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Old 02-27-2011, 08:07 PM   #1
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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.
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Old 02-27-2011, 08:08 PM   #2
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Wow!! Just realized that I've graduated to "Junior Member." WooHooo.
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Old 02-27-2011, 08:09 PM   #3
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It could go to the 5V in the proto area. Same thing.
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Old 02-27-2011, 09:33 PM   #4
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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?
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Old 02-27-2011, 09:34 PM   #5
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Correction . . . at most 4.4VDC. The switch to ground through the other diode would drain to 0.6VDC.
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Old 02-28-2011, 04:07 AM   #6
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oops wrong thread.

Last edited by WestfieldMX5; 02-28-2011 at 04:17 AM.
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Old 02-28-2011, 11:31 AM   #7
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You have the diode drop backwards, but yes, it's there to clamp the voltage.
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Old 02-28-2011, 11:52 AM   #8
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Quote:
Originally Posted by Matt Cramer View Post
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.
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Old 03-01-2011, 11:37 AM   #9
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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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