PWm driver circuit

[gspannelljr]

I have a problem I cant seem to solve. I have converted a computor power supply to run a pwm driver circuit I built to test my servos as I build my gauges. The circuit I used for the pwm driver is http://bellsouthpwp.net/b/o/bobpaige...Schematic.html
The problem is, when I connect both supplys, 5V and 12V, to the circuit, the power supply shuts down and goes into a protection mode. It seems the probem is the shared ground. If either supply is connected independently, then it stays on. I cant seem to figure out how to isolate the grounds from each other. In the back of my mind I am thinking I need a diode somewhere but my electronic building skills, which I haven't used since 1978, are a bit rusty. Any suggestions would be greatly appreciated.

[deering]

Hello GS.

There's only one supply in your diagram; i.e 9-12V. The 0-5VDC is analog input of the desired rotation. The supply and input signal must have a common ground.

Jim.

[RadarBob]

...So if you use the 12V from your computer supply to provide the supply rail, you could just connect a pot between 5V regulator output and ground and use the wiper to provide the analogue input.


Cheers,

Rob

[gspannelljr]

ohhhhhhh. that makes more sense. I did kind of wonder why I would need a 5V regulator and then add a 5V supply. This is why I love you guys. YOu keep me from "Letting the smoke out" of my circuits. You do know they run on smoke, right? If you let the smoke out, they stop working. The bigger the circuit, the more smoke it needs.NEVER let the smoke out!

[RadarBob]

Actually, thinking about it, you should be able to do without the 7805 regulator as you have +5/+12V available on your computer supply.

As far as I am aware, all of the outputs on a computer supply are with respect to a common ground (the black wires).

There is lots of info on the net with regard to pinouts of computer PSU's - But for ATX +12V is on the Yellow wire, and +5V on the larger Red.

So you could ditch the 7805 and run the LM324 off the +12V output
and the rest of the circuit from the +5V output.

Rob