J K Flip Flopon 24/2/2012 & Updated on Thursday 17th of May 2018 at 04:03:18 PM
In order to have an insight over the working of JK flip-flop, it has to be realized interms of basic gates similar to that in Figure 2 which expresses a positive-edge triggered JK flip-flop using AND gates and NOR gates. Here it is seen that the output Q is logically anded with input K and the clock pulse (using AND gate 1, A1) while the output Q̅ is anded with the input J and the clock pulse (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 Q̅. Similarly NOR gate 2, N2 has its two inputs (X2 and Y2) as output of A2 and output Q (respectively).
Initially let J = K = 0, Q = 0 and Q̅ = 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 Q̅ = 1; while the output of N2 will become 1 as X2 = 0 and Q = 0. Thus one gets Q = 0 and Q̅ = 1. However if one considers the initial states to be J = K = 0, Q = 1 and Q̅ = 0, then X1 = X2 = 0 which results in Q = 1 and Q̅ = 0. This indicates that the state of flip-flop outputs Q and Q̅ remains unchanged for the case of J = K = 0.
Now assume that J = 0, K = 1, Q = 0 and Q̅ = 1. Analyzing on the same grounds, one gets X1 = X2 = 0 which further results in Q = 0 (and hence Q̅ = 1). For the same case if Q and Q̅ were 1 and 0, respectively, then X1 = 1 and X2 = 0 which would result in Q = 0 (and hence Q̅ = 1). This implies that if J = 0 and K = 1, then the flip-flop resets (Q = 0 and Q̅ = 1). Next if J = 1, K = 0, Q = 1 and Q̅ = 0, then X1 = X2 = 0 which results in Q = 1 (and thus Q̅ = 0). For the same case if Q = 0 and Q̅ = 1, then X1 = 0, X2 = 1 which leads to Q̅ = 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 Q̅ = 0.
Similarly for J = 1, K = 1, Q = 1 and Q̅ = 0 one gets X1 = 1, X2 = 0 and Q = 0 (and hence Q̅ = 1); and if Q changes to 0 and Q̅ to 1, then X1 = 0, X2 = 1 which forces Q̅ 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 Q̅ accordingly.
However it is to be noted that the state of the flip-flops remain unaltered if there is no rising-edge of the clock at its input. All these details can be summarized as in Table I. The wave forms pertaining to the same are presented by Figure 3. Moreover it is to be noted that the working of negative edge triggered flip-flop is similar to that of positive-edge triggered one except that the changes occur at the trailing edge of the clock pulse instead of its leading edge. From the truth table above one can arrive at the equation for the output of the J K flip-flop as (Table II) In addition to the basic input-output pins shown in Figure 1, J K flip-flop can also have special inputs like clear (CLR) and preset (PR) (Figure 4). These can be used to bring the flip-flop to a definite state from its current state. For example, the output can be made equal to 0 using CLR pin while it can set to 1 using PR pin. However these pins can be either active high (Figure 4a) or active low (Figure 4b) operated. The waveforms pertaining to positive-edge triggered JK flip-flop with active high preset and clear pins are shown in Figure 5. Moreover it is to be noted that these pins can be either synchronous or asynchronous in nature meaning which the clear and set operations occur either depending on the clock (shown by green lines) or no (shown by red lines), respectively. Further if the preset and clear pins are active low, then the changes observed in the diagram occur at the instant when clear and preset go low instead of high.
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