I recently "inherited" tech support for a CCTV system. A camera went out recently, and water had damaged the power plug on the Siamese cable. I spliced on a new end and I was getting approx 12V reading, but the camera wouldn't work. After some troubleshooting I determined that it would work only if I plugged it into a different DC line from the power distribution box. Upon opening the box, I discovered that the positive terminal on that particular DC out path was burnt and blackened. Yet it was still supplying power?
The real question is this: There are two positive (red) and two negative (black) lines coming from the power supply that is plugged in to AC. The two positive lines are screwed into the +VIN terminal on the top left of the board. The two negative lines not attached to the same terminal. One negative line is attached to the -V terminal on the bottom left of the board, while the other negative line from the power supply is attached to the negative terminal of the 9th DC out path. Is this right?
I was planning on just moving the wires from the fried DC out path 2 to the open 9th path, but with the negative line coming from the power supply, I'm not sure. I just don't know enough about this stuff. One of my colleagues is an electrician and he told me that the negative terminals are bussed so I can just move the wire currently connected to the 2nd positive terminal over to the 9th positive terminal and everything will work. It's not that I doubt, but I would like a better understanding of what is going on.
Why is it wired like this? I would assume that both negative lines from the power supply should be hooked to the same terminal. But it has been functioning for several years as it is currently wired, so I must be wrong.
So just move the positive DC out wire from terminal 2 over to terminal 9 and that's it?
I will be posting a picture shortly.