Изучение
дешифраторов и мультиплексоров (Лабораторная работа)
Цель работы: изучить принцип действия, схемные реализации и особенности применения микросхем дешифраторов и мультиплексоров.
Ход работы
Рис. Схема подключения дешифратора
1.
Сняли таблицу истинности для элемента D10
|
№ |
8 |
4 |
2 |
1 |
|
|
|
|
|
|
|
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
|
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
1 |
1 |
1 |
|
2 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
|
3 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
|
4 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
|
5 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
|
6 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
|
7 |
0 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
|
8 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
9 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
|
10 |
1 |
0 |
1 |
0 |
X |
X |
X |
X |
X |
X |
X |
|
11 |
1 |
0 |
1 |
1 |
X |
X |
X |
X |
X |
X |
X |
|
12 |
1 |
1 |
0 |
0 |
X |
X |
X |
X |
X |
X |
X |
|
13 |
1 |
1 |
0 |
1 |
X |
X |
X |
X |
X |
X |
X |
|
14 |
1 |
1 |
1 |
0 |
X |
X |
X |
X |
X |
X |
X |
|
15 |
1 |
1 |
1 |
1 |
X |
X |
X |
X |
X |
X |
X |
2. Сняли таблицу истинности для элемента D10
Рис. Схема подключения мультиплексора
|
X1 |
X2 |
X3 |
X4 |
X5 |
X6 |
X7 |
X8 |
V1 |
V2 |
V3 |
|
|
1 |
X |
X |
X |
X |
X |
X |
X |
0 |
0 |
0 |
0 |
|
X |
1 |
X |
X |
X |
X |
X |
X |
0 |
0 |
1 |
0 |
|
X |
X |
1 |
X |
X |
X |
X |
X |
0 |
1 |
0 |
0 |
|
X |
X |
X |
1 |
X |
X |
X |
X |
0 |
1 |
1 |
0 |
|
X |
X |
X |
X |
0 |
X |
X |
X |
1 |
0 |
0 |
1 |
|
X |
X |
X |
X |
X |
0 |
X |
X |
1 |
0 |
1 |
1 |
|
X |
X |
X |
X |
X |
X |
0 |
X |
1 |
1 |
0 |
1 |
|
X |
X |
X |
X |
X |
X |
X |
0 |
1 |
1 |
1 |
1 |
3. Сняли таблицу истинности для элемента D13:
|
X1 |
X2 |
X3 |
X4 |
X5 |
X6 |
X7 |
X8 |
|
V1 |
V2 |
V3 |
HG7 |
|
|
X |
X |
X |
X |
X |
X |
X |
X |
1 |
X |
X |
X |
0 |
1 |
|
1 |
X |
X |
X |
X |
X |
X |
X |
0 |
0 |
0 |
0 |
1 |
0 |
|
X |
1 |
X |
X |
X |
X |
X |
X |
0 |
0 |
0 |
1 |
1 |
0 |
|
X |
X |
1 |
X |
X |
X |
X |
X |
0 |
0 |
1 |
0 |
1 |
0 |
|
X |
X |
X |
1 |
X |
X |
X |
X |
0 |
0 |
1 |
1 |
1 |
0 |
|
X |
X |
X |
X |
1 |
X |
X |
X |
0 |
1 |
0 |
0 |
1 |
0 |
|
X |
X |
X |
X |
X |
1 |
X |
X |
0 |
1 |
0 |
1 |
1 |
0 |
|
X |
X |
X |
X |
X |
X |
1 |
X |
0 |
1 |
1 |
0 |
1 |
0 |
|
X |
X |
X |
X |
X |
X |
X |
1 |
0 |
1 |
1 |
1 |
1 |
0 |
4.
Сняли таблицу истинности для элементов D2 и D3:
Рис. Схема преобразования кодов
|
Входы |
D2 |
D3 |
|||||||||||||
|
X5 |
X4 |
X3 |
X2 |
X1 |
Y5 |
Y4 |
Y3 |
Y2 |
Y1 |
Y6 |
Y5 |
Y4 |
Y3 |
Y2 |
Y1 |
|
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
|
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
|
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
|
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
|
0 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
|
0 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
|
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
|
0 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
|
0 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
1 |
|
0 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
|
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
|
0 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
|
0 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
0 |
|
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
|
0 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
|
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
0 |
|
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
|
1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
|
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
|
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
|
1 |
0 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
|
1 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
0 |
|
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
|
1 |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
|
1 |
1 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
|
1 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
1 |
|
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
|
1 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
1 |
0 |
1 |
1 |
|
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
|
1 |
1 |
1 |
1 |
0 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
0 |
|
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
0 |
0 |
0 |
1 |
Элемент D2 представляет собой преобразователь двоично-десятичного позиционного кода в двоичный, элемент D3 – преобразователь двоичного кода в двоично-десятичный.
Определим вес разрядов двоично-десятичного кода:
|
Элемент |
6 |
5 |
4 |
3 |
2 |
1 |
|
D2, X |
- |
10 |
5 |
4 |
2 |
1 |
|
D3, Y |
20 |
10 |
5 |
4 |
2 |
1 |
Выводы по работе: изучили принцип действия,
схемные реализации и особенности применения микросхем дешифраторов и
мультиплексоров.