單片機4*4矩陣鍵盤接口電路程序 Proteus仿真圖

ID:449849 · 發(fā)表于 2022-4-8 10:00

設(shè)計一個單片機鍵盤接口電路，實現(xiàn)4*4矩陣鍵盤，用數(shù)碼管顯示相應按鍵內(nèi)容，當指定某一個按鍵連續(xù)被按三次時，顯示“口”字符號，按鍵有去抖動功能，數(shù)碼管可以通過清零鍵清零。

仿真原理圖如下（proteus仿真工程文件可到本帖附件中下載）

單片機源程序如下:

//0x3F, 0x06, 0x5B, 0x4F, 0x66, 0x6D, 0x7D, 0x07, 0x7F, 0x6F, 0x77, 0x7C, 0x39, 0x5E, 0x79, 0x71 共陰
//0xC0, 0xF9, 0xA4, 0xB0, 0x99, 0x92, 0x82, 0xF8, 0x80, 0x90, 0x88, 0x83, 0xC6, 0xA1, 0x86, 0x8E 共楊
#include <reg52.h>
#define GPIO_DIG P0 //段選數(shù)碼管
#define GPIO_KEY P2 //矩陣按鍵
sbit led = P1^0;
typedef unsigned char uchar;
typedef unsigned int uint;
uint keynum = 0;
uint count=0;
uchar code table[]={0x3F, 0x06, 0x5B, 0x4F, 0x66, 0x6D, 0x7D, 0x07, 0x7F, 0x6F, 0x77, 0x7C, 0x39, 0x5E, 0x79, 0x71};
uchar code table0[]={0x5c};
void delay( uchar i ){
while( i-- );
}
void key44scan(void)//鍵盤掃描函數(shù)
{
// uint count=0;
uint t;
P2=0xfe;
t=P2;
t=t&0xf0;
if(t!=0xf0){
delay(100);
if(t!=0xf0)
{
t=P2;
count++;
switch(t)
{
case 0xee:keynum=1; break;
case 0xde:keynum=5; break;
case 0xbe:keynum=9; break;
case 0x7e:keynum=13; break;
}
P0=table[keynum];
while(t!=0xf0)
{
t=P2;
t=t&0xf0;
}
}
}
P2=0xfd;
t=P2;
t=t&0xf0;
if(t!=0xf0){
delay(100);
if(t!=0xf0)
{
t=P2;
count++;
switch(t)
{
case 0xed:keynum=2; break;
case 0xdd:keynum=6; break;
case 0xbd:keynum=10; break;
case 0x7d:keynum=14; break;
}
P0=table[keynum];
while(t!=0xf0)
{
t=P2;
t=t&0xf0;
}
}
}
P2=0xfb;
t=P2;
t=t&0xf0;
if(t!=0xf0){
delay(100);
if(t!=0xf0)
{
t=P2;
count++;
switch(t)
{
case 0xeb:keynum=3; break;
case 0xdb:keynum=7; break;
case 0xbb:keynum=11; break;
case 0x7b:keynum=15; break;
}
P0=table[keynum];
while(t!=0xf0)
{
t=P2;
t=t&0xf0;
}
}
}
P2=0xf7;
t=P2;
t=t&0xf0;
if(t!=0xf0){
delay(100);
if(t!=0xf0)
{
t=P2;
count++;
switch(t)
{
case 0xe7:keynum=4; break;
case 0xd7:keynum=8; break;
case 0xb7:keynum=12; break;
case 0x77:keynum=0; break;
}
P0=table[keynum];
while(t!=0xf0)
{
t=P2;
t=t&0xf0;
}
}
}
// if(count==3){
// count = 0;
// led =~ led;
// P0=table0[0];
// delay(1000);
// }
}
void main (){
led = 0;
while(1) {
key44scan();
if(count==3){
//count = 0;
led =~ led;
P0=table0[0];
// delay(1000);
}
if(count==4){
count = 0;
led =~ led;
P0=table[0];
// delay(1000);
}
}
}

復制代碼

Keil代碼與Proteus仿真下載:

實驗2.7z (65.26 KB, 下載次數(shù): 48)

ID:930389 · 發(fā)表于 2022-7-1 00:00

請問一下那個排阻需要多大的？10k 還是 1k

ID:278457 · 發(fā)表于 2025-5-20 03:11

kiki2314 發(fā)表于 2022-7-1 00:00
請問一下那個排阻需要多大的？10k 還是 1k

1K就行

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單片機4*4矩陣鍵盤接口電路程序 Proteus仿真圖

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