Power Supply rail shorts to ground when loaded

Question

Just to preface: this is probably a really dumb question.

I recently purchased a desktop power supply to work on circuit projects at home (I'm an hobbyist). I recently had issues with an op amp circuit. While troubleshooting, I figured out that the positive rail of my power supply was shorting to ground when I tried to build a voltage divider. For example: If my rails were -10 to 10 and I wanted to reduce the positive voltage to 5 volts, put 2 1K resistors in series to ground and use the voltage from between the two resistors. But instead of my 5 volts I just got earth. Please reference the schematic below.

So I started doing some research and I understand that the 'ground' terminal on power supplies sets the zero voltage reference for the two rails. My understanding is that if you connect the ground terminal and one of the other terminals you can change the voltage reference. For example, if my setting is constant voltage for 20 V and I connect the negative terminal and ground, the positive terminal will read 20V relative to ground.

That all makes sense to me. What I don't understand is when I connect a resistance across the ground and another terminal the voltage reference still changes. I know that the ground terminal is connected to earth through the bottom prong and splits the rails symmetrically when I don't connect a resistance across to ground.

I'm pretty sure this is how the power supply is supposed to work and I'm just missing something fundamental. Is it possible to build a voltage divider the way I described above? If not, how would I approach this problem?

Answer 1

By default the outputs of your supply are floating. That is they have no electrical connection to earth ground. It's like a big battery connected to nothing.

^{simulate this circuit – Schematic created using CircuitLab}

The green ground connector goes to the earth ground. If you connect either side of the battery to ground it references that side as ground and the other connection becomes either a positive voltage relative to ground or a negative voltage if you attach the red side to green.

However, if everything is wired correctly there should be no current flowing through that shunt to ground. As such, it does not really matter if it is shunted with a wire or with a resistor since without current there will be no voltage drop across the resistor.

---

As for your voltage divider idea. It is not at all clear what you have tried to do here.

I suspect you are trying to do something like this...

^{simulate this circuit}

Unfortunately that does not work since the load changes the bottom resistance of the divider and thus the voltage. But you may well be doing something else... add a schematic to the question.

Answer 2

Assuming your provided schematic is the correct description of the device you have created, you need to set resistor values that will allow R_LOAD to have a 5v difference across its terminals.

Your math indicates that would be R1 = R2 This is incorrect as R_LOAD is introduced.

I would do this instead, Remove R2 altogether and match R_LOAD with R1.

If R_LOAD is 10kohm, R1 is 10kohm. This will allow R_LOAD to have a voltage of 5v.

Answer 3

I have met the similar situation like him. But I used op amp for my virtual ground as well. At the beginning, I refer this schematic and conduct on the PCB and It works very well. But, when I add more LOAD at one side of the double rail, my dual source does not symmetric as well.

I am thinking that my virtual ground with op amp is based on the impedance of LOAD. It means that I need to put the similar load between my two rails and the source will also the sample +-Vdd/2.

Is it my thinking true? And below is my schematic