Maximum value of ADCIN

[fratello]

I try to find myself the answer on this question, but I am not sure if I understand correct. Please, help me ! I read, for 7 seconds, the value of ADCIN, using 12F675 ; I need to memorize
the maximum (the highest) value of reading. It's like this OK ?

for cnt=1 to 70
adcin 3, v_cal
v_cal=v_cal * 5000 ' I use VDD as Vref
v_cal=div32 1023
' vmax=0 max v_cal ' it's this
or
' vmax=vmax max v_cal ' or this correct ?
pause 100
next cnt

[aratti]

Why not using a simpler way?



for cnt=1 to 70
adcin 3, v_cal
v_cal=v_cal * 5000 ' I use VDD as Vref
v_cal=div32 1023

If cnt = 1 then
vmax = v_cal
else
If v_cal > vmax then vmax = v_cal
endif

pause 100
next cnt


Cheers

Al.

[rsocor01]

Fratello,

The result should be in mVolts.

[Darrel Taylor]

You almost had it right Fratello.
And the timing will be more accurate if you don't do the math on each loop.


vmax = 0
for cnt=1 to 70
adcin 3, v_cal
vmax=vmax max v_cal
pause 100
next cnt
v_cal=vmax * 5000 ' I use VDD as Vref
v_cal=div32 1023

[fratello]

Thank You all for reply !
Both variant (mine = 0 max v_cal and variant of Mr.Robert ) give me the same results : 24 .
According of my math, I think this is the value of 24 mVolts. I am right ?
Now I try the variant of Mr.Darrel (remember ? My guardian angel in PBP problems :) ...).
I post the results.
LE : Something strange...with this variant, I read only 5 mV ?!
Maybe the schematic help...

[rsocor01]

Oops, disregard my comment. Yes, you should get your result in millivolts.

Robert

[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 ?

[Darrel Taylor]

So, if eeprom it's like in picture :
75 (hex) = 117 (dec)
v_cal = 117 * 4,8875 = 572 mVolts ?
Yes, if you saved the MAX A/D value instead of the converted value.

In that case, be sure to use the previous formula to convert it to voltage.

v_cal=v_cal * 5000 ' I use VDD as Vref
v_cal=div32 1023.

[fratello]

I use this code :

vmax = 0
for cnt =1 to 70
adcin 3, v_cal
vmax =vmax max v_cal
pause 100
next cnt

v_cal=vmax * 5000 ' I use VDD as Vref
v_cal=div32 1023

write 0, v_cal.LowByte
pause 10
write 1, v_cal.HighByte
pause 10

[Darrel Taylor]

Then you have saved the value as a Voltage.
And 117 means 117mV.

Since I = E / R ... with a 0.02 ohm resistor ...

0.117 / 0.02 = 5.85 Amps

Dividing by 0.02 is the same as multiplying times 50.

117 * 50 = 5850 or 5.850 Amps.

[fratello]

One truck with e-beer for Mr.Darrel ! With thanks for fratello !
Now another worm walk to my convolutions ! If at start of colecting samples, my motor give me one peek (or "n" peeks) of voltage , but the real v_cal ( from the end of route) it's smaller than peek(s) , it's possible to reject them ? Something like here : http://www.picbasic.co.uk/forum/showthread.php?t=12183
It's possible to use :

AvgCount CON 70 ' Number of samples to average
FAspread CON 50 ' Fast Average threshold +/-
Reject CON 2 ' Spurious Rejection level
...etc - see post

[fratello]

Since the whole code don't fit into 12F675, I give up...till hw upgrade...
So, I put in my code one little delay (pause 500) before starting ADC reading...I hope it's enough.

[fratello]

It's ok with "pause'... I have though just one question...
If I use this code:

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

for cnt = 1 to 3
adcin 3, adval
vt=adval * 5000
vt=div32 1023
grup=vt+grup
pause 20
next cnt
vs=grup / 3
how small can be "pause' for correct reading of ADC ?
Thanks in advance !

[mackrackit]

It can be zero.

[Darrel Taylor]

You will still get correct A/D readings without any pause at all in there.

I guess it depends on how much the signal is changing, as to whether the average will be accurate or not.