# Next step electronics 1; Transistor-based thermometer

Throughout the years of my occupation with analog electronics I have realized that there is a considerable gap between intermediate and advanced analog knowledge. That is, even if you have read every known book regarding analog electronics, either for undergraduate students (The Art of Electronics, Microelectronic Circuits) or for graduate students (Analysis and Design of Analog Integrated Circuits), you will probably have no idea how to analyze monstrous circuits like the LM10 or an fm receiver. It’s not easy to analyze even the LM101 that got out decades ago.

The purpose of this line of posts is just that. To introduce and explain circuits relatively unknown to the average electronics enthusiast. Some of them will be simple (like the one that is going to be presented in this post) and some of them will be a little bit more complicated. But all of them have one common trace; when I stumbled across them for the first time, I felt like making a step towards advancing my expertise on the subject of analog electronics.

Ok, enough with the talking. Let’s find out how we can make a thermometer using off the shelf transistors.

Believe it or not, this is a thermometer. The output voltage, V(out), is proportional to the temperature of the transistors (we assume that all of the transistors are close enough, so their respective temperatures do not deviate much). In general, the relationship between the difference of a BJT’s base and emitter voltage ($V_{be}$) and its collector current ($I_C$) is given by the following formula

(1) $V_{be}=V_t*ln({I_C/I_S})$

where $V_t$ is the thermal voltage, $I_C$ is the collector current and $I_S$ is the saturation current. What’s important about $I_S$ is that it is a constant for each transistor and that it is directly proportional to its emitter area. The formula of the thermal voltage is

(2) $V_t=T k/q=86.6*10^{-6} T$

where $k$ is Boltzmann’s constant, $q$ is the charge of an electron and $T$ is the temperature in Kelvins.

From equation (1) it is apparent that if two identical transistors (same $I_S$) have the same base emitter voltage and are close enough to share the same temperature, then they must have the same collector current ($I_C$). This is the case with Q4 and Q5, as well as Q3 and Q6, assuming that all PNP transistors are identical with each other and all NPN transistors are also identical with each other.

The measurement of the temperature is taking place in the Q1-Q3 transistors. Since transistors Q1 and Q2 have all three of their terminals connected, they form a PNP transistor with twice as much the $I_S$, because it is proportional to its emitter area. Now, let us perform Kirchoff’s voltage law around the base emitter junctions of the compound Q12 transistor and Q3 transistor. The equation that we get is

(3) $V_{be3}=V_{be12}+I_{C12}*R_1$

assuming that the current is going into the emitter. Substituting equation (1) in (3), we take

(4) $V_t*ln({I_{C3}/I_{S3}})=V_t*ln({I_{C12}/I_{S12}}) + I_{C12}*R_1$

which can be written as

(5) $I_{C12}*R_1=V_t*ln(\frac{I_{C3}*I_{S12}}{I_{C12}*I_{S3}})$

As we mentioned earlier, the compound transistor Q12 has twice the emitter are of Q3, so

(6) $I_{S12}=2*I_{S3}$

and the collector currents of Q4-Q5 are identical, which means that, if we neglect the base currents as relatively small, we get

(7) $I_{C12}=I_{C4}=I_{C5}=I_{C3}$

Combining (2), (6) and (7) with (5), we derive the following formula

(8) $I_{C12}=\frac{ln(2)*86.6*10^{-6}}{R_1} T$

Now it is obvious how we get the temperature dependency. We have created a current which is directly proportional to the absolute temperature. All Q6 does is replicating this current and feeding it to resistor $R_2$ in order to create the output voltage which can be easily measured using a voltmeter. Therefore, the output voltage is

(9) $V_{(out)}= \frac{R_2}{R_1}*ln(2)*86.6*10^{-6}* T$

By properly selecting the ratio of $R_2$ to $R_1$ we can adjust the rate of change or output voltage for a particular temperature.

[IMPORTANT NOTE!] If you try to build this circuit, it is not going to work as it is. The reason behind this is that, equation (5) can also be satisfied with really small collector currents (which in real devices exist and we call them “leakage currents”). What one must do is to add a startup circuit that will ensure that the currents in the device are not negligible. We will show that in future posts.