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servo pwm problem, what´s wrong, where´s the errors?

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Lost_whit_servo
Guest

servo pwm problem, what´s wrong, where´s the errors?

Posted: Sun Apr 20, 2008 12:22 pm

I want to drive one or more rc servo but...I can´t find where´s the error(s).
Can some one help me?
thanks in advance

Code:

#include <16F628.h>
#use delay(clock=20000000) // Fosc
#fuses HS,NOWDT,NOPUT
#use fast_io(B)
#bit RB1 = 0x06.1
#bit RB2 =0x06.2
static int16 k=0;
#INT_TIMER0
void tmr0_int()
{

RB2 = !RB2; //int is working?
k++;
set_timer0(206); // 10useg
}

void main(void)
{
int i;
set_tris_b(0);
output_low(PIN_B1);
setup_timer_0(RTCC_INTERNAL | RTCC_DIV_1);
enable_interrupts(INT_RTCC);
enable_interrupts(GLOBAL);
set_timer0(206); // 10useg int
while(1)
{
for(i=90;i<=150;i=i+30) //900useg, 1200useg, 1500useg
{
if(k<=i)
{
bit_set(*RB1,1);
}
while(k<=2000) //itxaron 200mseg bete arte
{
bit_clear(*RB1,0);
k=0;
}
}
}
}

ckielstra

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

Posted: Sun Apr 20, 2008 1:41 pm

I'm not really sure what you want your code to do but there are several errors:

Code:

while(k<=2000) //itxaron 200mseg bete arte
{
bit_clear(*RB1,0);
k=0;
}

1) This loop will never exit. Even with the timer interrupt increasing the value of K it's value will soon be reset to 0 and never reach 2000.
2) This code for changing the value of an I/O-port is completely wrong: *RB1 means 'take the contents of the variable pointed to by RB1', with RB1 being a 1-bit variable this will be either the contents of address 0 or 1. Completely not what you intend to do.
Possible methods to clear an I/O-port are:

Code:

Method 1:
#byte PORT_B = 0x06
#bit RB1 = PORT_B.1
RB1 = 0;

Method 2:
bit_clear(PORT_B, 1);

Methode 3:
output_low(PIN_B1);

I like method_3 because you don't have to manually specify the port's memory address which makes it very easy to change the program to another processor.

Code:

void tmr0_int()
{

RB2 = !RB2; //int is working?
k++;
set_timer0(206); // 10useg
}

An interrupt routine that has to execute every 10us is extremely fast and will consume all your processor's execution time.
To understand this you have to know that an interrupt has a lot of overhead for saving all registers on entering the interrupt and restoring the registers on exit. Depending on the number of registers this takes about 45 instructions on a PIC16 and 80 on a PIC18.

Your timer interrupt routine takes about 10 instructions, a total of: 45 + 10 = 55 instructions.
The processor is running at 20MHz, or 5 instructions per us. Meaning it takes 11us to execute 1 interrupt and leaving no time to execute your main program.

Servo programs have been discussed many times before on this forum, I suggest you use the search function to find those discussions. Here is one link: http://www.ccsinfo.com/forum/viewtopic.php?t=30139

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