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Message |
saeidjabbari
Joined: 25 Mar 2010
Posts: 8
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| TIMER0 Interrupt Problem |
 Posted: Sun May 09, 2010 12:25 pm |
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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);
}
}
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This link is CODE and SCH:
http://www.4shared.com/file/bTwzl32U/TIMER0INTRRUPT.html |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Sun May 09, 2010 1:25 pm |
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| Tell us the purpose of your project. What do you want to do ? |
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saeidjabbari
Joined: 25 Mar 2010
Posts: 8
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| Posted: Sun May 09, 2010 10:42 pm |
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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 |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19349
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| Posted: Mon May 10, 2010 1:55 am |
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| 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);
}
}
}
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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 |
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ckielstra
Joined: 18 Mar 2004
Posts: 3680
Location: The Netherlands
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| Posted: Mon May 10, 2010 3:26 am |
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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. |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19349
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| Posted: Mon May 10, 2010 3:37 am |
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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 |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19349
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| Posted: Mon May 10, 2010 3:38 am |
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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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