# How does temperature compensation work for pellistors in Wheatstone bridge?

Pellistors are used to detect combustible gases. Basically, the gas is burned inside the pellistor and the heat released will change the resistance of the detection element. The concentration of the gas is therefore transduced into the variation of the resistance.

Several datasheets/articles (eg. VQ548MP Datasheet, Understanding catalytic LEL combustible gas sensor performance) mention a compensator (a compensator in this case is simply an inert detection element) should be used to eliminate the effects of ambient temperature changes, since the variation of ambient temperature would also change the resistance. The article says:

Because the two beads are strung on opposite arms of the Wheatstone Bridge circuit, the difference in temperature between the beads is registered by the instrument as a change in electrical resistance.

However, I can't see how it measures the difference in temperature. According to the circuit below (suppose $$\ R_1 = R_2 \$$, and $$\\Delta R\$$ is the resistance change caused by gas combustion, $$\ R_D = R_C + \Delta R\$$):

$$V_{out} = V_{DC} \cdot (\frac{1}{2} - \frac{R_C}{2R_C+\Delta R})$$

$$\R_C\$$ can be affected by the ambient temperature and is not known. So the compensation is not working in this case.

Or was the circuit drawn wrong? Because it says the two beads are strung on opposite arms of the Wheatstone Bridge circuit. If we put the compensator in the place of $$\ R_2\$$:

$$V_{out} = V_{DC} \cdot (\frac{R_C}{R_C+R_1}-\frac{R_1}{R_1+R_C+\Delta R})$$

I still can't see how it compensates.

• What makes you think that "the compensation is not working in this case"? You've defined the resistance of the detector RD = RC+ΔR. If the ambient temperature changes, RD changes along with it, as does the compensator RC, and they change by the same amount. Commented Mar 19, 2022 at 1:34
• The two components should be "side by side" at the "same" measurement location. In fact, it is a "ratiometric" measurement. Commented Mar 19, 2022 at 11:17
• @Antonio51 I'm confused because the ratio($\frac{R_D}{R_C}$) will change with temperature if $R_D \neq R_C$. I don't know how $\Delta R$ is calculated in a practical situation (especially how to determine the value of $R_C$). Commented Mar 19, 2022 at 11:36
• Simplify a little. No gas. The ratio of the two "resistances" should be almost "one", and this ratio should not change with temperature, so one can "calibrate" the W-bridge. Commented Mar 19, 2022 at 11:44
• Note also that voltage Vout measured is not a "linear" function. Commented Mar 19, 2022 at 13:16

$$\R_C\$$ can be affected by the ambient temperature and is not known.
• Thanks for your answer!! I can understand the point of compensation now. But still a bit confused about $R_C$. Do we need to know $R_C$ at different temperatures? Because according to the first equation, to obtain $\Delta R$, the value of $R_C$ is needed (which changes with the temperature). Commented Mar 19, 2022 at 8:38