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Quick question: why does Q of spiral inductor have a peak?

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Nonsense

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Hi, all,

Why does the quality factor of the on-chip spiral inductor have a peak? I mean what dominate the loss at low freq. and what at high freq.?

Thanks.
 

wlcsp

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Re: Quick question: why does Q of spiral inductor have a pea

Nonsense said:
Hi, all,

Why does the quality factor of the on-chip spiral inductor have a peak? I mean what dominate the loss at low freq. and what at high freq.?

Thanks.

Of course it has a peak since it's not a perfect inductor, instead it has many parasitic effects. At low freq, substrate loss and ohmic loss dominate, while at high frequency major loss mainly comes from the eddy current loss i.e. skin effect and proximity effect.

wlcsp
 

Nonsense

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Re: Quick question: why does Q of spiral inductor have a pea

wlcsp said:
Of course it has a peak since it's not a perfect inductor, instead it has many parasitic effects. At low freq, substrate loss and ohmic loss dominate, while at high frequency major loss mainly comes from the eddy current loss i.e. skin effect and proximity effect.

wlcsp

Hi, wlcsp,

Thanks for reply.

I know the loss when freq. get high, this could be the explaination for the quality factor to go down, but why Q rises at first and gets a peak?
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

At low frequency, resistance is a constant and inductive reactance goes to zero, so Q goes to zero.

At high frequency, there is inter-turn capacitance and capacitance to the substrate. This capacitance is in parallel with the inductor. At the frist self resonance (S sub RF) we have a parallel resonant circuit (C in parallel with a series RL). If you sit down and calculate the imaginary part of that impedance and divide it by the real part of that impedance, you find that the capacitance acts to decrease the Q. For R small, Q goes to zero at close to (but not quite equal to) the self resonant frequency.

With Q zero at low frequency and zero near the self resonant frequency and larger than zero in between, it must see a maximum at some frequency below the self resonant frequency.

The equation for Q in this case is:

Q = ωL/R - (ω²L²+R²)ωC/R

Note that a non-zero C decreases Q even though there is no resistance at all associated with the C in this case.
 

    Nonsense

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Nonsense

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Re: Quick question: why does Q of spiral inductor have a pea

rautio said:
At low frequency, resistance is a constant and inductive reactance goes to zero, so Q goes to zero.

At high frequency, there is inter-turn capacitance and capacitance to the substrate. This capacitance is in parallel with the inductor. At the frist self resonance (S sub RF) we have a parallel resonant circuit (C in parallel with a series RL). If you sit down and calculate the imaginary part of that impedance and divide it by the real part of that impedance, you find that the capacitance acts to decrease the Q. For R small, Q goes to zero at close to (but not quite equal to) the self resonant frequency.

With Q zero at low frequency and zero near the self resonant frequency and larger than zero in between, it must see a maximum at some frequency below the self resonant frequency.

The equation for Q in this case is:

Q = ωL/R - (ω²L²+R²)ωC/R

Note that a non-zero C decreases Q even though there is no resistance at all associated with the C in this case.

Thank you, Dr. Rautio,

You said that at low frequency, the inductance goes to zero, but if you use L=-1/2/pi/freq/imag(Y11) to extract the inductance, you'll find that the inductance at low frequency is actually very large.

And, with the Q equation you gave, the peak Q reaches at the freq. of ω=√((R²C+L)/3L²C). While doing EM simulation, it is not uncommon to see that the peak of Q drifts from the measured one (the peak value is almost the same, only the peak frequency is different), what may cause such drift? it seems hard to say with the given equation.

Thanks.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Nonsense -- Actually, you slightly mis-read my comments just a little. I said that inductive reactance (ωL) goes to zero as ω goes to zero. Indeed the inductance does change at low frequency. However, I don't think it goes to extremely large values. I would think this happens as 1) current penetrates the entire volume of the metal conductor, and 2) when the ground plane underneath the inductor becomes thin with respect to skin depth.

