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Dumb Question on Low Input Table Switching Circuit

Old Feb 27, 2011 | 07:07 PM
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Default Dumb Question on Low Input Table Switching Circuit

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



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.
Old Feb 27, 2011 | 07:08 PM
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Wow!! Just realized that I've graduated to "Junior Member." WooHooo.
Old Feb 27, 2011 | 07:09 PM
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It could go to the 5V in the proto area. Same thing.
Old Feb 27, 2011 | 08:33 PM
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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?
Old Feb 27, 2011 | 08:34 PM
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Correction . . . at most 4.4VDC. The switch to ground through the other diode would drain to 0.6VDC.
Old Feb 28, 2011 | 03:07 AM
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oops wrong thread.

Last edited by WestfieldMX5; Feb 28, 2011 at 03:17 AM.
Old Feb 28, 2011 | 10:31 AM
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You have the diode drop backwards, but yes, it's there to clamp the voltage.
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Old Feb 28, 2011 | 10:52 AM
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Originally Posted by Matt Cramer
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.
Old Mar 1, 2011 | 10:37 AM
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Actually, it's normally a little below 5 volts - the diode makes sure it won't go over 5.6.
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