| T O P I C R E V I E W |
| shahriar |
Posted - Aug 01 2011 : 22:40:32
Hi Dave and fellow FastCap users
How do you relate the capacitance matrix to measurements made by an impedance analyzer?
I understand that FastCap models infinity as a ground terminal. This explains why the C matrix is not symmetrical - since there is some coupling to infinite ground from the source terminal. But how do you take this C matrix and transform it to the "circuit" capacitance one would measure using an impedance analyzer.
I have started to work on it but Im not sure of my results. Essentially the two plates are at symmetric potentials (from infinity). So [Q]=[C][V - Vref] where Vref is the potential of the plates from infinity.
Help on this matter will really help many people out there struggling with this problem.
Best
Shahriar |
| 15 L A T E S T R E P L I E S (Newest First) |
| rebekasm |
Posted - Jan 16 2022 : 23:34:50
Hello Shahriar,
The whole discussion is very good one to understand the capacitance matrix. However, I have some questions regarding to COMSOL capacitance calculation. I have a modelled a system with six electrode and consider a sphere with infinite element as ground node. So, from COMSOL I can get the result for Maxwell/Mutual matrix with ground at infinity. Now, I need to replace the ground node with float at infinity to reduce the matrix, so the potential is constant and I can derive a mutual capacitance matrix with diagonal zero element. In COMSOL AC/DC module it is not possible to do study with ground node. So, do you have any idea how could I approach the problem? |
| rebekasm |
Posted - Jan 16 2022 : 23:31:16
Hello Shahriar,
The whole discussion is very good one to understand the capacitance matrix. However, I have some questions regarding to COMSOL capacitance calculation. I have a modelled a system with six electrode and consider a sphere with infinite element as ground node. So, from COMSOL I can get the result for Maxwell/Mutual matrix with ground at infinity. Now, I need to replace the ground node with float at infinity to reduce the matrix, so the potential is constant and I can derive a mutual capacitance matrix with diagonal zero element. In COMSOL AC/DC module it is not possible to do study with ground node. So, do you have any idea how could I approach the problem? |
| dave_royle |
Posted - Sep 02 2011 : 06:09:28
shahriar,
Sorry for leaving you hanging, I have been out.
I am not sure you have it yet, let me give you a worked out example:
I have two 0.2m square plates side by side (coplanar) separated by 0.1m. The matrix is:
CAPACITANCE MATRIX, picofarads
1 2
1%GROUP1 1 8.804 -2.285
1%GROUP2 2 -2.285 8.804
The equivalent circuit is:
2.285pf between plates
6.519pf between each plate and infinity/earth (sum of row).
Draw this and you can see that if I measure the capacitance between the plates I will get 5.545pf (all three caps contribute).
Next I add a 10m square plate under the small plates spaced 0.8m under. The matrix is:
CAPACITANCE MATRIX, picofarads
1 2 3
1%GROUP1 1 9.07 -2.043 -6.394
1%GROUP2 2 -2.043 9.07 -6.394
1%GROUP3 3 -6.394 -6.394 419.8
The equivalent circuit is:
2.043pf between small plates.
6.394pf between each small plate and big plate.
0.633pf between each small plate and infinity/earth (sum of group1/group2 rows).
406.212pf between big plate and infinity/earth (sum of group3 row).
Comparing these two examples we can see that:
1 - the capacitance between small plates is very similar.
2 - the relationship between the small plates and infinity/earth in the first example is similar to the relationship between the small plates and the big plate in the second example.
3 - the capacitance between the small plates and infinity/earth is very small in the second example.
The big plate has taken over the role of infinity/earth from the perspective of the small plates. Of course if I make the big plate 1Km square and move it 100m away it will more perfectly replace earth/infinity.
This is why we say that infinity = planet earth. Infinity is somewhat metaphysical, but we can all imagine parasitic capacitance to earth. It is the same as adding a really big plate reasonably far away from our device and considering parasitc capacitance to it.
In other words if a row in a capacitance matrix is non-zero than there is capacitive coupling between that element and the outside world (earth). Consider coplanar traces over ground plane in a PWB (two conductors and ground plane, three elements in matrix). The traces better have zero sum rows in the matrix, the ground plane should be capturing all of their flux, no leakage flux from traces to earth is expected. On the other hand can we expect the ground plane itself to have capactance to earth (this comes up when quantifying EMI problems for example), so that conductor's row will not have a zero sum.
Hope this helps.
Dave |
| shahriar |
Posted - Aug 12 2011 : 18:17:47
Hi Dave
Yes it is quite a pain. But I actually did some measurements and when you solve the three equations with three unknowns it gives consistent results between FastCap and experimental results for C12. My work right now involves understanding capacitive sensing on a much larger scale of a room.
