Skip to main content

SR to JK flip-flop conversion

Short answer

Combinational logic drives S and R of an SR flip-flop from J, K, Qn. Excitation table yields S = J·Qn', R = K·Qn. This prevents SR=11 and implements JK behavior: set when J=1, reset when K=1, hold when both 0, toggle when both 1.

With the help of RS flip-flop, realize JK flip-flop using excitation table and required expressions.

To make a JK flip-flop from an SR flip-flop, the JK inputs and the present state QnQ_n are fed to a small combinational circuit whose outputs drive S and R. The circuit is found from the JK characteristic table and the SR excitation table.

SR excitation table

QnQ_nQn+1Q_{n+1}SR
000X
0110
1001
11X0

Conversion table

JKQnQ_nQn+1Q_{n+1}SR
00000X
0011X0
01000X
011001
100110
1011X0
110110
111001

K-maps

 S:        KQn=00 01 11 10       R:      KQn=00 01 11 10
   J=0        0   X  0  0          J=0      X   0  1  X
   J=1        1   X  0  1          J=1      0   0  1  0
  • S: the two 1s at (J=1J=1, Qn=0Q_n=0), cells 100 and 110, form a pair giving S=JQn′S = JQ_n' (the X cells are not needed).
  • R: the 1s at (K=1K=1, Qn=1Q_n=1), cells 011 and 111, form a pair giving R=KQnR = KQ_n.
S=J Qn′,R=K QnS = J\,Q_n', \qquad R = K\,Q_n

Circuit

J ---[AND]----- S +---------+
Q'---[   ]        |  SR FF  |--+-- Q
                  |   CLK   |  |
K ---[AND]----- R |         |--|-+-- Q'
Q ---[   ]        +---------+  | |
  (Q and Q' fed back to the AND gates)

Check: J=K=1J = K = 1: if Qn=0Q_n = 0, S=1,R=0S = 1, R = 0 so QQ sets to 1; if Qn=1Q_n = 1, S=0,R=1S = 0, R = 1 so QQ resets. The output toggles, and the forbidden S=R=1S = R = 1 never occurs, because SS and RR need opposite values of QnQ_n.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Digital Logic chapter 6 questions →