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Negative edge JK flip-flop; T to JK conversion

Short answer

A negative edge-triggered JK flip-flop changes state only at the falling clock edge. Truth table: J=0,K=0 holds; J=0,K=1 resets; J=1,K=0 sets; J=1,K=1 toggles. To convert a T flip-flop into a JK flip-flop, derive T = J Q_n' + K Q_n using excitation tables and K-map, then feed this expression to the T input.

Explain briefly the operation of negative edge-triggered JK flip flop with its truth table. Convert T flip-flop into JK flip-flop.

Negative edge-triggered JK flip-flop

A negative edge-triggered JK flip-flop changes state only at the falling (HIGH-to-LOW) edge of the clock. An edge detector produces a narrow pulse at the falling edge, and only during that pulse are J and K allowed to act. Its symbol has a triangle with a bubble at CLK. Since the inputs act for a very short time, the output can change only once per clock, so there is no race-around.

CLKJKQn+1Q_{n+1}Action
Falling edge00QnQ_nNo change
Falling edge010Reset
Falling edge101Set
Falling edge11Qn′Q_n'Toggle
0, 1, risingXXQnQ_nNo change

Converting T flip-flop into JK flip-flop

We need a circuit with inputs J, K, QnQ_n whose output drives T. T excitation: T=1T = 1 when the state must change.

JKQnQ_nQn+1Q_{n+1}T
00000
00110
01000
01101
10011
10110
11011
11101

K-map for T

        KQn=00 01 11 10
J=0        0   0  1  0
J=1        1   0  1  1
  • Cells 011, 111 (K=1K=1, Qn=1Q_n=1) give KQnKQ_n.
  • Cells 100, 110 (J=1J=1, Qn=0Q_n=0) give JQn′JQ_n'.
T=J Qn′+K QnT = J\,Q_n' + K\,Q_n

Circuit

J --[AND]--+
Q'--[   ]  +--[OR]-- T +--------+--- Q
K --[AND]--+           | T FF   |
Q --[   ]              |  CLK   |--- Q'
                       +--------+
 (Q and Q' fed back to the AND gates)

Check: J=K=1J = K = 1 gives T=Qn′+Qn=1T = Q_n' + Q_n = 1, so it toggles; J=K=0J = K = 0 gives T=0T = 0, no change.

✓ Answer checked by Claude Opus 5.5 (high effort)

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