2 Основная часть
2.1 Постановка задачи
Задачей данного курсового проекта было проектирование и реализация синхронного счетчика на JK-триггерах с К=1287.
Следующей задачей работы стала разработка этого же делителя частоты на стандартных схема счетчиков: К155ИЕ2 (7490), К155ИЕ4 (7492), 155ИЕ5 (7493).
Существует несколько методов разработки счетчиков, которые формируют произвольную последовательность. Рассмотрим детали простого способа. перед тем как начать разработку необходимо проанализировать работу JK-триггера.
Переход |
Q(t) |
Q(t+1) |
J |
K |
0 -> 0 |
0 |
0 |
0 |
Х |
0 -> 1 |
0 |
1 |
1 |
Х |
1 -> 0 |
1 |
0 |
Х |
1 |
1 -> 1 |
1 |
1 |
Х |
0 |
Таблица 1. Таблица возбуждений JK-триггера
Разобьем счетчик на модули 9, 11, 13.
2.2 Расчет модуля с К=9
Рассчитаем счетчик К=9:
|
Q(t) |
Q(t+1) |
|||||||
№ |
D |
C |
B |
A |
d |
c |
b |
a |
|
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
|
2 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
|
3 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
|
4 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
|
5 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
|
6 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
|
7 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
|
8 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
|
9 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
10 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
|
11 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
|
12 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
|
13 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
|
14 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
|
15 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
|
16 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
|
Таблица 2. Таблица истинности счетчика в моменты t и t+1
Jd |
Kd |
Jc |
Kc |
Jb |
Kb |
Ja |
Ka |
0 |
X |
0 |
X |
0 |
X |
1 |
X |
0 |
X |
0 |
X |
1 |
X |
X |
1 |
0 |
X |
0 |
X |
X |
0 |
X |
1 |
0 |
X |
1 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
0 |
X |
X |
0 |
1 |
X |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
1 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
0 |
X |
0 |
X |
X |
1 |
0 |
X |
0 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
0 |
X |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
0 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
Таблица 3. Таблица состояний счетчика
Составим карты Карно:
D |
00 |
01 |
11 |
10 |
00 |
1 |
X |
X |
1 |
01 |
1 |
X |
X |
1 |
11 |
0 |
X |
X |
0 |
10 |
0 |
X |
X |
0 |
C\BAJ
a=D
D |
00 |
01 |
11 |
10 |
00 |
Х |
1 |
1 |
Х |
01 |
Х |
1 |
1 |
Х |
11 |
Х |
1 |
1 |
Х |
10 |
Х |
1 |
1 |
Х |
C\BAKa=1
DC\BA |
0 |
01 |
11 |
10 |
00 |
0 |
1 |
Х |
Х |
01 |
0 |
1 |
Х |
Х |
11 |
0 |
0 |
Х |
Х |
10 |
0 |
0 |
Х |
Х |
0J
b=A*D=A+D
DC\BA |
00 |
01 |
11 |
10 |
00 |
Х |
Х |
1 |
0 |
01 |
Х |
Х |
1 |
0 |
1 |
Х |
Х |
1 |
1 |
10 |
Х |
Х |
1 |
1 |
1
K
b=D+A=D*A
DC\BA |
00 |
0 |
11 |
10 |
00 |
0 |
0 |
1 |
0 |
01 |
X |
X |
X |
X |
11 |
X |
X |
X |
X |
10 |
0 |
0 |
0 |
0 |
1
Jc= D*B*A=D+A*B
DC\BA |
00 |
0 |
11 |
10 |
00 |
X |
X |
X |
X |
0 |
0 |
0 |
1 |
0 |
11 |
1 |
1 |
1 |
1 |
10 |
X |
X |
X |
X |
1
1
Kc=B*A+D =A*B*D
DC\BA |
00 |
01 |
11 |
10 |
00 |
0 |
0 |
0 |
0 |
01 |
0 |
0 |
1 |
0 |
11 |
X |
X |
X |
X |
10 |
X |
X |
X |
X |
J
d=B*A*C=A*B+C
D |
00 |
01 |
11 |
10 |
00 |
X |
X |
X |
X |
01 |
X |
X |
X |
X |
11 |
1 |
1 |
1 |
1 |
10 |
1 |
1 |
1 |
1 |
C\BA
Kd=1
Проведем анализ на сбои:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1111
1110
1101
1100
Рис 2. Граф состояний
2.3 Расчет модуля К=11
Рассчитаем счетчик К=11:
|
Q(t) |
Q(t+1) |
||||||
№ |
D |
C |
B |
A |
d |
c |
b |
a |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
2 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
3 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
4 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
5 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
6 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
7 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
8 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
9 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
10 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
11 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
12 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
13 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
14 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
15 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
