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c - 使用 beaglebone Black SPI 在 DIP203-6 LCD 屏幕上显示字符

转载 作者:行者123 更新时间:2023-11-30 16:19:54 25 4
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我正在尝试使用 Beagle Bone Black SPI 将一个字符写入我的 DIP203-6 LCD。当光标在正确的位置闪烁时,我已经初始化了 LCD 屏幕。我面临的问题是编写 the data sheet 中的字符字符的写入需要10位,而spi只需要8位。 LCD的RS需要按照datasheet中的命令设为1。我倾向于得到一些垃圾值。

在 int main() 部分,最后你可以看到我正在使用 main 上面的传输函数传输命令

#include <stdint.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
#include <fcntl.h>
#include <sys/ioctl.h>
#include <linux/types.h>
#include <linux/spi/spidev.h>

#define ARRAY_SIZE(a) (sizeof(a) / sizeof((a)[0]))
static void transfer(int fd, uint16_t* tx, uint8_t cs);
void init_display(int fd);

static void pabort(const char *s)
{
perror(s);
abort();
}

static const char *device = "/dev/spidev2.1";
static uint8_t mode=3;
static uint8_t bits = 8;
static uint32_t speed = 40000;
static uint16_t delay;

//initialization of display

void init_display(int fd)
{
int ret;
uint16_t tx[] = {
0x0F8, 0x000, 0x0C0,
0x0F8, 0x060, 0x000,
0x0F8, 0x060, 0x0C0,
0x0FC, 0x090, 0x000,
0x0FC, 0x000, 0x020,
0x0FE, 0x000, 0x000,
0x0F8, 0x000, 0x0C0,
0x0F8, 0x0F0, 0x000,
0x0F8, 0x080, 0x000,

};

for (ret=0; ret<27; ret++)
{
transfer(fd,&tx[ret], 0);
usleep(2000);
}
}
static void transfer(int fd, uint16_t* tx, uint8_t cs){
int ret;
uint16_t rx[ARRAY_SIZE(tx)] = {0, };
struct spi_ioc_transfer tr = {
.tx_buf = (unsigned long)tx,
.rx_buf = 0,
.len = 1,
.delay_usecs = delay,
.speed_hz = speed,
.bits_per_word = bits,
.cs_change=cs,
};
printf("%u", cs);

ret = ioctl(fd, SPI_IOC_MESSAGE(1), &tr);
if (ret < 1)
pabort("can't send spi message");

// for (ret = 0; ret < ARRAY_SIZE(tx); ret++) {
// if (!(ret % 6))
puts("");
// printf("%.2X ", rx[ret]);
}
// puts("");


static void print_usage(const char *prog)
{
printf("Usage: %s [-DsbdlHOLC3]\n", prog);
puts(" -D --device device to use (default /dev/spidev1.1)\n"
" -s --speed max speed (Hz)\n"
" -d --delay delay (usec)\n"
" -b --bpw bits per word \n"
" -l --loop loopback\n"
" -H --cpha clock phase\n"
" -O --cpol clock polarity\n"
" -L --lsb least significant bit first\n"
" -C --cs-high chip select active high\n"
" -3 --3wire SI/SO signals shared\n");
exit(1);
}

static void parse_opts(int argc, char *argv[])
{
while (1) {
static const struct option lopts[] = {
{ "device", 1, 0, 'D' },
{ "speed", 1, 0, 's' },
{ "delay", 1, 0, 'd' },
{ "bpw", 1, 0, 'b' },
{ "loop", 0, 0, 'l' },
{ "cpha", 0, 0, 'H' },
{ "cpol", 0, 0, 'O' },
{ "lsb", 0, 0, 'L' },
{ "cs-high", 0, 0, 'C' },
{ "3wire", 0, 0, '3' },
{ "no-cs", 0, 0, 'N' },
{ "ready", 0, 0, 'R' },
{ NULL, 0, 0, 0 },
};
int c;

