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Sam_40
Joined: 07 Jan 2015
Posts: 127
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| How to accurately set timer? [SOLVED] |
 Posted: Fri Nov 06, 2015 8:18 am |
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Hello guys,
I am trying to understand how to set the timer tick event. I have the PIC18F2685 mounted on PCB. I got TIMER1 using external 32768 crystal to generate a very accurate 1Hz (thanks to PCM Programmer for the TIMER1 code). I also got the RBs interrupt, the 7 segment display to work and now I can but the PIC to sleep.
However, I think I can improve my project by using a timer to handle the display instead of using the delay function.
I have been trying to use TIMER0 to generate a 1mS tick with no luck. Would you please help me with this? I want to know how to calculate the tick in case I want to use more or less than 1mS.
FYI; I did search the forum, google, The compiler manual, the examples and the PIC header file. I also tried to change the PIC clock to other than 8M and the T0_DIV_128 to other values in the header file. However the information I have is for trial and error :( I need to understand it so I can use it correctly.
This is a simple test code:
| Code: | #include <18F4520.h>
#fuses INTRC_IO, NOWDT, BROWNOUT
#use delay(clock=8M)
#int_timer0
void timer0_isr(void)
{
output_high(A7);
output_high(C2);
output_low(A7);
output_low(C2);
}
void main()
{
setup_timer_0(T0_INTERNAL|T0_DIV_128||T0_8_BIT);
delay_ms(500);
set_timer0(0x00);
clear_interrupt(INT_TIMER0);
enable_interrupts(INT_TIMER0);
enable_interrupts(GLOBAL);
while(TRUE);
} |
I want it to do what this code does inside the main():
| Code: | while(TRUE);
{
output_high(A7);
delay_ms(1);
output_high(C2);
delay_ms(1);
output_low(A7);
delay_ms(1);
output_low(C2);
delay_ms(1);
} |
Thanks in advance,
Last edited by Sam_40 on Wed Nov 11, 2015 11:29 am; edited 1 time in total |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19496
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| Posted: Fri Nov 06, 2015 8:33 am |
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The easiest timer to use to actually develop 1mSec, is timer2. Timer2 offers a programmable count, so can give an exact tick. Timer0 does not.
All the timers are fed at the start from Fosc/4. So for 1mSec (1000* per second), you need a count of 2000000/1000 = 2000.
So for timer2:
setup_timer_2(T2_DIV_BY_16,124,1);
This gives /16, and then /125 (the actual count is one greater than the 'PR2' value (second number - read the data sheet), so gives 2000 exactly. 
For timer0, the timer always divides intrinsically by either 65536, or 256 (16bit or 8bit). 2000/256, gives /7.8 required, and the nearest prescaler to this is /8. So:
setup_timer_0(T0_INTERNAL|T0_DIV_8||T0_8_BIT);
Will give 2000000/(8*256) = 976.5625/second (1.024mSec) which should be close enough. |
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Sam_40
Joined: 07 Jan 2015
Posts: 127
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| Posted: Fri Nov 06, 2015 9:13 am |
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Ttelmah,
Thanks a lot for the detailed information. You have a great weekend! |
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Sam_40
Joined: 07 Jan 2015
Posts: 127
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| Posted: Wed Nov 11, 2015 9:09 am |
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Ttelmah,
I am so confused in regards to timer2 set up to generate the 1mS. This is what I got:
For 1ms at 8Mhz internal osc and 1/16 prescaler for timer2 should be:
The math to calculate the period register
PR2 = (second * Fosc/4 ) / Prescaler
PR2 = (0.001 * 2000000)/16
PR2 = 125 (Why did you use 124?) the datasheet shows the PR2 should match TMR2?
/* Timer 2 Prototype */
/* void setup_timer_2(int8 mode, int8 period, int8 postscale); */
So:
1:16 Prescaler
125 PR2 Register I also tried your value of 124
1:1 Postscaler
Should yield 1000Hz per second or 1mS.
However I am not getting that?
the output result does not match:
| Code: | while(TRUE);
{
output_high(A7);
delay_ms(1);
output_high(C2);
delay_ms(1);
output_low(A7);
delay_ms(1);
output_low(C2);
delay_ms(1);
} |
It seems as timer2 ticks faster than 1mS. The LEDs are very dim which suggest that the LEDs ON/OFF for less than 1mS?
would you please edit my test code to make timer2 tick for 1mS. I edited the #use delay(clock=8M) to #use delay(internal=8M). I also tried to setup the osc inside the main without any luck. I am also aware that the internal osc is not accurate, However it is working fine if use the above code! I want to see how would you do it for this processor to setup and use timer2.
Regards, |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19496
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| Posted: Wed Nov 11, 2015 10:11 am |
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The timer counts from 0.....
0 to 124 = 125 counts.
This is common with PIC timers.
Think about it. You are waiting 1mSec for each change, not doing all the changes just 1mSec apart....
| Code: |
#INT_TIMER2
void t2_tick(void)
{
static unsigned int8 state=0;
switch (state)
{
case 0:
output_high(A7);
state++;
break;
case 1:
output_high(C2);
state++;
break;
case 3:
output low(A7); //2ticks from when A7 went high.....
state++;
break;
case 4:
output_low(C2);//2 ticks from when C2 went high.
state=0;
break;
}
}
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This takes 4 interrupts to perform the four changes. |
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Sam_40
Joined: 07 Jan 2015
Posts: 127
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| Posted: Wed Nov 11, 2015 11:25 am |
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Ttelmah,
Thanks a lot. Problem is solved.
My test program without using delay to help others:
| Code: | #include <18F2685.h>
#fuses INTRC_IO, NOWDT, BROWNOUT
#INT_TIMER2
void t2_tick(void)
{
static unsigned int8 state=0;
switch (state)
{
case 0:
output_high(A7);
state++;
break;
case 1:
output_high(C2);
state++;
break;
case 3:
output low(A7); //2ticks from when A7 went high.....
state++;
break;
case 4:
output_low(C2);//2 ticks from when C2 went high.
state=0;
break;
}
}
void main()
{
setup_oscillator(OSC_8MHZ);
setup_timer_2(T2_DIV_BY_16,124,1);
enable_interrupts(INT_TIMER2);
enable_interrupts(GLOBAL);
while(true)
{
}
} |
it also works great with my 7 segments project.
Thanks, |
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