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digital clock using 18F4550

 
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akhter900



Joined: 07 Jun 2009
Posts: 8

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digital clock using 18F4550
PostPosted: Sun Jul 12, 2009 3:57 am     Reply with quote
hie....

I tried to count the second but its not working.
I just take a look on some examples that is given in the CCS compiler but my program does not count the second properly. It becomes slow.

Anybody can give me some tips about this?
Code:

#define INTS_PER_SECOND 76
.......
#int_rtcc                         
void clock_isr() {             
                                   
    if(--int_count==0)
    { 
      ++seconds;
      int_count=INTS_PER_SECOND;
     
         TIME[S_]++;
         if(TIME[S_] >= 60)
         {
            TIME[S_]=0;
            TIME[M_]++;
         }
         if(TIME[M_] >= 60)
         {
            TIME[M_]=0;
            TIME[H_]++;
         }
         if(TIME[H_] >= 12)
         {
            TIME[H_]=0;
         }
    }
}

void main()
{
..........
int_count=INTS_PER_SECOND;
   set_timer0(0);
   setup_counters( RTCC_INTERNAL, RTCC_DIV_256 | RTCC_8_BIT);
   enable_interrupts(INT_RTCC);
   enable_interrupts(GLOBAL);
......
}
akhter900



Joined: 07 Jun 2009
Posts: 8

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PostPosted: Sun Jul 12, 2009 5:07 am     Reply with quote
How can I count the exact value for ...
INTS_PER_SECOND
???
ckielstra



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

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PostPosted: Sun Jul 12, 2009 7:26 am     Reply with quote
You didn't mention which CCS example program you used as a starting point but searching the directory I found only ex_stwt.c to have INTS_PER_SECOND defined as 76.
Now, if you look into the program you will see the following line:
Code:
#define INTS_PER_SECOND 76     // (20000000/(4*256*256))
Where:
- 20000000 is the clock speed of the processor
- 4 is the number of clocks to execute a single instruction. Always 4 for the PIC10 to PIC18 processors.
- The last two numbers 256 come from the line:
Code:
setup_counters( RTCC_INTERNAL, RTCC_DIV_256 | RTCC_8_BIT);
- An 8 bit counter counts up to 256 before restarting at 0.
- A post divider of 256 decreases the number of interrupts by 256.

Assuming you have an 8MHz clock your calculation for the number of interrupts would become:
8000000 / (4 * 256 * 256) = 30.51
Set the number of interrupts to 30 or 31. You will get an error of 1.6% or 23 minutes/day. For most applications this is no problem.
For higher accuracy you could change the post divider to 128, this will give you:
8000000 / (4 * 256 * 128) = 61.04
Setting the number of interrupts per second to 61 reduces the error to 0.04% = 56seconds/day.
Depending on your actual clock rate you can play a bit with the numbers to get a more exact clock but I recommend to keep the number of interrupts below 200 or the interrupt overhead will become too high.

Another approach for a really accurate clock can be found in: http://www.ccsinfo.com/forum/viewtopic.php?t=26177
akhter900



Joined: 07 Jun 2009
Posts: 8

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PostPosted: Wed Jul 15, 2009 11:17 pm     Reply with quote
Thanks a lot for the details.

I use the line.... and its working well.
Code:

#define INTS_PER_SECOND 183

Thanks
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