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I have an old Nokia phone charger lying around which I want to adapt for powering my breadboard/PCBs.

I was trying to find out how much current the charger was able to supply, so I connected a DMM across its leads and measured the current.

Now, I know that one is not supposed to measure a battery/power supply current by just connecting the leads (since the ammeter setting in a DMM effectively shorts out the battery/power supply) but use a load, what other method is there to measure maximum current?

Do I just use a small resistor(like a 10 ohm one) in series? I'm worried that that'll burn up the resistor and/or give a wrong reading.

I also have a USB cable that I could use with my current phone charger to power the breadboard, but as asked earlier - is there a risk of damage to the charger/DMM by directly measuring the current?

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  • \$\begingroup\$ Check what ratings are printed on the Nokia phone charger :-) \$\endgroup\$ – Huisman Oct 10 at 21:22
  • \$\begingroup\$ It's inaccurate. It says 5.7V 780mA on the charger, but actually gives 6.06V and 0.8A using the above method. I'm not even sure if the readings I have are accurate, hence the question. \$\endgroup\$ – cst1992 Oct 10 at 21:25
  • \$\begingroup\$ There is no guaranteed-safe way to determine how many amps a power supply can provide. If the power supply has a fuse, your may find out that you've reached the maximum when you blow the fuse and have to replace it. If it does not, you may find out that you've reached the maximum when it catches on fire. (This is unlikely for a consumer device; they are generally meant to be safe. But that doesn't make it a good idea!) \$\endgroup\$ – Glenn Willen Oct 10 at 22:15
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    \$\begingroup\$ @cst1992 If you believe that 5.7V at 780mA is the maximum ratings, then I believe you're being way too strict. Manufacturers will always have some tolerance that will go above what that typical rating says. I don't think they really share this to the customers because they probably want to prevent people from blowing up their device. \$\endgroup\$ – KingDuken Oct 11 at 15:10
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    \$\begingroup\$ A well-designed device should be able to exceed its ratings - but anything beyond is not guaranteed. So you know that this charger can deliver its rated current of 780mA (at 5.7V or higher) - and that is all you should expect from it. \$\endgroup\$ – Bruce Abbott Oct 11 at 18:04
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You should gradually decrease the value of the load resistance while monitoring the supply output voltage. When the voltage starts to droop you have reached the maximum current the PSU can supply.

Don't short out power supplies with your ammeter. It gives you little, if any, useful information. At the current you measure the voltage is zero and since P = VI then the delivered power is zero.

From the comments:

Good thing I bought a 2 Mohm potentiometer then :P – cst1992

A 2 MΩ potentiometer is useless for testing your power supply. If it is a 1/4 W potentiometer then the maximum current the resistance track can handle is \$ I = \sqrt {\frac {P}{R} } = \sqrt {\frac {0.25}{2M} } = 0.35 \ \text {mA} \$ when you are expecting 2,200 times that current. Even if it was powerful enough, 10 Ω would be \$ \frac {10 \times 330}{2000000} = 0.00165 \$° of rotation on a 330° potentiometer. Have you got a steady hand?

Note that a potentiometer's power rating is across the whole length of the track. If you are only using a certain percentage of it then derate the power dissipation by that factor as well.

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  • \$\begingroup\$ It gives me 6.06V consistent when measured directly. How much drop should I look for to know I've reached the limit? \$\endgroup\$ – cst1992 Oct 10 at 21:27
  • \$\begingroup\$ Good thing I bought a 2M ohm potentiometer then :P \$\endgroup\$ – cst1992 Oct 10 at 21:28
  • \$\begingroup\$ If it measures 6.06 volts with no load, then the nameplate rating of 5.7 volts at 780 ma seems quite reasonable as these devices generally are not highly regulated. By the way, a resistor of 7.3 ohms will draw the full rated current. You will have a lot of trouble obtaining that low a resistance reliably with a 2M potentiometer. \$\endgroup\$ – Barry Oct 10 at 23:21
  • \$\begingroup\$ @cst1992: See the update. \$\endgroup\$ – Transistor Oct 11 at 16:42
  • \$\begingroup\$ @Transistor It was a joke. At the 780mA that the DMM is showing, the power dissipated through the resistance is ~4.76W. That'll need a large resistor or it'll burn up. Same for a potentiometer. \$\endgroup\$ – cst1992 Oct 14 at 16:05

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