OK. I'm pretty sure my inexperience is going to show here. I had the task of finding out how much current a microprocessor was drawing. I was going to measure, with a multimeter, between the point after the power (using a AC to DC 5V) and ground. I didn't break the circuit and the microprocessor started smoking at the red probe. Damn.. was I shorting out the point after the power when I did this? But since it is current and not voltage, it shouldn't be. Does anyone have an explanation for why the microprocessor started smoking? Sorry, this description may be vague.

Back to my original task, I used a power supply to measure how much current the microprocessor was drawing. Is there any alternative way to measure the current without a power supply?

Thanks.... OTL

  • 1
    \$\begingroup\$ It's not totally clear what you did. A block diagram of how your uC is powered, and where you contacted it for the measurement would help. Also, to be clear, did you set the multimeter for voltage or current measurement? And for current measurement, did you rearrange the probes into the correct ports of the meter? \$\endgroup\$
    – The Photon
    Jan 4, 2013 at 2:12
  • \$\begingroup\$ I'm pretty sure I used the current measurement on the multimeter with the correct ports. It was just my method of measurement was incorrect. I don't have a block diagram at the moment, but the answers below do depict what I should have done.. \$\endgroup\$
    – O_O
    Jan 4, 2013 at 9:00
  • \$\begingroup\$ Most adjustable power supplies display current and voltage. Also, they usually have adjustable current limiting which really helps keep the smoke in. I have two of these: cheap but sufficient for most needs: mpja.com/0-18VDC-0-3-A-Variable-Benchtop-Power-Supply/… \$\endgroup\$
    – Phil Frost
    Oct 16, 2015 at 15:04

2 Answers 2


To measure current with a basic multimeter, you have to break the circuit and insert the ammeter in series. That, is you must a series connection the 5V -> circuit board -> ammeter -> ground. If you used the ammeter in parallel, which your description suggests, then low resistance ammeter would draw as much current as the power supply can provide. This certainly could overheat the probe, although ammeters usually contain a fuse that should prevent this.

It's also possible you accidentally shorted power to ground with the probe somehow, or if a second power supply is involved (quite likely for microprocessors), that you created a connection between the power supplies.

By contrast, voltage is measured in parallel, and voltmeters are high resistance so they do not (usually) interfere with a circuit's operation.

It is often easier to place a "shunt" resistor in series with your load and measure the voltage as an indirect way of measuring current, because you do with this without breaking a circuit on a PCB.


Measuring current is a little different to measuring voltage. This is why you (usually) have separate sockets on your multimeter for current/voltage.

Inside the multimeter there is a series "current shunt" resistor, of a very low value (e.g. 100mΩ) This is because the shunt resistor itself limits current and affects the reading, so it needs to be as low as possible.
This is pretty much the opposite scenario to the voltage range where you have a parallel resistor for a very high value (again to minimise errors caused by it drawing current from the source)

Anyway, the short story is never put a multimeter on current range across (i.e. in parallel) a voltage source (the more powerful the source, the more serious this could be - especially with a cheap multimeter not designed to handle severe overloads - they can explode)

What you need to do is wire in in series as jbarlow says in his answer. So you would do something like:

  • Meter on DC amps range and probes plugged into correct sockets
  • Red lead to positive supply
  • Black lead to uC power pin (not ground)

Now the meter is in series between the supply and power pin, and any current will pass through the aforementioned shunt resistor. The meter measures the voltage drop across this (again, it should be very small to avoid affecting the circuit too much) and calculates the current according to Ohm's law.


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