Im curious about your idea of creating a resonant circuit. Im not sure I understand it very well. Is the inductance of the inductor known and then you just sweep the frequency to find the appropriate C that creates resonance with w=1/Sqrt(LC). Just curious - and its an interesting idea.
Also to be sure about your 100'x100' GND example - is this what you mean - C matrix below
(1st row) C11+C12+C13 -C12 -C13
(2nd row) -C21 C21+C22+C23 -C23
(3rd row) -C31 -C32 C31+C32+C33
So adding the first row should give C11 but that should be very close to 0 because most of the return flux is in C13. Same applies to the second row C22=0. I think the third row could add up to 0 but C33 might not make it 0 because that is the return flux with respect to infinite GND. Just want to be sure my logic is right and want to outline it for future FASTCAP/Numerical Field Solver people.
Regarding my question about FastCAP's boundary condition for the domain of the problem being solved - There can be two boundary conditions - One being V=0 (GND at Infinity) , or dV/dx=0 (Conservation of Charge). So if you have a large Ground (100'x100') nearby, and charge is conserved (Since the matrix row adds up to 0) then it looks like over the entire domain (ground and conductors of interest) the conservation of charge boundary is being used.
The reason why I talk about this is because I am well versed with COMSOL and by default they have the Zero charge boundary condition. Ive spent two weeks trying to understand the difference between FastCAP and COMSOL results and your explanation has helped greatly.
Thank you Dave!
Shahriar
quote:
Originally posted by dave_royle
shahriar,
Wow, what you are describing sounds like it would be tough to measure. The impedance analyzer would have to be several feet away to avoid influencing the fringing fields. The long hookup wires would require a low frequency impedance analyzer measurement to avoid wire inductance. But you are talking about a very small capacitance (tens of pf I am guessing), so it would be difficult to measure at low frequency. I would be tempted to run wire right down the centerline between the plates and put an inductor in the center. Choose L value to resonate with your capacitor at something like 10MHz (low enought that everything is lumped) and sniff for resonance with a near field probe.
What is the application for this gem?
By the way, Fastcap does not define the infinity boundary condition as ground. We invent "earth ground = infinity" because it is easy to visualize and accurate so long as earth is sufficiently distant from your conductors (not always the case!). For example you can eliminate the need to think about infinity or earth altogether if you add a third, really big, conductor to your FastCap input. In your case, add a 100' x 100' plate, 10' distant from your capacitor, and call it node 3. Your matrix will now have C13 and C23 (negative numbers) whose magnitude will equal the cooresponding leakage capacitances before the plate was there. And your new leakage capacitance (sum of matrix row) will now be zero in each row (no flux going to infinity). Now if you move that big plate separation to 9' none of the matrix values will change, the plate is acting like infinity. If you move the big plate separation to 1' then your matrix values will change, the plate is no longer a good model of infinity.
Dave
|
| dave_royle |
Posted - Aug 11 2011 : 22:30:26
chromatik,
More precisely, all of the *flux* is kept between shahriar's capacitor and the big plate. The leakage flux is close to zero. I am using leakage flux to refer to flux that leaves one of the conductors defined in FastCap and does not return to any FastCap defined conductor (goes to infinity).
BTW, RF tools like Momentum will assume that there is a large/close reference ground that prevents significant flux from escaping from your circuit.
FastCap is more general in this sense, I can use it to calculate the capacitance between the Earth and the Moon.
Dave |
| chromatik |
Posted - Aug 11 2011 : 16:57:05
@shahriar
@dave_royle
As the mail form isn't working properly, if we could discuss by mail it would be very interesting.
Here's my mail: chromatik_fr at yahoo.com
Regards |
| chromatik |
Posted - Aug 11 2011 : 11:06:04
shahriar, dave,
Thanks for your help!
The lack of "ground/reference" or "ports" was a disturbing thing at first sight with FastCap.
I was using Agilent's Momentum previously, and "ground" and "port" were key parameters.
Previously, in dave's message (Aug 11 2011, 03:21:07), dave was using a 100x100 plate, and compared distance variation effects on the matrix result, 10' gave a sum of matrix row equal to zero, meaning all the leakage capacitance is 0.
