Stereo jack power switch AND Bipolar Power Supply

[Razoumihine]

Hello,

On stompboxes, when one plug the input jack, this switches on the power supply. An example in a circuit is here:
http://www.electrosmash.com/images/tech/microamp/mxr-microamp-schematic.png
Principle is that when pluging a Jack in the guitar input, the sleeve closes the power circuit to earth, switching on the power. In this case, the signal earth (jack sleeve) is common to power supply negative.

Now, If we change the above circuit with a bipolar power supply like this one: http://sessionville.com/assets/images/articles/bipolar-power-supply.jpg, the power supply negative is no more earth, it is -(V/2). The  sleeve of the guitar input Jack cannot be connected to -(V/2), because it is already connected to the floating ground of this bipolar supply.

So, in case of such bipolar power supply, how to use the jack as a power supply switch?

Lionel

[Kipper4]

Follow the same principle as the first schematic and use the ground for switching not v/2.
When the input has no plug in it the ground is cut off therefore no power.

[Seljer]

Well heres the concept from here http://www.geofex.com/FX_images/PNP_power_switching.pdf expanded so that switched positive side with the PNP Q1 then also switches a seperate negative side through a NPN Q2 (and Q3 and Q4 in between doing some other stuff.... it may be possible with a transistor less but it's 1am right now and I can't exactly figure it out). The base current required for Q1 and Q2 to turn on and saturate add a couple of mA to the pedal's overall consumption so thats a drawback if you're using it with batteries. The quiescent current when turned off is still a couple of 10uA so enough to empty up a 9V battery in a year or so. It also might be prone to breaking the balance of supply with a 'virtual' ground (in reference to the supply)...so only suitable for a proper bipolar supply (i.e. you already have a ground on the power supply side like the linked example with two 9v batteries)

Follow the same principle as the first schematic and use the ground for switching not v/2.
When the input has no plug in it the ground is cut off therefore no power.

Not really, in the the linked example with the battery and the 2x 10kohm voltage divider grounded in the middle even if you broke the ground connection, you'd still be left with 9V flowing over 20kilohms + the rest of the circuit.

[electrosonic]

Maybe this from the Geo transformer splitter.

Andrew

[Seljer]

Hah, I knew I had already seen it on Geofex! That setup of transistors makes much more sense than what I was trying to do 

[Razoumihine]

Ok very good, especially the last schematic, with only one transistor in between! Works perfectly!
Thank you
Lionel

[Razoumihine]

Also I found that another option is possible and very simple, but uses an IC (MAX1044) :
http://gaussmarkov.net/wordpress/circuits/bipolar-9v-power-supply/