数码管: 根据STC89C52的数码管规划详解

一. 硬件规划详解

阐明:
选通一个数码管分两步,
第一步是位选,即挑选哪个数码管亮, 这儿位选是经过锁存器U2的WE1~WE6来挑选第一个数码管到第六个数码管,因为数码管为共阴极(相对于8位数码管),那么这儿WE端为0代表位选选通,
第二步是段选,即挑选数码管的哪位亮, 这儿段选是经过锁存器U1的a~h来挑选数码管的8位管子,这儿DUAN端为1代表段选选通.

二. 程序规划详解
1. 让第一个数码管显现8
#include

sbit duan = P2 ^ 6;
sbit wela = P2 ^ 7;

void main()
{
wela = 1;
P0 = 0xfe;
wela = 0;

duan = 1;
P0 = 0x7f;
duan = 0;

while (1);
}

2. 让6个数码管显现重复显现0~F
#include
#define uchar unsigned char
#define uint unsigned int

sbit duan = P2 ^ 6;
sbit wela = P2 ^ 7;

uchar code table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

void delayms(uint xms);

void main()
{
uchar num;

wela = 1;
P0 = 0xc0;
wela = 0;

while (1)
{
for (num = 0; num < 16; num++)
{
duan = 1;
P0 = table[num];
duan = 0;

delayms(500);
}
}

}

void delayms(uint xms)
{
uint i, j;

for (i = xms; i > 0; i–)
for (j = 110; j > 0; j–)
;
}

3.
第一个数码管显现1,时刻为0.5s,然后封闭它,
第二个数码管显现2,时刻为0.5s,然后封闭它,
…
第六个数码管显现6,时刻为0.5s,然后封闭它,
顺次循环.
#include
#define uchar unsigned char
#define uint unsigned int

sbit duan = P2 ^ 6;
sbit wela = P2 ^ 7;

uchar code we_table[] = {
0xfe,0xfd,0xfb,0xf7,0xef,0xdf
};

uchar code du_table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

void delayms(uint xms);

void main()
{
uchar num;

while (1)
{
for (num = 0; num < 6; num++)
{
wela = 1;
P0 = we_table[num];
wela = 0;

P0 = 0xff;

duan = 1;
P0 = du_table[num+1];
duan = 0;

delayms(1);
}
}
}

void delayms(uint xms)
{
uint i, j;

for (i = xms; i > 0; i–)
for (j = 110; j > 0; j–)
;
}

阐明:
只要将延时时刻调整为1s,咱们能够看到数码管一同显现123456.
P0 = 0xff; 是用来消影, 即消除段选与位选之间的影响.



4.
守时器0的办法1完成第一个发光二极管以200ms距离闪耀,
守时器1的办法1完成数码管前两位59s循环计时
#include
#define uchar unsigned char
#define uint unsigned int

sbit dula = P2 ^ 6;
sbit wela = P2 ^ 7;
sbit led1 = P1 ^ 0;

uchar code table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

void delayms(uint);
void display(uint);

uchar num, num1, num2;

void main()
{
TMOD = 0x11;
TH0 = (65536 – 45872) / 256;
TL0 = (65536 – 45872) % 256;
TH1 = (65536 – 45872) / 256;
TL1 = (65536 – 45872) % 256;
EA = 1;
ET0 = 1;
ET1 = 1;
TR0 = 1;
TR1 = 1;
while (1)
{
display(num);
}
}

void display(uint num)
{
dula = 1;
P0 = table[num / 10];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = 0xfe;
wela = 0;

delayms(5);

dula = 1;
P0 = table[num % 10];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = 0xfd;
wela = 0;

delayms(5);
}

void delayms(uint xms)
{
uint i, j;

for (i = xms; i > 0; i–)
for (j = 110; j > 0; j–)
;
}

void T0_time() interrupt 1
{

TH0 = (65536 – 45872) / 256;
TL0 = (65536 – 45872) % 256;
num1++;
if (num1 == 4)
{
num1 = 0;
led1 = ~led1;
}
}

void T1_time() interrupt 3
{

TH1 = (65536 – 45872) / 256;
TL1 = (65536 – 45872) % 256;
num2++;
if (num2 == 20)
{
num2 = 0;
num++;
if (num == 60)
num = 0;
}
}

5. 用守时器以距离500MS在6位数码管上顺次显现0、1、2、3….C、D、E、F，重复
#include
#include

#define uint unsigned int
#define uchar unsigned char

sbit dula = P2 ^ 6;
sbit wela = P2 ^ 7;

uchar num, tt;
uchar code table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

void main()
{
num = 0;
tt = 0;

TMOD = 0x01;
TH0 = (65536 – 45872) / 256;
TL0 = (65536 – 45872) % 256;
EA = 1;
ET0 = 1;
TR0 = 1;

dula = 1;
P0 = 0x3f;
dula = 0;

wela = 1;
P0 = 0xc0;
wela = 0;

while (1)
{
if (tt == 10)
{
tt = 0;
num++;
if (num == 16)
num = 0;

dula = 1;
P0 = table[num];
dula = 0;
}
}
}

void T0_timer() interrupt 1
{
TH0 = (65536 – 45872) / 256;
TL0 = (65536 – 45872) % 256;
tt++;
}

