两个单片机之间的简单异步串行通信

1、两个单片机之间的简单异步串行通信2009-9-1 21:13提问者：feitian_001|悬赏分：10|浏览次数：898 次我在 proteus 中用两个AT89C52 单片机做了一个小设计，想以来来验证两个单片之间异步串行通信TXD 和上位机的RXD 0 x07 0 x07 P1 口（P1 口连着八个发光二极管（共阴极），按理说如果是正0 x07P1 口连接着的八个发光二极管会点亮后三个，但是我编写程序加载到proteus 下位机（发送数据的单片机）：#include #define uchar unsigned char #define uint unsigned int voiduar

2、t_init()TMOD=0 x20;/T12TH1=0 xFD;/baud rate9600bit/s TL1=0 xFD;TR1=1;REN=1;/允许接收SM0=0;/SM1=1;TI=0;void delay(uint z)uintx,y; for(x=z;x0;x-)for(y=125;y0;y-);void main()uart_init(); while(1)SBUF=0 x07;while(!TI); TI=0;delay(5000);上位机（接收发过来的数据的单片机）：#include #define uchar unsigned char#defineuintunsigne

3、dvoiduart_init()TMOD=0 x20;TH1=0 xFD;/baud rate9600bit/s TL1=0 xFD;TR1=1; REN=1;SM0=0;/SM1=1;void delay(uint z)uintx,y; for(x=z;x0;x-)for(y=125;y0;y-);void main()uart_init(); while(1)if (RI)RI=0; P1=SBUF;delay(3000);问题补充：AT89C52 P1 个发光二极管，就这些东西最佳答案不可能固定不变的。接受机发送机开始发送地位数据，接受机才开始接受数据。#include #define

4、uchar unsigned char #define uint unsigned int voiduart_init()TMOD=0 x20; TH1=0 xFD; TL1=0 xFD; PCON=0 x00; TR1=1; SCON=0 x50;void delay(uint z)uintx,y; for(x=z;x0;x-)for(y=125;y0;y-);void main()uart_init();doSBUF=0 x01;while(TI=0); TI=0;while(RI=0); REN=0;while(SBUF!=0 x02) ;while(1) SBUF=0 x07;whil

5、e(TI=0); TI=0;delay(500) ; SBUF=0 x00;while(TI=0); TI=0;delay(500) ;#include #define uchar unsigned char #define uint unsigned int voiduart_init()TMOD=0 x20; TH1=0 xFD; TL1=0 xFD; PCON=0 x00; TR1=1;SM0=0;/SM1=1;REN=1;void main()uart_init(); do SBUF=0 x02;while(TI=0); TI=0;while(RI=0); RI=0; P2=SBUF;

6、while(P2!=0 x01);while(1)while(RI=0); P1=SBUF;RI=0;2010-11-18 16:46串行通信控制2010-11-18 16:46串行通信控制P1. 0 0 x01P1. 0 P1. 1 0 x03P1. 1 ，只有下位机两个口都变成高电平，才发送 0 x07，上位机在 P1. 0 和 P1. 1 口都输出低电平悬赏分：10 -解决时间：2010-11-19 09:51/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

7、- - - - - - - - - - - - - - - -最 佳 答 案 ：根 据 要 求 ， 绘 制 P R O T E U S 仿 真 电 路 图 如 下 ：http/hi. baidu. com/D7F6B6F8C2DBB5C0/album/item/d9581a5108d990171238图 中 的 上 位 机 ， 是 完 成 串 行 接 收 、 并 根 据 收 到 的 数 据 ， 输 出 高 低 电 平 。 使 用 的 程 序 如 下 ：/ / = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

8、= = = = = = = = = = = = = =# i n c l u d e define XTAL 11059200/ CPU define baudrate 9600/#define uchar unsigned char #define uint unsigned ints b i t K 1P 1 0/s b i t K 2 = P 1 1 ;v o i d i n i t ( v o i d )S C O N0 x 5 0/PCON & = 0 x00/T M O D =0 x 2 0;/定时器 1 工作于 8 位自动重载模式,用于产生波特率T H 1 = ( u n s i

9、 g n e d c h a r ) ( 2 5 6 - ( X T A L / ( 3 2 L * 1 2 L * b a u d r a t e ) ) ) ;T L 1 = ( u n s i g n e d c h a r ) ( 2 5 6 - ( X T A L / ( 3 2 L * 1 2 L * b a u d r a t e ) ) ) ; / / T 1 赋 初 值T R 1 = 1/1I E = 0 x 00/E A = 1/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

10、- - - - - - - - - - - - - - - - - - - - - - - -v o i d m a i n ( v o i d )i n i t ( ) ;w h i l e ( 1 ) w h i l e ( R I = = 0 ) ;R I = 0 ;if (SBUF = 0 x01) K1 = 1; if (SBUF = 0 x03) K2 = 1; if (SBUF = 0 x07) K 1 = 0 ;K 2 = 0 ; / / = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

11、= = = = = = = = = = = = =图 中 的 下 位 机 ， 要 完 成 输 入 的 检 测 ， 并 发 送 不 同 的 串 行 数 据 。 使 用 的 程 序 如 下 ：/ / = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =# i n c l u d e define XTAL 11059200/ CPU define baudrate 9600/#define uchar unsigned char #define uint uns

12、igned ints b i t K 1P 1 0/s b i t K 2 = P 1 1 ;/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -v o i d i n i t ( v o i d )S C O N0 x 5 0/PCON & = 0 x00/T M O D =0 x 2 0;/定时器 1 工作于 8 位自动重载模式,用于产生波特率T H 1 = ( u n s i g n e d c h

13、a r ) ( 2 5 6 - ( X T A L / ( 3 2 L * 1 2 L * b a u d r a t e ) ) ) ;T L 1 = ( u n s i g n e d c h a r ) ( 2 5 6 - ( X T A L / ( 3 2 L * 1 2 L * b a u d r a t e ) ) ) ; / / T 1 赋 初 值T R 1 = 1/1I E = 0 x 00/E A = 1/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

14、 - - - - - - - - - - - - - - - - -v o i d m a i n ( v o i d )b i t T _ 1 = 0 , T _ 2 = 0 , T _ 3 = 0 ;i n i t ( ) ;/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -w h i l e ( 1 ) i f ( ( K 1 = = 0 ) & & ( T _ 1 = = 0 ) ) T _ 1 = 1;T _ 3 = 0 ;S B

15、 U F = 0 x 0 1 ;w h i l e ( T I = = 0 ) T I = 0 ;/- i f ( ( K 2 = = 0 ) & & ( T _ 2 = = 0 ) )T _ 2 = 1 ;T _ 3 = 0 ;S B U F = 0 x 0 3 ;w h i l e ( T I = = 0 ) T I = 0 ;/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -if (K1 = 1) & (K2 = 1) & (T_3 = 0) T _ 1 = 0;T _ 2 = 0;T _ 3 = 1 ;S B U F = 0 x 0 7 ;w h i l e ( T I = = 0 ) T I = 0 ; / / = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =