Hello from a newbie, both to the forum and (relatively) to electronics.

I'm trying to find a method of replacing a mechanical on-off switch with an electronic version. A much simplified version of my circuit is:

A connects to B via PL1/J1 and passes low frequency (possibly down to 0.01Hz) signals by taking B's input low via the diode.

My aim is to replace S1 with an electronic equivalent that switches on when PL1 is plugged into J1 and off when the plug is removed. Mechanical switches in the plug/socket are not acceptable!

The only requirements are that in the off state virtually no current should be drawn from the battery and in the on state, since A itself only draws about 6-7ma, any additional load required by the extra circuitry should ideally be 1ma or less. Obviously this is to conserve battery life.

Am I seeking the impossible?

Regards

richard.

 

Regulator with on/off control

http://rocky.digikey.com/WebLib/Sharp/Web Data/PQ05DZ51_11 Series, PQ3DZ53_13.pdf

You gets to figure out how to power it on/of. If you need more power just search digikey for a regulator with an on/off control.

Or you could insert a transistor in the ground of the regulator and use it as an on/off switch.

For either you'd still need a circuit to pull power from the battery and some means of controlling the transistor switching action. Without a mechanical switch that's going to be fun.

Hrm....

Ah, use a stereo jack rather than a mono one and use one of the 'unused' pair as your switch. Just put a low signal on it to run either the transistor or the on/off on the IC.

 

Regulator with on/off control

http://rocky.digikey.com/WebLib/Shar...PQ3DZ53_13.pdf

That looks useful!

For either you'd still need a circuit to pull power from the battery and some means of controlling the transistor switching action. Without a mechanical switch that's going to be fun.

Yup, that's the problem! As far as I can see, with the circuit as-is and the diode pulled low, there would not be enough voltage at point 'P' to switch a transistor on.

Unless... perhaps I can somehow raise the ground of B above that of A thus raising P (relative to B ground) above the 0.7v forward drop of the diode? Dunno, could be talking beginners rubbish.

Ah, use a stereo jack rather than a mono one and use one of the 'unused' pair as your switch.

Sorry can't to that. The actual connector is a RJ45 with all signals used, I just didn't have that symbol available!

 

what's the duty cycle of the signal? Are the frequency and duty cycle fairly constant?

 

No, the frequency /duty cycle are essentially random.