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znieh
1 Posts |
 Posted - Jul 07 2011 : 19:18:27
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Hey!
I want to solve a simple Capacitor.
C=e0*er*l*w/d
l=365um
w=2.89um
d=1.28
C= 2.845672141406249e-014
CODE FASTCAP
0 Title
Q cappa1 0.000000 0.000000 0.000000 0.365000 0.000000 0.000000 0.365000 0.002890 0.000000 0.000000 0.002890 0.000000
Q cappa2 0.000000 0.000000 0.001280 0.365000 0.000000 0.001280 0.365000 0.002890 0.001280 0.000000 0.002890 0.001280
RESULT
cappa1%GROUP1 1 41.41 -34.82
cappa2%GROUP1 2 -34.82 41.41
What is wrong? Thanks in advance. |
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dave_royle
USA
13 Posts |
Posted - Jul 30 2011 : 05:25:59
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Znieh,
Dude -
1 - Inputs to fastcap are in meters. 365um is not .365
2 - It took me some time to figure out that you had er=3.9, you could have mentioned that.
Fixing your input deck we get:
CAPACITANCE MATRIX, femtofarads
1 2
cappa1%GROUP1 1 41.62 -35
cappa2%GROUP1 2 -35 41.62
Note the unit is femtofarads. Your formula is producing 28 femtofarads.
If your question is why formula is producing smaller values than fastcap output I will attempt to explain that here. There are actually two ways to view this, physicist way and EE way. I will speak in EE terms for you.
Imagine this capacitor is physically on a table in your lab.
There is capacitance between the plates of the capacitor (call it C1)and there is also capacitance between each plate and ground (call them C2 and C3). If I try to measure the capacitance between the plates with a capacitance meter, I am actually going to measure C1 in parallel with (C2 in series with C3). In the field solver the C1 = |-35ff| (if you really want to know why it is negative in the capacitance matrix reply back and I can elaborate). 41.62 - 35 = 6.62ff is the capacitance between the conductor and ground (or think of it as the electric flux not captured by the other conductor).
Your formula underestimated C1 (28ff vs. 35ff) because it is assuming that all of the electric flux between the plates is in the normal direction and constant. In your case, because w=2.89um is not much larger than d=1.28um you have a significant amount of fringing E-field not accounted for in formula. Capacitance is flux per volt, so not accounting for the fringing flux will underestimate the capacitance.
Your formula is attempting to approximate the value of C1, but assuming that all of the electric flux between plates is in the normal direction.
Dave |
Edited by - dave_royle on Aug 12 2011 02:54:51 |
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chromatik
France
29 Posts |
Posted - Aug 08 2011 : 14:24:49
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Dear Dave,
Thanks for your answer, I tried to work on it, but I didn't noticed the "er=3.9"...
And BTW, thanks for the clear explanation :) |
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Problem with simple Capacitor
