Problems with ADC

Hell again all... seems like I get one thing fixed and something else goes wrong. I am having a problem with my ADC on the PIC16F877A. Here is the code that I am using:

Code:

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DEFINE OSC 20
DEFINE ADC_BITS 10
DEFINE ADC_CLOCK 3
DEFINE ADC_SAMPLEUS 1
PAUSE 1000

LIGHT VAR WORD

ADCON1=%10000010
TRISA=%11111111
TRISB=%00000000
TRISD=%00000000

MAIN:
    ADCIN 1, LIGHT
        LIGHT.0=PORTD.6
        LIGHT.1=PORTD.7
        LIGHT.2=PORTB.0
        LIGHT.3=PORTB.1
        LIGHT.4=PORTB.2
        LIGHT.5=PORTB.3
        LIGHT.6=PORTB.4
        LIGHT.7=PORTB.5
        LIGHT.8=PORTB.6
        LIGHT.9=PORTB.7
GOTO MAIN
END

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This measures the voltage coming from a CDS Photocell circuit. I have Vcc to photocell to 10k resistor to ground. ADC is connected between the photocell and 10k resistor, so as the resistance of the photocell changes, as does the voltage. I can measure this with a meter but when I try to output to light up the corresponding LEDs it doesn't work right. Any help with this will be greatly appreciated.

2 things

1. CDS usually have high impedance, but the PIC require a max of 10K. You'll need to use a buffer in between (op-amp, transistor, FET, ...)
2. The way you have it now, you read PORTB and D and place it in your LIGHT variable

try

Code:

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        PORTB = (LIGHT >>2)
        PORTD.6 = LIGHT.0
        PORTD.7 = LIGHT.1

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Quote:

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Originally Posted by mister_e

Code:

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        PORTB = (LIGHT >>2)
        PORTD.6 = LIGHT.0
        PORTD.7 = LIGHT.1

---

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Doesn't this line just shift everything two bits to the right:

Code:

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portb = (light >>2)

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By doing this I'm actually losing two bits right?

LIGHT is a Word, PORT is a BYTE... this is always hard to fit 16bits in a 8 bit register huh? ;)

LIGHT>>2 - yes it shift two position to the right... yes you miss bit0 and bit1, but the next lines
PORTD.6 = LIGHT.0
PORTD.7 = LIGHT.1

place them on PORTD.

no lost at all.

None of your bit are lost LIGHT>>2 as long as you dont use something like
LIGHT=LIGHT>>2

ok i was thinking that bit 0 and bit 1 would just be moved to where bits 2 and bits 3 are... thus showing bits 0 and 1 twice and losing bits 8 and 9 with the code you provided me with... i'll give it a shot and see what happens...