$I\$ as \$V_{drop} = IR\$. This doesn't damage the transformer, only affects our load, and whether the voltage is sufficient at the load.

Temperature rise is more difficult, and can damage the transformer. We can see how warm the transformer gets to the touch, but that doesn't tell us whether one particular winding is getting too hot or not.

If we assume that all windings are cooled to ambient in the same way (which is not too wrong, especially for a first stab, and especially for a toroidal) then the VA of a winding is proportional to the mass \$m\$ of copper used in it, regardless of the number of turns \$N\$, length \$L\$ or wire area \$A\$.

As \$V\propto N \propto L\$, and \$R\propto \frac{L}{A}\$, we can see that the mass varies as \$m \propto \frac{V^2}{R}\$

I'll leave it as an exercise for you to prove this, hint, dimensional analysis helps (don't forget the dimensions of the constants of proportionality). Note that the expression itself has units of power, which we would expect as it's supposed to help estimate VA.

So, for each secondary, calculate \$V^2/R\$, and apportion your total 20 or 30VA in proportion, this is your estimate of VA for each winding.

Having got initial estimates for the VA of individual windings, fire the transformer up from a fused supply. Load each winding with a resistor to draw half your calculated VA, and measure the voltage drop. Make sure this is reasonable (a few percent for big transformers, possibly 10% for small ones) for all windings before you continue the test.

Load all windings with half your calculated VA, and let it run for 30 mins to reach thermal equilibrium. Now disconnect everything, and quickly measure the resistance of each winding before they have had a chance to change temperature. You can estimate the temperature rise of each winding by knowing that copper has a tempco of 0.4% per degree C at room temperature. For example, if your 2.5ohm winding went to 2.75ohm (+10%), that indicates a 25C rise above ambient. You may need to do a four-terminal measurement to get differences accurate enough to be worth using at the ohm level.

If any winding is particularly hot or cool, you can vary the proportional of the total VA going to it, and try again. The maximum temperature a winding can reach is governed by the insulation used on the transformer. I don't like to go above 70C (remember if you box the transformer up, its ambient will increase) without knowing more about the specific wire used.

Before you finally test the transformer at full VA, remember the temperature rise of the transformer is proportional to \$I^2\$, so the half VA temperature rise you measured was one quarter of the final, not half!

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