[PIC] 18F452 ADCON1 PORTAbits.RA0 & PORTAbits.RA1 as input Digital

[ppperez]

Hello MASTERS.

Someone can tell me what I am doing wrong....

I am trying to understand the PIC18F452, I did a couple of programs, I am trying to understand the Digital input signal.

I connected a couple of push bottoms on RA0 & RA1 to simulate a digital signal, the output will be the PORTB, I am doing a counter. With RA0 add 1 and RA1 menus 1.

According to the Datasheet in order to get all inputs on PORTA as Digital PCFG3CFG0 011X and ADCON0 I understood it's not necessary.

This is my easy code and I don't have any output, I don't know if I need to read more or my PIC is damaged.

char salida = 0;
void main(int argc, char** argv) {
    TRISB = 0;    //Puerto B output
    TRISA = 1;    //Puerto A input
    ADCON1 = 0x10000111;
    //       0x1-------    Justificacion a la derecha
    //       0x-0------    Fosc/2
    //       0x--00----    No implemented
    //       0x----0110    AN0 -> AN7 = Digitals

    
    while(1)
    {
        if (PORTAbits.RA0 == 1)
        {
            salida = salida + 1;
            while(PORTAbits.RA0 == 1)
            {}
        }
        if (PORTAbits.RA1 == 1)
        {
            salida = salida - 1;
            while(PORTAbits.RA1 == 1)
            {}
        }
        PORTB = salida;
        }
  
            }

I hope that someone can give me some ideas.

Thanks in advance.

 

[esp1]

hi p,
The bit pattern appears to be set for Analog inputs for ADCON1 xxxx0111
E

 

[ppperez]

Hello esp1.

Thank you for your prompt answer. I found my error, the problem was my definition of ADCON1, x means hexadecimal and d decimal, in this case I have to use decimal so

wrong definition:
ADCON1 = 0x10000111;

Correct definition of ADCON1
ADCON1 = 0b10000111;

I am very happy, I am understanding more about microcontrollers.

Thanks again...