# What is the minimum supply voltage for an LM335?

The answer to this question is probably staring me in the face, but I just can't see it. I'm attempting to interface with an LM335AZ temperature sensor (datasheet) at 3.3V. The datasheet clearly states that the output voltage is 10mV/K, and that at 25°C and 1mV it should output between 2.92V and 3.04V. What I do not see specified is the minimum voltage it will operate at, and the sensor I have in hand doesn't appear to work correctly at 3.3V.

With a 3.3KΩ resistor the current to the LM335 should be 1mA at 3.3V. If I directly measure the output voltage when supplying 3.3V, I get 2.46V, which correlates to −27°C; if I supply 5V I get 2.93V, or about 20°C. Suffice it to say that I'm not performing these tests outside at the North Pole. :-)

Does the LM335AZ only work on 5V, or 5V and up?

With 3.3K you are not getting 1mA. You are getting around 100uA.

At 25DegC the LM335 is at 3V. With a Vin of 3.3V you getting 0.3V across the resistor. 0.3V / 3.3K is apx 100uA. If you use a 300Ohm resistor you should be at about 1mA.

• Stupid decimal point. :-) Thanks, John! I'd missed that I need to be looking at the voltage drop, not Vss. Commented Jan 3, 2010 at 3:26
• @blalor - HTML doesn't work in comments, you need to use mini-markdown: [link](http://example.com) _italic_ **bold** code Commented Jun 27, 2011 at 16:28

The LM335 works a bit like a variable zener diode, as the symbol in the datasheet shows.

That means that it will have a voltage over it when a current flows through it. So you don't supply a voltage, you supply a current. That's why you can't find the voltage in the datasheet.
The datasheet specifies that the current must be between 400$\mu$A and 5mA. You have to calculate the value of R1 depending on V+ and the maximum and minimum temperatures you want to measure.

In this answer I provide a detailed calculation of the series resistor R1.

I know this is an old thread, but just wanted to supply my findings for using LM335 at 3.3v using Arduino Pro Mini.

This article here suggests using a 2kΩ resistor at 5v, so using the R = V/I formula we can change to 1.3kΩ resistor. I had a 1.2kΩ one handy and this worked.