In my book, A Programmer’s Introduction to Mathematics, I describe the Taylor Series as a “hammer for every nail.” I learned about another nail in the design of modern smartphone accelerometers from “Eight Amazing Engineering Stories” by Hammack, Ryan, and Ziech, which I’ll share here.

These accelerometers are designed using a system involving three plates, which correspond to two capacitors. A quick recap on my (limited) understanding of how capacitors work. A capacitor involving two conductive plates looks like this:

The voltage provided by the battery pushes electrons along the negative direction, or equivalently pushing “charge” along the positive direction (see the difference between charge flow and election flow). These elections build up in the plate labeled $ -Q$, and the difference in charge across the two plates generates an electric field. If that electric field is strong enough, the electrons can jump the gap to the positive plate and complete the circuit. Otherwise, the plate reaches “capacity” and current stops flowing. Whether the jump happens or the current stops depends on the area of the plate $ A$, the distance between the plates $ d$, and the properties of the material between the plates, the last one is called the “dielectric constant” $ \varepsilon$. (Nb., I’m not sure why it doesn’t depend on the material the plate is composed of, but I imagine it’s smooshed into the dielectric constant if necessary) This relationship is summarized by the formula

$ \displaystyle C = \frac{\varepsilon A}{d}$

Then, an external event can cause the plates to move close enough together so that the electrons can jump the gap and current can begin to flow. This discharges the negatively charged plate.

A naive, Taylor-series-free accelerometer could work as follows:

- Allow the negatively charged plate to wobble a little bit by fixing just one end of the plate, pictured like a diving board (a cantilever).
- The amount of wobble will be proportional to the force of acceleration due to Hooke’s law for springs.
- When displaced by a distance of $ \delta$, the capacitance in the plate changes to $ C = \frac{\varepsilon A}{d – \delta}$.
- Use the amount of discharge to tell how much the plate displaced.

This is able to measure the force of acceleration in one dimension, and so thee of these devices are arranged in perpendicular axes to allow one to measure acceleration in 3-dimensional space.

The problem with this design is that $ C = \frac{\varepsilon A}{d – \delta}$ is a nonlinear change in capacitance with respect to the amount of displacement. To see how nonlinear, expand this as a Taylor series:

$ \displaystyle \begin{aligned} C &= \frac{\varepsilon A}{d – \delta} \\ &= \frac{\varepsilon A}{d} \left ( \frac{1}{1 – \frac{\delta}{d}} \right ) \\ &= \frac{\varepsilon A}{d} \left ( 1 + \frac{\delta}{d} + \left ( \frac{\delta}{d} \right )^2 + O_{\delta \to 0}(\delta^3) \right ) \end{aligned}$

I’m using the big-O notation $ O_{\delta \to 0}(\delta^3)$ to more rigorously say that I’m “ignoring” all cubic and higher terms. I can do this because in these engineering systems (I’m taking Hammack at his word here), the quantity $ (\delta / d)^2$ is meaningfully large, but later terms like $ (\delta / d)^3$ are negligibly small. Of course, this is only true when the displacement $ \delta$ is very small compared to $ d$, which is why the big-O has a subscript $ \delta \to 0$.

Apparently, working backwards through the nonlinearity in the capacitance change is difficult enough to warrant changing the design of the system. (I don’t know why this is difficult, but I imagine it has to do with the engineering constraints of measurement devices; please do chime in if you know more)

The system design that avoids this is a three-plate system instead of a two-plate system.

In this system, the middle plate moves back and forth between two stationary plates that are connected to a voltage source. As it moves away from one and closer to the other, the increased capacitance on one side is balanced by the decreased capacitance on the other. The Taylor series shows how these two changes cancel out on the squared term only.

If $ C_1 = \frac{\varepsilon A}{d – \delta}$ represents the changed capacitance of the left plate (the middle plate moves closer to it), and $ C_2 = \frac{\varepsilon A}{d + \delta}$ represents the right plate (the middle plate moves farther from it), then we expand the difference in capacitances via Taylor series (using the Taylor series for $ 1/(1-x)$ for both, but in the $ 1 + \delta/d$ case it’s $ 1 / (1 – (-x))$).

$ \displaystyle \begin{aligned} C_1 – C_2 &= \frac{\varepsilon A}{d – \delta} – \frac{\varepsilon A}{d + \delta} \\ &= \frac{\varepsilon A}{d} \left ( \frac{1}{1 – \frac{\delta}{d}} – \frac{1}{1 + \frac{\delta}{d}} \right ) \\ &= \frac{\varepsilon A}{d} \left ( 1 + \frac{\delta}{d} + \left ( \frac{\delta}{d} \right )^2 + O_{\delta \to 0}(\delta^3) – 1 + \frac{\delta}{d} – \left ( \frac{\delta}{d} \right )^2 + O_{\delta \to 0}(\delta^3) \right ) \\ &= \frac{\varepsilon A}{d} \left ( \frac{2\delta}{d} + O_{\delta \to 0}(\delta^3) \right ) \end{aligned}$

Again, since the cubic and higher terms are negligibly small, we can “ignore” those parts. What remains is a linear response to the change in the middle plate’s displacement. This makes it significantly easier to measure. Because we’re measuring the difference in capacitance, this design is called a “differential capacitor.”

Though the math is tidy in retrospect, I marvel at how one might have conceived of this design from scratch. Did the inventor notice the symmetries in the Taylor series approximations could be arranged to negate each other? Was there some other sort of “physical intuition” at play?

Until next time!

“Did the inventor notice the symmetries in the Taylor series approximations could be arranged to negate each other? Was there some other sort of “physical intuition” at play?”

Most likely simply the engineer decided to use a bridge measuring method, which is a standard approach whenever small values with a lot of possible common-mode noise are measured. The bridge will cancel the common mode noise out. Look up Wheatstone bridge for capacitance measurement.

I am not a mathematician or hardware engineer, but this makes me think of forward difference vs central difference approximation error in finite approximations – 1st order vs 2nd order error due to very similar cancellation.

This makes it sound like charge jumping the gap if the field strength is high enough is the normal operation mode, but for a real capacitor this internal short-circuiting is almost always fatal and I think that this device works differently, too. Moreover, the capacitance C is not related to the short-circuiting, but determines how much the voltage over the capacitor changes for a given current into the capacitor. A real capacitor has a maximum voltage rating and should not short-circuit below that voltage (of course there are leak currents, which is why a real capacitor is better modeled by a capacitance in parallel to a resistor).

Not trying to show off, but these types of approximations are not so uncommon in senior high school without the use of taylor series. I also came across some of the questions pertaining to differential capacitors and the harmonic nature of the plates in high school text books.

It is true, most physics books use invalid/bad math to avoid a calculus prerequisite. IMO interpreting that as “this result is basic” is lame.

Seconding what tomlx said: I don’t think any charge is jumping between plates here. I think the change in capacitance is being measured.

If you can assume the charge, Q, on the plates is constant then that reduces to measuring the voltage. But that change in voltage probably now causes a current to flow, so I expect it’s more complicated than that.