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And I mean, how does it work internally, how does it output a voltage when temperature changes, etc. When I see explanations online, I don't understand much of how it works (supposedly something called a PTAT is involved).

Can someone explain it to me like I'm a total noob when it comes to electronics?

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    \$\begingroup\$ The datasheet has a "detailed description", did you read that? At which level of understanding are you stuck? \$\endgroup\$ – Arsenal Dec 14 '17 at 15:37
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    \$\begingroup\$ I'm stuck with the PTAT thing. Our professor requires us to have a thorough explanation of how they work. I understand some parts of the internal circuit in the datasheet, but I can't seem to find how it seems to work. Like, how exactly does having different current densities for the emitter cause a voltage for the resistor. \$\endgroup\$ – RenV Dec 14 '17 at 15:41
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    \$\begingroup\$ Yes, the datasheet merely details how to use it, not how it works fundamentally, as you say. @Arsenal, have you found the information the OP's question asks in the datasheet you linked to then? \$\endgroup\$ – TonyM Dec 14 '17 at 15:52
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    \$\begingroup\$ \$1.38 V_{PTAT}\$ on Functional Block Diagram - Figure 12 (p13) is the only reference in datasheet. No real explanation. \$\endgroup\$ – StainlessSteelRat Dec 14 '17 at 16:08
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    \$\begingroup\$ The key phrase is "The temperature-sensing element is comprised of a delta-V BE architecture", which is a bit heavy on the jargon. Have you read en.wikipedia.org/wiki/Silicon_bandgap_temperature_sensor ? How much semiconductor theory do you already know? \$\endgroup\$ – pjc50 Dec 14 '17 at 16:23
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So to understand how you can build something which is Proportional To Absolute Temperature (PTAT) you have to dive a bit deeper into semiconductor basics, or rather principles.

Let's start with a simple diode (which is not how it is done in the LM35). The current through a diode is given by the Shockley equation (which neglects the series resistance which is fine for our use): $$I=I_S \left(e^{\frac{V_D}{nV_T}}-1\right)$$

with:

  • \$I\$ the diode current
  • \$I_S\$ the reverse bias saturation current
  • \$V_D\$ the voltage across the diode
  • \$V_T\$ the thermal voltage
  • \$n\$ the ideality factor

Now \$I_S\$ and \$n\$ are properties of a specific diode, so they depend on the model you have and you can't influence them.

\$V_T\$ is actually what is really interesting. It is the thermal voltage which is given by this formula: \$V_T=\frac{kT}{q}\$ with \$k\$ being the Boltzmann constant, \$T\$ the Absolute Temperature of the diode junction and \$q\$ the elementary charge.

Now what you usually do with diodes as a temperature sensor is to force a constant small current through it and measure the voltage over the diode.

So you can solve the above equation to read: $$ V_D = n\ \frac{kT}{q} \ \ln\left(1+\frac{I}{I_S}\right) $$

\$n\$,\$I\$,\$k\$,\$q\$ and \$I_S\$ are all constant, so you can see the voltage across the diode will be varying only with the absolute temperature of the diode.

You will find that often the 1 inside the logarithm is neglected as \$\frac{I}{I_S}\$ is much greater than 1. To get a feeling for it: you might use a constant current of 30 µA as \$I\$ but \$I_S\$ is in the order of pA, so \$\frac{I}{I_S}\$ is in the order of a million and a million + 1 is still pretty close to a million.


Now the fun part is, that the base-emitter voltage of a transistor behaves just the same way as a diode.

The method used is called \$\Delta V_{BE}\$ because you use two transistors with different currents and you measure the difference of the two base-emitter voltages. The key is here to use two transistors which are essentially the same (so their properties are the same and can be cancelled out).

$$V_{BE1} = n\ \frac{kT}{q} \ \ln\left(\frac{I_{C1}}{I_S}\right) \\ V_{BE2} = n\ \frac{kT}{q} \ \ln\left(\frac{I_{C2}}{I_S}\right) \\ \Delta V_{BE} = V_{BE2} - V_{BE1} = n\ \frac{kT}{q} \ \left(\ln\left(\frac{I_{C1}}{I_S}\right) - \ln\left(\frac{I_{C2}}{I_S}\right)\right) \\ \Delta V_{BE} = n\ \frac{kT}{q} \ \ln\left(\frac{I_{C1}}{I_{C2}}\right) $$

So what is the benefit of this compared to the diode above? You eliminated the process variation of \$I_S\$ and replaced it with two constant currents you have to provide, but you might get that under better control than the process variation.

