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我自己用8×8连16×16，为什么不能用啊，只有第八行和第16行亮

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ID:453993 发表于 2018-12-25 01:33 | 显示全部楼层 |阅读模式


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ID:446674 发表于 2018-12-25 09:16 | 显示全部楼层
请说清楚问题，发照片、程序、电路图
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ID:155507 发表于 2018-12-25 09:25 | 显示全部楼层
程序呢?线路图呢?什么都没有,人家想帮你也帮不上呀

我给你来个试试
  1. /*-----------------------------------------------
  2.   名称：595控制4汉字点阵屏
  3.   修改：无 51单片机
  4.   内容：屏+5V、GND连接电源5V
  5.         P0.0-P0.3分别连接A、B、C、D，用于控制16行，通过2路HC138控制
  6.         SCK时钟连接P2.5
  7.         LATCH锁存连接P2.6
  8.         R1红色数据信号连接至P0.4
  9.         G1绿色数据信号连接至P2.4，如果使用单红色屏此信号无需连接
  10.         OE使能端一般接GND，可以用IO引脚PWM控制亮度
  11.         或者使用此引脚控制拖尾现象。
  12.   现象：使用普通速度51系列单片机。单个汉字屏出现红色“德”
  13. ------------------------------------------------*/
  14. #include<reg52.h>

  16. sbit T_STR = P2^6;       //锁存引脚，屏上标识位STR
  17. sbit R1    = P0^4;       //数据引脚，屏上标识R1
  18. sbit G1    = P2^4;       //数据引脚，屏上标识G1
  19. sbit T_CLK = P2^5;       //时钟引脚，屏上标识位CLK
  20. sbit OE    = P2^1;       //使能引脚，屏上标识EN/OE

  22. unsigned char tab[8];
  23. unsigned char DIS[]={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};//行数据

  25. /*-----------------------------------------------
  26.              16x16汉字取模数据
  27. ------------------------------------------------*/
  28. unsigned char code hztest[][32]=    //二维数组，目的是减少运算，可以用最基本的一维数组，每个汉字有32个字节数据
  29. {
  30. /*--  文字:  德  --*/
  31. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  32. 0x08,0x02,0x08,0x02,0xF4,0x7F,0x02,0x02,0xE9,0x3F,0x28,0x25,0x24,0x25,0xE6,0x3F,
  33. 0x05,0x00,0xF4,0x7F,0x04,0x02,0x04,0x24,0xA4,0x44,0xA4,0x50,0x94,0x10,0x04,0x1F,

  35. /*--  文字:  飞  --*/
  36. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  37. 0x00,0x00,0xFF,0x03,0x00,0x02,0x00,0x22,0x00,0x12,0x00,0x0A,0x00,0x06,0x00,0x0A,
  38. 0x00,0x12,0x00,0x22,0x00,0x04,0x00,0x04,0x00,0x48,0x00,0x50,0x00,0x60,0x00,0x40,

  40. /*--  文字:  莱  --*/
  41. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  42. 0x10,0x04,0x10,0x04,0xFF,0x7F,0x90,0x04,0x80,0x00,0xFC,0x1F,0x80,0x00,0x88,0x08,
  43. 0x90,0x04,0xFE,0x3F,0xC0,0x01,0xA0,0x02,0x90,0x04,0x8C,0x18,0x83,0x60,0x80,0x00,
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led16x16_2.jpg
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ID:155507 发表于 2018-12-25 10:03 | 显示全部楼层

  2. /*-----------------------------------------------
  3.   名称：595控制4汉字点阵屏
  4.   修改：无 51单片机
  5.   内容：屏+5V、GND连接电源5V
  6.   现象：使用普通速度51系列单片机。单个汉字屏出现红色“德”
  7. ------------------------------------------------*/
  8. #include<reg52.h>

  10. sbit T_STR = P2^6;       //锁存引脚，屏上标识位STR
  11. sbit R1    = P0^4;       //数据引脚，屏上标识R1
  12. sbit G1    = P2^4;       //数据引脚，屏上标识G1
  13. sbit T_CLK = P2^5;       //时钟引脚，屏上标识位CLK
  14. sbit OE    = P2^1;       //使能引脚，屏上标识EN/OE

  16. unsigned char tab[8];
  17. unsigned char DIS[]={0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};//行数据

  19. /*-----------------------------------------------
  20.              16x16汉字取模数据
  21. ------------------------------------------------*/
  22. unsigned char code hztest[][32]=    //二维数组，目的是减少运算，可以用最基本的一维数组，每个汉字有32个字节数据
  23. {
  24. /*--  文字:  德  --*/
  25. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  26. 0x08,0x02,0x08,0x02,0xF4,0x7F,0x02,0x02,0xE9,0x3F,0x28,0x25,0x24,0x25,0xE6,0x3F,
  27. 0x05,0x00,0xF4,0x7F,0x04,0x02,0x04,0x24,0xA4,0x44,0xA4,0x50,0x94,0x10,0x04,0x1F,