Am I right if I say all the leakage flux is kept inside this "capacitor + plate" ?
|
| dave_royle |
Posted - Aug 11 2011 : 04:48:09
chromatik,
Expanding on shahriar's response;
FastCap does not know anything about "grounds" or "references". All it knows about are conductors. If you have 2 conductors without a ground plane you will have some leakage flux that will escape from this 2 conductor structure and go elsewhere (earth if you do it in your lab, infinity if you do it in the middle of an empty universe). This leakage flux can be very small if the two conductors are intimately close to each other (as in an actual component), and FastCap will give you this leakage flux. If you include a ground plane you now have three conductors and you will still have leakage flux escaping this three conductor structure and going to earth/infinity. So now your circuit will have four nodes: conductor1, conductor2, groundplane, earth(infinity). In practice if your ground plane is large and close to your other two conductors the leakage flux is insignificant, and FastCap will show you that.
By the way if you are using a piece of equipment like a network analyzer than you better not have a significant amount of leakage flux coupling to earth or your measurement will be sensitive to the environment. For example standing near the fixture could effect the measurement.
Dave |
| dave_royle |
Posted - Aug 11 2011 : 03:23:11
shahriar,
I forgot to answer yor question, I am not familiar with COMSOL but if it can produce a Maxwell capacitance matrix then everything we have discussed will apply.
Dave |
| dave_royle |
Posted - Aug 11 2011 : 03:21:07
shahriar,
Wow, what you are describing sounds like it would be tough to measure. The impedance analyzer would have to be several feet away to avoid influencing the fringing fields. The long hookup wires would require a low frequency impedance analyzer measurement to avoid wire inductance. But you are talking about a very small capacitance (tens of pf I am guessing), so it would be difficult to measure at low frequency. I would be tempted to run wire right down the centerline between the plates and put an inductor in the center. Choose L value to resonate with your capacitor at something like 10MHz (low enought that everything is lumped) and sniff for resonance with a near field probe.
What is the application for this gem?
By the way, Fastcap does not define the infinity boundary condition as ground. We invent "earth ground = infinity" because it is easy to visualize and accurate so long as earth is sufficiently distant from your conductors (not always the case!). For example you can eliminate the need to think about infinity or earth altogether if you add a third, really big, conductor to your FastCap input. In your case, add a 100' x 100' plate, 10' distant from your capacitor, and call it node 3. Your matrix will now have C13 and C23 (negative numbers) whose magnitude will equal the cooresponding leakage capacitances before the plate was there. And your new leakage capacitance (sum of matrix row) will now be zero in each row (no flux going to infinity). Now if you move that big plate separation to 9' none of the matrix values will change, the plate is acting like infinity. If you move the big plate separation to 1' then your matrix values will change, the plate is no longer a good model of infinity.
Dave |
| shahriar |
Posted - Aug 10 2011 : 17:33:41
Hi Chromatik
With regards to your two wire capacitors. Ground is usually some reference potential - it could be Earth Ground but doesnt have to be. My equations are with respect to an "Earth GND" which you can measure with respect to the earth jack of your three prong terminal. These equations are more with respect to a field's understanding of what is going on - but does translate into designing circuits. BTW, network analyzers usually operate at very high frequencies - FASTCAP is a quasi-static solver so the dimensions of your device should be much smaller than the wavelength of operation.
With regards to your second question.
If they are capacitors there wont be a resistance because there is not current - open circuit = infinite resistance. So the resistance has no effect on your voltage drop. Inductance is a measure of the magnetic flux per unit current. Magnetic flux is the integral of perpendicular magnetic flux density (B) over some cross sectional area. Magnetic flux is generated by a current - no current no magnetic flux no inductance.
This is all for quasi-statics where you have no displacement current - and that is what FASTCAP solves
Best
Shahriar
quote:
Originally posted by chromatik
Hi shahriar, hi Dave,
I am also interested in your discussion, because there is not so many threads in here talking about comparing simulation and measurements.
If you're ok, I would like to point out a misunderstanding I have about capacitors, and more generally about capacitance extraction.
At the beginning, the problem is basically the same as shahriar: few conductors facing each other, but with a little difference, these conductors are wires.
Let's talk about 2 conductors
Capacitances are likely:
- C12: mutual capacitance between conductor 1 and conductor 2
- C1G: capacitance of conductor 1 to ground
- C2G: capacitance of conductor 2 to ground
First question: if I do not define a ground plane in my schematic, does the reference ground would be the Earth? So If I use a network analyzer, the capacitance measured will follow shahriar's equations
Cm12 = C12 + C1G||C2G
Cm1 = C1G + C12||C2G
Cm2 = C2G + C12||C1G
Second question: considering the length of the wires is small compared to wavelength, considering a one volt source connected across the capacitor, due to resistance and inductance of the wires, the voltage at the end of the wires would be different, so charges, so... capacitances would be different, too?
Thanks for your help.