16 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
Таблица 4. Таблица истинности счетчика в моменты t и t+1
Jd |
Kd |
Jc |
Kc |
Jb |
Kb |
Ja |
Ka |
0 |
X |
0 |
X |
0 |
X |
1 |
X |
0 |
X |
0 |
X |
1 |
X |
X |
1 |
0 |
X |
0 |
X |
X |
0 |
X |
1 |
0 |
X |
1 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
0 |
X |
X |
0 |
1 |
X |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
1 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
0 |
0 |
X |
0 |
X |
1 |
X |
X |
0 |
0 |
X |
1 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
0 |
X |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
0 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
Таблица 5. Таблица состояний счетчика
DC\BA |
00 |
01 |
11 |
10 |
0 |
1 |
X |
X |
1 |
01 |
1 |
X |
X |
1 |
11 |
0 |
X |
X |
0 |
1 |
1 |
X |
X |
0 |
0
0
Ja=D*C*B
DC\BA |
00 |
01 |
11 |
10 |
0 |
Х |
1 |
1 |
Х |
01 |
Х |
1 |
1 |
Х |
11 |
Х |
1 |
1 |
Х |
10 |
Х |
1 |
1 |
Х |
0
Ka=1
DC\BA |
0 |
01 |
11 |
10 |
00 |
0 |
1 |
Х |
Х |
01 |
0 |
1 |
Х |
Х |
11 |
0 |
0 |
Х |
Х |
10 |
0 |
1 |
Х |
Х |
0
Jb=A*C+A*D=A+D*C
DC\BA |
00 |
01 |
11 |
10 |
00 |
Х |
Х |
1 |
0 |
0 |
Х |
Х |
1 |
0 |
11 |
Х |
Х |
1 |
1 |
10 |
Х |
Х |
1 |
1 |
1
Kb=D+A=D*A
DC\BA |
00 |
0 |
11 |
10 |
00 |
0 |
0 |
1 |
0 |
01 |
X |
X |
X |
X |
11 |
X |
X |
X |
X |
10 |
0 |
0 |
0 |
0 |
1
Jc= B*A*D=B*A+D
DC\BA |
00 |
0 |
11 |
10 |
00 |
X |
X |
X |
X |
0 |
0 |
0 |
1 |
0 |
11 |
1 |
1 |
1 |
1 |
10 |
X |
X |
X |
X |
1
1
Kc=B*A+D=D*A*B
DC\BA |
00 |
01 |
11 |
10 |
00 |
0 |
0 |
0 |
0 |
01 |
0 |
0 |
1 |
0 |
11 |
X |
X |
X |
X |
10 |
X |
X |
X |
X |
J
d=B*A*C=B*A+C
DC\BA |
00 |
0 |
11 |
10 |
0 |
X |
X |
X |
X |
01 |
X |
X |
X |
X |
11 |
1 |
1 |
1 |
1 |
10 |
0 |
0 |
1 |
1 |
1
0
K
d=C+B=C*B
Проведем анализ на сбои:
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1111
1110
1101
1100
Рис. 3. Граф состояний
2.4 Расчет модуля К=13
№ |
D |
C |
B |
A |
d |
c |
b |
a |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
2 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
3 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
4 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
5 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
6 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
7 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
8 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
9 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
10 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
11 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
12 |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
0 |
13 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
14 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
15 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
16 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
Таблица 7. Таблица истинности счетчика в моменты t и t+1
Jd |
Kd |
Jc |
Kc |
Jb |
Kb |
Ja |
Ka |
0 |
X |
0 |
X |
0 |
X |
1 |
X |
0 |
X |
0 |
X |
1 |
X |
X |
1 |
0 |
X |
0 |
X |
X |
0 |
X |
1 |
0 |
X |
1 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
0 |
X |
X |
0 |
1 |
X |
X |
1 |
0 |
X |
X |
0 |
X |
0 |
1 |
X |
1 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
0 |
0 |
X |
0 |
X |
1 |
X |
X |
0 |
0 |
X |
1 |
X |
X |
1 |
X |
0 |
0 |
X |
X |
0 |
1 |
X |
X |
0 |
1 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
0 |
X |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
0 |
X |
X |
1 |
X |
1 |
X |
1 |
X |
1 |
Таблица 8. Таблица состояний счетчика
D |
00 |
01 |
11 |
10 |
0 |
1 |
X |
X |
1 |
01 |
1 |
X |
X |
1 |
11 |
0 |
X |
X |
0 |
1 |
1 |
X |
X |
1 |
C\BA
0
0
J
a=D+C=D*C
D |
00 |
01 |
11 |
10 |
00 |
Х |
1 |
1 |
Х |
01 |
Х |
1 |
1 |
Х |
11 |
Х |
1 |
1 |
Х |
10 |
Х |
1 |
1 |
Х |
C\BA
Ka=1
DC\BA |
00 |
01 |
11 |
10 |
00 |
0 |
1 |
Х |
Х |
01 |
0 |
1 |
Х |
Х |
11 |
0 |
0 |
Х |
Х |
10 |
0 |
1 |
Х |
Х |
J
b=A*D+A*C=A+D*C
DC\BA |
0 |
01 |
11 |
10 |
00 |
Х |
Х |
1 |
0 |
0 |
Х |
Х |
1 |
0 |
11 |
Х |
Х |
1 |
1 |
10 |
Х |
Х |
1 |
0 |
0
1
K
b=D*C+A=D*C*A
DC\BA |
00 |
0 |
11 |
10 |
00 |
0 |
0 |
1 |
0 |
01 |
X |
X |
X |
X |
11 |
X |
X |
X |
X |
10 |
0 |
0 |
1 |
0 |
1
Jc=B*A
DC\BA |
00 |
0 |
11 |
10 |
00 |
X |
X |
X |
X |
0 |
0 |
0 |
1 |
0 |
11 |
1 |
1 |
1 |
1 |
10 |
X |
X |
X |
X |
1
1
K c=D+A*B=D*A*B
DC\BA |
00 |
01 |
11 |
10 |
00 |
0 |
0 |
0 |
0 |
01 |
0 |
0 |
1 |
0 |
11 |
X |
X |
X |
X |
10 |
X |
X |
X |
X |
J
d=A*B+C
DC\BA |
00 |
01 |
11 |
10 |
0 |
X |
X |
X |
X |
01 |
X |
X |
X |
X |
11 |
1 |
1 |
1 |
1 |
10 |
0 |
0 |
0 |
0 |
0Kd=C
Проведем анализ на сбои:
Рис. 4. Граф состояний