# variation of thickness of dielectric layer for different textile-based capacitive pressure sensors

For my thesis I am working with textile-based capacitive pressure sensors and I am struggling to see how much variation in thickness of the dielectric layer you can have before you have noticeable impact on sensor output. If I wanted to have my sensors somewhere between 2x2in and 4x4in for the surface area, and I was using nylon 6,6 for my dielectric layer (dielectric constant of 3.4), and I was testing different construction methods for the nylon (different knits and wovens) and I wanted to keep them as close in thickness as possible, how much leeway is there before the variation in thickness would result in a measureable difference? Since I am using different construction methods for the dielectric layer there is no way I can make them EXACTLY the same thickness, so I am trying to figure out my wiggle room. If I have a dielectric layer that is 900µm thick and another that is 400µm thick, is that going to be way too much variation or is that relatively close enough? I've done the capacitance calculation for each thickness using a 3x3in sized sensor and got a capacitance of 194.318pF for 900µm thickness and 435.924pF for 400µm thickness. I just don't know what is considered a significant difference for capacitance.

• Welcome to EE.SE. You have no location information in your user profile so it is difficult to know what dimension units you are using. 2 x 2 what? mm? What is "nylon 6,6"? It's probably better to use standard SI units for "µm" (or even "um") rather than micron as this is an international site. Since we know that $C = k \frac {A}{d}$ separation will be critical. – Transistor Oct 20 '18 at 9:45
• Hey! Sorry, I will edit now. There are different kinds of nylon, commonly for fabric it will either be nylon 6 or nylon 6,6, and they have slightly different dielectric constants (like a .2 difference), so I specified. The dimensions of the sensors are in inches, I didn't realize I forgot to include that. Thanks for the tips. – darriage Oct 21 '18 at 18:44
• Ah, inches. You must be from the colonies then. Why are you mixing your units? Use SI for all. – Transistor Oct 21 '18 at 19:50

The capacitance of a parallel plate capacitor is given by

$$C = \kappa \epsilon_0 \frac {A}{d}$$

where $$\ \kappa \$$ is the dielectric constant, $$\ \epsilon \$$ is the permittivity of free space, 8.85 × 10−12 F/m, A is the area and d is the distance between the plates. Here we'll be using metric measurements rather than the quaint inches.

Checking your calculations for your 3" x 3" at 900 µm capactior with 3.4 dielectric we get

$$C_{900} = 3.4 \times 8.85 \times 10^{-12} \frac {0.0762^2}{900 \times 10^{-6}} = 1.88 \times 10^{-10} = 194 \ \text {pF}$$

If we decrease $$\ d\$$ from 900 µm to 400 µm the formula says that the capacance should increase by $$\ \frac {9}{4} \$$. Checking this we get $$\ C_{400} = 188 \frac {9}{4} = 436.5 \ \text {pF}\$$.

Only you can decide that. If you can tolerate a $$\ \frac {9}{4} \$$ variation then yes, otherwise no. I suspect not.