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variable pic16f628a

 
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alex
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variable pic16f628a
PostPosted: Sat Feb 16, 2008 1:20 pm     Reply with quote
Hi, i need help for a little problem.

My source code:
Code:

#include<16f628a.h>
#fuses NOWDT,NOPROTECT,PUT,NOLVP,MCLR,INTRC_IO
#use delay(clock=48000)

#use fast_io(b)
#use fast_io(a)


void main()
{
int duty;
 
duty= 1;

setup_oscillator(OSC_48KHZ);

set_tris_b(0xF7);
output_low(PIN_b3);

while(1)
  {
    output_high(pin_b3);
    delay_ms(duty);
    output_low(pin_b3);
    delay_ms(20);
  }
}

Easy? No, because I have a problem. when I use duty=1 I have a rectangular wave of 1.5 ms Hi and 20 ms low.
If i use delay_ms(1) I have 1ms Hi, if I use delay_ms(2) I have 2ms Hi., but with delay_ms(duty) I can´t create the same value of duty=x.
So, a variable int is wrong?
PCM programmer



Joined: 06 Sep 2003
Posts: 21708

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PostPosted: Sat Feb 16, 2008 3:26 pm     Reply with quote
The problem is likely caused by your very low clock speed of 48 KHz.
The instruction cycle frequency is 1/4 of that, which is only 12 kHz.

CCS uses software delay loops to create the delays in the delay_ms()
function. With a 12 KHz instruction clock, there are 12 instructions per
millisecond. "Branch" instructions take two cycles. I haven't looked at
the .LST file, but I suspect that the CCS library code for delay_ms() just
doesn't work very well at low clock speeds.
alex
Guest
PostPosted: Sat Feb 16, 2008 3:50 pm     Reply with quote
But, PCM programmer, the instructions delay_ms(1) e delay_ms(2) are working well. I see 1.1 ms hi e 2.1 hi.
The instruction delay_ms(duty) , where duty is a variable int that is not working.
Is it not possible that i am using a wrong variable?
Ttelmah
Guest
PostPosted: Sat Feb 16, 2008 4:00 pm     Reply with quote
The problem is that the delay_ms, using a _variable_, is much more complex code, than that using a fixed value. For the constant, the compiler inserts the instructions needed to get as close as it can the the specified time. For the variable version, it has to have a loop, and count down the variable. The minimum loop time is about 10 instruction times, and the 'step size', is about 15 instruction times. The delay_ms code, for variables, is just not able to cope with times where there are so few instructions per loop. I suspect you will find that the arithmetic, actually goes 'haywire' internally, if one mSec, is less than the shortest available loop time...
At this clock rate, a variable loop, is beyond the ability of the code to deal with.

Best Wishes
alex
Guest
PostPosted: Mon Feb 18, 2008 8:29 am     Reply with quote
Well, I gave up of delay. I used a instruction "for" and I did 1ms hi and 20ms low.
The problem is that I can´t change the values of duty cycle.
The variable int of instruction "for" doesn't increase in decimal.. I can´t increase or decrease the duty cycle.
Are there solutions for that?
Wayne_



Joined: 10 Oct 2007
Posts: 681

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PostPosted: Mon Feb 18, 2008 10:46 am     Reply with quote
Something like

Code:

#include<16f628a.h>
#fuses NOWDT,NOPROTECT,PUT,NOLVP,MCLR,INTRC_IO
#use delay(clock=48000)

#use fast_io(b)
#use fast_io(a)

#define PERIOD 20

void main()
{
int duty;
 
duty= 1;

setup_oscillator(OSC_48KHZ);

set_tris_b(0xF7);
output_low(PIN_b3);

int n,i;

while(1)
  {
    output_high(pin_b3);
    for(n=0; n<duty; n++) {
        // Insert dummy code to create required delay or use another loop
        for(i=0; i<10; i++) {}
    }
    output_low(pin_b3);
    for(n=0; n<(PERIOD  - duty); n++) {
        // Insert dummy code to create required delay or use another loop
        for(i=0; i<10; i++) {}
    }
  }
}


duty = 0 to PERIOD

the low period will be slightly out due to the PERIOD - duty calculation. This can be corrected by changing the internal for(i= loop.

There is still a problem with this.
If duty = 0 you will still get a low pulse.
If duty = PERIOD you will still get a high pulse.

If the delay (for) routines are too slow then you will not be able to get the desired 1ms duty. You do the math.
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