Third try? (without the schematic)

I am trying to use the circuit diagramed in the associated schematic as the source of power in a project I am building, but am apparently missing something obvious. It provides +5V when the switch is depressed, but the -5V is always on regardless of the switch position. I have checked and rechecked, and don?t believe I have anything hooked up incorrectly. Have also replaced the 7905, without effect. If no one can find anything wrong with the provided schematic, I am back to square one, and will simply disassemble and reassemble from scratch to see if that provides any results. The basic idea for this schematic came from the 7905 datasheet.

Many thanks,


PS: schematic posted to ABSE under title "+/-5V schematic".

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source of power in a

provides +5V when the

position. I have checked

also replaced the

schematic, I am back to

that provides any


Your switch simply disconnects to the common point between the two batteries from ground. There are still plenty of alternate current paths between the positive side of B1 and the negative side of B2.

You need to use a double pole switch to disconnect both B1 and B2 from the inputs to the voltage regulators. Leave the common point B1 and B2 connected to ground.

Reply to
Dan Coby

Agreed, with the switch open you effectively have an 18V supply with a floating ground. The charged capacitors will tend to put the ground around

9V so the 7905 will look like it has -9V supply and the 7805 a +9v supply. Not sure why the 7805 is working with the switch open - perhaps the virtual ground is sitting higher than 9v so the supply on the 7805 is insufficient but that on the 7905 is fine.

As previous poster said you want a dual pole switch to take out both the batteries.



Reply to
Nigel Heather

Thank you, Dan. Now that you point it out I can see it. And I had it as you describe once upon a time, not even sure why I changed it anymore. Quick to fix. :) I'm still in the breadboard phase...

Much appreciated.


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