# Capacitance formula

1. Mar 14, 2012

can somebody give formula to calculate value of capacitance at co planar comb electrode as at the attachment..

#### Attached Files:

• ###### Untitled.png
File size:
5.9 KB
Views:
514
2. Mar 15, 2012

### Staff: Mentor

Welcome to the PF.

That is a pretty non-trivial geometry to get the capacitance for. Do you have any FEA computational tools available to you (like ANSYS for example)?

3. Mar 15, 2012

from where I can I get the theoritical formula for designing I Capacitor? i don't want to do the simulation...i just want to calculate the theoretical value..

4. Mar 16, 2012

### rbj

$$\ C= \epsilon \frac{A}{d}$$

where $\epsilon$ is the permittivity of the dielectric, $A$ is the mean area of the facing surfaces and $d$ is the spacing between the surfaces. it looks like you have 8 pair of facing surfaces.

you say this is all coplaner, then it really depends on how thick this coplaner comb is. if it's thin enough, all it will be is edge effects. if the thickness is much larger than $d$, it should approximate what it would be for facing plates.

5. Mar 16, 2012

the formula you give for parallel plate capacitor but not for co plnar capacitor i think..

Last edited by a moderator: Mar 16, 2012
6. Mar 16, 2012

### rbj

of course. and if the thickness of your coplanar capacitor is zero the capacitance is zero. if the thickness of your coplanar capacitor is enough greater than $d$, then the plate formula will become approximately correct for your coplaner capacitor. but you haven't said a word about how thick this thing is (relative to the spacing you have between the teeth).

Last edited by a moderator: Mar 16, 2012
7. Mar 16, 2012

### rbj

i can't remember how we dealt with edge effects and i don't know where the old textbook is. here is something i got from http://www.phenix.bnl.gov/phenix/WWW/muon/phnotes/PN125/node1.html :

"The capacitance between adjacent strips having a thickness, t, width, w, and separation, s, laying on a dielectric with constant, k, is approximately given by,

C(pf/cm) = 0.12t/w + 0.09(1+k)log_10(1 + 2w/s + w^2/s^2)."

$$C \approx 0.12\frac{t}{w} + 0.18(1+k)\log_{10}\left( 1 + \frac{w}{s} \right)$$

judging from the units (pf/cm), i think this formula comes from capacitance per unit length is

$$C \approx (1.35) \epsilon_0 \frac{t}{w} + (0.87) (\epsilon_0 + \epsilon) \log \left( 1 + \frac{w}{s} \right)$$

in whatever units. length should be much longer than $t$ or $s$ or even $w$.

it's obviously heuristic and not a theoretical formula.

Last edited: Mar 16, 2012
8. Mar 16, 2012

### f95toli

There is no single formula for interdigital capacitors (which is what you are showing in the picutre) and if you really want accurate results you need to use simulation software (and even then it is not trivial to get accurate results).

That said, there are a bunch of heuristic formulas that can work quite well. rbj is showing one, but if that is not applicable you can find them in books on microwave transmission lines and elements.
I mostly use coplanar waveguides, and a good source of information about the use of interdigital capacitors in coplanar geometries can is Simons book ("Coplanar waveguides, circuits..." you can find it on Amazon).

9. Mar 16, 2012

### Bob S

10. Mar 17, 2012

### f95toli

No, it is not.
What the OP is asking about is an on-chip capacitor made in a single layer of metal on a substrate (i,e, it is "2-dimensional").
Capacitors of this type are used in microwave circuits, and will typically have a C of tens of fF.

11. Mar 17, 2012

### Staff: Mentor

Agreed. There is a lot more fringe capacitance in this coplanar capacitor geometry. That's what makes it so hard to calculate analytically...

12. Mar 18, 2012

can i use this formula to calculate the capacitor :

c = 2ϵw(L-x)/ g

13. Mar 18, 2012

### Staff: Mentor

Sure. Bet your job on it.

Where did you get that?

14. Mar 18, 2012