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iVoVa
Joined: 12 Mar 2011 Posts: 7
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Explain about code timer0, help me ! |
Posted: Sat Mar 12, 2011 11:39 pm |
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Code: |
#include <18F4520.h>
#fuses NOWDT,PUT,XT,NOPROTECT
#use delay(clock=4000000)
#byte PORTB = 0xF81
int16 count = 0;
int8 a;
//Chuong trinh ngat TMR0
#int_timer0
void interrupt_timer0()
{
set_timer0(0x23);
++count;
if(count == 8)// 8*131ms ~ 1s
{
count=0;
rotate_left(&a,1);
}
}//
void main(void)
{
set_tris_b(0);
setup_timer_0(RTCC_INTERNAL|RTCC_DIV_2);
set_timer0(0x23);// T_out = 2*(65535 - 35)*1us = 131ms
enable_interrupts(global);
enable_interrupts(int_timer0);
a = 0x01;
while(true)
{
PORTB = a;
}
}
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I use Timer to set PORTB = 0x01,0x02,0x04........delay within time is 1 sec.
I have imitated with Proteus and it run correct. But I do not understand how timer0 can be recount with this code. I think that PORTB = 0x01, when timer is overflowed, PORTB = 0x02, Program will stop here. Help me explain why Timer0 recount without enable_interrupt(init_timer0) .
I am Vietnamese, so my english skill is so bad. Please try to understand what I said. Thanks for reading.
Last edited by iVoVa on Sun Mar 13, 2011 6:54 am; edited 1 time in total |
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Ttelmah
Joined: 11 Mar 2010 Posts: 19513
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Posted: Sun Mar 13, 2011 4:00 am |
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OK.
Think of the timer, like a sort of clock face. It is numbered from 0 (at the top), round to 65535.
Behind it is a gearbox, so it takes 65536 turns of the input shaft to go round once.
Then behind this is another gearbox, with a number of output shafts fed from the motor. The motor goes through a 4:1 reduction, then feeds the first shaft. There are four output shafts. One at the motor speed/4, one at half this, one at 1/4 this, and one at one eighth this.
Then there is one final thing. At the top of the face, is a bell. This rings when the hand goes from 65535, to 0.
When you start, the motor is spinning, but none of the four shafts connect to the main 'face' gearbox. Then:
setup_timer_0(RTCC_INTERNAL|RTCC_DIV_2);
Connects the shaft at motor/8 to the face gearbox.
Then:
set_timer0(0x23);
Moves the hand forwards to '35'.
The hand now moves, and counts round to the top, where the bell rings.
enable_interrupts(global);
enable_interrupts(int_timer0);
says 'when the bell rings, execute the int_timer0 code'.
At no point is the gearbox disconnected (setup_timer_0(RTCC_OFF)), or the bell stopped (disable_interrupts(INT_TIMER0)), so the bell keeps ringing at regular intervals (close to one second).
The timing will not be perfect, since the hand moves forwards a little, between the bell going off (timer interrupt triggering), and the code actually reaching the instruction to move the hand forward a little in the interrupt code. These counts will be 'extra', So the code will take this little amount extra each second, and will run about 0.1% slow.
The point is that once started, the hand keeps going round, and so the interrupt code will keep being called.
Best Wishes |
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iVoVa
Joined: 12 Mar 2011 Posts: 7
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Posted: Sun Mar 13, 2011 6:58 am |
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@Ttelmah : Thanks for helping ! |
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