110/220V AC to 12V DC switching mode power supply (metal enclosed, input is 220V with separate L,N and ground.)

  • Measured voltage difference with a multimeter between V- and V+ 12V DC as expected
  • Measured voltage difference between ground and V- or ground and V+ reads practically 0V DC

According to this text about floating power supplies:

For floating DC power supplies, the voltage potential appears from the positive output terminal to the negative output terminal. There is no voltage potential (at least, none with any power behind it) from either the positive terminal to earth ground or from the negative output terminal to earth ground.

So that is in accordance with my measurement, but how can this be? If there is a voltage different between point A and B in any electrical network, how can the voltage C-A and C-B both be zero? The only explanation I can think of is a measuring error (multimeter ties the measured line to the ground?), but if so, the quoted text above is wrong?


3 Answers 3


Because the supply is floating, and you only made one measurement at once.

Floating means isolated, which means the resistance between the supply and ground is very very high. So high that the 10 Mohm input resistance of your multimeter is almost a short circuit in comparison.


simulate this circuit – Schematic created using CircuitLab

If you were to make both measurements at once, with identical meters, you would find V- at -6V and V+ at +6V with respect to ground, and 12V of course between V- and V+.

  • \$\begingroup\$ Thanks, nice schematic. So in essence I did tie the measured line to the ground? Similar to a pulldown transistor? \$\endgroup\$
    – Sebastian
    Commented Jan 7, 2015 at 12:58
  • \$\begingroup\$ Yes, more like a pulldown resistor than transistor. It's worth remembering that the act of measuring something always modifies it - in this context, by the fairly major addition of 10 Megohms or whatever your multimeter is. Also note Andy's comment re: AC measurements - you may get a surprise there. \$\endgroup\$
    – user16324
    Commented Jan 7, 2015 at 13:38

Did you measure using DC on the multimeter? You need to measure the AC voltage that might be present between output and ground. Then you could see several tens rising to over one hundred volts AC - this of course doesn't mean it's dangerous - it's just the EMI decoupling capacitors (if any) on the output.

For DC measurements your meter's input impedance will likely reduce the voltage to near zero.

  • \$\begingroup\$ Thanks for the extra info, on this specific power supply your suggested AC measurement yields 0V. \$\endgroup\$
    – Sebastian
    Commented Jan 7, 2015 at 16:41

The 12V power supply is galvanically isolated from the ground via a transformer so theoretically there's an infinite impedance between the primary (and ground of course) and the secondary. That's why you measure 0V. In reality there's always a leakage impedance (Several Mega ohms though)between both. If the input impedance of the multimeter is high enough (10MOhm DMM for example) you sometimes measure a certain AC voltage.


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