Saving last two samples of ADC readings....

[armghan11]

Saving latest two samples of ADC readings....

I am working on a non intrusive load monitoring system i want my micro controller to reads the analogue input of the ADC then this input is normalized to remove the values in points because the micro controller cannot store values which are after decimal point. So to save important information being discarded the values are multiplied by 100. after that i want micro controller keeps the latest two samples. After normalization of the sample, it is subtracted from the previous sample. If the difference is positive, it show it on LCD.

I am attaching a flow chart below it might help you... i want help in saving samples of adc... i work on ccs c compiler

 

[keith1200rs]

I am not quite sure exactly what your problem is.

You need to have two variables - new_ADC and old_ADC for example. Before entering your loop you need to read the ADC and store it in old_ADC.

Once in the loop you need to read the ADC - this is new_ADC - and do your comparison. Then store new_ADC into old_ADC ready for the next time round the loop.

Keith.

 

[armghan11]

I am not quite sure exactly what your problem is.

You need to have two variables - new_ADC and old_ADC for example. Before entering your loop you need to read the ADC and store it in old_ADC.

Once in the loop you need to read the ADC - this is new_ADC - and do your comparison. Then store new_ADC into old_ADC ready for the next time round the loop.

Keith.

i just want to store latest two adc samples and do a comparison and after that it again take latest two sample that how the loop continues
check out the flowchart it might help

 

[keith1200rs]

I did look at the flowchart and the description I gave above is the parts you are missing, I believe.

Keith

 

[armghan11]

here is the c code i made... but it isnt working properly

#include <16F877a.H> 
#device adc=10 
#fuses HS,NOWDT,NOPROTECT,NOLVP
#use delay(clock=4000000)

#include <lcd2.c>

char look(int a)
{
switch(a)
{
case 0:
return '0';
case 1:
return '1';
case 2:
return '2';
case 3:
return '3';
case 4:
return '4';
case 5:
return '5';
case 6:
return '6';
case 7:
return '7';
case 8:
return '8';
case 9:
return '9';
default:
return '.';
}
}

void main()
{
unsigned int16 new_adc,old_adc;
lcd_init(); 
setup_adc_ports(ALL_Analog); 
setup_adc(ADC_CLOCK_DIV_8); 
set_adc_channel(3);
delay_us(20);
old_adc= read_ADC();
old_adc = (old_adc * 4.89)/0.47;
Char *current = "0.00";
Char *current1= "0.00";
do
{
delay_us(20);
new_adc = read_ADC();
new_adc = (new_adc * 4.89)/0.47;
if ( new_adc  > Old_adc) {
int16 k;
k = new_adc - old_adc;
current1[0] = look(k/1000);
current1[2] = look((k/100)%10);
current1[3] = look((k/10)%10);
lcd_gotoxy(1,2);
printf(LCD_PUTC,"L1=%s",current);
lcd_gotoxy(9,2);
printf(LCD_PUTC,"L2=%s",current1);
        }
else
        {
current[0] = look(new_adc/1000);
current[2] = look((new_adc/100)%10);
current[3] = look((new_adc/10)%10);
lcd_gotoxy(1,2);
printf(LCD_PUTC, "L1=%s",current);
        }
Delay_ms(250);
old_adc = new_adc;
} while(1);
}

 

[keith1200rs]

Does it compile? You have Old_adc instead of old_adc on one line if ( new_adc > Old_adc) {.

Keith

 

[armghan11]

Does it compile? You have Old_adc instead of old_adc on one line if ( new_adc > Old_adc) {.

Keith

changed it to old but still not working properly on protues.... some logical error in my code

 

[jayanth.devarayanadurga]

Instead of this current1[0] = look(k/1000); you can use current1[0] = k/1000 + 48;