兩個單片機之間的簡單異步串行通信

1、兩個單片機之間的簡單異步串行通信2009-9-1 21:13提問者：feitian_001|懸賞分：10|瀏覽次數(shù)：898 次我在 proteus 中用兩個AT89C52 單片機做了一個小設(shè)計，想以來來驗證兩個單片之間異步串行通信TXD 和上位機的RXD 0 x07 0 x07 P1 口（P1 口連著八個發(fā)光二極管（共陰極），按理說如果是正0 x07P1 口連接著的八個發(fā)光二極管會點亮后三個，但是我編寫程序加載到proteus 下位機（發(fā)送數(shù)據(jù)的單片機）：#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;/允許接收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);上位機（接收發(fā)過來的數(shù)據(jù)的單片機）：#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);問題補充：AT89C52 P1 個發(fā)光二極管，就這些東西最佳答案不可能固定不變的。接受機發(fā)送機開始發(fā)送地位數(shù)據(jù)，接受機才開始接受數(shù)據(jù)。#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 ，只有下位機兩個口都變成高電平，才發(fā)送 0 x07，上位機在 P1. 0 和 P1. 1 口都輸出低電平懸賞分：10 -解決時間：2010-11-19 09:51/ / - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

7、- - - - - - - - - - - - - - - -最 佳 答 案 ：根 據(jù) 要 求 ， 繪 制 P R O T E U S 仿 真 電 路 圖 如 下 ：http/hi. baidu. com/D7F6B6F8C2DBB5C0/album/item/d9581a5108d990171238圖 中 的 上 位 機 ， 是 完 成 串 行 接 收 、 并 根 據(jù) 收 到 的 數(shù) 據(jù) ， 輸 出 高 低 電 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;/定時器 1 工作于 8 位自動重載模式,用于產(chǎn)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 賦 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、= = = = = = = = = = = = =圖 中 的 下 位 機 ， 要 完 成 輸 入 的 檢 測 ， 并 發(fā) 送 不 同 的 串 行 數(shù) 據(jù) 。 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;/定時器 1 工作于 8 位自動重載模式,用于產(chǎn)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 賦 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 ; / / = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =