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How can I calculate interrupt time?? (TMR0)

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fuzzy

Joined: 26 Feb 2005
Posts: 64

How can I calculate interrupt time?? (TMR0)

Posted: Sat Apr 29, 2006 6:02 am

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

Posted: Sat Apr 29, 2006 7:22 am

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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