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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.

enter image description here

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  • \$\begingroup\$ 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. \$\endgroup\$
    – brhans
    Mar 19, 2022 at 1:34
  • \$\begingroup\$ The two components should be "side by side" at the "same" measurement location. In fact, it is a "ratiometric" measurement. \$\endgroup\$
    – Antonio51
    Mar 19, 2022 at 11:17
  • \$\begingroup\$ @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\$). \$\endgroup\$
    – Jack Black
    Mar 19, 2022 at 11:36
  • \$\begingroup\$ 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. \$\endgroup\$
    – Antonio51
    Mar 19, 2022 at 11:44
  • \$\begingroup\$ Note also that voltage Vout measured is not a "linear" function. \$\endgroup\$
    – Antonio51
    Mar 19, 2022 at 13:16

1 Answer 1

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\$R_C\$ can be affected by the ambient temperature and is not known.

It is known: its purpose is to be at the same ambient temperature, and thus have the same resistance.

The idea is that you have two identical sensors. Both are exposed to the same ambient temperature, and installed so they have same thermal conductance to ambient. But only one of them is reacts to gas (e.g. is exposed to gas, or is coated with the catalyst, etc.).

The ambient temperature changes will not destabilize the bridge, since both beads will change their resistance the same due to ambient changes. Only the temperature difference between the two beads will be measured, and that temperature difference will be due to additional heat provided by the oxidation of the combustible gas on one arm of the bridge. And that's what you want to measure :)

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  • \$\begingroup\$ 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). \$\endgroup\$
    – Jack Black
    Mar 19, 2022 at 8:38
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    \$\begingroup\$ @GeorgeGuo Rc is not something that you provide. The entire left side of the bridge, on grey background, is the sensor you buy. It has two beads, ie. two resistances inside. The only resistors you provide are on the right arm. \$\endgroup\$ Mar 19, 2022 at 16:29
  • \$\begingroup\$ And no, you generalny don’t care about exact resistance values, since the calibration is not usually done according to any equations, but by measuring the voltage change in the output relative to excitation voltage or excitation current, while feeding the sensor various calibrated gas mixtures. \$\endgroup\$ Mar 19, 2022 at 16:33
  • \$\begingroup\$ As for equations: all resistances in the bridge have the same nominal value. Each of them has a thermal coefficient. The coefficients of the beads are the same, but one is hotter. So any voltage change will be proportional to temperature difference if the drive scheme is constant current. With constant voltage things are a bit more nonlinear. But then the sensor’s temperature delta vs gas concentration is nonlinear anyway. So you are curve fitting the thing during calibration. \$\endgroup\$ Mar 19, 2022 at 16:35

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