16×16點(diǎn)陣LED顯示器漢字顯示—基于74HC595和74HC154驅(qū)動(dòng)

ID:268172 · 發(fā)表于 2018-1-16 10:06

點(diǎn)陣LED顯示器結(jié)構(gòu)及漢字字模。設(shè)計(jì)16×16點(diǎn)陣LED顯示器的驅(qū)動(dòng)電路，編寫程序?qū)崿F(xiàn)在16×16點(diǎn)陣LED顯示器上的漢字顯示。16×16點(diǎn)陣LED顯示器可由4塊8×8點(diǎn)陣LED顯示器構(gòu)成。可采用單片機(jī)并行端口、移位寄存器、譯碼器或鎖存器等4種驅(qū)動(dòng)方式中的組合實(shí)現(xiàn)對16×16點(diǎn)陣LED顯示器的驅(qū)動(dòng)
電路方面主要包括以下3部分。
（a）設(shè)計(jì)單片機(jī)的最小系統(tǒng)（包括復(fù)位電路和外接的晶振電路），并確定相關(guān)元器件參數(shù)。
（b）采用動(dòng)態(tài)驅(qū)動(dòng)的方式，設(shè)計(jì)單片機(jī)并行端口、移位寄存器、譯碼器或鎖存器等與16×16點(diǎn)陣LED顯示器的驅(qū)動(dòng)電路，主要包括控制點(diǎn)陣LED行和列的連線。
（c）采用移位寄存器、譯碼器或鎖存器驅(qū)動(dòng)點(diǎn)陣LED顯示器時(shí)，設(shè)計(jì)單片機(jī)與移位寄存器、譯碼器或鎖存器間的控制連線。

仿真原理圖如下（proteus仿真工程文件可到本帖附件中下載）

單片機(jī)源程序如下:

#include<regx51.h>
#include<intrins.h>
#define uchar unsigned char
#define clk P2_5
#define sclk P2_6
#define en P2_4
#define dat P2_7
void hang(uchar a)
{
uchar k;
//p2=0;
//en=1;
for(k=0;k<8;k++)
{
dat=a&0x80;
clk=1;
clk=0;
a<<=1;
}
}
delay1ms(unsigned char m)
{
unsigned char n;
while(m--)
{
for(n=0;n<122;n++);
}
}
void main(void)
{
uchar i,j,q;
uchar code a[32]=
{0x04,0x00,0x04,0x00,0x0F,0xE0,0x10,0x20,0x20,0x40,0x7F,0xF8,0xA1,0x08,0x21,0x08,
0x3F,0xF8,0x21,0x08,0x21,0x08,0x3F,0xF8,0x21,0x08,0x01,0x02,0x01,0x02,0x00,0xFE,};
uchar code b[32]=
{0x00,0x08,0x20,0x3C,0x17,0xC0,0x14,0x00,0x84,0x1C,0x45,0xE0,0x45,0x20,0x15,0x22,
0x15,0x24,0x25,0x28,0xE5,0x10,0x25,0x10,0x25,0x08,0x29,0x44,0x29,0x82,0x11,0x00,};
uchar code c[32]=
{0x10,0x00,0x10,0x00,0x3F,0xFC,0x20,0x00,0x4F,0xF0,0x80,0x00,0x3F,0xF0,0x00,0x10,
0x00,0x10,0x00,0x10,0x00,0x10,0x00,0x10,0x00,0x0A,0x00,0x0A,0x00,0x06,0x00,0x02,};
uchar code d[32]=
{0x00,0x40,0x00,0x40,0x00,0x40,0xFE,0x40,0x11,0xFC,0x10,0x44,0x10,0x44,0x10,0x44,
0x10,0x44,0x10,0x84,0x10,0x84,0x1E,0x84,0xF1,0x04,0x41,0x04,0x02,0x28,0x04,0x10,};
while(1)
{
P2=0;
en=1;
for(j=0;j<62;j++)
{
for(i=0;i<32;i=i+2)
{
hang(a[i]);
hang(a[i+1]);
q=(31-i)/2;
P2=P2&0xf0;
P2=P2|q;
sclk=1;
sclk=0;
delay1ms(1);
}
}
for(j=0;j<62;j++)
{
for(i=0;i<32;i=i+2)
{
hang(b[i]);
hang(b[i+1]);
q=(31-i)/2;
P2=P2&0xf0;
P2=P2|q;
sclk=1;
sclk=0;
delay1ms(1);
}
}
for(j=0;j<62;j++)
{
for(i=0;i<32;i=i+2)
{
hang(c[i]);
hang(c[i+1]);
q=(31-i)/2;
P2=P2&0xf0;
P2=P2|q;
sclk=1;
sclk=0;
delay1ms(1);
}
}
for(j=0;j<62;j++)
……………………
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