STM32F103C8T6无法与HD44780通信

Question

我试图用HD44780控制16X2液晶显示器(4位通信)使用STM32F103C8T6。我用这样的方式连接LCD：

RS => PA0 EN => PA2 RW地面 D7 => PB7 D6 => PB6 D5 => PB5 D4 => PB4

液晶显示器不显示任何东西。问题可能在哪里？有人知道这个问题吗？

这是我的代码：

#include "delay.h"
#include "stm32f10x.h" // Device header

void lcd_command(unsigned char command);
void lcd_init(void);
void lcdPosition(unsigned int row, unsigned int column);
void lcd_data(unsigned char data);

int main() {
  delay_init();
  RCC->APB2ENR |= 1 << 2; // Port A Enabled.
  RCC->APB2ENR |= 1 << 3; // Port B Enabled.

  GPIOA->CRL = 0x22222222; // A0 and A2 Output.
  GPIOB->CRL = 0x22222222; // B7,B6,B5,B4 Output.
  GPIOB->ODR = 0x00;       // Port B clear.

  delay_ms(20);
  lcd_command(0x30); // Datasheet says.
  delay_ms(5);
  lcd_command(0x30); // Datasheet says.
  delay_ms(1);
  lcd_command(0x30); // Datasheet says.
  delay_ms(1);
  lcd_init();
  lcdPosition(1, 1); // first row first column
  delay_ms(1);
  lcd_data('A'); // Letter A
  while (1)
    ;
}

void lcd_command(unsigned char command) {

  GPIOA->BRR |= 1 << 0; // RS reset.

  GPIOA->BSRR |= 1 << 2;   // E set.
  GPIOB->ODR = command;    // upper nibble
  delay_ms(2);             // delay
  GPIOA->BRR |= 1 << 2;    // E reset.
  GPIOB->BRR = 0x000000F0; // clear data bits
  delay_ms(2);             // delay

  command = command << 4; // lower nibble

  GPIOA->BSRR |= 1 << 2;   // E set.
  GPIOB->ODR = command;    // lower nibble
  delay_ms(2);             // delay
  GPIOA->BRR |= 1 << 2;    // E reset.
  GPIOB->BRR = 0x000000FF; // clear data bits
}
void lcd_init() {
  lcd_command(0x02); // Return
  delay_ms(2);       // delay
  lcd_command(0x28);
  set 4 - bit data, 2 - line, 5x7 font delay_ms(2); // delay
  lcd_command(0x0C);
  turn on display, cursor off.delay_ms(2); // delay
  lcd_command(0x01);                       // Clear.
  delay_ms(2);                             // delay
  lcd_command(0x06);
  move cursor right delay_ms(4); // delay
}
void lcdPosition(unsigned int row, unsigned int column) {
  if (row == 1) {
    column--;
    lcd_command(0x80 + column); // Define row

  } else if (row == 2) {
    column--;
    lcd_command(0xC0 + column); // Define column
  }
}

void lcd_data(unsigned char data) {
  GPIOA->BSRR |= 1 << 0; // RS reset.
  GPIOA->BSRR |= 1 << 2; // E set.
  GPIOB->ODR = data;
  upper nibble first delay_ms(4); // delay
  GPIOA->BRR |= 1 << 2;           // E reset.
  GPIOB->BRR = 0x000000F0;        // clear data bits
  delay_ms(4);                    // delay

  data = data << 4; // lower nibble

  GPIOA->BSRR |= 1 << 2;   // E set.
  GPIOB->ODR = data;       // lower nibble
  delay_ms(4);             // delay
  GPIOA->BRR |= 1 << 2;    // E reset.
  GPIOB->BRR = 0x000000FF; // clear data bits
}

Answer 1

另外，PA0和PA2不是5V容错引脚，通常HD44780是5V设备，如果它在+5V上运行，请参考表5中的数据表。“中等密度STM32F103xx引脚定义”，如果该引脚在I/O列级别上为5V，则应用FT表示！

Answer 2

请仔细阅读如何在数据表中初始化4位接口，见图24第46页.

简而言之:您的lcd_command一个接一个地发送两个小块，但是您只需要为前几个命令发送一个咬口(因为显示控制器仍然处于8位模式)。您需要有一个单独的函数才能只发送一条信息。