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jampani
USA
5 Posts |
 Posted - Jul 16 2007 : 20:06:50
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Hi,
Can I apply specific voltage to a conductor in fastcap2. I wanted to measure the capacitance of a conductor(interconnect) by applying a 0V on all the surrounding conductors and 1V on the target conductor.
Is there any way to do this is fastcap2???
jampani |
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Enrico
550 Posts |
Posted - Jul 18 2007 : 23:37:04
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Your question touches a fundamental point, that is how to interpret the results from FastCap2.
The capacitance matrix you get from FastCap2 is, as for most field solvers, a Maxwell capacitance matrix.
This topic is well explained in any good electromagnetic text book, but since this is a widely asked question, let's recap briefly the theory, with some examples.
A Maxwell capacitance matrix provides the relation between voltages on a set of conductors to charges on the conductors. For example, for a condutor set:
Q = C * V
where C is the Maxwell capacitance matrix, and V and Q are the voltage and charge vectors respectively.
Now:
1. What is the relation between the elements of C, the Maxwell capacitance matrix, and the capacitances between the conductors?
Consider for example a 4 conductors case, as shown in the below picture. Here the mutual and auto capacitances (capacitances towards infinity) are explicitly drawn.
What is then the charge Q1 on the conductor #1, given the voltages V1, V2, V3, V4 on the four conductors? Physics tells us
Q1 = C11 * V1 + C12 * (V1 - V2) + C13 * (V1 - V3) + C14 * (V1 - V4)
which can be arranged as
Q1 = (C11 + C12 + C13 + C14) * V1 - C12 * V2 - C13 * V3 - C14 * V4
So the first row of the Maxwell capacitance matrix C is
| C11+C12+C13+C14 -C12 -C13 -C14 |
Completing the exercise, the full matrix has the form
| C11+C12+C13+C14 -C12 -C13 -C14 |
| -C21 C21+C22+C23+C24 -C23 -C24 |
| -C31 -C32 C31+C32+C33+C34 -C34 |
| -C41 -C42 -C43 C41+C42+C43+C44 |
Extending to the general case, the Maxwell capacitance matrix has the form
| C11+C12+..+C1n -C12 .. -C1n |
| -C21 C21+C22+..+C2n .. -C2n |
| .. .. .. .. |
| -Cn1 -Cn2 .. Cn1+Cn2+..+Cnn |
Note by the way that, when using a SPICE simulator, you must use the Cxy values, NOT directly the values appearing in the Maxwell capacitance matrix. So you have to perform some subtractions to find the auto capacitance.
For instance, if your matrix is the following, for a two conductor problem:
| 4nF -1nF |
| -1nF 4nF |
then C11 = 4nf - 1nf = 3nF
and C12 = C21 = - (-1nF) = 1nF
Your SPICE netlist will then report something like this:
C11 node1 0 3n
C22 node2 0 3n
C12 node1 node2 1n
2. How FastCap2 calculates the Maxwell capacitance matrix?
Since in general there is the relation
Q = C * V
to calculate the k-th column of C, you can set the potential of the k-th conductor to one (i.e. the k-th element of V is set to 1) and the potential of all remaining conductors to zero (i.e. all n-1 remaining elements of V are set to 0).
The resulting Q' vector will be therefore equal to the k-th column of C.
In general, knowledge of the C matrix gives you all the elements needed to calculate the Q-V relation (i.e. the capacitance) for any set of voltages V.
In the end, here you are the simple answer: for your specific case, the result is already in your hands, with no further calculation. It is the element on the diagonal of the capacitance matrix resulting from FastCap2, taken from the row corresponding to the conductor you are interested in.
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jampani
USA
5 Posts |
Posted - Jul 23 2007 : 18:27:47
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Hi Enrico,
Thank you very much for your reply. It was very informative.
regards,
jampani
jampani |
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anand_jog
USA
2 Posts |
Posted - Jul 28 2007 : 01:02:06
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So if in an electronic circuit one of the plates of the capacitor is connected to ground, does the C11 term become zero?
In this case, does the Maxwell capacitance matrix become identical to the 'balanced charge' capacitance matrix?
Basically, I'm confused as to which of the two to use while computing the capacitance matrix of a multi-terminal capacitor which is then implemented in an electronic circuit. Ideally some of the terminals would be connected to ground and some would be connected to some driving potential.
Please clarify.
Anand Jog |
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fayefan
7 Posts |
Posted - Dec 11 2007 : 06:18:04
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quote:
The capacitance matrix you get from FastCap2 is, as for most field solvers, a Maxwell capacitance matrix. That is as follows:
| C11+C12+..+C1n -C12 .. -C1n |
| -C21 C21+C22+..+C2n .. -C2n |
| .. .. .. .. |
| -Cn1 -Cn2 .. Cn1+Cn2+..+Cnn |
Note by the way that, when using a SPICE simulator, you must use the Cxy values, NOT directly the values appearing in the Maxwell capacitance matrix. So you have to perform some subtractions to find the auto capacitance.
For instance, if your matrix is the following, for a two conductor problem:
| 4nF -1nF |
| -1nF 4nF |
then C11 = 4nf - 1nf = 3nF
and C12 = C21 = - (-1nF) = 1nF
Your SPICE netlist will then report something like this:
C11 node1 0 3n
C22 node2 0 3n
C12 node1 node2 1n
It means that in the following result:
CAPACITANCE MATRIX, picofarads
1 2
CONDUCTOR%GROUP1 1 1.95 -1.69
CONDUCTOR%GROUP2 2 -1.69 1.89
C11+C12=1.95pf
C12=C21=1.69pf
C21+C22=1.89pf
C11=0.26pf
C22=0.20pf
Am I right?
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Edited by - fayefan on Dec 11 2007 06:19:39 |
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Enrico
550 Posts |
Posted - Feb 12 2008 : 12:29:12
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quote:
CAPACITANCE MATRIX, picofarads
1 2
CONDUCTOR%GROUP1 1 1.95 -1.69
CONDUCTOR%GROUP2 2 -1.69 1.89
C11+C12=1.95pf
C12=C21=1.69pf
C21+C22=1.89pf
C11=0.26pf
C22=0.20pf
Am I right?
Yes
Enrico
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How to make a conductor
