Electronic – Voltage divider for Xbee’s ADC – what resistor values

Have a look at the examples on page 6 of the datasheet. It shows a basic application and a calibrated sensor. This is the calibrated one:

Set the potentiometer so that at 25°C you have 2.982V at the output. If you don't need the calibration just leave out the potentiometer. Your reading will be less precise. Uncalibrated error is for the LM335 typically 2K @25°C, maximum 9K(!) over the full range (table on page 2), so that you might prefer to calibrate it after all.
The value of R1 must be chosen in function of the power supply; on the first page you can read that it needs between 400\$\mu\$A and 5mA. Suppose the maximum temperature you want to measure is 30°C (that's 303.15K). The output will then be 3.03V. The current through the potentiometer will be \$\frac{3.03V}{10k\Omega}=303\mu A\$. You have a 3.3V power supply, then \$R1=\frac{3.3V-3.03V}{400\mu A + 303\mu A}=384\Omega\$ maximum.

Being good designers :-) we now check if the current isn't going to be too high at the lowest temperature. Suppose we want to measure temperatures as low as 0°C, that's 273K, giving \$V_{OUT}=2.73V\$. The resistor current will then be \$I_{R1}=\frac{3.3V-2.73V}{384\Omega}=1.5mA\$, subtract the 273\$\mu\$A through the potentiometer, and we have 1.2mA through the LM335, way below the allowable 5mA, so that's OK.

R1 has to be smaller than \$384\Omega\$ if you want to measure higher temperatures. With a 3.3V supply your theoretically maximum measurable is 330K, or 57°C. In practice somewhat less, because you will always need some voltage over the resistor. The resistor value will be so low, however, that at lower temperatures you almost certainly will exceed the allowable 5mA.

Connect \$V_{REF}\$ to \$V_{CC}\$.

National has also a sensor giving you the temperature directly in °C (LM35) and I thought this would be a solution for measuring higher temperatures (at 57°C it will only output 570mV, so you'd have enough headroom), but that needs at least 4V, so that wouldn't help us.

By your values, I have to assume that you are measuring off of the 10kOhm resistor on the bottom. I would hazard a guess that your 10kOhm resistor is a little high and the 100kOhm resistor is a little low. This would mean if your 10kOhm resistor is taking more of the voltage in the voltage division than expected.

There are a few problems at work here.

1. The first is the resistor value error. Ohm's Law: V=I*R. So if there is 5% error in R, then for a constant I, there will be a %5 error in the expected value of V. The %error from resistor tolerances are the maximum +/- error. So in reality, there could be a resistor with +5% error and other with -5% error. So there is a much larger possible range in output voltages for a constant current.

   • This can be eliminated by either hard-coding the real resistances in software or adding a reference voltage that is more accurate than the resistor or A/D quantization error. This reference voltage would be used to get the real resistance values as the A/D sees them.

2. A/Ds can have an offset error. This really has to be calibrated out. Some A/Ds have this built in.

3. A/Ds also have quantization error. So if a voltage falls between two consecutive quantization levels, it will have to be rounded to one of those two, introducing error. There are ways of increasing the number of bits for an A/D by oversampling and averaging blocks of samples into one sample.

There are other errors that could be at work, but those three are the main ones I have run into. If the A/D is fairly linear, then measuring two accurate voltages with the A/D circuitry would let you build an affine linear equation to correct the data. This builds off of the problems in 1 and 2.

A/Ds can appear linear, but end up being very nonlinear at a particular range. I have heard of some implementations using a look-up table to correct the nonlinear behavior. But that is getting a little beyond your current problems.

Edit: One more item. It is a good idea to buffer your analog signals to the A/D. It does a few things, like add another device to protect the microcontroller, and limit any possible transient sampling behavior from the A/D go into the analog signal.