The Analysis of A JK-Flip-Flop With Deeds

Abstract

In order to have an insight over the working of JK flip-flop, it has to be realized in terms of basic gates similar to that in Diagram 2 which expresses a positive-edge triggered JK flip-flop using AND gates and NOR gates. Here, it can be observed that the output Q and the clock pulse are logically anded using the AND gate 1, A1, whereas the output   Q?  is anded using the clock pulse and the input J. (using AND gate 2, A2). Further the output of A1 is fed as one of the inputs (X1) to the NOR gate 1, N1 whose other input (Y1) is connected to output Q?. Similarly NOR gate 2, N2 has its two inputs (X2 and Y2) as the output of A2 and output Q (respectively). Initially let J = K = 0, Q = 0 and Q? = 1. Now consider the appearance of positive-edge of the first clock pulse at the CLK pin of the flip-flop. This results in X1 = 0 and X2 = 0. Then the output of N1 will become 0 as X1 = 0 and Q? = 1; while the output of N2 will become 1 as X2 = 0 and Q = 0. Thus one gets Q = 0 and Q? = 1. However if one considers the initial states to be J = K = 0, Q = 1 and Q? = 0, then X1 = X2 = 0 which results in Q = 1 and Q? = 0. This indicates that the state of flip-flop outputs Q and Q? remains unchanged for the case of J = K = 0.

Now assume that J = 0, K = 1, Q = 0 and Q? = 1. Analyzing on the same grounds, one gets X1 = X2 = 0 which further results in Q = 0 (and hence Q? = 1). For the same case if Q and Q? were 1 and 0, respectively, then X1 = 1 and X2 = 0 which would result in Q = 0 (and hence Q? = 1).

This implies that if J = 0 and K = 1, then the flip-flop resets (Q = 0 and Q? = 1).
Next if J = 1, K = 0, Q = 1 and Q? = 0, then X1 = X2 = 0 which results in Q = 1 (and thus Q? = 0). For the same case if Q = 0 and Q? = 1, then X1 = 0, X2 = 1 which leads to Q? = 0 and hence Q is forced to value 1. This means that for the case of J = 1 and K = 0, flip-flop output will always be set i.e. Q = 1 and Q? = 0.

Similarly for J = 1, K = 1, Q = 1 and Q? = 0 one gets X1 = 1, X2 = 0 and Q = 0 (and hence Q? = 1); and if Q changes to 0 and Q? to 1, then X1 = 0, X2 = 1 which forces Q? to 0 and hence Q to 1. This indicates that for J = K = 1, flip-flop outputs toggle meaning which Q changes from 0 to 1 or from 1 to 0, and these changes are reflected at the output pin Q? accordingly.