Another thing you eliminate is the complex temperature dependency of the saturation current, which limits the linearity of your voltage output.

If you dive deep enough into these things you will encounter many properties which are not quite as ideal as the above equations might suggest. Some of these effects are cancelled out by the \$\Delta V_{BE}\$ method.

Also the typical diode has an ideality factor which is not as close to 1 as you would like it to be. Transistors base-emitter diodes have one much closer to 1.

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  • \$\begingroup\$ Excellent answer, upvoted. \$\endgroup\$ – TonyM Dec 14 '17 at 18:09
  • \$\begingroup\$ It's not only important to cancel out Is because it varies with the process but also because it varies with temperature (and not in the simple way kT/q does). \$\endgroup\$ – τεκ Dec 14 '17 at 18:27
  • \$\begingroup\$ @τεκ that is actually a very good point to add, thanks. At that level a lot is going on and it is hard to keep the balance between how deep to go and sometimes I forget the most important parts. \$\endgroup\$ – Arsenal Dec 14 '17 at 18:33
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    \$\begingroup\$ +1 Another approach is to the use the same transistor at two (or even three) different currents (time shared). The improvements increase the linearity and unadjusted accuracy but decrease the signal level. Three currents can be used to compensate for ohmic resistance. \$\endgroup\$ – Spehro Pefhany Dec 14 '17 at 21:06
  • \$\begingroup\$ Explanation is good and thorough, I get it now. Thanks for all the answers \$\endgroup\$ – RenV Dec 14 '17 at 22:51
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After following links through Quora and Edaboard, it basically comes down to Wikipedia:

Basically, it boils down to the fact that the voltage across a PN junction (the BE junction of a transistor) varies both with current density and with temperature. This can be exploited by the surrounding circuit that compares the responses of two transistors operating at different current densities to either cancel out the temperature dependency (to create a voltage reference) or to linearize the transfer function (to create a temperature sensor).

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  • \$\begingroup\$ Okay this helps very much, but if you don't mind, I have one last question, why does the voltage across the junctions vary with temperature and current density? Is there an underlying physics explanation to it? It doesn't seem ti be in the wikipedia articles \$\endgroup\$ – RenV Dec 14 '17 at 16:50
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    \$\begingroup\$ That's a solid-state physics question that starts to get into the quantum-mechanical properties of the charge carriers. The basic idea is that the energy of the carriers is dependent on temperature, and the statistics of large numbers of carriers result in the various terms shown in the Wikipedia equations. \$\endgroup\$ – Dave Tweed Dec 14 '17 at 16:59
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    \$\begingroup\$ The question comes from a staggering inability to use Internet search. Quick googling for the datasheet keyphrase "delta-VBE architecture" gives a lot of information, that this kind of circuit was patented in 1981/1984 by Motorola, google.com/patents/US4450367 , and a full explanation is given here, iascaled.com/delta-vbe-temperature-measurement It takes 30 seconds to figure all out. \$\endgroup\$ – Ale..chenski Dec 14 '17 at 17:15
  • \$\begingroup\$ @AliChen, seems a pointlessly harsh comment, OP had their answers already and is just trying to learn, as we all do. \$\endgroup\$ – TonyM Dec 14 '17 at 21:18
  • \$\begingroup\$ @TonyM, I have even a stronger point to make: a student must learn how to use modern informational tools. Internet has many dumb-down levels of explanations for everything. If a student in EE can't spiral it down to his level of comprehension and learn basics of necessary elements involved on his own, he/she has "no use in engineering". \$\endgroup\$ – Ale..chenski Dec 14 '17 at 23:47

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