我在尝试使用两个595移位寄存器在4位7seg显示屏上输出数字时遇到了麻烦。
我已经到了正确显示数字的地步,但我现在遇到的问题是输出正在显示的数字之间闪烁一些垃圾。我该如何防止这种情况发生?
我非常确定问题在于,因为我使用字节发送到寄存器,所以它在显示的字节之间进行锁存。
这是我的代码
int latchPin = 5;
int clockPin = 6;
int dataPin = 4;
int i = 0;
int waitTime = 500;
// digits from the right
byte colDig[4] =
{
B00001000, // digit 1
B00000100, // digit 2
B00000010, // digit 3
B00000001, // digit 4
};
const byte digit[10] = //seven segment digits in bits
{
B11000000, // 0
B11111001, // 1
B10100100, // 2
B10110000, // 3
B10011001, // 4
B10010010, // 5
B10000010, // 6
B11111000, // 7
B10000000, // 8
B10011000, // 9
};
void setup()
{
pinMode(latchPin, OUTPUT);
pinMode(dataPin, OUTPUT);
pinMode(clockPin, OUTPUT);
}
void loop()
{
// step through each digit then increment
// the counter by one, until nine
for(int j = 0;j<9;j++){
updateShiftRegister(0, j);
delay(waitTime);
updateShiftRegister(1, j);
delay(waitTime);
updateShiftRegister(2, j);
delay(waitTime);
updateShiftRegister(3, j);
delay(waitTime);
}
}
void updateShiftRegister(int col, int num)
{
digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST, colDig[col]);
shiftOut(dataPin, clockPin, MSBFIRST, digit[num]);
digitalWrite(latchPin, HIGH);
}
答案 0 :(得分:1)
所以看起来我有点正确,
shiftOut功能在功能结束时将时钟引脚设置为低电平,有效地强制锁存。
通过略微修改此页面上的代码,我能够阻止它,现在它完美无缺。 http://arduino.cc/en/Tutorial/ShftOut23
// the heart of the program
void shiftItOut(int myDataPin, int myClockPin, byte myDataOut) {
// This shifts 8 bits out MSB first,
//on the rising edge of the clock,
//clock idles low
//internal function setup
int i=0;
int pinState;
pinMode(myClockPin, OUTPUT);
pinMode(myDataPin, OUTPUT);
//clear everything out just in case to
//prepare shift register for bit shifting
digitalWrite(myDataPin, 0);
digitalWrite(myClockPin, 0);
//for each bit in the byte myDataOut
//NOTICE THAT WE ARE COUNTING DOWN in our for loop
//This means that %00000001 or "1" will go through such
//that it will be pin Q0 that lights.
for (i=7; i>=0; i--) {
digitalWrite(myClockPin, 0);
//if the value passed to myDataOut and a bitmask result
// true then... so if we are at i=6 and our value is
// %11010100 it would the code compares it to %01000000
// and proceeds to set pinState to 1.
if ( myDataOut & (1<<i) ) {
pinState= 1;
}
else {
pinState= 0;
}
//Sets the pin to HIGH or LOW depending on pinState
digitalWrite(myDataPin, pinState);
//register shifts bits on upstroke of clock pin
digitalWrite(myClockPin, 1);
//zero the data pin after shift to prevent bleed through
digitalWrite(myDataPin, 0);
}
//stop shifting
//digitalWrite(myClockPin, 0);
}
简单地评论说最后一个digitalWrite修复了它。