74HC14 issues please HELP!

[SAWULA]

I have two built circuits and I am trying to integrate them. In one ciruit there is 74HC14 IC and there is a 2.2k resister between two gates (pin2 and pin 3) and in other circuit they are directly connected (pin1 and 2). I want to know why is 2.2k resister used? I know the purpose is to invert the output of first gate but why we can't directly connect those pins in both cases?

 

[FvM]

in other circuit they are directly connected (pin1 and 2)

You won't want to directly connect pin1 and 2 of HC14... (unless designing an unreliable oscillator)

 

[alexan_e]

You won't want to directly connect pin1 and 2 of HC14... (unless designing an unreliable oscillator)

Actually he said pins 2 and 3 (output to input)

I don't know the exact circuit but you can connect the gates directly without a problem, the resistor doesn't change the state between the gates.

Alex

 

[joelalfuerto]

nice... It helps to my assign..

---------- Post added at 16:13 ---------- Previous post was at 16:09 ----------

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[SAWULA]

Sorry for the mistake. I wanted to say pin2 and pin3. The two circuits are x10 transmitter and a x10 receiver. 74HC14 is used to detect the zero crossings of the AC power lines. (x10 is a protocol that is used to communicate over power lines) And I want to combine both transmitter and the receiver together and create a transceicver. In the transmitter ciruit pin2 and pin3 of 74HC14 is directly connected and in the receiver circuit they are connected through a 2.2k resister. I want know the difference inbetween so that I can use either directly connection or the connection through 2.2k resister for my combined transceicver circuit.

 

[betwixt]

The only possible reason for using the resistor is to introduce a tiny delay in the signal but it would be so insignificant at power line frequency as to be useless. In some designs a resistor is used to isolate a driving signal from the pin it drives so test signals can be injected or the logic gate can be tested in-circuit.

Your schematic seems to be missing a pull-up resistor on pin 1 and a ground on the opto-couplers. Wouldn't it be a better and cheaper solution to use a bridge rectifier and one opto?

Brian.