- Asked 2/22
- 2081 Baisakh · 6 marks
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)
| Q | Q+ | T |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Step 2: conversion table. For each J, K and Q, take Q+ from the J-K characteristic table and write the T needed:
| J | K | Q | Q+ | T |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
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: .
- Group Q = 1, JK = 01 and 11: .
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 (toggle); J = K = 0 gives T = 0 (hold); J = 1, K = 0 gives , 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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