Serin serout problem

[lerameur]

yes actually, i went and read that , its like a diode, loosing 0.7 volt, so you need a bit more to compensate..
i have been working on the a/d conversion, again this is my first time, I am reading section 12 of he pic16f88 datasheet.

i wrote this program , bit it do not display the right inotmation:


'RECEIVE PIC
INCLUDE "modedefs.bas"
DEFINE OSC 20 'use external 20mhz crystal
CMCON = 7
ANSEL = %00001100 'Put port 2 and 3 analog and otehr digital
ADCON1 = %11000001 ' Set PORTA analog and RIGHT justify result
ADCON0 = %00010001 ' Configure and turn on A/D Module

DEFINE LCD_DREG PORTB ' Set LCD Data port
DEFINE LCD_DBIT 4 ' Set starting Data bit (0 or 4) if 4-bit bus
DEFINE LCD_RSREG PORTB ' Set LCD Register Select port
DEFINE LCD_RSBIT 1 ' Set LCD Register Select bit
DEFINE LCD_EREG PORTB ' Set LCD Enable port
DEFINE LCD_EBIT 0 ' Set LCD Enable bit
DEFINE LCD_BITS 4 ' Set LCD bus size (4 or 8 bits)
DEFINE LCD_LINES 2 ' Set number of lines on LCD
DEFINE LCD_COMMANDUS 2500
DEFINE LCD_DATAUS 250
DEFINE CHAR_PACING 2000
pause 1000

' Define ADCIN parameters
Define ADC_BITS 10 ' Set number of bits in result
Define ADC_CLOCK 3 ' Set clock source (3=rc)
Define ADC_SAMPLEUS 50 ' Set sampling time in uS

input1 var word : input2 var word

TRISA = %11111111 ' Set PORTA to all input

Pause 500

loop: ADCIN portA.2, input1
ADCIN portA.3, input2

Lcdout $fe, 1 ' Clear LCD
Lcdout "Value1: ", DEC input1 ' Display the decimal value
lcdout $FE,$C0, "Value2: ", dec input2
Pause 200

Goto loop
End

[skimask]

yes actually, i went and read that , its like a diode, loosing 0.7 volt, so you need a bit more to compensate..
i have been working on the a/d conversion, again this is my first time, I am reading section 12 of he pic16f88 datasheet.

i wrote this program , bit it do not display the right inotmation:


'RECEIVE PIC
INCLUDE "modedefs.bas"
DEFINE OSC 20 'use external 20mhz crystal
CMCON = 7
ANSEL = %00001100 'Put port 2 and 3 analog and otehr digital
ADCON1 = %11000001 ' Set PORTA analog and RIGHT justify result
ADCON0 = %00010001 ' Configure and turn on A/D Module

DEFINE LCD_DREG PORTB ' Set LCD Data port
DEFINE LCD_DBIT 4 ' Set starting Data bit (0 or 4) if 4-bit bus
DEFINE LCD_RSREG PORTB ' Set LCD Register Select port
DEFINE LCD_RSBIT 1 ' Set LCD Register Select bit
DEFINE LCD_EREG PORTB ' Set LCD Enable port
DEFINE LCD_EBIT 0 ' Set LCD Enable bit
DEFINE LCD_BITS 4 ' Set LCD bus size (4 or 8 bits)
DEFINE LCD_LINES 2 ' Set number of lines on LCD
DEFINE LCD_COMMANDUS 2500
DEFINE LCD_DATAUS 250
DEFINE CHAR_PACING 2000
pause 1000

' Define ADCIN parameters
Define ADC_BITS 10 ' Set number of bits in result
Define ADC_CLOCK 3 ' Set clock source (3=rc)
Define ADC_SAMPLEUS 50 ' Set sampling time in uS

input1 var word : input2 var word

TRISA = %11111111 ' Set PORTA to all input

Pause 500

loop: ADCIN portA.2, input1
ADCIN portA.3, input2

Lcdout $fe, 1 ' Clear LCD
Lcdout "Value1: ", DEC input1 ' Display the decimal value
lcdout $FE,$C0, "Value2: ", dec input2
Pause 200

Goto loop
End

What did it display? How are you hooked up?

The thing I had in mind is keeping Vref+ and Vref- on Vdd and Vss (battery power). One A/D input is connected to the diode, .7v, which should always stay the same (give or take a bit depending on temperature). The output number from the A/D will be a ratio of that diode (.7v) to the battery voltage., the lower the number, the higher the battery voltage.

For example, if the battery voltage was actually .7v (just for arguments sake, I know it would never be that low), and the diode voltage was .7v, the A/D output would be 1023.
If the battery voltage was 4.8v and the diode was .7, .7 divided by 4.8 multiplied by 1024 (full scale A/D reading) would be 149 (149.3). So, 149 would be full charge.
Taking that same train of thought, if the battery voltage was 3.6v (.9v/cell, battery dead), .7 divided by 3.6 multplied 1024 would be 199 (199.1) which would mean dead battery.

Are ya following me here?