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Author Topic: What does this circuit do?  (Read 2903 times)

Offline Geezer

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What does this circuit do?
« on: 25/08/2009 22:52:44 »



If the input and the output of the buffer are shorted together, it can't be doing much - right?



 

Offline RD

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What does this circuit do?
« Reply #1 on: 26/08/2009 03:41:57 »
Looks like an inverter circuit without the resistors drawn in




http://www.play-hookey.com/analog/experiments/basic_op_amp_inverter.html
« Last Edit: 26/08/2009 03:44:50 by RD »
 

Offline Geezer

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What does this circuit do?
« Reply #2 on: 26/08/2009 04:06:48 »
Nope. I should have mentioned  :D  It's a digital non-inverting buffer.
 

Offline syhprum

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What does this circuit do?
« Reply #3 on: 26/08/2009 06:18:10 »
The output goes directly to the switch contact hence the output voltage must follow the voltage there, as the chip is non inverting its output will follow its input after a few nano seconds so there will be a current forced into the chip until it settles down but this will do no damage.
 

Offline RD

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« Last Edit: 26/08/2009 06:40:14 by RD »
 

Offline Geezer

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What does this circuit do?
« Reply #5 on: 26/08/2009 06:41:33 »
RD is correct. Yes, it's a de-bounce circuit. Well done!

The buffer acts as a memory while the switch is open circuit. The feedback loop is rather "extreme", most people would prfer to see a resistor there, but it's not really necessary for the reasons that Syhprum points out. Further, board tester circuits frequently drive the outputs of gates against their will. It's not a problem as long as you only do it for short intervals.

The set and reset from the switch do their job with extreme prejudice, so they are bound to overcome the extreme feedback path. BTW, the feedback path is deliberately drawn to not look like a feedback path!
 

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What does this circuit do?
« Reply #5 on: 26/08/2009 06:41:33 »

 

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