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I am using a dual power supply model CPX200 from TTI in my school lab, and I am trying use it to supply 1.8V and 3.3V to a PCB, so I figured it was a good idea to check whether the ground of the dual port is coupled.

However, when I use multimeter DC voltage mode to measure the voltage difference between two ground terminal, it shows that the voltage difference is 0.3V and it starts decreasing slowly. I am having difficulties understanding this phenomenon.

So is the ground of dual power supply coupled? Why would my measurement decrease slowly? Thanks a lot.

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The spec sheet for that supply says it has two isolated outputs. If you want to use them as two positive supplies, with the same ground, you MUST connect the two negative terminals together, and to the ground of your circuit.

Since the two supplies are isolated, they could also be used to provide positive and negative supplies (as often needed for audio amplifiers and other analog circuits) by connecting the negative terminal of one supply to the positive terminal of the other, and calling that connection "ground".

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That sounds like they are not connected, and what you are reading is slowly varying leakage into their capacitance.

You must connect the grounds together if you want to be sure they are both system ground.

I would fully expect that the two supplies are independent, for best flexibility. The fact that you measured any significant voltage difference says that they are not hard coupled together. But sometimes, grounds are connected with diodes so that they can float a little to cope with small differences, but are still safe and able to blow fuses if live shorts to ground.

As an experiement, you could connect the ground of one to the output of the other. With a current limited bench power supply like that, the worst that could happen if they are connected with diodes internally is that the current limit comes on. Then measure the voltages at all terminals.

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The CPX200 has two completely independent and isolated outputs. This means that the negative terminal of the output is not connected to ground. If you want it grounded, you must provide ground connection externally. Anyways it is very probable that there is a capacitor between the negative output and the ground, and this capacitor can be charged for different reasons, so when you measure the voltage between the terminals you are discharging this capacitor via the input resistance of your voltmeter. You can verify that the output are isolated with an ohmeter.

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