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TIMER0 Interrupt Problem

 
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saeidjabbari



Joined: 25 Mar 2010
Posts: 8

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TIMER0 Interrupt Problem
PostPosted: Sun May 09, 2010 12:25 pm     Reply with quote

Hi for all
First of all excuse me about my English.
I want to use timer0 interrupt but it isn't work well!
where is the problem?
Code:

#include <16f877a.h>             // Define PIC Microcontroller Header
#fuses   HS                      // Set Oscillator On "High Speed" Mode
#use  delay(clock=20000000)   // Set 20MHz Oscillator For Microcontroller

#include <LCD.c>                 // Define LCD Header
#define  use_portd_lcd TRUE      // Set PORTD For LCD Input Pin

#int_timer0
void Timer0_isr()
{
 int i=0;
 
 lcd_init();
 while(i<11){
 printf(lcd_putc,"\fCOUNT=%u",i);
 delay_ms(100);
 i++;}
 return;
}

//----------------     PIN CONFIG     ----------------
#byte PORTA=0x05                 // Call PORTA Register From BANK0
#byte PORTB=0x06                 // Call PORTB Register From BANK0
#byte PORTC=0x07                 // Call PORTC Register From BANK0
#byte PORTD=0x08                 // Call PORTD Register From BANK0
#byte PORTE=0x09                 // Call PORTE Register From BANK0
#byte TRISA=0x85                 // Call TRISA Register From BANK0
#byte TRISB=0x86                 // Call TRISB Register From BANK0
#byte TRISC=0x87                 // Call TRISC Register From BANK0
#byte TRISD=0x88                 // Call TRISD Register From BANK0
#byte TRISE=0x89                 // Call TRISE Register From BANK0
//----------------------------------------------------
#bit  TMR0IF=0x0B.2



void main()
 {
   TRISA=0;
   TRISB=0;
   TRISC=0;
   TRISE=0;
   PORTC=0;
   setup_timer_0 (RTCC_INTERNAL|RTCC_DIV_256);
   interrupt_active(INT_TIMER0);
   enable_interrupts(INT_TIMER0);
   enable_interrupts(GLOBAL);

 for(;;)
  {
   
   
   PORTB=0b00000001;
   delay_ms(1000);
   
   if(get_timer0()==0){set_timer0(0);};
   
   PORTB=0b00000000;
   delay_ms(1000);

   
   

  }
 }


This link is CODE and SCH:
http://www.4shared.com/file/bTwzl32U/TIMER0INTRRUPT.html
PCM programmer



Joined: 06 Sep 2003
Posts: 21708

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PostPosted: Sun May 09, 2010 1:25 pm     Reply with quote

Tell us the purpose of your project. What do you want to do ?
saeidjabbari



Joined: 25 Mar 2010
Posts: 8

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PostPosted: Sun May 09, 2010 10:42 pm     Reply with quote

Hi my friend
It's easy! I want when timer0 is overflow, jump to interrupt loop and count 0 to 10, and go back to main loop and turn off led. But it's not done correctly.
If download schematic and code you can understand my opinion.

Thanks For Advice
Ttelmah



Joined: 11 Mar 2010
Posts: 19552

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PostPosted: Mon May 10, 2010 1:55 am     Reply with quote

Code:

int1 timer_overflow=false;

#int_timer0
void Timer0_isr() {
   timer_overflow=true;
}

void main()  {
   int i;
   output_b(0);
   output_c(0);
   output_d(0);
   output_e(0);
   setup_timer_0 (RTCC_INTERNAL|RTCC_DIV_256);
   enable_interrupts(INT_TIMER0);
   enable_interrupts(GLOBAL);
   for(;;) {
     if (timer_overflow) {
        timer_overflow=false;
        output_high(PIN_B0); 
        lcd_init();
        for (i=0;i<11;i++){
           printf(lcd_putc,"\fCOUNT=%u",i);
           delay_ms(100);
        }
        output_low(PIN_B0);
     }
  }
}

As shown, this will wait for the timer to overflow, turn on pin B0, count for 0 to 10 on the LCD, with 100mSec between each count, then turn off the LED. Repeating whenever the timer overflows.

Best Wishes
ckielstra



Joined: 18 Mar 2004
Posts: 3680
Location: The Netherlands

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PostPosted: Mon May 10, 2010 3:26 am     Reply with quote

I still don't get the goal of the project.
At the current settings the timer will overflow in 13ms. Keeping this in mind the 100ms delay makes no sense.
Ttelmah



Joined: 11 Mar 2010
Posts: 19552

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PostPosted: Mon May 10, 2010 3:37 am     Reply with quote

Yes.
I'm so used to 18F chips, where Timer0, is 16bit by default, that the 'old' 8bit version didn't occur to me. 13.1mSec.....

Best Wishes
Ttelmah



Joined: 11 Mar 2010
Posts: 19552

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PostPosted: Mon May 10, 2010 3:38 am     Reply with quote

and (of course), in the original code, it'll never get out of the interrupt routine, for the same reason.

Best Wishes
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