Third try? (without the schematic)

[Dave]

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,



Dave





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

 

[Dan Coby]

"Dave" <db5151@hotmail.com> wrote in message
news:8fadnaJvc_z9AGLanZ2dnUVZ_saknZ2d@internetamerica...

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.

snip

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

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.

 

[Nigel Heather]

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.

Cheers,

Nigel

 

[Dave]

"Dan Coby" <adcoby@earthlink.net> wrote in message
news:BdWdnekfPsXxNGLanZ2dnUVZ_gWdnZ2d@earthlink.com...

"Dave" <db5151@hotmail.com> wrote in message
news:8fadnaJvc_z9AGLanZ2dnUVZ_saknZ2d@internetamerica...
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.

snip

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

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.

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. <shaking head>

Dave