To see (1) occur, you need a volume current flowing in your metal. This can be approximated in Sonnet with the multi-sheet model, it can be approximated in volume meshers by setting "Solve inside" and using a sufficiently refined volume mesh. If you use a tube-like model with surface currents only for thickness, I think it would be a problem.

To see what happens for (2), just remember that at zero frequency, it is magneto-static, and conductors in the vicinity (except super-conductors) are transparent to magnetic field. Transitions into (1) and (2) would occur at different (low) frequencies.

I have not studied this matter in detail, but I'm sure fully understanding the low frequency behavior of inductors would be quite interesting.

I have not studied how peak Q shifts in frequency and can not answer your question on that matter. I do know that the precise value of peak Q is very sensitive to the exact S-parameters and that peak Q is also the most sensitive to subsection size error.
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, All:

I consider that tube like model is much better than multi-sheet model and volume meshing model for modeling the field inside a metal trace. The reasons are the following:

Tube like model is trying to solve the tangential field (or surface current) on the surfaces of a metallic trace. From the field on the surface, we know how the current is penetrating into the trace. This behaviour can be solved analytically with very high accuracy. In fact, it is even better than using volume meshing. As you know, the field is changing very fast when it is penetrating into a high conductiivty metal. Using a volume meshing, you need to have extremely fine meshing for the region close to the surfaces in order to capture the fast field changing in it. Numerically, this is a very big burden and it may not even capture the field variation as good as the analytical formula. Analytically, we know the wave is perpendicular to the surface and it is decaying exponentially and this analytical behaviour is welll modeled and it is extremely accurate. Actually, such a model is very elegant because its frequency dependency is built-in. Numerically modeling of the skin effect using volume meshing may also have accuracy problem due to the fact that the field variation is not the same at different frequency range. Regards.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian -- The equivalent surface impedance model is exact only in the case of a plane wave normally incident on an infinite plane conductor. In all other cases it is an approximation. I shall demonstrate.

For a microstrip line much thicker than a skin depth, I would think the tube-like model is a very good approximation for most any application, provided subsections size is small enough (on all four sides) to realize the desired error level.

However, for the case where the line is about a skin depth or less thick, the approximation degrades. It degrades most for square cross section lines, it degrades less for lines with w >> t or t >> w.

To see why, just go to DC. The tube-like model gives the exact answer for DC only if you leave off the side current. But this is just the multi-sheet model with N=2. The nice thing about the multi-sheet model is that it gives the exact answer at DC no matter how many sheets are used. (This makes it much easier to do convergence analysis at high frequency.)

Of course, the trade-off is that the multi-sheet model models the actual side current with filaments of current (along the edges of the interior sheets). The tube-like model models it with a sheet of current. I think it is interesting how each model has advantages where the other has disadvantages. Nature seems to like symmetry.

P.S. There should be no problem using the multi-sheet model in IE3D if desired.
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

I think your understanding of tube model is quite different from my understanding of tube model. In IE3D, the thickness model (tube model) doesn't behave like what you described. IE3D users can give it a try and they will see both the DC and AC resistances are behaving like expected (at DC, it is approaching the ohms law while it is approaching the high frequency behavior at the high frequency).

I think the elegance of tube model used in IE3D, is that all the current distribution (no matter on the top or bottom or the side surfaces) are taken into full consideration while the current inside is described completely using the surface current as long as the conductivity of the metal is not very low. I understand that surface impedance which is used in all MOM simulators in modeling loss will fail when the conductivity of a metal is very low. The question is what is "low". Let's analyze some example.

Assume there is a conductor with sigma = 1000 s/m and we are considering the frequency at 10 GHz. Assume the permittivity (real part) is 1 since it is not an important factor for conductor loss. The Er_complex = Er - j sigma / (omega*8.86e-12) = 1 - j 1797. Then, the refraction coefficient |n| = 42 (sqrt of 1797). If a plane wave is incident onto such a metal with an incident angle of Theta0, the diffraction angle is Theta1. Then, you will see Sin(Theta1) = Sin(Theta0)/n. As you can see, even Theta0 is approaching 90 degrees, Sin(Theta1) is 0.024 and Theta1 is just 1.36 degrees. Basically, the field penetrating into the conductor is almost normal to the conductor surface even for such a big incident angle and such a poor conductive metal. Typical good conductor's sigma is in the range of 1.0e7 s/m or 4 orders higher than the assumed value here. In some sense, for good conductors, the field inside a metal is only 1e-4 degree away from the normal line. Again, every body can follow what I discussed here to verify the values I put in this posting and you can see that the tube model with surface impedance is very accurate theoretically.

