Naked Science Forum

Non Life Sciences => Geek Speak => Topic started by: Geezer on 25/08/2009 22:52:44

Title: What does this circuit do?
Post by: Geezer on 25/08/2009 22:52:44

[diagram=501_0]


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

Title: What does this circuit do?
Post by: RD on 26/08/2009 03:41:57

Looks like an inverter circuit without the resistors drawn in

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http://www.play-hookey.com/analog/experiments/basic_op_amp_inverter.html

Title: What does this circuit do?
Post by: Geezer on 26/08/2009 04:06:48

Nope. I should have mentioned  [:D]  It's a digital non-inverting buffer.

Title: What does this circuit do?
Post by: syhprum 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.

Title: What does this circuit do?
Post by: RD on 26/08/2009 06:22:43

De-bounce ? (http://en.wikipedia.org/wiki/Switch#Contact_bounce)

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http://books.google.co.uk/books?id=TU8h7whAKOYC&pg=PT181&lpg#v=onepage&q=&f=false

Title: What does this circuit do?
Post by: Geezer 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!