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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 ?
Code: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
Why not using a simpler way?
CheersCode: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
Al.
Fratello,
The result should be in mVolts.
You almost had it right Fratello.
And the timing will be more accurate if you don't do the math on each loop.
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
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...
Oops, disregard my comment. Yes, you should get your result in millivolts.
Robert
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...
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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Wow, Thank You again Mr.Darrel !
Maybe my code it's wrong ; I use
insteadCode:DEFINE OSC 4
DEFINE ADC_BITS 10
DEFINE ADC_CLOCK 3
DEFINE ADC_SAMPLEUS 50
This is the explanation, don't ?Code:DEFINE ADC_BITS 10 ' ADCIN resolution (Bits)
DEFINE ADC_CLOCK 1 ' ADC clock source (Fosc/8)
DEFINE ADC_SAMPLEUS 11 ' ADC sampling time (uSec)
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 :(
No, that won't matter.
Can you post your Proteus files (zipped)?
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...
If you have PBP 2.60, then you could ...Quote:
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 don't have 2.60 yet, you'll need to ...Code:write 0, WORD v_cal
read 0, WORD v_cal
Code:write 0, v_cal.LowByte
write 1, v_cal.HighByte
read 0, v_cal.LowByte
read 1, v_cal.HighByte
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 ?
Yes, if you saved the MAX A/D value instead of the converted value.Quote:
So, if eeprom it's like in picture :
75 (hex) = 117 (dec)
v_cal = 117 * 4,8875 = 572 mVolts ?
In that case, be sure to use the previous formula to convert it to voltage.
Code:v_cal=v_cal * 5000 ' I use VDD as Vref
v_cal=div32 1023.
I use this code :
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
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.
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 :
Code:AvgCount CON 70 ' Number of samples to average
FAspread CON 50 ' Fast Average threshold +/-
Reject CON 2 ' Spurious Rejection level
...etc - see post
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.
It's ok with "pause'... I have though just one question...
If I use this code:
....Code:DEFINE OSC 4
DEFINE ADC_BITS 10
DEFINE ADC_CLOCK 3
DEFINE ADC_SAMPLEUS 50
how small can be "pause' for correct reading of ADC ?Code:for cnt = 1 to 3
adcin 3, adval
vt=adval * 5000
vt=div32 1023
grup=vt+grup
pause 20
next cnt
vs=grup / 3
Thanks in advance !
It can be zero.
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.