🚀 go-pugleaf

"Antony N. Lord" wrote:
> 
> OK, for starters I made a twin typo : its "temperature" not "temporator"
> and the input board measures 0-4095 mV (that's voltage, not current.)
> 
> Here's the CORRECTED original post along with some replies (e-mail) and
> further questions which should help bring us all up to date!
> 
> ------------------------------------------------------------
> 
> I am wondering if someone could help me interpret the information on the
> LM35 Precision Centigrade Temperature Sensor and assist in design of
> suitable electronics around it.
> 
> I need to install several remote temperature sensors (LM35), run over
> some spare Cat 5 cable to a PC interface (Weeder Technologies Analog
> Input Board.)
> 
> Notes
> 
> * Runs to sensors are less than 10m over Cat 5 ethernet cable (4 - 8
> conductors available)
> * Interface has 5V, 12V and a variable 3-12V supply rail available
> * Interface board input range 0-4095 mV (source impedance 8k ohm max)
> * My temperature measuring range is from -5�C (fridge) to about 45�
> (hottest day here) perhaps even 80� for measuring some electrical items
> 
> My greatest problem is correct selection of component values for the
> right temperature range. The ranges shown in several figures of the
> "Typical Applications" section of the data sheet are not adequate.
> Figures 6 & 8 appear to be good starts.
> 
> If anyone can offer advice as to the "best approach" perhaps with some
> working so I might glean an understanding as to the calculations it
> would be most appreciated!
> 
> ------------------------------------------------------------
> 
> LM 35 Data Sheet :
> 
> http://www.national.com/ds/LM/LM35.pdf
> 
> LM 35 Application Notes
> 
> http://www.national.com/an/AN/AN-460.pdf
> 
> Weeder Tech Analog Input Module
> 
> http://www.weedtech.com/wtadc-m.pdf
> 
> ------------------------------------------------------------
> 
> nico@nctdevpuntnl (punt=.) wrote :
> 
> > I think the problem lies in your interface board. The LM35 is a device
> > with a variable voltage output and your interface board expects a
> > current. That's not going to match.
> 
> As I wrote above I made a typo. It's 0 - 4095 mV input.
> 
> ------------------------------------------------------------
> 
> Robert (e-mail omitted for privacy) wrote :
> 
> > Hello Antony
> >
> > Figure 8 is the one to go for. Imagine that R1 and R2 are not connected.
> >
> > The IC generates a change in voltage of 10mV for every degree centigrade
> > change between Vout and GND. This results in a current flow through the 200R
> > resistor. This current must flow through the other 200R resistor and since
> > they are of equal value, the voltage across each resistor must be equal. So
> > the voltage across the other resistor changes by 10mV for every degree
> > centigrade.
> >  When the temperature is zero centigrade, output voltage is 0mV with respect
> >  to the GND terminal. Actually minimum is about 20mV, so minimum temperature
> >  is +2C.
> > As the output cannot go any lower than the GND terminal, negative
> > temperatures cannot be measured with just the two 200R resistors.
> > The IC itself consumes around 50uA of current. If a 2K resistor (R1) is put
> > in series with the ground terminal, this develops 100mV across R1. This
> > represents a temperature of +10C. This raises the internal reference within
> > the IC by 100mV and so adds 100mV to the voltage across the 200R resistor
> > between Vout and GND. R2 is added to compensate for the additional current
> > flow (50uA), which would otherwise flow through the 200R resistor at the
> > input of the analog card.
> > So when the temperature is 0C, the voltage across the 200R resistor will be
> > 100mV, therefore the voltage seen by the analog card will also be 100mV
> > When the temperature is -5C the analog input voltage will be +50mV (-50mV +
> > 100mV).
> > So when the temperature is say +20C, the analog voltage will be 20*10mV +
> > 100mV = 300mV
> > and at +45C the voltage will be 45*10mV + 100mV = 550mV
> > The software will need to make the 10 degree correction so that the correct
> > temperature is displayed.
> >
> > Regards
> > Robert
> 
> (see attached image for clarification)
> 
> ------------------------------------------------------------
> 
> Thanks to Robert I now have a much better understanding of the theory
> behind the circuit.
> 
> I would however appreciate it if someone could tell me how to configure
> it so that the maximum readable temperature is closer to 4095mV (the
> input board maximum) as this should increase the accuracy (and number of
> increments) of the reading. Roberts solution only outputs 550mV at 45�.
> 
> I will also add that any calculation / correction of the voltage reading
> is easily done in the software. I normally calibrate new temperature
> sensors by using a "cold" and "hot" heat source with a know good
> thermometer and then calculate a "line of best fit" for calculating the
> temperature (from the given voltage readings.)
> 
> Regards,
> Antony.
> 
> ------------------------------------------------------------
> 
> *** Take the letter 'X' out of my e-mail address if replying via e-mail
> ***
> 
>                        Name: LM35 circuit.bmp
>    LM35 circuit.bmp    Type: image/bmp
>                    Encoding: x-uuencode

  Actually, the LM35 and variants are more accurate than most commercial
thermometers; one has to shell out a fairly good chunk of change to get
a calibrated thermometer as good as or better than these ICs.
  No calculations, lines of best fit, or any other trash like that.