JK flip flop question.

[SeanHatch]

We're analyzing JK flip flop circuits and I don't understand it.

So, Given that J = x, and K = xB' , my notes say that Q+ = JQ' + K'Q. I understand that. Then, substituting, we get xA' + x'A + xB.

I have no idea how that result was obtained. Can anyone help me with this?

 

[ljcox]

Have you searched this forum for JK Flip Flop? There have been several posts in the past.

I don't understand your Boolean expressions. A JK is very simple (and very useful).

J K Q+
0 0 Qo ie. no change
1 0 1
0 1 0
1 1 Qo' ie. it toggles

Where
Qo is the initial state of Q
Q+ is the state of Q after the clock pulse transition. (some trigger on the positive edge, others on the negative)

Qo' = Qo bar.

 

[mkh]

what is that A and B? i dont understand...

 

[Dean Huster]

"(some trigger on the positive edge, others on the negative)"

In TTL logic, all JK flip flops and ripple counters clock on the negative edge; all D flip flops and synchronous counters clock on the positive edge.

Dean

 

[jtallent]

JK flip flop analysis

"what is that A and B?"

The A's and B's usually refer to the output of the specific flip flop. In this analysis case, there are two JK flip flops with arbitrary names "A" and "B". The output's are also sometimes referred to as Q(A) and Q(B), where Q is the output and the respective subscript refers to the specific flip flop.

In the initial analysis question J=x and K=xB are the two excitation equations (the input equations) for the flip flop in question.

Q*=JQ' + K'Q is the characteristic equation for JK flip flops which describes how the flip flop arrives at its next state in algebraic terms. (Could be re-written as A*=JA' + K'A to clarify).

The initial post is missing the excitation equations for the second flip flop (assuming there are two since A and B appear in the equations). However, assuming the given excitation equations are for flip flop "A", we substitute J(A)=x and K(A)=xB into the characteristic equation which results in the next state equation for flip flop A:

A*= xA' + (XB)'A by DeMorgan's law, expand to get ...
A* = xA' + (x' + B')A expand ...
A* = xA' + x'A + B'A

It looks as if that answer: xA' + x'A + xB is wrong. I think that I am looking at this correctly, but without the schematic or more information, it's hard to know for sure if I'm missing something else.

-Justin