  29. /*--  文字:  飞  --*/
  30. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  31. 0x00,0x00,0xFF,0x03,0x00,0x02,0x00,0x22,0x00,0x12,0x00,0x0A,0x00,0x06,0x00,0x0A,
  32. 0x00,0x12,0x00,0x22,0x00,0x04,0x00,0x04,0x00,0x48,0x00,0x50,0x00,0x60,0x00,0x40,

  34. /*--  文字:  莱  --*/
  35. /*--  宋体12;  此字体下对应的点阵为：宽x高=16x16   --*/
  36. 0x10,0x04,0x10,0x04,0xFF,0x7F,0x90,0x04,0x80,0x00,0xFC,0x1F,0x80,0x00,0x88,0x08,
  37. 0x90,0x04,0xFE,0x3F,0xC0,0x01,0xA0,0x02,0x90,0x04,0x8C,0x18,0x83,0x60,0x80,0x00,

  39. };
  40. /*-----------------------------------------------
  41.              向595写入一个字节 单红色
  42. ------------------------------------------------*/
  43. void InputByte( unsigned  char DataR1) //写一个字节
  44. {
  45. unsigned  char i;
  46. for(i=8; i>0; i--)
  47. {
  48. R1 = !(DataR1&0x01);                  
  49. T_CLK = 0;
  50. T_CLK = 1;
  51. DataR1 = DataR1 >> 1;
  52. }
  53. }
  54. /*-----------------------------------------------
  55.              向595写入两个字节 双色
  56. ------------------------------------------------*/
  57. /*
  58. void Input2Byte( unsigned  char DataR1,unsigned  char DataG1) //写一个字节
  59. {
  60. unsigned  char i;
  61. for(i=8; i>0; i--)
  62. {
  63. R1 = !(DataR1&0x01);   
  64. G1 = !(DataG1&0x01);               
  65. T_CLK = 0;
  66. T_CLK = 1;
  67. DataR1 = DataR1 >> 1;
  68. DataG1 = DataG1 >> 1;
  69. }
  70. }
  71. */

  73. /*-----------------------------------------------
  74.             初始化定时器，做为动态扫描
  75. ------------------------------------------------*/
  76. void Init_Timer1(void)
  77. {
  78. TMOD |= 0x10;                             
  79. EA=1;                      //总中断使能
  80. ET1=1;                     //定时中断使能
  81. TR1=1;                     //打开定时器开关
  82. }


  85. /*-----------------------------------------------
  86.                       主程序
  87. ------------------------------------------------*/
  88. main()
  89. {
  90. Init_Timer1(); //初始化定时器

  92. while(1)
  93. {
  94.   //可以加入其它的控制程序
  95. }
  96. }

  98. /*-----------------------------------------------
  99.              定时器执行动态扫描
  100. ------------------------------------------------*/
  101. void Timer1_isr(void) interrupt 3
  102. {

  104. static unsigned char count,j;
  105. unsigned char i;
  106. TH1=0xff;                //重装初值
  107. TL1=0;


  110. T_STR=0;       //锁存释放
  111.    
  112. for(j=0;j<1;j++) //取每个汉字的前2个字节，数据传输完成后锁存输出即可 j表示汉字个数，如果是3个汉字屏就可以修改成3
  113.         {
  114.     InputByte(~hztest[j][i]);
  115.     InputByte(~hztest[j][i+1]);
  116.         }
  117. OE=1;  //暂时关闭屏幕，防止出现鬼影，如果没有使用该引脚控制则会出现拖影现象。
  118. T_STR=1;      //锁存有效，此时一行的数据显示到屏上
  119. P0=DIS[15-count]|0xf0;//用P0口的前4位控制16行，屏内部通过4-16译码器工作，循环扫描其他15行
  120. OE=0;//使能
  121. count++;
  122. i+=2;              //累加调用下一行数据
  123. if(count==16)
  124.     count=0;
  125. if(i==32)
  126.     i=0;

  128. }
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ID:453993 发表于 2018-12-25 10:50 来自手机 | 显示全部楼层
Arvin0011 发表于 2018-12-25 09:16
请说清楚问题，发照片、程序、电路图

