Substituting a 74HCT74 for a 4013

--- ___ _____ The HC74 needs a logic low asserted on the SET and RESET inputs in order to make them work, while the 4013 needs a logic high to do the same thing. Therefore, if you want to substitute an HC74 for a 4013, you'll need to invert the signals driving SET and RESET on the 4013.

I guess I wasn't clear, but that is what I am trying to do. After more checking, I think the inverter I have isn't working right. With no connections the inputs all read 1.6v and it takes a resistor no larger than 100 ohms to take pull it down below .1v. Having a resistor in there is obviously going to interferer with my capacitor charging. Shouldn't the capacitor pull an input down to 0 volts or is this thing defective? The markings on the chip say DM7404N.

You don't really need an inverter. The chip has both Q and /Q logic outputs for both flipflops. Thus, just connect the timeout circuit to /Q on the first flipflop and tie the unused sets and reset to 5V instead of ground. Everything else should be the same.

That worked, thanks John. I don't know why I didn't think of that, as an experienced computer programmer I don't know why it is taking so much to get my head wrapped around these logic circuits. I think I am just trying to make it more complex than it really is. Thanks again.

Being a programmer you're not accustomed to have the value of a variable and its inverted value available at the same time. So you implemented an "invert" operation. Always wondered why some people tend to split up the world in hardware and software with hardly any appreciation for each other. From the first time I touched a computer I've always been aware one needs both hard- and software to run that machine. You apparently experienced the same :)

petrus bitbyter