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            51單片機(jī)-數(shù)碼管

            作者: 時(shí)間:2016-12-02 來(lái)源:網(wǎng)絡(luò) 收藏


            #include <reg52.h>

            void delay(){
            int i,j;
            for(i = 0; i < 0xff; i++)
            for(j = 0; j < 0xff; j++)
            ;
            }
            unsigned char code duan[]={
            0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D,0x07,0x7F,0x6F,
            0x77,0x7C,0x39,0x5E,0x79,0x71};
            unsigned char code wei[]={
            0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07};

            void show(unsigned char we,unsigned char du){
            P1 = wei[we];
            P2 = duan[du];
            delay();
            }
            void main(){
            while(1){
            show(1,5);
            }
            }



            動(dòng)態(tài)顯示: 3個(gè)數(shù)碼管從0"999計(jì)數(shù)。





            #include <reg52.h>

            unsigned int code duan[]={
            0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D,0x07,0x7f,0x6f};

            unsigned int code wei[]={
            0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,0xfe,0xff};

            unsigned int num,sum;

            void delay(int k){
            int i,j;
            for(i = 0; i < k; i++)
            for(j = 0; j < 0xff; j++)
            ;
            }

            void display(unsigned int bai, unsigned int shi, unsigned int ge){
            P1 = 0x0;
            P2 = duan[bai];
            delay(1); //少延時(shí)一些數(shù)碼管更亮,如果延時(shí)很大就
            P1 = 0x1; //變成靜態(tài)的了
            P2 = duan[shi];
            delay(1);
            P1 = 0x2;
            P2 = duan[ge];
            delay(1);
            }
            void main(){
            EA = 1;
            ET1 = 1;
            TMOD = 0x10;
            TR1 = 1;
            num = 0;
            sum = 0;
            TH1 = (65535 - 50000) / 256;
            TL1 = (65535 - 50000) % 256;
            while(1){
            if(num == 20){ //每一秒計(jì)數(shù)增加1
            num = 0;
            if(sum == 1000)
            sum = 0; //用3位數(shù)碼管顯示,當(dāng)達(dá)到1000時(shí)從頭顯示
            sum++;
            }
            display(sum/100,sum%100/10,sum%10);
            }
            }

            void time1() interrupt 3{
            TH1 = (65535 - 50000) / 256;
            TL1 = (65535 - 50000) % 256;
            num++;
            }








            #include <reg52.h>

            unsigned char code duan[]={
            0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D};

            void delay(int k){
            int i,j;
            for(i = 0; i < k; i++)
            for(j = 0; j < 0xff; j++)
            ;
            }

            void display(){
            int i;
            for(i = 1; i < 7; i++){
            P1 = i-1;
            P2 = duan[7-i];
            delay(1);
            }
            }

            void main(){
            while(1){
            display();
            }
            }







            #include <reg52.h>

            unsigned int num,sum;

            unsigned int code duan[]={
            0x3F,0x06,0x5B,0x4F,0x66,0x6D,0x7D,0x07,0x7f,0x6f};

            void delay(int m){
            int i,j;
            for(i = 0; i < m; i++)
            for(j = 0; j < 0xff; j++)
            ;
            }

            void display(int n1,int n2,int n3){
            P1 = 0;
            P2 = duan[n1];
            delay(1);
            P1 = 1;
            P2 = duan[n2];
            delay(1);
            P1 = 2;
            P2 = duan[n3];
            delay(1);
            }

            void main(){
            EA = 1;
            ET1 = 1;
            TMOD = 0x10;
            TR1 = 1;
            sum = 0;
            num = 0;
            TH1 = (65535 - 50000) / 256;
            TL1 = (65535 - 50000) % 256;
            while(1){
            if(num == 1){
            num = 0;
            sum++;
            }
            if(sum == 1000)
            sum = 0;
            display(sum/100,sum%100/10,sum%10);
            }
            }

            void time1() interrupt 3{
            TH1 = (65535 - 10000) / 256;
            TL1 = (65535 - 10000) % 256;
            num++;
            }

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