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JK flip-flop characteristic and excitation tables

Short answer

A JK flip-flop eliminates SR's invalid state by feeding Q and Q' back to input NANDs. J,K=00 hold, 01 reset, 10 set, 11 toggle. Characteristic equation: Qn+1 = J·Qn' + K'·Qn. Excitation table: 0→0: J=0,K=X; 0→1: J=1,K=X; 1→0: J=X,K=1; 1→1: J=X,K=0.

Explain the operation of JK flip flop showing its logic diagram, characteristic table and then derive its characteristic equation and excitation table.

A JK flip-flop is a clocked flip-flop with inputs J (like set) and K (like reset). It removes the invalid state of the SR flip-flop: when J=K=1J = K = 1 the output toggles.

Logic diagram

It is an SR flip-flop with QQ and Q′Q' fed back to the input NAND gates.

J ---[NAND1]--S'-->[NAND3]---> Q
CLK -[     ]   Q'-->[     ]
Q' --[     ]
K ---[NAND2]--R'-->[NAND4]---> Q'
CLK -[     ]   Q--->[     ]
Q  --[     ]

NAND1 gets J, CLK and Q′Q'; NAND2 gets K, CLK and QQ; NAND3 and NAND4 form the cross-coupled latch.

Operation (clock active)

  • J=0,K=0J = 0, K = 0: both input NANDs give 1, latch holds: no change.
  • J=0,K=1J = 0, K = 1: if Q=1Q = 1, NAND2 gives 0 and resets: Q = 0.
  • J=1,K=0J = 1, K = 0: if Q=0Q = 0, NAND1 gives 0 and sets: Q = 1.
  • J=1,K=1J = 1, K = 1: the gate enabled by feedback changes the state: toggle.

Characteristic table

JKQnQ_nQn+1Q_{n+1}Action
0000No change
0011No change
0100Reset
0110Reset
1001Set
1011Set
1101Toggle
1110Toggle

Characteristic equation

        KQn=00 01 11 10
J=0        0   1  0  0
J=1        1   1  0  1
  • J=1,Qn=0J=1, Q_n=0 (cells 100, 110) gives JQn′JQ_n'.
  • K=0,Qn=1K=0, Q_n=1 (cells 001, 101) gives K′QnK'Q_n.
Qn+1=J Qn′+K′ QnQ_{n+1} = J\,Q_n' + K'\,Q_n

Excitation table

Shows the inputs needed for each required transition:

QnQ_nQn+1Q_{n+1}JK
000X
011X
10X1
11X0

For example, for 0→10 \to 1 we may use J=1,K=0J=1, K=0 (set) or J=1,K=1J=1, K=1 (toggle), so KK is a don't care.

Note: with a level-triggered JK and J=K=1J = K = 1, the output can toggle many times while the clock is high (race-around condition). It is removed by using an edge-triggered or master-slave JK flip-flop.

✓ Answer checked by Claude Opus 5.5 (high effort)

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