#include <reg51.h>
unsigned char i;
sbit DATA1=P3^7;
sbit DATA2=P3^5;
sbit CLCK1=P3^6;
sbit CLCK2=P3^4;
#define uchar unsigned char
#define uint unsigned int
void shuru_1();
void shuru_2();
unsigned char num,k,temp1,temp2,zuo=0;
unsigned int timecount;
unsigned char code lie[16]={0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,};/*74hc154显示代码从左到右16列依次选通*/
unsigned char code hang[288]={
0x00,0x00,0x00,0x00,0x1F,0xF8,0x11,0x10,0x11,0x10,0x11,0x10,0x11,0x10,0xFF,0xFE,
0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x1F,0xF9,0x00,0x01,0x00,0x0F,0x00,0x00,

0x02,0x02,0x0A,0xEA,0xAA,0xAA,0xAA,0xAF,0xAA,0xAA,0xAA,0xEA,0x0A,0x02,0x02,0x00,
0x82,0x4A,0x2A,0x1E,0x0B,0x4A,0x8A,0x7A,0x02,0xFA,0x4B,0x4A,0x4A,0xFA,0x02,0x00,

0x00,0x80,0x60,0xF8,0x07,0x40,0x20,0x18,0x0F,0x08,0xC8,0x08,0x08,0x28,0x18,0x00,
0x01,0x00,0x00,0xFF,0x00,0x10,0x0C,0x03,0x40,0x80,0x7F,0x00,0x01,0x06,0x18,0x00,

0x00,0x00,0x00,0x7F,0x49,0x49,0x49,0x49,0x49,0x49,0x49,0x7F,0x00,0x00,0x00,0x00,
0x81,0x41,0x21,0x1D,0x21,0x41,0x81,0xFF,0x89,0x89,0x89,0x89,0x89,0x81,0x81,0x00,

0x40,0x40,0xC0,0x5F,0x55,0x55,0xD5,0x55,0x55,0x55,0x55,0x5F,0x40,0x40,0x40,0x00,
0x20,0x60,0x3F,0x25,0x15,0x15,0xFF,0x90,0x47,0x29,0x11,0x2D,0x43,0x80,0x80,0x00,

0x10,0x10,0xD0,0xFF,0x90,0x10,0x44,0x54,0xD4,0x74,0x5F,0x54,0xD4,0x54,0x44,0x00,
0x04,0x03,0x00,0xFF,0x00,0x01,0x12,0x11,0x14,0x14,0xFF,0x14,0x14,0x11,0x12,0x00,


0x00,0xF8,0x0C,0x0B,0x08,0x08,0xF8,0x40,0x30,0x8F,0x08,0x08,0x08,0xF8,0x00,0x00,
0x00,0x7F,0x21,0x21,0x21,0x21,0x7F,0x00,0x00,0x00,0x43,0x80,0x40,0x3F,0x00,0x00,


0x10,0x10,0x10,0xFF,0x10,0x10,0xF0,0x00,0x00,0xF8,0x08,0x08,0x08,0xF8,0x00,0x00,
0x80,0x40,0x30,0x0F,0x40,0x80,0x7F,0x00,0x00,0x7F,0x20,0x20,0x20,0x7F,0x00,0x00,

0x10,0x60,0x02,0x8C,0x00,0xF0,0x10,0x10,0x10,0xFF,0x10,0x10,0x10,0xF0,0x00,0x00,
0x04,0x04,0x7E,0x01,0x00,0xFF,0x42,0x42,0x42,0x7F,0x42,0x42,0x42,0xFF,0x00,0x00,

};
void shuru_1()
{uchar t;
for(t=0;t<8;t++)
{
temp1<<=1;
DATA1=CY;
CLCK1=1;
CLCK1=0;
}
}
void shuru_2()
{
uchar t;
for(t=0;t<8;t++)
{
temp2<<=1;
DATA2=CY;
CLCK2=1;
CLCK2=0;
}
}
void main(void)
{
TMOD=0x01;
TH0=(65536-2000)/256;
TL0=(65536-2000)%256;
TR0=1;
ET0=1;
EA=1;
while(1);
}
void serve_led()interrupt 1 using 0
{ k=2;
TH0=(65536-2000)/256;
TL0=(65536-2000)%256;
P2=0xff;
P2 = lie[num];
temp1=hang[num*2+zuo];
shuru_1();
P2 = lie[num];
temp2=hang[zuo+(num*2+1)];
shuru_2();
num++;
if(num==16)
{
num=0;
}
timecount++;
if(timecount==100)
{num=0;
timecount=0;
zuo=zuo+2;
if(zuo==272)
{
zuo=0;
}
}
}
75b1c773b64caa50.png
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ID:453993 发表于 2018-12-25 11:01 来自手机 | 显示全部楼层
angmall 发表于 2018-12-25 09:25
程序呢?线路图呢?什么都没有,人家想帮你也帮不上呀

