Capacitors and resonant frequency

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meyousikmann - 29 Oct 2006 21:44 GMT

I have a problem that asks about the movement of the second plate of a
capacitor in order to change the resonant frequency by 10 ppm in an LC
circuit. I am getting stuck in terminology.

Here is the information given:

Area of each capacitor plate (one attached to solid nonmoving post while
other can move small distances perpendicular to the plates)
.01 m^2

Initial plate separation
100 micometers

Inductance
120 nH

I have found the initial capacitance and frequency, but the second part of
the question says "How far must the second plate move to change the resonant
frequency by 10ppm". I am not entirely sure what this question is asking
for. Where does ppm fit into any of the calculations. I don't want
answers, but I would appreciate if somebody can translate the question into
something I might understand.

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Baron - 29 Oct 2006 21:50 GMT

> I have a problem that asks about the movement of the second plate of a
> capacitor in order to change the resonant frequency by 10 ppm in an LC
[quoted text clipped - 20 lines]
> answers, but I would appreciate if somebody can translate the question
> into something I might understand.

Think about the definition of a capacitor ! The value of capacitance
for a fixed plate area will vary in inverse proportion to the
separation !

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Baron:

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John Popelish - 29 Oct 2006 21:52 GMT

> I have a problem that asks about the movement of the second plate of a
> capacitor in order to change the resonant frequency by 10 ppm in an LC
[quoted text clipped - 18 lines]
> answers, but I would appreciate if somebody can translate the question into
> something I might understand.

First you need to derive the relationship between
capacitance and frequency (the power of proportionality).

Then you derive the relationship between plate spacing and
capacitance (another power of proportionality). Then
combine these two things to get 10 ppm frequency change.
10ppm is just a fraction 10/1,000,000 times the original
frequency. So the original frequency must become either 1 +
10/1,000,000 of its original value, or 1 - 10/1,000,000 of
it (depending on whether the plates are moved apart or
together).

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Charles Schuler - 29 Oct 2006 21:52 GMT

>I have a problem that asks about the movement of the second plate of a
>capacitor in order to change the resonant frequency by 10 ppm in an LC
[quoted text clipped - 18 lines]
> want answers, but I would appreciate if somebody can translate the
> question into something I might understand.

delta f = 10/1e6 x f1

fx = f1 - delta f

solve for new value of C based on fx

solve for new plate spacing based on C

(If I understand the question ;>)

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Tom Biasi - 29 Oct 2006 21:52 GMT

I believe you are being asked to determine the capacitance needed for that
frequency change and then determine the plate areas that must face each
other at the given distance.

Tom

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Eeyore - 29 Oct 2006 22:44 GMT

> Where does ppm fit into any of the calculations

It's another way of representing a small number ( as in parts per million ).
Compare with per_cent for example which is parts per hundred.

Graham

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mphillipps2@yahoo.com - 31 Oct 2006 18:38 GMT

Off the top of my head, I think it's:

100um (plate spacing) * (10 ^ -6) = 10^ -9 m = 1 nanometer

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mphillipps2@yahoo.com - 31 Oct 2006 18:42 GMT

Wait, my last post was wrong.

f = 1/(2 * pi * sqrt(LC))

so I get 2 nanometers.

mphillip...@yahoo.com wrote:

> Off the top of my head, I think it's:
>
> 100um (plate spacing) * (10 ^ -6) = 10^ -9 m = 1 nanometer

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Electronics Forum / Basics / October 2006

Capacitors and resonant frequency