c = getopt_long(argc, argv, "D:s:d:b:lHOLC3NR", lopts,
NULL);

if (c == -1)
break;

switch (c) {
case 'D':
device = optarg;
break;
case 's':
speed = atoi(optarg);
break;
case 'd':
delay = atoi(optarg);
break;
case 'b':
bits = atoi(optarg);
break;
case 'l':
mode |= SPI_LOOP;
break;
case 'H':
mode |= SPI_CPHA;
break;
case 'O':
mode |= SPI_CPOL;
break;
case 'L':
mode |= SPI_LSB_FIRST;
break;
case 'C':
mode |= SPI_CS_HIGH;
break;
case '3':
mode |= SPI_3WIRE;
break;
case 'N':
mode |= SPI_NO_CS;
break;
case 'R':
mode |= SPI_READY;
break;
default:
print_usage(argv[0]);
break;
}
}
}

int main(int argc, char *argv[])
{
int ret = 0;
int fd;

parse_opts(argc, argv);

fd = open(device, O_RDWR);
if (fd < 0)
pabort("can't open device");

/*
* spi mode
*/
ret = ioctl(fd, SPI_IOC_WR_MODE, &mode);
if (ret == -1)
pabort("can't set spi mode");

ret = ioctl(fd, SPI_IOC_RD_MODE, &mode);
if (ret == -1)
pabort("can't get spi mode");

/*
* bits per word
*/
ret = ioctl(fd, SPI_IOC_WR_BITS_PER_WORD, &bits);
if (ret == -1)
pabort("can't set bits per word");

ret = ioctl(fd, SPI_IOC_RD_BITS_PER_WORD, &bits);
if (ret == -1)
pabort("can't get bits per word");

/*
* max speed hz
*/
ret = ioctl(fd, SPI_IOC_WR_MAX_SPEED_HZ, &speed);
if (ret == -1)
pabort("can't set max speed hz");

ret = ioctl(fd, SPI_IOC_RD_MAX_SPEED_HZ, &speed);
if (ret == -1)
pabort("can't get max speed hz");

printf("spi mode: %d\n", mode);
printf("bits per word: %d\n", bits);
printf("max speed: %d Hz (%d KHz)\n", speed, speed/1000);

init_display(fd);
usleep(20000);
uint16_t hl[]={0X38,0X0E,0X06,0XF8,0X30,0x50,0x02};
trnsfer(fd,hl,0);

close(fd);

return ret;
}

最佳答案

您的液晶模块使用SSD1803 Controller 。根据the data sheet ,写操作需要传输3个字节。

第一个字节是起始字节,由 5 位“1”组成,后跟 R/W、RS 和结束位(“0”)。

接下来的 2 个字节是数据字节。低 4 位数据后面跟着四个“0”。然后,高 4 位数据后面跟着四个“0”。

  • 第一个字节:“1” “1” “1” “1” “1” R/W("0") RS("1") “0”
  • 第二个字节:D0 D1 D2 D3 “0” “0” “0” “0”
  • 第三个字节:D4 D5 D6 D7 “0” “0” “0” “0”

数据表中的图 7-11 说明了此操作。

enter image description here

此外,看起来您正在传递一个数组,但我认为传输函数仅传输一个字节,因为 .len = 1;

struct spi_ioc_transfer tr = {
.tx_buf = (unsigned long)tx,
.rx_buf = 0,
.len = 1, // <- HERE
.delay_usecs = delay,
.speed_hz = speed,
.bits_per_word = bits,
.cs_change=cs,
};

您想要传输 3 个字节。您可以修改该函数并根据数组的大小指定长度。或者,如果在传输之间保持 CS 为低电平,则可以循环遍历数组并一个接一个地发送字节。

关于c - 使用 beaglebone Black SPI 在 DIP203-6 LCD 屏幕上显示字符,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55443681/

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