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How can I calculate interrupt time?? (TMR0)

 
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fuzzy



Joined: 26 Feb 2005
Posts: 64

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How can I calculate interrupt time?? (TMR0)
PostPosted: Sat Apr 29, 2006 6:02 am     Reply with quote

Code:

#define INTS_PER_SECOND 76     // (20000000/(4*256*256))

BYTE seconds;      // A running seconds counter
BYTE int_count;    // Number of interrupts left before a second has elapsed


#int_rtcc                          // This function is called every time
void clock_isr() {                 // the RTCC (timer0) overflows (255->0).
                                   // For this program this is apx 76 times
    if(--int_count==0) {           // per second.
      ++seconds;
      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);


this code is a compiler's example.

What does it mean this expression: (20000000/(4*256*256)) ?

As i can understand 20000000 is clock frequency.
256 is prescaler division. What are 4 and the second 256??

Thanks
ckielstra



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

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PostPosted: Sat Apr 29, 2006 7:22 am     Reply with quote

Quote:
As i can understand 20000000 is clock frequency.
256 is prescaler division.
True.

Quote:
What are 4 and the second 256??
The 4 comes from the timers being driven at Fosc / 4. This is related to the PIC using 4 clock cycles to execute a single instruction.

The 256 is the maximum value your timer can count up to before overflowing, at overflow an interrupt is generated and the timer starts counting at 0 again. For 8-bit timers use 256 and for 16 bit timers 65536.
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