I wanted to measure the current of a generic lithium ion battery charger. I set I plug my red lead in the A 10A max fused and black into COM. The charger is rated at 600 mA and my other lead plug is fused to 250 mA so obviously that wouldn't work.

So I put my leads on either side of the battery and there's a small spark. Did I break something?

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    \$\begingroup\$ Can you draw it please? Sounds that you have connected your MM in parallel to the battery, effectively shorting it through your device getting max current the battery can give. I guess it is not working now... \$\endgroup\$ – Eugene Sh. Jun 7 '17 at 18:10
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    \$\begingroup\$ Your just shorted the battery, which is not nice, but not a disaster. And probably burned the fuse. Again, not a disaster. \$\endgroup\$ – Gregory Kornblum Jun 7 '17 at 18:10
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    \$\begingroup\$ @EugeneSh, OP says it was connected to the 10A fused socket. \$\endgroup\$ – Chris M. Jun 7 '17 at 18:13
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    \$\begingroup\$ Oh. Right. Then just blown the fuse. \$\endgroup\$ – Eugene Sh. Jun 7 '17 at 18:13
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    \$\begingroup\$ Never connect a meter in current mode directly across a power source. You are basically shorting out the source through the fuse... \$\endgroup\$ – Trevor_G Jun 7 '17 at 18:20

You placed an ammeter in parallel to a voltage supply, which created a short circuit through the meter and most likely blew the 10A fuse in the ammeter socket.

What you want to do is put the meter between the charger and the battery, and then connect the other end of the battery to the charger. Diagram will help:


simulate this circuit – Schematic created using CircuitLab

Edit: If everything still works fine, it's likely you tripped some kind of over-current protection in the charger before the fuse in the meter blew.


You used an ammeter in parallel to a voltage supply, and not in series. This resulted in more current than the ammeter could handle. You have thus blown some fuse.


The principles behind your idea were OK. You wondered what the maximum current that the PSU could deliver. So you connected a very low resistance across it (the DMM) to draw the maximum load and to observe the measurement. It's the execution of it that's wrong, I'm afraid.

A simple linear supply, commonplace decades ago, could be shorted like you tried to. They consisted of a bridge rectifier and a smoothing capacitor. Very shortly after though, the high current would overheat the diodes and/or draw a relatively very high current from the supply and probably take out the supply fuse.

But a modern switch-mode PSU is more sophisticated than that and contains protection, such as output current limit and 'hiccup mode'. The latter will switch the PSU off if its output is low voltage while the current is high. It then waits a short while and switches on to try again. It repeats this cycle for as long as the output overload persists.

The practical way to determine your PSU's maximum output current is to connect your PSU output as (V+)->ammeter->Radj->(V-) where Radj is an adjustable load resistance i.e. a high-power rheostat. You also connect a voltmeter across (V+) and (V-). You set the rheostat to draw (say) 75 % max current, so 450 mA here, and switch the PSU on. You then decrease the rheostat resistance to draw increasing current and observe the output current as the output voltage starts to fall. You will be able to see when the voltage goes outside of its specified voltage range and therefore when the PSU can no longer provide the increase in load current.

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    \$\begingroup\$ The OP was testing a Li-Ion battery charger, and not a PSU. Also, the problem was really shorting the battery, which usually can supply many amps. The charger, if the least bit intelligent, should have some short-circuit protection. \$\endgroup\$ – Ronan Paixão Jun 8 '17 at 13:14

If the meter still works you didn't break anything.

A lithium ion charger has both a constant voltage mode and a constant current mode. So it acts sometimes as a voltage supply and sometimes as a current supply. It decides which mode to use at a given time by monitoring the battery voltage, current, or, in some cases, the temperature (battery has a sensor built in, which the Samsung phone that caught fire apparently did not).

In either mode, connecting the ammeter in parallel with the operating charger+battery circuit will pull current from both the battery and the charger (depending on, as one poster said, its output current limit). So if you get a reading, it doesn't tell you anything specific about either the battery or the charger.

If you put the meter in series with the battery, as Chris M. said, you see how much charging current is being supplied to the battery, which may tell you about the state of the battery (low current means it's mostly charged and being sustained, high current means it's mostly discharged) but not really anything about the charger's maximum current.


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