Floating power supply: voltage difference fixed but separate voltages not?
If your supply is truly floating output then the act of connecting the meter will ground that terminal through the (high) internal resistance of the meter.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. (a) Your first test. (b) Your second test. (c) The test you didn't do.

If you try a second meter simultaneously with the first - or a resistor with the same impedance as your meter input (10 MΩ is typical for DMMs) - you should get close to the readings of Figure 1c.


  That's confusing, because \$\Delta v = v_+ - v_- =(v_+ - v_{GND}) - (v_- - v_{GND}) = 0V \ne 5V\$.


The fallacy with your maths is that the two conditions \$v_+ - v_{GND} \$ and \$ v_- - v_{GND} \$ were not measured simultaneously. The act of measuring one altered the other and vice versa. If you measure both simultaneously the maths will work out. In Figure 1c this would be \$\Delta v =(v_+ - v_{GND}) - (v_- - v_{GND}) = 2.5 - (-2.5) = 5V\$.


  So the + and - terminals of the supply are constantly floating and not fixed, but their difference is fixed?


Correct.


  Also, is there any benefit (e.g. protection from leakage current) to short the GND and - terminal and get my voltage off the + and GND connections rather than +/-? I'm concerned about damaging my device.


No, but it's not recommended. The return wire should go to the V- terminal as that's where it ultimately needs to return to. The problem arises not because of any real electrical reason but because you may cause confusion when you or another has to maintain it later. In addition, you have to consider what will happen if the V- to GND link becomes disconnected.