Hello everyone. Long time lurker, first time poster here.

I'm looking to use the Texas Instruments LM35 temperature sensor to (yes, you guessed it) monitor ambient temperature in a refrigerator.

The overall goal is to eliminate the existing Temperature Control Knob and replace it with an Arduino controlled/triggered Solid State Relay. This will also allow me to have an LCD readout of the temperature and be able to "dial in" a Set/Desired Temperature via programming or some other source such as a potentiometer.

I'm looking to operate the LM35 in Full-Range (-55°C to 150°C). Figure 2 on page 1 of the datasheet (found here: http://www.ti.com/lit/ds/symlink/lm35.pdf) shows that a resistor must be added to the output of the LM35 in order for it to operate in Full-Range.

I'm confused as to what the other side of the resistor (labeled -Vs) should be connected to and how to figure out what size resistor I'll need being that I'm operating from a single supply. Also, what is the purpose of listing the different Vout ratings in this figure?

Here is Figure 2 from the datasheet:

Lastly, I just noticed Figure 18 on page 11 of the datasheet. Now I'm even more confused as to whether I should wire it up as shown in Figure 2 or Figure 18??? What's the difference between the two?

Figure 18:

Any help would be greatly appreciated.

 

Since you are in the USA you may want to look at the LM34 instead which is calibrated for 10mV/°F.

 

In order to get the full range of the sensor, you need to be able to have the output negative to GND. Figure 2 just provides some information necessary to get the correct voltage for temperature less than 0°C. Figure 18 shows how you can achieve a negative voltage using just a single voltage supply by raising the GND of the sensor about 0.9V above the voltage return. With negative temperatures, the voltage on Vout will be negative with respect to GND. For better accuracy, I would replace the diodes with a zener.

 

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I'm looking to operate the LM35 in Full-Range (-55°C to 150°C). Figure 2 on page 1 of the datasheet (found here: http://www.ti.com/lit/ds/symlink/lm35.pdf) shows that a resistor must be added to the output of the LM35 in order for it to operate in Full-Range.

I'm confused as to what the other side of the resistor (labeled -Vs) should be connected to and how to figure out what size resistor I'll need being that I'm operating from a single supply. Also, what is the purpose of listing the different Vout ratings in this figure?

Here is Figure 2 from the datasheet:
.....................................
Lastly, I just noticed Figure 18 on page 11 of the datasheet. Now I'm even more confused as to whether I should wire it up as shown in Figure 2 or Figure 18??? What's the difference between the two?
.....................................
Any help would be greatly appreciated.

The difference is that the circuit in Figure 1 requires a negative voltage (-Vs) with the output being referenced to ground, while the circuit in Figure 18 requires only a positive voltage but the output is referenced to the diode voltage (in other words you need a differential input to properly see the LM35 output voltage.

If you don't need to go below 0F then you may want to use an LM34 as suggested by MrChips since 0V output is 0F (with no negative voltage or added parts).

 

You guys are the best! That's why I love coming here

With your collective help I now notice two things.

1 - In Figure 18 The output is a differential output, so I would need an ADC with a differential input or a Differential-to-Single chip in order to read the output. Looks like I would also be better suited using a potentiometer in place of a 10% resistor so I could calibrate the output to be a little more precise.

2 - In Figure 2 they list one of the outputs as being negative voltage for a negative Celsius reading. This should have been my first indicator that the Output pin (-Vs) needs to be below ground via two power supplies.

Originally I wanted to go with the LM35 because of it's accuracy of about ±0.5°C. However, the LM34 has an accuracy of about ±1.0°F. At first it's a little off-putting but 1 degree should be fine for what I'm using it for.

Thanks again fellas, I really appreciate the help!

 

hi,
There are LM135 and LM335 types that operate minus thru plus zero, 10mV/Cent

E

 

±0.5°C and ±1.0°F are almost the same. You can adjust the scale by changing the scaling factor.
LM34 will allow you to go down to 0°F without the need for negative voltages.

 

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1 - In Figure 18 The output is a differential output, so I would need an ADC with a differential input or a Differential-to-Single chip in order to read the output. Looks like I would also be better suited using a potentiometer in place of a 10% resistor so I could calibrate the output to be a little more precise.
.........

The output voltage is determined by the LM35 and is largely independent of the resistor tolerance, the same as a load resistor change causing little output change. The resistor is just to provide a negative bias so the output can go below 0V.

 

The output voltage is determined by the LM35 and is largely independent of the resistor tolerance, the same as a load resistor change causing little output change. The resistor is just to provide a negative bias so the output can go below 0V.

Gotcha. That makes a lot more sense. Thank you

 

Hello everyone. Long time lurker, first time poster here.

I'm looking to use the Texas Instruments LM35 temperature sensor to (yes, you guessed it) monitor ambient temperature in a refrigerator.

The overall goal is to eliminate the existing Temperature Control Knob and replace it with an Arduino controlled/triggered Solid State Relay. This will also allow me to have an LCD readout of the temperature and be able to "dial in" a Set/Desired Temperature via programming or some other source such as a potentiometer.

I'm looking to operate the LM35 in Full-Range (-55°C to 150°C). Figure 2 on page 1 of the datasheet (found here: http://www.ti.com/lit/ds/symlink/lm35.pdf) shows that a resistor must be added to the output of the LM35 in order for it to operate in Full-Range.

I'm confused as to what the other side of the resistor (labeled -Vs) should be connected to and how to figure out what size resistor I'll need being that I'm operating from a single supply. Also, what is the purpose of listing the different Vout ratings in this figure?

Here is Figure 2 from the datasheet:

​

Lastly, I just noticed Figure 18 on page 11 of the datasheet. Now I'm even more confused as to whether I should wire it up as shown in Figure 2 or Figure 18??? What's the difference between the two?

Figure 18:

​

Any help would be greatly appreciated.

While this is somewhat redundant to what others have already said, I wanted to show you where the information you were looking for is given/implied in the figures you were looking at.

The purpose of listing the Vout values for different temperatures is to give you the output voltage at the extreme limits of the operating temperature range (as well as the room temperature value). So you know that your output voltage, relative to the bottom terminal of the device, will be between -0.55V and +1.50V. The first line tells you how to size the resistor, namely to choose a value that results in about 50uA of current flowing downward to negative supply. The implication is that whatever voltage you use for +Vs, you will use the negative of that for -Vs (hence why the used Vs for both and not, say, Vp for one and Vn for the other). For many applications using symmetric supplies is probably very simple to do, but if you have a single supply then you can use the circuit in Figure 18. With a quiescent current of only about 100uA or so, the voltage drop across a 1N914 is only about 0.5V, so your effective -Vs is about -1V and with an 18kΩ resistor you would expect about 55uA of current. Since the quiescent current could be smaller, that gives a bit of cushion, though the exact current doesn't appear to be at all too critical since they are using a 10% tolerance resistor.

As for the notion that you are giving up something in accuracy by using a device that has a rated accuracy of 1°F instead of 0.5°C, you need to take into account that 0.5°C is 0.9°F, so they are very similar -- in fact, they are almost certainly the same and it is merely a matter of rounding the specs to something convenient in the two scales.

Since this is for a refrigerator, you might ask yourself why you want to operate over the full temperature scale of the device at all. Don't add in capabilities that you don't need if they increase the cost or complexity -- if it's good enough, then it's good enough. Now, if you want to do so just for grins and giggles, then by all means go for it.

 

That's one hell of an explenation and exactly what I was looking for. Thank you so much WBahn! Nothing vague anymore