# How do I test how many amps a transformer actually puts out

I have a 120v <-> 9v ac transformer rated at 3.34 amps, but my multimeter suggests that (after running it through a diode bridge), I can get 7+ amps out of it @ 9.4 volts ac.

Can I?

I have a device that draws 4 amps DC at 6-10 volts,a nd I'd like to power it from this device.

• How does your multimeter suggest you that? Feb 18, 2013 at 19:24
• There is an 'unfused' 10A setting -- I tried it at the outputs of my diode bridge. Not sure I was doing it right. Feb 18, 2013 at 19:25
• I'm pretty sure you didn't do it right if you had no load in series with the Amp meter. Always make sure there is a load when measuring ampère. Feb 18, 2013 at 19:40
• what he did was practically use the Multimeter's ( should i say ammeter's ) resistance as the series resistance for measuring Current. NOT RECOMMENDED UNLESS you know what you are doing. Feb 19, 2013 at 1:39

If it's rated at 3.3A and you draw much more than that from it for an extended time, it will exceed its rated temperature and eventually its insulation is likely to break down. At which point, smoke ensues.

Whether 20% overload (4A) is too much more I can't say. I would guess short periods of operation (up to an hour, then cooling down for an hour) would be OK for my own use, but I wouldn't sell anything made that way.

7 amps is definitely too much.

• Reading his question again, i have a feeling that @gbronner had made mistake in assessing his observation. He measured voltage and Current at two different load condition. So practically his claim that he gets 7+ amps out of it @ 9.4 volts ac is invalid. Nonetheless your answer is very helpful and shows true experience :). Feb 19, 2013 at 1:45
• It is possible that he read 7A at 9VDC after the bridge. Now 9V * 1.4 - 1.4V(bridge) = 11.2V so 9.4V DC (not AC) at 7A would be plausible, if he measured both with the same load. I suspect a transformer rated for 3A would have greatly exceeded 7A short circuit directly into a multimeter! But in any case, 7A is not recommended for operation. Feb 19, 2013 at 8:58

You have to measure voltage while you are measuring current to get power rating. The transformer will not be giving out the same voltage when shorted out thru a shunt resistor in a multimeter.

Take another multimeter, set it up as voltmeter and measure the voltage while measuring current. Then power delivered would be P = I * V and you can roughly expect I = P/V of current in normal operating conditions.

Note that the voltage measured won't be accurate, rather some approximation your multimeter comes out with as you will not be getting constant voltage, rather something like that:

Try putting the biggest electrolytic capacitor you can find accross the bridge output while measuring (watch polarity and voltage rating).

BTW, you are lucky you didn't pop the fuse in your multimeter, never measure current while shorting the power source with multimeter, put some load in series, like 1ohm power resistor.