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I have a 9v AC to DC wall adapter that I use to power my Arduino boards. The label says it gives 9v at 1000ma but I wanted to test the validity of the label to troubleshoot.

When I connected my probes to the terminals however it said 0 amps and then the green power light on the adapter turned off. After that the adapter would not power anything. I opened it and could not identify any blown or otherwise damaged pieces.

So, before I fry something else like the USB port on my computer :(, does anyone know what I was doing wrong? The multimeter still works and gives valid readings on other things and I have measured chargers before without issues. I have this multimeter from radioshack.

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    \$\begingroup\$ If you simply connected the two multimeter probes to the + and - on the supply in ampere mode you effectively shorted the power supply and probably burned it \$\endgroup\$
    – crasic
    Commented Jun 29, 2015 at 20:38
  • \$\begingroup\$ And to add to @crasic, if nothing is actually burned looking, it may well be the transformer itself. \$\endgroup\$
    – Asmyldof
    Commented Jun 29, 2015 at 20:39
  • \$\begingroup\$ Ampere readings go in series with the load. The 1000ma number is the largest current you can draw from that power supply since the vast majority of commercial power supplies are designed for constant voltage output. Your multimeter will read current by placing a small resistor and measuring the voltage drop across it. If it was a 1 ohm resistor you just tried to pull 5A from your supply and something blew up. \$\endgroup\$
    – crasic
    Commented Jun 29, 2015 at 20:42
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    \$\begingroup\$ What was wrong with my question? \$\endgroup\$
    – NULL
    Commented Jun 29, 2015 at 21:16
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    \$\begingroup\$ Nothing is wrong with your question - the problem is with your technique. You made a fundamental (but common) beginner's error in misusing your ammater, and while there's nothing wrong with that as a learning experience it makes the actual content of your question a bit irrelevant, and some of the crusty regulars here are reacting to that in a characteristic, less-than-ideally-educational manner. \$\endgroup\$
    – Chris Stratton
    Commented Jun 30, 2015 at 3:15

4 Answers 4

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As others have said, connecting a multimeter set to measure Amps directly across a power supply effectively puts a short circuit on the supply, due to the very low resistance of the multimeter when set to measure current. The short-circuit current you measure this way will probably be far above the supply's rated output, unless the supply has overcurrent protection.

I don't think it is practical to try to verify the rated output current of a power supply by a direct measurement - you just have to trust the maker's claim.

You could, however, connect a load of the correct resistance to draw the rated current to the supply, and monitor the output voltage and power supply temperature over an hour or two to see if the supply survives, and continues to output the advertised voltage, and doesn't get too hot.

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  • \$\begingroup\$ Indeed. If you know what the load impedance (in simply resistive cases, resistance) is, then there's actually little reason to measure current at all - rather, measure voltage across the load and then calculate the current from the voltage under load and the known load impedance. A simple unregulated supply will show noticeable droop of the output voltage as the current drawn approaches the rating, which is presumably where the transformer starts to saturate... \$\endgroup\$
    – Chris Stratton
    Commented Jun 30, 2015 at 3:21
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How did you test the current draw? Like you would voltage or by placing the meter inline with the power? Ideally in voltage measurements the meter represents infinite resistance, and in current measurements represents zero resistance. If you have the meter in current mode and short the leads across the output, you're effectively shorting the circuit with zero resistance, which can damage the meter and/or the circuit being tested. In current measurements you place the meter inline with the circuit so that current flows through the meter to measure it.

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  • \$\begingroup\$ Also note that it's 1A rated, but the device determines if that much current is pulled. \$\endgroup\$
    – Jarrod Christman
    Commented Jun 29, 2015 at 20:40
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    \$\begingroup\$ I measured it like this: Multimeter + ══════ power supply +; multimeter - ══════ power supply -. In this case I don't see the difference between in series and parallel because attaching the respective lead to each end is the same as attaching the leads together and attaching each end to a terminal? What is the correct way to measure it then? \$\endgroup\$
    – NULL
    Commented Jun 29, 2015 at 21:10
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    \$\begingroup\$ @NULL - that's exactly the wrong way to use an ammeter. You need to connect it in series with a suitable load, not as a near-short across the supply with no other load. \$\endgroup\$
    – Chris Stratton
    Commented Jun 30, 2015 at 3:18
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you made a short, the internal resistance of the multimeter is under/around 1 ohms in ampmeter mode ( in most case). So if you have a 1 ohm res with 9 v apply to it, V = R*I , that is around 9 amps that you are requesting to your adapter.

Either, you burn a fuse or you broke your device. Normally, you should put a nominal charge connected to your adapter than you can measure the current.

Regards,MathieuL.

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  • \$\begingroup\$ I am still new to the technical part:); what do you mean by nominal charge? \$\endgroup\$
    – NULL
    Commented Jun 29, 2015 at 21:13
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    \$\begingroup\$ A nominal charge mean the charge that the device was made for, If your adapter is made for a 10k res as an example. You know the nominal current when you connect the 10K res. Nominal charge just represent the optimal/normal condition in which the device was made. In this case, you should just trust the manufacturer, if you live in NA, all power supply that end on the market respect the American Standard, so you can trust what the manufacturer said. How ever if you purchase a cheap supply from China it is different. \$\endgroup\$
    – MathieuL
    Commented Jun 30, 2015 at 1:20
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    \$\begingroup\$ Right idea, but there's a wording problem in this answer. It should be recommending a nominal "load" (measured in ohms), not a nominal "charge" (measured in coulombs) \$\endgroup\$
    – Chris Stratton
    Commented Jun 30, 2015 at 3:16
  • \$\begingroup\$ Thanks you, Chris for the correction, my main language is French and Load = Charge. \$\endgroup\$
    – MathieuL
    Commented Jun 30, 2015 at 12:11
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Further to the other answers, if you want to avoid shorting out supplies it's better if you get a resistor, which you calculate the resistance for by using V=IR, which rearranges to get R=V/I

You say the voltage is 9v and the current is 1000mA. There is 1000mA in 1 Amp. Therefore the current is 1A.

The resistance is therefore 9/1 = 9 ohms.

Therefore, If you connect a 9 Ohm resistor in series with the ammeter, you can measure the maximum theoretical current. That means you won't burn out another supply.

Of course, you have internal resistance to worry about, but you can always measure this by putting the meter to resistance mode, and then probing another power supply.

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    \$\begingroup\$ "+1" thank you for the clarification! I said the voltage was 9v though, so it would be 9 ohms, correct? What would happen if I used a slightly higher value resistor(I would hate to have to buy a 9 ohm when I could use a 10ohm on hand!)? The measurement would be slightly incorrect though right? \$\endgroup\$
    – NULL
    Commented Jul 1, 2015 at 13:31
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    \$\begingroup\$ Oops, I can't read! :P If you had a 100 ohm resistor, the current would drop significantly. From V=IR -> I = V/R, you can see 9/100 is 0.09A, which is a tiny amount of current. Remember the higher the resistance the lower the current, as the two are inversely proportional. You can buy resistors on ebay for about 99p. I bought a set of about 30 types for £2 from China, and they are accurate. They are useful. Here is an example: ebay.co.uk/itm/291399638610. I would NOT buy this though, as the lowest value is 10 ohms, which is slightly too high. Just have a look round on ebay :) \$\endgroup\$
    – George
    Commented Jul 1, 2015 at 17:11

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