Regards.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian -- Yes, I guess I do not understand what you are doing at low frequency, let's explore this a little bit:

Let's take a resistor that is a cube one CM on a side. There is a perfect conductor one CM square cross section line attached to one side and another to the opposite side. These lines go to the ports.

Let's say it is a 10 Ohm resistor. At low frequency, the surface impedance Zs = 1/σt (t=metal thickness, this is the low frequency case of my eq. 5 on pg. 916 in my paper on microstrip loss in March 2003 MTT Trans.). For the 2-sheet model, each t is 0.5 CM and 1/σt gives us 20 Ohms of resistance in the top sheet and 20 Ohms of resistance in the bottom sheet. The two connected in parallel make 10 Ohms.

So, if you get the same 10 Ohms from 4 sheets, then you must make each one 40 Ohms. That would give you 10 Ohms when all four are connected in parallel. Is this what you actually do?
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

Thank you for your interests in knowing what we do? However, company policy does not allow me to reveal detail that are not published. Sorry for not being able to answer your questions on the detail.

My point is the tube model of thickness is matching the physics and mathematics. It is very solid if it is implemented properly. I believe it is much more accurate than multi-sheet model because the tube model is closer to reality. Also, the tube model is much more practical than volume meshing because volume meshing has to use very tiny cells in order to capture the fast field and current change inside a metal. Don't you agree?

Anyway, you can try to explore it. Thanks!

Best regards,
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian -- You are the company, and you set your company policy. If it is your policy to not discuss details that are not published (actually, I don't recall seeing any of your publications recently), that is your choice. We keep some things proprietary also, but not very much. Most of our work is published. Please feel free to enjoy it.

We publish the details so that no one must take our word for what we say. They can view the entire theory and judge for themselves. Perhaps they will copy it. Perhaps they will improve on it. Sometimes they find mistakes, too. But that is OK. That is how science works. You have choosen to not take part in this process, at least for this matter, and that is your choice.

You seem to be suggesting that the tube model is more accurate than the multi-sheet model, but you provide no data or specific line of reasoning to show that this is the case.

Somehow (and you won't tell us how) you get the tube model to match at DC for resistance. OK, maybe so. But then you say your tube model is "closer to reality" and therefore it is more accurate. But it is not clear what you mean by "closer to reality".

To me, "closer to reality" is achieved by doing a convergence analysis. For the multi-sheet model, you just add more sheets (and make cell size smaller), and this converges uniformly and asympotically to the exact answer. The multi-sheet model can always get "closer to reality" as much as is desired (until you run out of computer). This is verified in extreme detail in the MTT Trans. microstrip loss paper I mentioned above.

It seems your "closer to reality" at low frequency is modeling a uniform volume current as a tube composed of four infinitly thin sheets of current. Even with the right DC resistance, you can not possibly get the DC inductance right. Confining the current to just the surface when the real current is over the entire volume over estimates the true inductance. Your DC inductance will be too high.

"How much higher?" Now there is a problem. I see no practical way to do a convergence analysis for the tube model and find out how much the error is. Any suggestions, that you can share with us? If you need "closer to reality" than the four sheets provides, how do you do it?

Bottom line: All models give the wrong answer. To do science and engineering, we must ask, "How much wrong are the results," and provide a solid number for the answer. If we say, "It is accurate," without providing a number (or a way to get the number), we are just doing sales (which I do do myself sometimes too), not science and engineering.

