mod-6 synchronous counter using T f/f

mod - 6 counter has 6 states.

Hence, the number of f/f's is,

$2^n \gt N \gt 2^n-1$

$\therefore 2^3 \gt 6$

$\therefore$ Number of flip-flops required = 3

**Step 1:** State Diagram.

**Step 2:** State table.

present state (Ps) | Next state (Ns) |
---|---|

a | b |

b | c |

c | d |

d | e |

e | f |

f | a |

**Step 3:** State assignment.

a = 000, b = 001, c = 010, d = 011, e = 100, f = 101.

**Step 4:** Excitation table of Tf/f and truth table of T f/f

**Truth table.**

T | $Q_n+1$ |
---|---|

0 | $Q_n$ |

1 | $Q_n$ |

**Excitation table.**

$Q_n$ | $Q_n+1$ | T |
---|---|---|

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 0 |

**Step 5:** Excitation table and maps.

| Present state (Ps) | Next state (Ns) | Excitation inputs | |----|----|----|

q2 | q1 | q0 | Q2 | Q1 | Q0 | T2 | T1 | T0 |
---|---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |

0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |

0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |

0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 |

1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |

1 | 1 | 0 | x | x | x | x | x | x |

1 | 1 | 1 | x | x | x | x | x | x |

**Step 6:** Simplifying logic expressions.

$T_2 = q_1 q_0 + q_2 q_0$

$T_1 = q_2 q_0$

$\therefore T_o = 1$