Regards
|
| shahriar |
Posted - Aug 10 2011 : 17:22:41
Sorry Dave for the confusion. I was being sloppy and cutting corners
This is what I meant
Cm1= C1G + (C12*C2G)/(C12+C2G) - which is C1G parallel with series combination of C12 and C2G
Cm2= C2G + (C12*C1G)/(C12+C1G) - C2G parallel with series combination of C12 and C1G
Cm12 = C12 +(C1G*C2G)/(C1G+C2G) - C12 in parallel with series combination of C1G and C2G
What you said is correct - but Ive outlined it above for others to be clear as well.
What I mean by non-ideal capacitor - is one where the capacitance is not typically given by C=epsilon0 *plateArea/plateseparationdistance
This is because the plate separation distance is on the order of or greater than the actual dimensions of the plate. So the fringe field in this case will be dominant and the only way to determine its capacitance is by simulating it and also verifying it with impedance measurements.
Hence the reason why I need to make sense of the measurements and the simulations - they need to be equal in some capacity.
I will be using an impedance analyzer to make these measurements - and the capacitor are just "non-ideal" dominated by fringe fields so difficult to make hand calculations of the mutual capacitance between the two plates. The capacitor plates are on the order of half a foot in either dimension so they are not on the scale of a room. But I am investigating the induced charge on other objects in a room so it is difficult to put all the metal in a faraday cage. The method I proposed was to just measure (using an impedance analyzer) Cm1, Cm2 and Cm12 (wrt Earth GND) and then determine C12 from solving 3 eqns with 3 unknowns. I think this works and should match up with results given by FASTCAP/COMSOL.
On another note - are you familiar with COMSOL? COMSOL doesnt define the infinity boundary condition as GND. It just has a zero charge boundary condition. From what I understand one cannot actually determine the mutual capacitance C12 in the [C] matrix using the zero charge boundary condition because all the charge is converged on all the plates - even if you put 1V on one terminal and GND the other terminal - what you end up getting is the same amount of charge on both terminals so you really end up with the diagonal terms of the [C] matrix. Have you had any experience/insights with this boundary condition and the resulting Capacitance matrix?
Thanks for all your help Dave.
Shahriar
quote:
Originally posted by dave_royle
Shahriar,
Your formulas look correct so long your "||" operator means:
C1G||C2G = C1G*C2G/(C1G+C2G).
This symbol means "in parallel with" which would be (C1G + C2G) for capacitors, but I think you are using it to refer to the "resistors in parallel" formula right?. I am noticing that I said "C1 in parallel with (C2 + C3) in an earlier post where it should have been "C1 in parallel with (C2 in series with C3).
Anyway, you have me curious. What are you building? What is a non-ideal capacitor and how big is it? Are you going to be using an impedance analyzer or is it just theoretical? If it is physically large compared to your room, then you may not be able to consider the room to be infinity.
Dave
|
| chromatik |
Posted - Aug 10 2011 : 10:03:08
Hi shahriar, hi Dave,
I am also interested in your discussion, because there is not so many threads in here talking about comparing simulation and measurements.
If you're ok, I would like to point out a misunderstanding I have about capacitors, and more generally about capacitance extraction.
At the beginning, the problem is basically the same as shahriar: few conductors facing each other, but with a little difference, these conductors are wires.
Let's talk about 2 conductors
Capacitances are likely:
- C12: mutual capacitance between conductor 1 and conductor 2
- C1G: capacitance of conductor 1 to ground
- C2G: capacitance of conductor 2 to ground
First question: if I do not define a ground plane in my schematic, does the reference ground would be the Earth? So If I use a network analyzer, the capacitance measured will follow shahriar's equations
Cm12 = C12 + C1G||C2G
Cm1 = C1G + C12||C2G
Cm2 = C2G + C12||C1G
Second question: considering the length of the wires is small compared to wavelength, considering a one volt source connected across the capacitor, due to resistance and inductance of the wires, the voltage at the end of the wires would be different, so charges, so... capacitances would be different, too?
Thanks for your help.
Regards |
| dave_royle |
Posted - Aug 10 2011 : 05:48:50
Shahriar,
Your formulas look correct so long your "||" operator means:
C1G||C2G = C1G*C2G/(C1G+C2G).
This symbol means "in parallel with" which would be (C1G + C2G) for capacitors, but I think you are using it to refer to the "resistors in parallel" formula right?. I am noticing that I said "C1 in parallel with (C2 + C3) in an earlier post where it should have been "C1 in parallel with (C2 in series with C3).
Anyway, you have me curious. What are you building? What is a non-ideal capacitor and how big is it? Are you going to be using an impedance analyzer or is it just theoretical? If it is physically large compared to your room, then you may not be able to consider the room to be infinity.