6. 使用动态扫描办法在六位数码管上显现出安稳的654321
#include
#include

#define uint unsigned int
#define uchar unsigned char

sbit dula = P2 ^ 6;
sbit wela = P2 ^ 7;

uchar code du_table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

uchar code we_table[] = {
0xfe,0xfd,0xfb,0xf7,0xef,0xdf
};

void delayms(uint);

void main()
{
uchar num;

for (num = 6; num > 0; num–)
{
dula = 1;
P0 = du_table[num];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = we_table[6 – num];
wela = 0;

delayms(1);
}
}

void delayms(uint xms)
{
uint i, j;

for (i = xms; i > 0; i–)
for (j = 110; j > 0; j–)
;
}



7. 用动态扫描办法和守时器1在数码管的前三位显现出秒表，准确到1%秒，即后两位显现1%秒，一向循环下去
#include
#include

#define uint unsigned int
#define uchar unsigned char

sbit dula = P2 ^ 6;
sbit wela = P2 ^ 7;

uint tt;
uchar code du_table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71
};

void display(uint);
void delayms(uint);

void main()
{
TMOD = 0x10;
TH1 = (65536 – 10000) / 256;
TL1 = (65536 – 10000) % 256;
EA = 1;
ET1 = 1;
TR1 = 1;
while (1)
{
display(tt);
}
}

void T1_timer() interrupt 3
{
TH1 = (65536 – 10000) / 256;
TL1 = (65536 – 10000) % 256;
tt++;
if (tt == 1000)
tt = 0;
}

void display(uint num)
{
dula = 1;
P0 = du_table[num / 100];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = 0xfe;
wela = 0;

delayms(1);

dula = 1;
P0 = du_table[num % 100 / 10];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = 0xfd;
wela = 0;

delayms(1);

dula = 1;
P0 = du_table[num % 100 % 10];
dula = 0;

P0 = 0xff;

wela = 1;
P0 = 0xfb;
wela = 0;

delayms(1);
}

void delayms(uint xms)
{
uint i, j;

for (i = xms; i > 0; i–)
for (j = 110; j > 0; j–)
;
}

8. 使用动态扫描和守时器1在数码管上显现出从765432开端以1/10秒的速度往下递减直至765398并坚持显现此数，与此同时使用守时器0以500MS速度进行流水灯从上至下移动，当数码管上数减到中止时，试验板上流水灯也中止然后悉数开端闪耀，3秒后（用T0守时）流水灯悉数封闭、数码管上显现出“HELLO”。到此坚持住
#include
#include

#define uint unsigned int
#define uchar unsigned char

sbit dula = P2 ^ 6;
sbit wela = P2 ^ 7;

uchar code table[] = {
0x3f,0x06,0x5b,0x4f,
0x66,0x6d,0x7d,0x07,
0x7f,0x6f,0x77,0x7c,
0x39,0x5e,0x79,0x71,
0x76,0x79,0x38,0x3f,0
};

uchar temp, t0, t1, bai, shi, ge, flag, flag1;
uint shu;

void init();
void display(uchar, uchar, uchar, uchar, uchar, uchar);
void delayms(uint);

void main()
{
init();
while (1)
{
if (flag1 != 1)
display(7,6,5,bai,shi,ge);
else
display(16,17,18,18,19,20);
}
}

void init()
{
shu = 432;
temp = 0xfe;
P1 = temp;
TMOD = 0x11;
TH0 = (65536 – 50000) / 256;
TL0 = (65536 – 50000) % 256;
TH1 = (65536 – 50000) / 256;
TL1 = (65536 – 50000) % 256;
EA = 1;
ET0 = 1;
ET1 = 1;
TR0 = 1;

TR1 = 1;
}

void T0_timer() interrupt 1
{
TH0 = (65536 – 50000) / 256;
TL0 = (65536 – 50000) % 256;
t0++;
if (flag != 1)
{
if (t0 == 10)
{
t0 = 1;
temp = _crol_(temp,1);
P1 = temp;
}
}
else
{
if (t0 % 4 == 0)
{
P1 = ~P1;
}
if (t0 == 60)
{
TR0 = 0;
P1 = 0xff;
flag1 = 1;
}
}
}

void T1_timer() interrupt 3
{
TH1 = (65536 – 50000) / 256;
TL1 = (65536 – 50000) % 256;
t1++;
if (t1 == 2)
{
t1 = 0;
shu–;
bai = shu / 100;
shi = shu % 100 / 10;
ge = shu % 10;
if (shu == 398)
{
TR0 = 0;
TH0 = (65536 – 50000) / 256;
TL0 = (65536 – 50000) % 256;
TR0 = 1;
flag = 1;
t0 = 0;
P1 = 0xff;
TR1 = 0;
}
}
}

void display(uchar aa, uchar bb, uchar cc, uchar bai, uchar shi, uchar ge)
{
dula = 1;
P0 = table[aa];
dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xfe;
wela = 0;
delayms(1);

dula = 1;
P0 = table[bb];
dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xfd;
wela = 0;
delayms(1);

dula = 1;
P0 = table[cc];
dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xfb;
wela = 0;
delayms(1);

dula = 1;
P0 = table[bai];
dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xf7;
wela = 0;
delayms(1);

dula = 1;
P0 = table[shi];
dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xef;
wela = 0;
delayms(1);

dula = 1;
P0 = table[ge];

dula = 0;
P0 = 0xff;
wela = 1;
P0 = 0xdf;
wela = 0;
delayms(1);
}