The LM335 belongs to a range of a fairly accurate and simple to use temperature sensors.

If we want to use the calibrated configuration of the LM335 then this is how to calculate the series resistor for the LM335:

According to the datasheet, the potentiometer needs to set the voltage over the LM335 at 2.98 Volts at 25°C (¶ 6.4 in the datasheet). The LM335 needs between 400uA and 5 mA (¶ 6.2 in the datasheet). Suppose that the max temperature we want to measure is 30 C (=303.15 °K). The output then will be 3.0315 Volt (let’s say 3.03 V).

At 3.03V the current through the 10k potentiometer will be 3.03/10k=303uA. The minimum current that needs to flow through the LM335 is 400uA, so R1 needs to supply 703uA over a voltage of Vcc-3.03V.

So at 5 Volt that would be 1.97/0.703 = 2.8kΩ

At 3.3 Volt it would be 0.27/0.703 = 0.384kOhm =384Ω

(Note 703uA= 0.703mA = 0.000703A. I chose to divide by mA so I get outcome in kΩ).

Now we still have to check if the current will not be too high at the lowest temperature we want to measure:

Let’s say our minimum temperature is 0°C that is 273.15. At 10mV/°K that means an output voltage of 2.73 Volt.

At 5 Volt the resistor current will be (5-2.73)/2.8kΩ = 810uA. We still need to subtract 2.73V/10kOhm=273uA which flows through the potentiometer so we have 810-273=537uA, which is well below the 5mA max.

If we calculate that for 3.3Volt, we find the following:

The resistor current will be (3.3-2.73)/384=1.5mA. Subtract 273uA that flows through the potentiometer and that gives roughly 1.23mA, well within limits.

If you want to measure temperatures higher than 30°C then R1 has to be smaller. When using 3.3 Volt the theoretical maximum temperature measurable is 330°K which is about 57°C. In practice however that does not work as you need a drop voltage over R1 so the value of R1 cannot be zero, or, in other words: you cannot connect the LM335 directly to Vcc and still expect to get an output lower than Vcc.

Bet let’s see how that would be at 50°C (323.15°K) and 3.3V Vcc :

Well again you need 703uA but this time over 0.07V.

That gives a value for R1 of 100Ω.

But then at 0 degrees the current is 5.7mA. Still need to subtract 2.73uA which gives 5.4mA which is out of spec.

A program would be as follows:

//A0 is used to read the LM335 void setup() { Serial.begin(9600); } void loop() { int rawvoltage= analogRead(A0); float millivolts= (rawvoltage/1024.0) * 5000; float kelvin= (millivolts/10); Serial.print(kelvin); Serial.println(" degrees Kelvin"); float celsius= kelvin - 273.15; Serial.print(celsius); Serial.println(" degrees Celsius"); float fahrenheit= ((celsius * 9)/5 +32); Serial.print(fahrenheit); Serial.println(" degrees Fahrenheit"); delay(2000); }