Maximum value of ADCIN

[fratello]

Tested and re-tested today both variant : Mr.Darrel's and Mr.Aratti.
Still different results : 5mV vs. 24 mV. It's so strange for me...

[Darrel Taylor]

fratello,
I'm not sure what you've done in your code, but I've tested it here and it works.

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[fratello]

Wow, Thank You again Mr.Darrel !
Maybe my code it's wrong ; I use

DEFINE OSC  4
DEFINE ADC_BITS 10
DEFINE ADC_CLOCK 3
DEFINE ADC_SAMPLEUS 50

instead

DEFINE ADC_BITS 10     ' ADCIN resolution  (Bits)
DEFINE ADC_CLOCK 1     ' ADC clock source  (Fosc/8)
DEFINE ADC_SAMPLEUS 11 ' ADC sampling time (uSec)

This is the explanation, don't ?
NO, I think this is NOT the explanation ! My math teacher it's verry, verry angry ...
I learn more here : http://www.darreltaylor.com/DT_Analog/
1 step=0.0048875. So, if I have Max_result 450, Volts= 450 * 0.0048875 =2.199 Volts; for 696 --> Volts= 696 * 0.0048875 =3.401 Volts...
Stupid me, again

[Darrel Taylor]

No, that won't matter.

Can you post your Proteus files (zipped)?

[fratello]

Yes, all is OK now !
If v_cal is 1023 (maximum 5.000V) it's correct this command for memorise in EEPROM :
write 0, v_cal
I read the PBP help, but I find only this : write 5,B0 ' Send value in B0 to EEPROM location 5.
And I can read them when I need, with this command :
read 0, v_cal ???
It's enough like these or need more ?
Thanks in advance and sorry for my ignorance...

[Darrel Taylor]

If v_cal is 1023 (maximum 5.000V) it's correct this command for memorise in EEPROM :
write 0, v_cal
...
read 0, v_cal ???
It's enough like these or need more ?

If you have PBP 2.60, then you could ...

write 0, WORD v_cal

read  0, WORD v_cal

If you don't have 2.60 yet, you'll need to ...

write 0, v_cal.LowByte
write 1, v_cal.HighByte

read  0, v_cal.LowByte
read  1, v_cal.HighByte

[fratello]

I understand now ; I read 10 pages of search "eeprom".
Of course, the answer it's simple if somebody help ! Thanks, Mr.Darrel !
....
So, if eeprom it's like in picture :
75 (hex) = 117 (dec)
v_cal = 117 * 4,8875 = 572 mVolts ?