The LM335 and it’s serieresistor

Fig 1: LM335 in Calibrated mode

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

The LM335 operates from -40°C to 100°C. It’s output is lineair with 10mV/°Kelvin (¶ 6.6 in the datasheet). It can be used in two ways: Calibrated and uncalibrated mode. The difference is a  10kΩ potentiometer that is used to regulate the output voltage to 2.98 Volt at 25°C. With more and more devices running off of 3.3 Volt, the question is sometimes raised: will the LM335 run off of 3.3 Volt as well. Going through the datasheet you will not see a min or max value stated for the voltage to be used as all depends on the current through the LM335 and that current is determined by a series resistor (R1 in Fig. 1)

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);
}

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