problem with output RA2 and RA3

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- 8th December 2006, 07:32
savnik

problem with output RA2 and RA3

I have the below code.
My problem is that the output RA2 and RA3 not change if (value - P2) >= 1 and (value1 - P3) >= 1.
What wrong i have made;

Code:

'************************************************************************ @ DEVICE PIC16F88, MCLR_OFF ' Master Clear Options (Internal) '******************************** LCD *********************************** DEFINE LCD_DREG PORTB ' Selection of the port B DEFINE LCD_DBIT 4 DEFINE LCD_RSREG PORTB ' RS on port RB2 DEFINE LCD_RSBIT 2 DEFINE LCD_EREG PORTB ' E on port RB3 DEFINE LCD_EBIT 3 DEFINE LCD_BITS 4 ' Mode 4 bits DEFINE LCD_LINES 2 ' LCD 2 lines of 16 caracter '****************************Define ADCIN parameters ********************** DEFINE ADC_BITS 8 ' Set number of bits in result DEFINE ADC_CLOCK 3 ' Set clock source (3=rc) DEFINE ADC_SAMPLEUS 50 ' Set sampling time in uS '*********************************** VARIABLES ********************* TRISA = %00000011 CMCON = 7 ' PortA = digital I/O ANSEL = %00000011 ' Will set RA0-RA1 as analog and all others as digital ADCON0 = %11100001 ' Configure and turn on A/D Module ADCON1 = %00000010 ' Set PORTA analog and RIGHT justify result adval var Byte ' Create adval to store result value var Word adval1 var Byte ' Create adval1 to store result value1 var Word P VAR Byte P1 VAR Byte P2 var Word P3 var Word PAUSE 500 LCDOUT $FE,1," VOLTMETER " PAUSE 1000 P2 = P*50/255 P3 = P1*50/255 P2=28 P3=10 NORMAL: ADCIN 0, adval ' Read channel 0 to adval value = adval*50/255 ADCIN 1, adval1 ' Read channel 1 to adval1 value1 = adval1*50/255 LCDOUT $FE,1,"VOLT : ",DEC2 value LCDOUT $FE,$C0,"AMPER: ",DEC2 value1 if (value - P2) >= 1 then PORTA.2=1 else PORTA.2=0 endif if (value1 - P3) >= 1 then PORTA.3=1 else PORTA.3=0 endif PAUSE 500 GOTO NORMAL
- 8th December 2006, 08:25
mat janssen

When I read the datasheet of the 16F88 then I see other configurations than you filled in at adcon0 and adcon1.
You say:
ADCON0 = %11100001 ' Configure and turn on A/D Module
I read for this:
do nothing with adcon0
You say:
ADCON1 = %00000010 ' Set PORTA analog and RIGHT justify result
I read for this:
adcon1 = %10000000 'only right justify and nothing about set port a.
- 8th December 2006, 09:00
savnik

Quote:

Originally Posted by mat janssen

When I read the datasheet of the 16F88 then I see other configurations than you filled in at adcon0 and adcon1.
You say:
ADCON0 = %11100001 ' Configure and turn on A/D Module
I read for this:
do nothing with adcon0
You say:
ADCON1 = %00000010 ' Set PORTA analog and RIGHT justify result
I read for this:
adcon1 = %10000000 'only right justify and nothing about set port a.

I read this
ADCON0.0=1 ' A/D converter module is operating
- 8th December 2006, 09:48
mat janssen

The adcon0,0 is an internal bit to tell us what the condition is of the adc.
You need it if your programm is made in assembler.
All of this is solved for you in the picbasicpro compiler.
- 8th December 2006, 10:31
savnik

i change with this:

Code:

CMCON = 7 ' PortA = digital I/O ANSEL = %00000011 ' Will set RA0-RA1 as analog and all others as digital TRISA = %00000011 ADCON0 = %11000001 ' Configure and turn on A/D Module ADCON1 = %00000000 ' Left justified. Six Least Significant bits of ADRESL are read as ‘0’

and work but i want only the positive result and not the negative of command: if (value - P2) >= 1
23 - 25 = -2 (P2=25)
27 - 25 = 2 (P2=25)
the pin PORTA.2 turn on , and for -2 and for 2 , but i want only for 2.
How to make this;
- 8th December 2006, 10:55
mat janssen

Test if the first number is greater then the second number and then do the subtraction. If otherwise then do nothing.
- 8th December 2006, 13:49
savnik

Quote:

Originally Posted by mat janssen

Test if the first number is greater then the second number and then do the subtraction. If otherwise then do nothing.

Thank you.
I use this code:

Code:

If value >= P2 then If (value - P2) >= 1 then PORTB.1=1 else PORTB.1=0 endif endif