Interfacing LM335 to XBee ADC pin

Question

I'm a software engineer dipping my toe into hardware. I am trying to interface an LM335 temperature sensor to an XBee radio device. The XBee has an ADC built in. The XBee pins are:

• 1 = V_CC = 3.3V
• 10 = GND
• 14 = V_REF
• 19 = ADC

The specifications on the XBee say that V_REF "must be between 2.6V and V_CC".

The LM335 has (-) and (+) and (adj) pins, see LMx35, LMx35A Precision Temperature Sensors.

Could someone give me some guidance on what circuit I would need to make this work? Or at least a reference where I could learn more?

Answer 1

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.

Answer 2

As stevenvh has covered the LM335 issue with his excellent answer, I will not comment on that. However, I feel that neither the XBee V_REF issue was not fully explained, nor the XBee ADC input - with respect to Series 1 or 2.

You don't specify which series of XBee you are using, but if it be Series 2 then V_REF can be left disconnected.

From Common XBee Mistakes:

The XBee Series 2 hardware for ZigBee does not utilize this pin. However the Series 1 hardware for 802.15.4 radios does. It’s safe to hook it up for both, but on the ZigBee version it simply has no effect.

That said, whether you are using a Series 1 or 2 XBee, if you do as Steven suggests and connect V_REF to V_CC then no harm done...

In addition, the ADC on the XBee Series 2 can only sample up to 1.2V, anything beyond that will register as 1023 regardless. So a potential divider is also required at the output to map the 3.3V max down to 1.2V. A combination of 200kΩ and 100KΩ resistors would achieve this, with ADC connected to the point between the two, like so:

If you are using an XBee Series 1 module then the ADC can sample up to V_REF, which in this case is ~3.3V, and so no potential divider is required.

See Digital and analog sampling using XBee radios for the different maximum sampling voltages of the various XBee ranges:

Link to original table.