Question:

Show how D flip flop can be implement by J-K flip flop?

by Guest6717  |  earlier

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How D flip flop can be implement by J-K flip flop? Please share us about it. Thanks!

 Tags: Flip, Flop, implement, JK

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  1. James Augustus

    D flip-flop
    Now a day, the D flip-flop is the most common and widely in use flip-flop. It is mostly known as data or delay flip-flop because its output Q appearance likes a delay of input D.
    As they form the basis for shift registers these flip-flops are very useful. The D flip-flops are an essential part of various electronic devices. The signal on the D input pin is captured the moment the flip-flop is clocked, and subsequent changes on the D input will be ignored until the further clock event, which is the main advantage of the D flip-flop on the D-type "transparent latch". Some flip-flops have exception of "reset" signal input. It will reset Q to zero, and might be either synchronous or asynchronous with the clock.
    One bit position on each active transition of the clock above circuit shifts the contents of the register to the right. The input X is moved into the leftmost bit position.
    JK flip-flop
    The JK flip-flop augments the behavior of the SR flip-flop where J=Set, and K=Reset by interpreting the S = R = 1 condition as a "flip" or toggle command. The combination specifically, J = 1, K = 0 is a command to fix firmly the flip-flop; the combination J = 0, K = 1 is a command to set again the flip-flop; and the combination J = K = 1 is a command to toggle the flip-flop, such as change its output to the logical complement of its current value. Setting J = K = 0 does NOT result in a D flip-flop, but rather, will keep the current state. Simply fix K equal to the complement of J To synthesize a D flip-flop. The JK flip-flop is known as a universal flip-flop, because it can be configured to work as an SR flip-flop, a D flip-flop, or a T flip-flop.
     

  2. Guest4194600
    Cant you just put a NOT gate before the K pin, connect that to J and make that new input D? (ie D=0 then J=0,K=1 then Q = 0. D=1 then J=1, K=0 then Q=1. ie D=Q (after clock cycle)) change from a positive to negative edge by just applying a not gate before the clock (introduces minimal clock skew)

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