[英]Is there a possibility to glow the led in 5*5 matrix in a square shape with the common anode configuration
我有一个 LED 矩阵,它在行中是共阳极,在列中是共阴极。 我需要以正方形的形状发光矩阵(第一行 && 最后一行 && 第一列 && 最后一列),但我做不到。 我能够分别发光第一行和最后一行以及第一列和第二列)
void first_last_row();
void first_last_column();
int Led_Row_Pins[] = { 2 , 3 , 4 , 5 , 6 } ; // Anode pins are shorted in row_wise_manner
int Led_Column_Pins[] = {8 , 9 , 10 , 11 , 12} ; // Column Pins are shorted in column_wise_manner
int Loop_Count = 5 ;
int Wait_Time_On = 1000 ;
int Wait_Time_Off = 500 ;
int i = 0 ;
int j = 0 ;
int state = 1 ;
void setup() {
for( i = 0 ; i < Loop_Count ; i++ ){ // Anode Pins are connected in row_wise manner and are made LOW so that they dont conduct
pinMode(Led_Row_Pins[i],OUTPUT);
digitalWrite(Led_Row_Pins[i],LOW);
pinMode(Led_Column_Pins[i],OUTPUT); // Cathode Pins are connected in column_wise manner and are made HIGH so that they dont conduct
digitalWrite(Led_Column_Pins[i],HIGH);
}
}
void loop() {
first_last_row();
delay(1000);
first_last_column();
}
void first_last_row()
{
for( i = 0 ; i < Loop_Count ; i++ )// Led First And Last Row
{
for( j = 0 ; j < Loop_Count ; j++)
{
if( i == 0 || i == 4 )
{
digitalWrite(Led_Row_Pins[i],state); //Led_On_State
digitalWrite(Led_Column_Pins[j],!state);
}
}
}
}
void first_last_column()
{
for( j = 0 ; j < Loop_Count ; j++ )// Led First And Last Column
{
for( i = 0 ; i < Loop_Count ; i++)
{
if( j == 0 || j == 4 )
{
digitalWrite(Led_Row_Pins[i],state);// Led_On_State
digitalWrite(Led_Column_Pins[j],!state);
}
}
}
}
我需要将第一行和最后一行以及第一列和最后一列的 LED 一起发光,以便 forms 呈正方形,但我只能将第一行和最后一行以及第一列和最后一列分别发光。
在您的特定情况下,您可以打开第一行和最后一行,例如 1 毫秒,然后打开第一列和最后一列 1 毫秒,然后重复此操作。 我进一步建议在打开新模式之前关闭所有 LED。
void loop() {
static int show_rows = 0;
switch_all_off();
if(show_rows) {
first_last_row();
} else {
first_last_column();
}
show_rows = !show_rows;
delay(10);
}
void switch_all_off()
{
int off = 0;
for( j = 0 ; j < Loop_Count ; j++ )
{
digitalWrite(Led_Row_Pins[i],off);
digitalWrite(Led_Column_Pins[j],!off);
}
}
通过这第一步,行和列中包含的 LED 将更亮。 要解决此问题,您可以从行或列中排除角落 LED,例如
void first_last_column()
{
for( j = 0 ; j < Loop_Count ; j++ )// Led First And Last Column
{
for( i = 0 ; i < Loop_Count ; i++)
{
if((i != 0) && (i != 4) && ( j == 0 || j == 4 ))
{
digitalWrite(Led_Row_Pins[i],state);// Led_On_State
digitalWrite(Led_Column_Pins[j],!state);
}
}
}
}
对于更通用的解决方案,您必须遍历行,逐行激活并根据当前行的需要切换列引脚。
#define MAXLEDS 5
int states[MAXLEDS][MAXLEDS] = {
{ 1, 0, 0, 0, 1 },
{ 0, 1, 0, 1, 0 },
{ 0, 0, 1, 0, 0 },
{ 0, 1, 0, 1, 0 },
{ 1, 0, 0, 0, 1 }
};
void switch_leds(int row) {
int i;
/* switch off all rows */
for(i = 0; i < MAXLEDS, i++) {
digitalWrite(Led_Row_Pins[i], 0);
}
/* switch columns according to current row */
for(i = 0; i < MAXLEDS; i++) {
digitalWrite(Led_Column_Pins[i], !states[row][i]);
}
/* switch on current row */
digitalWrite(Led_Row_Pins[row], 1);
}
void loop() {
static int row = 0;
/* switch on LEDs in a single row */
switch_leds(row);
/* next row */
row++; row %= MAXLEDS;
/* The processing delay between calls to loop() is added to this delay. */
delay(5);
}
所有代码示例都未经测试,可能包含错误。
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