Hello,

sorry I do not understand ops:

 

It's fairly simple, you could just make a mirror image of the previous circuit I posted, but to turn it ON would require a -5V signal. The output of the PIC can be either 0V (LOW) or 5V (HIGH), both of these would turn a negative version OFF - you need to find a way of generating a 0V to -5V to switch a circuit in this way.

There are probably many ways of doing it (but not as simple as the positive switch), I suspect a number of people may post suggestions, but one fairly simple way would be to use an opto-isolator.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

OK I understand now.
I'm building the whole schematic of the control board and I will post it later on :wink:

 

Here is the schematic of the controller. 8)

There are the IR , the DSA (drive control) , the LCD as well as the button (A/D) interfaces.

There is as well the DB9F connector in order to program the pic "in situ".

You will notice that there is not the "switch" interface we are talking about, I'm still trying to inderstand and find the expected transistors.

Feel free to comment this board.

Regards

Attached Images

schemacontroller.jpg (100.9 KB, 171 views)

 

do you think I can use two diodes to provide the negative supply,
something like this ?

Attached Images

switch_780.jpg (6.0 KB, 157 views)

 

And how did you think that would work?

It will just short the pic's output to ground trough the first diode, and the second will block the 5V...