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單片機(jī)點(diǎn)陣式LED簡單圖形顯示技術(shù)

作者: 時(shí)間:2012-08-01 來源:網(wǎng)絡(luò) 收藏

1.實(shí)驗(yàn)任務(wù)

本文引用地址:http://butianyuan.cn/article/186014.htm

在8X8顯示“★”、“●”和心形圖,通過按鍵來選擇要顯示的圖形。

2.電路原理圖

點(diǎn)陣式LED簡單圖形顯示技術(shù)Character displayer

3.硬件系統(tǒng)連線

(1).把“系統(tǒng)”區(qū)域中的P1端口用8芯排芯連接到“點(diǎn)陣模塊”區(qū)域中的“DR1-DR8”端口上;

(2).把“系統(tǒng)”區(qū)域中的P3端口用8芯排芯連接到“點(diǎn)陣模塊”區(qū)域中的“DC1-DC8”端口上;

(3).把“系統(tǒng)”區(qū)域中的P2.0/A8端子用導(dǎo)線連接到“獨(dú)立式鍵盤”區(qū)域中的SP1端子上;

4.程序設(shè)計(jì)內(nèi)容

(1).“★”在8X8點(diǎn)陣上顯示圖如下圖所示

1.jpg

12H,14H,3CH,48H,3CH,14H,12H,00H

(2).“●”在8X8點(diǎn)陣上顯示圖如下圖所示

2.jpg

00H,00H,38H,44H,44H,44H,38H,00H

(3).心形圖在8X8LED點(diǎn)陣上顯示圖如下圖所示

3.jpg

30H,48H,44H,22H,44H,48H,30H,00H

5.匯編源程序

CNTAEQU 30H

COUNTEQU 31H

ORG 00H

LJMP START

ORG 0BH

LJMP T0X

ORG 30H

START:MOV CNTA,#00H

MOV COUNT,#00H

MOV TMOD,#01H

MOV TH0,#(65536-4000) / 256

MOV TL0,#(65536-4000) MOD 256

SETB TR0

SETB ET0

SETB EA

WT:JB P2.0,WT

MOV R6,#5

MOV R7,#248

D1:DJNZ R7,$

DJNZ R6,D1

JB P2.0,WT

INC COUNT

MOV A,COUNT

CJNE A,#03H,NEXT

MOV COUNT,#00H

NEXT:JNB P2.0,$

SJMP WT

T0X: NOP

MOV TH0,#(65536-4000) / 256

MOV TL0,#(65536-4000) MOD 256

MOV DPTR,#TAB

MOV A,CNTA

MOVC A,@A+DPTR

MOV P3,A

MOV DPTR,#GRAPH

MOV A,COUNT

MOV B,#8

MUL AB

ADD A,CNTA

MOVC A,@A+DPTR

MOV P1,A

INC CNTA

MOV A,CNTA

CJNE A,#8,NEX

MOV CNTA,#00H

NEX:RETI

TAB:DB 0FEH,0FDH,0FBH,0F7H,0EFH,0DFH,0BFH,07FH

GRAPH:DB 12H,14H,3CH,48H,3CH,14H,12H,00H

DB 00H,00H,38H,44H,44H,44H,38H,00H

DB 30H,48H,44H,22H,44H,48H,30H,00H

END

6.C語言源程序

#include

unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};

unsigned char code graph[3][8]={{0x12,0x14,0x3c,0x48,0x3c,0x14,0x12,0x00},

{0x00,0x00,0x38,0x44,0x44,0x44,0x38,0x00},

{0x30,0x48,0x44,0x22,0x44,0x48,0x30,0x00}

};

unsigned char count;

unsigned char cnta;

void main(void)

{

unsigned char i,j;

TMOD=0x01;

TH0=(65536-4000)/256;

TL0=(65536-4000)%6;

TR0=1;

ET0=1;

EA=1;

while(1)

{

if(P2_0==0)

{

for(i=5;i>0;i--)

for(j=248;j>0;j--);

if(P2_0==0)

{

count++;

if(count==3)

{

count=0;

}

while(P2_0==0);

}

}

}

}

void t0(void) interrupt 1 using 0

{

TH0=(65536-4000)/256;

TL0=(65536-4000)%6;

P3=tab[cnta];

P1=graph[count][cnta];

cnta++;

if(cnta==8)

{

cnta=0;

}

}

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