Design an 8421BCD asynchronous up counter with a ripple carry output.

Step-by-Step

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Step 1 Construct the state table.

Because an 8421BCD counter has ten states represented by ten fixed binary code, the coded state table can be listed directly, as shown in Table 6.3.7. Z is defined as the ripple carry output.

Step 2 Choose the type of flip-flops to be used. Derive the excitation equations and output equation.

The 8421BCD has four-bit binary codes and thus four flip-flops are required. Here, four J-K flipflops are chosen to construct the 8421BCD asynchronous up counter.

Since there is no global clock signal for all flip-flops in asynchronous circuit, the CP for each flipflop must be applied as long as the state of flip-flop produce a change .

According to Table 6.3.7, the output Q_1 of the first flip-flop changes state at each CP, so the input CP as its clock input. That is, {{C P}}_{1}={{C P}} and J_{1}=K_{1} = 1.

The output Q_{2} of the first flip-flop changes its state when Q_{1} changes from HIGH to LOW, that is, at the negative going edge of Q_{1} . Thus, Q_{1} is set as the input CP of the second flip-flop, CP_{2}=Q_{1}.

Note that at the 10th CP, although Q_{1} changes from HIGH to LOW, Q_{2} does not change. Thus, J_{2} and K_{2} input for the second flip-flop should be controlled to make sure that Q_{2} does not change at the 10th CP.

The output Q_{3} changes its state when Q_{2} changes from HIGH to LOW, so Q_{2} is set as the input CP of the third flip-flop, C P_{3} = Q_{2} and J_{3} = K_{3} = 1.

The output Q_{4} changes state when Q_{1} changes from HIGH to LOW, so Q_{1} is set as the input CP of the fourth flip-flop, CP_{4}=Q_{1}. Note that the J_{4} and K_{4} input still need to be controlled since Q_{4} does not change at its first three CP.

According to aforementioned analysis, Table 6.3.7 can be simplified as Table 6.3.8.

The state transition table is shown as Table 6.3.9.

According to Table 6.3.9, the Karnaughmaps of the inputs of J-K flip-flops are shown in Figure 6.3.10.

Excitation equations can be obtained by Karnaugh map simplification as follows:

\begin{cases} J_{4}=Q_{3}Q_{2} \\ K_{4}=1 \end{cases} , \quad \ \begin{cases} J_{2}=\bar{Q}_{4} \\ K_{2}=1 \end{cases} (6.3.6)

Only when Q_{4}\,Q_{3}\,Q_{2}\,Q_{1} = 1001, Z = 1, the output equation is

{{{Z}}}=Q_{4}\,\bar{Q}_{3}\,\bar{Q_{2}}\,Q_{1} (6.3.7)

Step 3 Check if the 8421BCD asynchronous up counter can be self-corrected.

There are six unused states from 1010 and 1111. Assume that the circuit is in state 1010, when next CP is applied to the clock input of FF1, the next state of the output Q_{{1}} changes from a 0 to a 1 since the values of J and K inputs are both 1. With Q_{{1}} from LOW to HIGH, there is no falling edge of CP applied to the clock input of FF2 and thus the output Q_{{2}} still keeps a 1. Since the output of Q_{{2}} does not change, there is no effective CP applied to the clock input of FF3 and thus the output Q_{{3}} also keeps a 0. With Q_{{1}} from LOW to HIGH, there is no falling edge of CP applied to the clock input of FF4 and thus the output Q_{{4}} does not change. Thus, after state 1010, the next state is state 1011.

Assume that the circuit is in state 1011; when next CP is applied to the clock input of FF1, the next state of the output Q_{{1}} changes from a 1 to a 0 since the values of J and K inputs are both 1. With Q_{{1}} from HIGH to LOW, there is a falling edge of CP applied to the clock input of FF2 and FF4. Therefore, the output Q_{{2}} changes from a 1 to a 0 due to J_{{2}} = 0 and K_{{2}} = 1 and the output Q_{{4}} changes from a 1 to a 0 since J_{{4}} = 0 and K_{{4}} = 1. With Q_{{2}} from HIGH to LOW, there is a falling edge of CP applied to the clock input of FF3 and thus the output Q_{{3}} changes from a 0 to a 1 since J_{{3}} = K_{{3}} = 1. It is clearly seen that the next state is state 0100 after state 1011. Similarly, you can check the other four unused states. The complete state diagram is shown in Figure 6.3.11. It can be seen from Figure 6.3.11 that the design circuit can be self-corrected.

Step 3 Draw a logic diagram according to excitation equations and output equation.

According the excitation eq. (6.3.6) and output eq. (6.3.7), the logic diagram of the 8421BCD asynchronous up counter is implemented with four J-K flip-flops, as shown in Figure 6.3.12.

Table 6.3.7: The state table. | |||||

CP | Q_{4} | Q_{3} | Q_{2} | Q_{1} | Z |

0 | 0 | 0 | 0 | 0 | 0 |

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

2 | 0 | 0 | 1 | 0 | 0 |

3 | 0 | 0 | 1 | 1 | 0 |

4 | 0 | 1 | 0 | 0 | 0 |

5 | 0 | 1 | 0 | 1 | 0 |

6 | 0 | 1 | 1 | 0 | 0 |

7 | 0 | 1 | 1 | 1 | 0 |

8 | 1 | 0 | 0 | 0 | 0 |

9 | 1 | 0 | 0 | 1 | 1 |

10 | 0 | 0 | 0 | 0 | 0 |

Table 6.3.8: The simplified state table. | |||

CP | Q_{4} | Q_{3} | Q_{2} |

0 | 0 | 0 | 0 |

2 | 0 | 0 | 1 |

4 | 0 | 1 | 0 |

6 | 0 | 1 | 1 |

8 | 1 | 0 | 0 |

Table 6.3.9: The state transition table. | ||||||||||

CP | Present state | Next state | Excitation inputs | |||||||

Q_{4n} | Q_{3n} | {Q_{2n}} | {Q_{4n+1}} | {Q_{3n+1}} | {Q_{2n+1}} | J_{4} | K_{4} | J_{2} | K_{2} | |

0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | × | 1 | × |

2 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | × | × | 1 |

4 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | × | 1 | × |

6 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | × | × | 1 |

8 | 1 | 0 | 0 | 0 | 0 | 0 | × | 1 | 0 | × |

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