# Math Help - Capacitance calculation (parallel plate type)

1. ## Capacitance calculation (parallel plate type)

Im just having a bit of trouble. Here is the question.

Calculate approximate capacitance a parallel plate type device would have on the wafer if the electrode dimensions where 150um by 50um and the thickness of the SiO2 (insulator) between the conductors was 1um.

I keep getting different (and stupid answers). Do i have to take into account the dielectric constant of the silicon dioxide (SiO2) or not?

Any help. I know it should be fairly straight forward. Im just getting confused by a few things.

2. Yeh, you need to take into account the permittivity, $\epsilon$ of the $\text{SiO}_2$ dielectric, the area of overlap of the plates, A, and the the distance between them, d, using the equation:

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

3. Ok thanks, so 3.9(150um*50um)/1um
=0.02925uf
=29.25nf

3.9 is the dielectric constant of SiO2. Does this seem right? i seem to be getting a different answer.

4. Well, with your numbers I got a different answer, 29.25mF, a fairly large capacitance. I'm not sure what the constant is for silica, but this page:

Dielectric Constants of Materials

thinks 4.5, giving an even bigger capacitance.