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T to JK flip-flop conversion

Short answer

Use the T excitation table and JK characteristic table to build a conversion table. K-map simplification gives T = J·Q' + K·Q. Implement with two AND gates (J with Q', K with Q) feeding an OR gate to the T input; this makes the T flip-flop hold, set, reset, or toggle as required.

Convert T flip-flop into JK flip-flop.

To make a T flip-flop behave as a J-K flip-flop, we add input logic that produces T from J, K and the present state Q. The required J-K behaviour: JK = 00 hold, 01 reset, 10 set, 11 toggle.

Step 1: excitation table of T flip-flop (T = 1 whenever the state must change)

QQ+T
000
011
101
110

Step 2: conversion table. For each J, K and Q, take Q+ from the J-K characteristic table and write the T needed:

JKQQ+T
00000
00110
01000
01101
10011
10110
11011
11101

Step 3: K-map for T

K-map for T
Q\JK   00  01  11  10
 0      0   0   1   1
 1      0   1   1   0
  • Group Q = 0, JK = 11 and 10: JQ‾J\overline{Q}.
  • Group Q = 1, JK = 01 and 11: KQKQ.
T=J Q‾+K QT = J\,\overline{Q} + K\,Q

Step 4: logic circuit

          +-----+
 J ------>|     |
 Q' ----->| AND |--+
          +-----+  |  +----+     +---------+
                   +->|    |     |         |
                      | OR |---->| T     Q |---> Q
                   +->|    |CLK->|>   T    |
          +-----+  |  +----+     |      Q' |---> Q'
 K ------>|     |  |             +---------+
 Q  ----->| AND |--+
          +-----+
 (Q and Q' are fed back from the outputs)

Check: J = K = 1 gives T=Q‾+Q=1T = \overline{Q} + Q = 1 (toggle); J = K = 0 gives T = 0 (hold); J = 1, K = 0 gives T=Q‾T = \overline{Q}, so a 0 toggles to 1 and a 1 holds (set); J = 0, K = 1 gives T = Q (reset).

✓ Answer checked by Claude Opus 5.5 (high effort)

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