Post by thread:[PIC] Converting A/D into Temperature

I've got an A/D circuit on my 16F874. I have a couple of OP Amps hooked to some resistors a few Pots and a thermistor so that I get a 5X gain on the voltage. (schematic available if needed) I have it giving me 2.5V at room temp. But for the LIFE of me I can't figure out how to convert the 0-1023 value into a usable temperature. I have an Excel spreadsheet that will tell me the readout if I type in the voltage example: 0v=0, 5V=1023, 2.5V=512, 2.813782991V=576 >From my chart that last Voltage should correspond to a reading of 68F (20C) Maybe it's late and it's really simple but I can't figure it out.. any help out there? Damon Hopkins

Damon wrote: > I've got an A/D circuit on my 16F874. > I have a couple of OP Amps hooked to some resistors a few Pots and a > thermistor so that I get a 5X gain on the voltage. (schematic available > if needed) > I have it giving me 2.5V at room temp. But for the LIFE of me I can't > figure out how to convert the 0-1023 value into a usable temperature. > I have an Excel spreadsheet that will tell me the readout if I type in > the voltage example: 0v=0, 5V=1023, 2.5V=512, 2.813782991V=576 > > From my chart that last Voltage should correspond to a reading of 68F > (20C) > Maybe it's late and it's really simple but I can't figure it out.. any > help out there? > > Well, using a thermistor, it's not exactly straightforward. The reason is the highly non-linear resistance to temperature characteristic of the thermistor. I have been working on something similar (in my copious spare time). This equation is from memory, i'll look it up tonight when I get home: T(Kelvin) = 1/(A + B*ln(R) + C*(ln(R)^3)) This equation is the Steinhart-Hart equation. R is the resistance of the thermistor. A, B, and C are material constants that you can figure out by measuring the resistance at three different temperatures. The constants are all less than 1 for my particular thermistor. Now you can see the difficulty with this equation in a PIC. While the fractional values can be taken care of by scaling everything, calculating the natural log function is a bit more difficult. On top of that, the temperature (in Kelvin) is actually the inverse. Of course, I admit I am not the world's greatest PIC programmer. Perhaps there is an easy way to realize the required mathematical operations on a PIC. In my circuit, I placed the thermistor in parallel with a fixed resistor in an attempt to linearize the resistance a little. Over a certain range of values, this works fairly well. A lookup table is one possible solution. You could put a lookup table for all possible values in external EEPROM. Or you could place a smaller lookup table in memory, and interpolate between points. If anyone has any other ideas, please jump in. Phil Eisermann H:(440) 284-3787 (.....mazerKILLspam@spam@ix.netcom.com) O:(440) 329-4680 (peisermaKILLspamridgid.com)

On Mon, 22 May 2000, Eisermann, Phil [Ridg/CO] wrote: > Well, using a thermistor, it's not exactly straightforward. The > reason > is the highly non-linear resistance to temperature characteristic of the > thermistor. I have been working on something similar (in my copious > spare time). This equation is from memory, i'll look it up tonight when > I get home: > > T(Kelvin) = 1/(A + B*ln(R) + C*(ln(R)^3)) > > This equation is the Steinhart-Hart equation. R is the resistance > of the thermistor. A, B, and C are material constants that you can figure > out by measuring the resistance at three different temperatures. The > constants are all less than 1 for my particular thermistor. > > Now you can see the difficulty with this equation in a PIC. While the > fractional values can be taken care of by scaling everything, > calculating the natural log function is a bit more difficult. On top > of that, the temperature (in Kelvin) is actually the inverse. If calculating logarithms is all that's holding you back, then check out: http://www.dattalo.com/technical/software/pic/piclog.html and http://www.dattalo.com/technical/theory/logs.html Scott

> In my circuit, I placed the thermistor in parallel with a fixed resistor > in an attempt to linearize the resistance a little. Over a certain > range of values, this works fairly well. > > If anyone has any other ideas, please jump in. This is one application where Excel is actually useful. Use it to build a table of thermistor values over the range you want to test. Then take that value to calculate the voltage you would get at the A/D input for your circuit topology. Graph the values. Try a simple voltage divider with a resistor at the top and the thermistor on the bottom. It works for me. >From the measured voltage values, have Excel calculate the factors for a y=mx+c function, generate another column with the result of that function and plot them on the same graph. Also calculate the power dissipation for each step and have Excel put the maximum value somewhere you can see it. Do the same for the error value. Now you can fiddle with the value of the series R to get the closest fit. If the error is still out of range, plot it. You may see a solution staring you in the face. Also try extending the temperature range that you are plotting to give the linear regression a bit more room to move. I did this recently and got <0.5 degree error over 0-100C with an algorithm that was no more than y=mx+c plus a test for values within two ranges. If they were, I added or subtracted half a degree. The circuit was just a voltage divider. Steve. ====================================================== Steve Baldwin Electronic Product Design TLA Microsystems Ltd Microcontroller Specialists PO Box 15-680, New Lynn http://www.tla.co.nz Auckland, New Zealand ph +64 9 820-2221 email: .....stevebKILLspam.....tla.co.nz fax +64 9 820-1929 ======================================================

On Mon, 22 May 2000 13:46:59 -0400, you wrote: {Quote hidden}>Damon wrote: >> I've got an A/D circuit on my 16F874. >> I have a couple of OP Amps hooked to some resistors a few Pots and a >> thermistor so that I get a 5X gain on the voltage. (schematic available >> if needed) >> I have it giving me 2.5V at room temp. But for the LIFE of me I can't >> figure out how to convert the 0-1023 value into a usable temperature. >> I have an Excel spreadsheet that will tell me the readout if I type in >> the voltage example: 0v=0, 5V=1023, 2.5V=512, 2.813782991V=576 >> >> From my chart that last Voltage should correspond to a reading of 68F >> (20C) >> Maybe it's late and it's really simple but I can't figure it out.. any >> help out there? >> >> > Well, using a thermistor, it's not exactly straightforward. The >reason >is the highly non-linear resistance to temperature characteristic of the >thermistor. I have been working on something similar (in my copious >spare time). This equation is from memory, i'll look it up tonight when >I get home: > > T(Kelvin) = 1/(A + B*ln(R) + C*(ln(R)^3)) > >This equation is the Steinhart-Hart equation. R is the resistance >of the thermistor. A, B, and C are material constants that you can figure >out by measuring the resistance at three different temperatures. The >constants are all less than 1 for my particular thermistor. > >Now you can see the difficulty with this equation in a PIC. While the >fractional values can be taken care of by scaling everything, >calculating the natural log function is a bit more difficult. On top >of that, the temperature (in Kelvin) is actually the inverse. Use a lookup table - you can compensate for the thermistor nonlinearity, and the nonlinearity of using a simple resistive divider in one go. You can generate the table using the equation from the thermistor databook data with a simple Basic/[insert your favorite language] program, or probably Excel. If you need a wide temperature range, use 2 switchable pullups to improve linearity at the hot end, otherwise select the pullup/pulldown to get a reasonable voltage swing over the range you're interested in.