Question: Why would one use a diode bridging a resistor?

Thread Starter

mmoerdijk

Joined Dec 18, 2023
2
I just found this in a schematic i was going through. This is part of the circuit filters a tacho input signal and then passes it on to a micro controller.

The 74HC2G14 is an dual inverting smith trigger (https://www.mouser.com/datasheet/2/302/74HC_HCT2G14-109587.pdf).

I understand why you would use a smith trigger to clean up the signal. I assume the dual trigger use is to negate the inversion of the first trigger. I also see that there is a rc filter between the first and the second smith trigger. But I do not understand why there is a diode bridging R407 there.

Could some one explain this to me?


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Thread Starter

mmoerdijk

Joined Dec 18, 2023
2
How does that work?

When the input is high, the resistor is bypassed. So the cap loads up fast. Then when the input signal is low, the cap discharges through the resistor slowly..

Would this result in more of a sawtooth wave?

edit:
Oh right, now I get it, i forgot that it is inverting. So the input will be high most of the time and go low when there is a pulse. That pulse is now filtered through the resistor and cap. That way it takes more time for the signal to go low, resulting in a shorter pulse, which is then inverted by the second smith trigger. :)

Thanks!
 
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