I've set up my PSU to use as a "lab power supply", I'm trying to obtain 12 Volts at 14 amps. I have the red, orange, yellow, black wires separated, and connected the green wire to ground to keep the PSU on.

That being said I feel that the problem is not necessarily with those first few steps. Every time I connect my multimeter between all the black and red, or black and yellow wires the entire PSU shuts off and won't restart until I touch the power wires to a grounding source (Some plates of metal I have).

I'm trying to figure out how I can test the connection with my multimeter without shutting off the entire PSU.

  • 1
    \$\begingroup\$ Is your multimeter set to measure current or a Tesla coil? \$\endgroup\$
    – Phil Frost
    Commented Mar 12, 2013 at 23:35
  • 1
    \$\begingroup\$ @PhilFrost I maybe misunderstanding your question, but I have the multimeter set to measure up to 10A DC. \$\endgroup\$
    – Mark Y.
    Commented Mar 12, 2013 at 23:37
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    \$\begingroup\$ Then you have set it to short circuit the power supply. Shutting down is the right thing for the PSU to do under those circumstances. \$\endgroup\$
    – user16324
    Commented Mar 12, 2013 at 23:39
  • \$\begingroup\$ @BrianDrummond I've connected it to a load without it shutting down. (Two stainless steel plates submersed in water, I'm just trying to get better output on a Hydrogen Fuel Cell). The PSU works fine in those circumstances, electrolysis takes place. \$\endgroup\$
    – Mark Y.
    Commented Mar 12, 2013 at 23:44
  • \$\begingroup\$ Here goes another question about a similar scenario electronics.stackexchange.com/q/26583/3552 \$\endgroup\$
    – sharptooth
    Commented Mar 13, 2013 at 9:40

1 Answer 1


A multimeter set to measure current (an ammeter) looks like a short circuit, or just a wire. When you connect your multimeter to the PSU, you are overloading it by presenting it with such a low resistance that it can not provide enough current to maintain its designed output voltage, so the protection circuits kick in and turn it off. Were it not for those protection circuits, a fuse in your meter would blow, or some wires would get hot and melt, or something similarly bad would happen.

More likely what you want to do is set your meter to measure voltage. In this mode, the meter looks like an open circuit, as if it's not there at all. Or, if you are interesting in measuring the current being supplied to a load, put the meter in series with the load, not in parallel with it.

  • \$\begingroup\$ Thanks that makes a lot of sense :P Is it the absence of a load or the absence of resistance that matters the most? Or are those acting the same? I would like to find the amperage being produced though, Voltage is measured correctly. \$\endgroup\$
    – Mark Y.
    Commented Mar 12, 2013 at 23:46
  • \$\begingroup\$ Thanks for pointing that out, the short circuit was definitely the problem. I haven't gotten full readings from the multimeter in series with the load, but the output from my fuel cell definitely indicates it is receiving the expected voltage and amperage. \$\endgroup\$
    – Mark Y.
    Commented Mar 13, 2013 at 0:02
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    \$\begingroup\$ General rule: to measure voltage, connect the meter in parallel with the power source. To measure current, connect in series with the power source. \$\endgroup\$
    – DoxyLover
    Commented Mar 13, 2013 at 0:51
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    \$\begingroup\$ When you say "I haven't gotten full readings from the multimeter in series with the load". If it's in amps or milli amps then that is the load your circuit is consuming. If you want to know what your PSU is capable of then you need a variable resistor rated 1 ohm (12V/14A = 0.9 ohms) and less and 200 Watts (12V x 14A). Once you exceed the 14 amps good supplies cutout to protect themselves and depending on the supply there may be a delay before it trips at 14 amps ( To allow for surges at power on and at ofther times). \$\endgroup\$
    – Spoon
    Commented Mar 13, 2013 at 8:50
  • \$\begingroup\$ @DoxyLover: He already has it connected in series, just there's no other components, so he gets a short. \$\endgroup\$
    – sharptooth
    Commented Mar 13, 2013 at 9:42

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