Dave |
| shahriar |
Posted - Aug 09 2011 : 23:37:22
Thank You Dave. Yes I understand what you mean.
I am trying to understand capacitance of non-ideal capacitors in a room. The value I get is significantly different from what FastCAP/COMSOL give me when I measure it with impedance analyzer. I could make a faraday cage (but it would have to be the size of a room) to have a reference GND. But I have another idea which builds on yours. Please comment if you think this is incorrect.
A capacitor in space is really a 2 conductor system with earth ground
One plate - called 1
Second plate - called 2
Cm12= Total Capacitance measured by the impedance analyzer between 1 and 2
Cm1 = Total Capacitance of plate 1 measured by impedance analyzer to earth GND
Cm2 = Total Capacitance of plate 2 measured by impedance analyzer to earth GND
C1G = Capacitance of plate 1 to GND
C2G = Capacitance of plate 2 to GND
C12 = Actual Mutual Capacitance of plate 1 and 2 (should be equal to FASTCAP/COMSOL predicted values)
Cm12 = C12 + C1G||C2G
Cm1 = C1G + C12||C2G
Cm2 = C2G + C12||C1G
If I make measurements of Cm1, Cm2 and Cm12 and solve the three equations (Cm12,Cm1,Cm2) for C12. I could calculate the real value of C12 - without worrying about the room sized faraday cage- and these results should match FASTCAP/COMSOL.
Is this logic correct?
Best
Shahriar
quote:
Originally posted by dave_royle
Shahriar,
I will give you the procedure by example, then a word of caution:
Given the following matrix:
1 2 3
1 14 -1.5 -8
2 -1.5 3 -1.2
3 -8 -1.2 14
Your circuit will have three nodes (1,2,3) and earth ground.
The capacitances among the three nodes (1,2,3) you read right out of the matrix, disregarding the sign. For example the capacitor that you will place between node 2&3 is 1.2pf (you find this in row1 column2 or row2 column1 both will be the same). This will give you 3 capacitors so far (1-2, 1-3, 2-3).
Next you draw a capacitor from node1 to earth ground. The value of this cap will be the sum (including signs) of row1 (14 -1.5 - 8 = 4.5pf). This sum will always be positive or zero. Do the same for the other two conductors. This will give you another 3 caps (1-gnd, 2-gnd, 3-gnd). This is your circuit. If you want the net capacitance between two nodes you will need to analyze the circuit, all of the capacitors will contribute to it. You could use PSPICE for example with the following deck:
C1 1 2 1.5pf
C2 1 3 8pf
C3 2 3 1.2pf
C4 1 0 4.5pf
C5 2 0 0.3pf
C6 3 0 4.8pf
Thats it!
Now that you know the procedure I need to caution you. When I am referring to ground in the above discussion I an not talking about the ground plane on you PCB, I mean literally the planet earth.
Imagine a charged tennis ball size metal sphere hovering in deep space. Electric flux will flow radially away from the sphere, and those lines will continue on to infinity. We have all seen this picture in our physics books. Next imagine two electrically neutral spheres separated by 1 meter (out in space). I pull some electrons off of one and put them onto the other and I now have electric flux lines that leave the positive sphere and terminate on the negative sphere. However, not all of the electric flux goes between the balls. Some still escapes the dipole and goes to infinity. The closer the balls are to each other, the less flux escapes to infinity. (I hope you can imagine this, I wish I had a blackboard). Now lets imagine repeating this experiment on your lab bench. The flux that escapes the dipole, that would have gone to infinity in free space, will virtually all terminate on the planet earth due to its sheer size. So, from a pragmatic engineers perspective, we can consider infinity to be the planet earth. Getting back to the capacitance matrix, when we do an integer sum of row1 in the capacitance matrix, the result represents the amount of flux leaving conductor1 and escaping the 3-conductor system and going to infinity (planet earth, your body, the big metal rack next to your bench ...). In other words it is leakage capacitance that is highly uncontrolled. If you are designing a device, it will include a controlled ground which needs to be one of the nodes in your FastCap (not the "infinity node"). For example if you are modeling a differential pair over ground plane you will have three nodes, conductor1, conductor2, and ground plane. Then when you sum the values in the matrix rows you better have a negligibly small number since this represents the electrix flux that is escaping from your PCB and coupling to whatever happens to be nearby. When I was in high school one of the many things that I tried to build that did not work well was an oscillator using a 12BE6 and a homemade capacitor. The problem was that whenever I put my hand near it to tune it the presence of my hand shifted the frequency so I could never get it tuned where I wanted. Way too much leakage flux!
Dave
Dave
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