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ex_patg question..

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xindy
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ex_patg question..

Posted: Tue Mar 04, 2008 3:49 am

just a question about the ex_patg example..

Code:

#include <16F877.h>
#fuses HS,NOWDT,NOPROTECT,NOLVP
#use delay(clock=20000000)
#use rs232(baud=9600,xmit=PIN_C6,rcv=PIN_C7)

#define NUM_OUTPUTS 7

//NOTE: periods MUST be multiples of 400
//Periods are in microseconds
#define PERIOD_0 400
#define PERIOD_1 800
#define PERIOD_2 1600
#define PERIOD_3 2000
#define PERIOD_4 20000
#define PERIOD_5 64000
#define PERIOD_6 2000000

const long wave_period[NUM_OUTPUTS] = {
PERIOD_0/400, PERIOD_1/400, PERIOD_2/400, PERIOD_3/400,
PERIOD_4/400, PERIOD_5/400, PERIOD_6/400};

long counter[NUM_OUTPUTS] = {0,0,0,0,0,0,0};

int port_b_image;

// This interrupt is used to output the waveforms. The interrupt
// is automatically called ever 200us.
#INT_TIMER1
void wave_timer() {
int i;

set_timer1(0xFC4F); // sets timer to interrupt in 200us
output_b(port_b_image); // outputs the waveform

for(i=0; i<NUM_OUTPUTS; i++) // sets up next output for each pin
{
if((++counter[i]) == wave_period[i]) // if counter is expired
{
counter[i] = 0; // reset counter
if(bit_test(port_b_image,i)) // and set pin as needed
bit_clear(port_b_image,i);
else
bit_set(port_b_image,i);

}
}
}

void main() {

setup_timer_1(T1_INTERNAL|T1_DIV_BY_1); // setup interrupts
enable_interrupts(INT_TIMER1);
enable_interrupts(GLOBAL);

port_b_image=0; // initialize variable
output_b(port_b_image);

while(TRUE); // loop forever
}

why is it that in the code above, it will interrupt every 200 us? i tried to do the math but i cant figure out how it arrives at 200 us.. please help.. thanks guys..

ckielstra

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

Posted: Tue Mar 04, 2008 5:52 am

Quote:

why is it that in the code above, it will interrupt every 200 us? i tried to do the math but i cant figure out how it arrives at 200 us..

At 20MHz every instruction clock cycle takes 20Mhz / 4 = 0.2us.

Timer1 in this program is setup to use no prescalers, i.e. it is dividing by 1

Code:

setup_timer_1(T1_INTERNAL|T1_DIV_BY_1); // setup interrupts

This results in timer1 ticking with a speed of 1 tick every 0.2us.
An interrupt will be generated when the 16-bit timer overflows from 0xFFFF to 0x0000. Without modifications this would take 0.2us * 0x10000 = 13.1ms.

Now have a look at the start of the interrupt routine:

Code:

set_timer1(0xFC4F); // sets timer to interrupt in 200us

The trick here is that timer1 is assigned a new start value which causes the next timer overflow to happen much sooner, after:
0x10000 - 0x0FC4F = 0x3B1 = 945 ticks

This is strange, I would have expected the math to come up with 1000. Most likely this deviation is there to compensate for the interrupt overhead; it takes about 40 instructions to save registers before your interrupt function is entered. I don't like the workaround here because the overhead is not a constant between processors and compiler versions, even another interrupt could cause large deviations. That's why I prefer to change this line to

Code:

set_timer1( 1000 - get_timer1()); // sets timer to interrupt in 200us

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