Dumb Question on Low Input Table Switching Circuit
 

OK, :noob: question time. I'm building the following on a daughter board:

http://i1114.photobucket.com/albums/...l/lowinput.jpg

The diode going to the switch is self-explanatory. What is the other diode doing, and why does is specify 5VDC from the TPS? I'm confused.

Wow!! Just realized that I've graduated to "Junior Member." WooHooo.

It could go to the 5V in the proto area. Same thing.

Thanks! That's what I thought, but wanted to confirm. Do you know what it does? It would seem to ensure that JS9 is always drained to at least 4.4VDC (assuming 0.6VDC diode drop). Is that the function?

Correction . . . at most 4.4VDC. The switch to ground through the other diode would drain to 0.6VDC.

oops wrong thread.

You have the diode drop backwards, but yes, it's there to clamp the voltage.

Duh!

OK. JS9 is 5.6VDC if switch is open and 0.6VDC if switch is closed. Thanks.

Actually, it's normally a little below 5 volts - the diode makes sure it won't go over 5.6.