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Math Help - Capacitance calculation (parallel plate type)

  1. #1
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    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.
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  2. #2
    Member Greengoblin's Avatar
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    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}
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  3. #3
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    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.
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  4. #4
    Member Greengoblin's Avatar
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    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.
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