我给你来个试试

#include <reg51.h>
unsigned char i;
sbit DATA1=P3^7;
sbit DATA2=P3^5;
sbit CLCK1=P3^6;
sbit CLCK2=P3^4;
#define uchar unsigned char
#define uint unsigned int
void shuru_1();
void shuru_2();
unsigned char num,k,temp1,temp2,zuo=0;
unsigned int timecount;
unsigned char code lie[16]={0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x0f,};/*74hc154显示代码从左到右16列依次选通*/
unsigned char code hang[288]={
0x00,0x00,0x00,0x00,0x1F,0xF8,0x11,0x10,0x11,0x10,0x11,0x10,0x11,0x10,0xFF,0xFE,
0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x1F,0xF9,0x00,0x01,0x00,0x0F,0x00,0x00,

0x02,0x02,0x0A,0xEA,0xAA,0xAA,0xAA,0xAF,0xAA,0xAA,0xAA,0xEA,0x0A,0x02,0x02,0x00,
0x82,0x4A,0x2A,0x1E,0x0B,0x4A,0x8A,0x7A,0x02,0xFA,0x4B,0x4A,0x4A,0xFA,0x02,0x00,

0x00,0x80,0x60,0xF8,0x07,0x40,0x20,0x18,0x0F,0x08,0xC8,0x08,0x08,0x28,0x18,0x00,
0x01,0x00,0x00,0xFF,0x00,0x10,0x0C,0x03,0x40,0x80,0x7F,0x00,0x01,0x06,0x18,0x00,

0x00,0x00,0x00,0x7F,0x49,0x49,0x49,0x49,0x49,0x49,0x49,0x7F,0x00,0x00,0x00,0x00,
0x81,0x41,0x21,0x1D,0x21,0x41,0x81,0xFF,0x89,0x89,0x89,0x89,0x89,0x81,0x81,0x00,

0x40,0x40,0xC0,0x5F,0x55,0x55,0xD5,0x55,0x55,0x55,0x55,0x5F,0x40,0x40,0x40,0x00,
0x20,0x60,0x3F,0x25,0x15,0x15,0xFF,0x90,0x47,0x29,0x11,0x2D,0x43,0x80,0x80,0x00,

0x10,0x10,0xD0,0xFF,0x90,0x10,0x44,0x54,0xD4,0x74,0x5F,0x54,0xD4,0x54,0x44,0x00,
0x04,0x03,0x00,0xFF,0x00,0x01,0x12,0x11,0x14,0x14,0xFF,0x14,0x14,0x11,0x12,0x00,


0x00,0xF8,0x0C,0x0B,0x08,0x08,0xF8,0x40,0x30,0x8F,0x08,0x08,0x08,0xF8,0x00,0x00,
0x00,0x7F,0x21,0x21,0x21,0x21,0x7F,0x00,0x00,0x00,0x43,0x80,0x40,0x3F,0x00,0x00,


0x10,0x10,0x10,0xFF,0x10,0x10,0xF0,0x00,0x00,0xF8,0x08,0x08,0x08,0xF8,0x00,0x00,
0x80,0x40,0x30,0x0F,0x40,0x80,0x7F,0x00,0x00,0x7F,0x20,0x20,0x20,0x7F,0x00,0x00,

0x10,0x60,0x02,0x8C,0x00,0xF0,0x10,0x10,0x10,0xFF,0x10,0x10,0x10,0xF0,0x00,0x00,
0x04,0x04,0x7E,0x01,0x00,0xFF,0x42,0x42,0x42,0x7F,0x42,0x42,0x42,0xFF,0x00,0x00,

};
void shuru_1()
{uchar t;
for(t=0;t<8;t++)
{
temp1<<=1;
DATA1=CY;
CLCK1=1;
CLCK1=0;
}
}
void shuru_2()
{
uchar t;
for(t=0;t<8;t++)
{
temp2<<=1;
DATA2=CY;
CLCK2=1;
CLCK2=0;
}
}
void main(void)
{
TMOD=0x01;
TH0=(65536-2000)/256;
TL0=(65536-2000)%256;
TR0=1;
ET0=1;
EA=1;
while(1);
}
void serve_led()interrupt 1 using 0
{ k=2;
TH0=(65536-2000)/256;
TL0=(65536-2000)%256;
P2=0xff;
P2 = lie[num];
temp1=hang[num*2+zuo];
shuru_1();
P2 = lie[num];
temp2=hang[zuo+(num*2+1)];
shuru_2();
num++;
if(num==16)
{
num=0;
}
timecount++;
if(timecount==100)
{num=0;
timecount=0;
zuo=zuo+2;
if(zuo==272)
{
zuo=0;
}
}
}
75b1c773b64caa50.png
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