AVR UART 接收到错误的字节

Question

经过长时间的停顿后，我再次开始对 AVR 进行编程。我能够在我的 Linux 计算机和 Atmega8 之间设置 UART 通信。不幸的是，当我发送大于 0x1f 的字节时，Atmega8 似乎接收到了错误的字节。 UART 是运行 9600 BAUD 和数据格式 8N1.

// clock frequency 1Mhz 
#define F_CPU 1000000UL

// baud rate
#define BAUD 9600
#define BAUDRATE ((F_CPU)/(BAUD*16UL)-1)

#include <avr/io.h>
#include <avr/interrupt.h>
#include <util/delay.h>
#include <stdlib.h>


void uart_init (void);
void uart_putc (unsigned char data);
void uart_puts (unsigned char * str, uint8_t size);

// interrupt service routine for UART receiver interupt
ISR(USART_RXC_vect) {

    // get received char
    unsigned char received_char = UDR;

    uart_puts("received=", 9);

    if((received_char & (1<<7)) == (1<<7)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<6)) == (1<<6)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<5)) == (1<<5)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<4)) == (1<<4)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<3)) == (1<<3)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<2)) == (1<<2)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<1)) == (1<<1)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    if((received_char & (1<<0)) == (1<<0)) {
        uart_putc('1');
    } else {
        uart_putc('0');
    }

    uart_puts("\n\r",2);
}


// function to initialize UART
// dataformat 8N1
void uart_init (void) {

    // shift the register right by 8 bits
    UBRRH = (BAUDRATE>>8);

    // set baud rate
    UBRRL = BAUDRATE;            

    // enable receiver, transmitter and receiver interrupt
    UCSRB|= (1<<TXEN)|(1<<RXEN)|(1<<RXCIE);

    // 8bit data format
    UCSRC|= (1<<URSEL)|(1<<UCSZ0)|(1<<UCSZ1);   

}

// sends a single char over the UART
void uart_putc (unsigned char data) {

    // wait while register is free
    while (!( UCSRA & (1<<UDRE)));
    // load data in the register
    UDR = data;
}

// sends a string over the UART
void uart_puts (unsigned char * str, uint8_t size) {

    uint8_t i; 

    for(i = 0; i < size; i++) {
        uart_putc(str[i]);
    }
}

// receives a single char over the UART
unsigned char uart_getc (void) {

    // wait while data is being received
    while(!(UCSRA) & (1<<RXC));
    // return 8-bit data
    return UDR;
}


uint8_t main (void) {

    // enable interrupts
    sei();

    // enable uart
    uart_init(); 

    uart_puts("ready\n\r", 7);

    while(1) {
    } 

    return 0; 
}

在 GtkTerm 中，我发送以下字节序列：0x01、0x02、0x03、0x1d、0x1e、0x1f、 0x20 中断服务例程 ISR 响应以下序列。

但是，我应该收到 received=00100000 0x20
怎么了？

Answer 1

来自我的评论：

可能是由于使用较慢的时钟速率/内部 RC 振荡器导致波特率关闭？波特率越慢，每个位与"middle"的偏移量就越大。这可能导致跳过位或控制器看到位两次。

AVR UART 接收到错误的字节

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