Probably should know this but.... When a Cd4013 flip-flop switches from 'Q' to 'not Q' does the 'Q' pin just go to a low(0V) or does it go to an open? By open I mean no signal, neither high or low.

If it changes to a 'low'(0V) can it be fed into an inverter input to make it a 'high' to control another part of a circuit?

Thanks for help again

 

CMOS = complementary metal-oxide semiconductor.

Complementary means the output is implemented with a pair of gates that complement each other.

After transitioning to low, they slap each other on the back saying "Great job!" (Oops, sorry, that's "compliment".)

When the gate pulling high turns off, its complement, a ground-connected gate, turns on to pull low.

That is the answer to your question - it pulls low.

 

Your second question is the easiest; yes, it can be fed to an inverter.

'Q' and 'not Q' are the inverse of each other. When 'Q' transitions low, "not Q' transitions to a high signal. Inversely, when 'Q' transitions from a low to high signal, 'not Q' changes from a high signal to a low signal (~0V). I don't believe there is a case where it will be neither high nor low.

 

I can, right after the blue smoke escapes.

It is a pretty standard definition of flip flops, Q and Q' are inverted.

 

CMOS = complementary metal-oxide semiconductor.

Complementary means the output is implemented with a pair of gates that complement each other.

After transitioning to low, they slap each other on the back saying "Great job!" (Oops, sorry, that's "compliment".)

When the gate pulling high turns off, its complement, a ground-connected gate, turns on to pull low.

That is the answer to your question - it pulls low.

Substitute "transistor" for "gate" and this will be accurate.
CMOS generally applies to the techniques used in gate (NAND, NOR, etc.) design, and not just to the output stages.

 

Thank you to all that answered! Thats what I thought happened with the gates but wasn't sure. Now on to the rest of my circuit!!