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Message |
kocakaya
Guest
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| RTCC Problem |
 Posted: Sun Feb 11, 2007 2:14 pm |
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Is this code cause deviation in RTCC time. If yes ? what can I do? My crystal is 4 Mhz internal
Waiting your help
| Code: |
#define high_start 1000
void akimgerilim()
{
int adcdeger;
set_adc_channel( 1 );
delay_us(1);
value = Read_ADC();
deger=(float)value/51.0;
value=deger;
delay_us(1);
set_adc_channel( 0 );
delay_us(1);
adcdeger=Read_ADC();
max=adcdeger*1.5;
delay_us(1);
}
void besli() {
OUTPUT_HIGH(PIN_D4);
delay_ms(200);
OUTPUT_LOW(PIN_D4);
delay_ms(200);
OUTPUT_HIGH(PIN_C5);
delay_ms(200);
OUTPUT_LOW(PIN_C5);
delay_ms(200);
OUTPUT_HIGH(PIN_D3);
delay_ms(200);
OUTPUT_LOW(PIN_D3);
delay_ms(200);
OUTPUT_HIGH(PIN_D2);
delay_ms(200);
OUTPUT_LOW(PIN_D2);
delay_ms(200);
OUTPUT_HIGH(PIN_C2);
delay_ms(200);
OUTPUT_LOW(PIN_C2);
delay_ms(200);
}
#INT_RTCC
void clock_isr() {
high_count -= 1;
akimgerilim();
if(high_count==0)
{
++seconds;
high_count=high_start;
ahtoplam=ahtoplam+sah;
ah=ahtoplam;
}
if(seconds>=60)
{
minute++;
seconds=0;
}
if(minute>=60)
{
hour++;
minute=0;
}
}
main()
{
high_count=high_start;
setup_timer_0(RTCC_INTERNAL|RTCC_DIV_8|RTCC_8_bit);
do
{
besli();
}while(true);
}
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ckielstra
Joined: 18 Mar 2004
Posts: 3680
Location: The Netherlands
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| Posted: Sun Feb 11, 2007 5:15 pm |
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With a 4MHz crystal and | Code: | | setup_timer_0(RTCC_INTERNAL|RTCC_DIV_8|RTCC_8_bit); |
you will get 488.28125 interrupts per second, not 1000 interrupts as you have specified.
Even when you set high_start to 488 you will get almost 50 seconds deviation per day because of the rounding. For a really accurate clock have a look at http://www.ccsinfo.com/forum/viewtopic.php?t=26177 |
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kocakaya
Guest
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| Thanks |
| Posted: Sun Feb 11, 2007 11:47 pm |
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How can I calcute this number? What is formula?
Do you think making floating point number division and calling akimgerilim() function in INT_RTCC make any problem? or make any delay in interrupt?
This system calculates amper hour per seconds so I must use this code in RTCC interrupt
When I use RTCC , Timer1 and Timer2 together the system works very slow.And RTCC count 1 minute in two minute .Why? do you think?
The code that I changed is below
| Code: |
#INT_RTCC
void clock_isr() {
high_count -= 1;
akimgerilim();
if(high_count==0)
{
++seconds;
high_count=high_start;
[b]ahtoplam=ahtoplam+max/3600; [/b] "Max is Floating point number"
}
if(seconds>=60)
{
minute++;
seconds=0;
}
if(minute>=60)
{
hour++;
minute=0;
}
}
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kocakaya
Guest
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| help |
| Posted: Mon Feb 12, 2007 4:27 pm |
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| anybody help me in this forum? |
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ckielstra
Joined: 18 Mar 2004
Posts: 3680
Location: The Netherlands
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| Posted: Mon Feb 12, 2007 5:06 pm |
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| Quote: | | How can I calcute this number? What is formula? |
Check the figure for timer0 in the datasheet of your PIC.
The timer is clocked with Fosc/4 divided by the number you set for the prescaler. An 8 bit timer will generate an interrupt every 256 clocks, a 16-bit timer every 65536 clocks.
| Code: | | setup_timer_0(RTCC_INTERNAL|RTCC_DIV_8|RTCC_8_bit); |
This will give an interrupt rate of: 4MHz / 4 / prescaler 8 / 256 = 488.something Hz
| Quote: | | Do you think making floating point number division and calling akimgerilim() function in INT_RTCC make any problem? or make any delay in interrupt? |
Interrupt routines should always be as short as possible. Floating point calculation takes a lot of computing power, but I think you are here still save. Search this forum or internet for an explanation of 'scaled integers' this is much quicker and might be an option for you.
| Quote: | | When I use RTCC , Timer1 and Timer2 together the system works very slow.And RTCC count 1 minute in two minute .Why? do you think? |
This is after you changed the 1000 to 488?
How should we know? You don't show the code for Timer1 and Timer 2. |
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