okey .... i understood .

The range between 2 to 60 degcent .... ok

i want to ask about the accuarcy of the a LM35 , from the datasheet it is a +0.5/-0.5 , knowing this value , in what that will help me ?

thank you

 

hi,
So your Tmax is 60Cdeg, the LM35 will output +0.6V for 60C.

If you set the LM358 to a gain of 4.17, this will increase the 0.6V to 2.5V.
So for Tmax the adc [with a Vref=2.5] will have an Vinp of 2.5V which will convert to 255 decimal.

In your PC program you will have to multiply the 255 by say,2.353 to make it equal to 600, which be equivalent to 60.0Cdeg

As an option, as you are amplifying the LM35 output to the ADC, you could just use the +V[5v] ref of the ADC.
This would mean amplifying the LM35 0.6V by 8.33, so the 0.6V would become 5V, which with a +5Vref would give 255.
This method for the Vref is simpler and cheaper.!

If you think about it, if you made the Tmax = 51Cdeg, so the LM35's 0.51V was multiplied by 9.8 to +5V, this would give an adc value of 255.
When multiplied by 2 in the PC program would give 51Cdeg.!

The +/-0.5C accuracy is over the temperature working range of the LM35.

OK.?

__________________
Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
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ok ..... allright .

???? i did not get it

thank you very much eric .

 

hi,
Look at this graph from the LM35 data.
Examine the Tempr plot over the full range of the LM35.

Attached Thumbnails

 

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Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
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thank you so much eric .

for any new doubts I'll DISCUSS it with you .

 

hello .. eric ! i want to ask about something

About the temperature . someone implement the project like me with these values :

Vref=1.28 v . applied on pin9 is 0.64 v
LM358 works as buffer .

he used these formulas :
To get the minimum span that the adc can identify . min-span=Vref/255
To get the Vin . Vin= Dout * min-span

then :
=1.28/255 = 5.01 mV

He that this the min-span according the formula. and we know that the adc will give us 10 mV / centi but here will give us 5mv/centi ..... which mean we will take the output of the adc and divide it by 2 to get the tempr .

example : if the output is 00000010= 2 in decimal then the program will read 2 and divide it by 2 to get the tempr ...... 2/2=1 c .

is that make any sense ? i asked him about why you divided by 2 ? but he did not respond till now ?

what do u think ?

note : the program is working fine as this person illustrated .

 

Hi,
I would still advise to use the LM358 as an amplifier for the LM35 output also use the ADC's internal 5Vref.

__________________
Eric " Good enough is Perfect "
I will NOT answer PM's requesting technical help, please use the Forum
PIC tutorials: Nigel's www.winpicprog.co.uk/ Bill's: www.blueroomelectronics.com/

 

okay .... i think this is convinced . thank you so much ...any new doubt i 'll DISCUSS it with you .