What does this do to the signal ?

Thread Starter

simmsdk

Joined Jul 16, 2011
2
Please look at the attached image for the schematic...

My question is:

What exactly does the schottky diode in parallel with the resister, and the capacitor do to the signal before it enters the last NAND.

What is it good for ?
 

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praondevou

Joined Jul 9, 2011
2,942
When pin 3 of IC1A goes HIGH the capacitor C1 is being charged rapidly through diode D1.
When pin 3 of IC1A goes LOW the capacitor C1 is being slowly (more slowly than it was charged)
discharged through R1.

You create a delay on only one edge of the signal.

If you join the inputs of the left gate and feed a digital signal to it, then the rising edge of the
output signal IC1B pin 4 will be delayed, the falling edge remains almost unchanged.
 

hgmjr

Joined Jan 28, 2005
9,029
A better description of the effect of this circuit would be that the original pulse would be stretched so that the output is wider that the input.

hgmjr
 
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