P.S.: Yes I do agree with your points on volume meshing for certain ranges of geometries.
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

I think at least a few of your comments are incorrect. I am a member of Zeland Software, Inc. and I am not the company. I am proud that I am a very important member of the company. I think every company has policy on its intellectual properties. I think Sonnet also has its policy on it.

I have to say that I have extremely good memory. I still remember the 1st time we met in 1990. We both had papers published on the MTT 1990 in Dallas. At that time, I was still a Ph.D. student. I still remember that we were in the same poster session. My paper was on a general MOM algorithm with non-uniform meshing (mixed rectangular and triangular meshing) and its applications in modeling MMIC circuits. You came to my poster and asked me how I handled the triangles and I explained to you how I did it. Later, I saw you in your booth in the show and I asked you how you did the de-embedding. You may not remember what you said to me. However, I still remember you said “I don’t want to tell you?” to me. It is understandable to me. In fact, I didn’t keep it in mind. However, as I mentioned that I have extremely good memory and I still remembered it when you asked me questions after 16 years.

It was interesting that I still heard that you claimed that triangular cells were causing current to zig-zag and it was not accurate in the MTT in late 1990s. I assume you had tried it and it did not work for you. However, I am quite confident triangular cells are working equally well compared to rectangles. I think it must be true because at least a few simulators are using triangular cells successfully.

It is funny to see so much has changed in the last 16 years. Yes. I have seldom published any papers in the recent years. I have been working hard to bring good things to designers. I really do not have time to write papers. I am glad to see so many designers are using the products Zeland has been developing. I feel honored when I see so many people publishing papers using IE3D in different conferences.

I understand everybody makes mistakes. However, I think smart persons should not make the same mistake repeatedly. As EM simulation is concerned, when a user gets more and more familiar with a simulator, the user should make few mistakes and they will get confidence on what they are doing.

Regarding the surface current, the tube model is using the surface current and it describes the volume current inside the metal analytically. As I showed in my earlier posting, it is very accurate for normal conductors. Nothing is missed in the model no matter DC or high frequency is concerned. It is a very elegant approach. Regards.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian -- In 1990, our de-embedding was still proprietary. I put it in the public domain in 1991, that was my choice. I think your de-embedding is still proprietary, and that is OK because that is your choice.

Your statements that seem to suggest the tube-like model is essentially exact are not supported by the information you have given (e.g., "it is elegant"). The statement that "Nothing is missed," is incorrect. A uniform volume current flowing through a rectangular cross-section does not give the same inductance as a surface current flowing on the volume's surface. Those are two different problems and they give two different inductances. If and when you can give specific information to support your claims and refute the specific counter points I have raised above, I think we will all be very happy to listen to it. Non-specfic generalizations (having the effect of, "I know it's accurate, take my word for it") are not helpful.

If you wish to support your claim that a pure triangle meshing is just as accurate as a rectangular meshing, I will very happy to see that as well. Performing the stripline standard using both types of meshing would be entirely adequate in that regard, if you wish to do so. If you can present solid data that supports your claim, I will immediately modify my viewpoint accordingly. Until then, my viewpoint remains determined by data I have already seen. You have published nice current distributions for rectangular meshings. You are still welcome to post a nice current dsitribution for a pure triangular meshing.
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

On my posting, what I said "It is elegant" means that we could use the surface current density to include the effects of the current inside a metal accurately without dividing the volume into many tiny cells. Dividing a volume into tiny cells to model the fast current change in a good conductor is not practical in a daily design. Also, dividing a volume into multiple sheets of traces also does not match the physics because it is limiting the current on the sheets and it does not take care of the current on the edge of the traces in both longitudinal direction and vertical direction. From this sense, I consider the tube model is much more elegant.

When I say “Nothing is missed”, I say that even we are modeling the surface current, it includes the current inside the trace and it does not miss the effects of the current inside a volume. If it was not clear enough to you, I would like to emphasize it here. As you know, no model is perfect when we are solving it numerically. However, I believe that the tube model is matching the physics very well and it is the best model when accuracy and efficiency are concerned. Every model needs to be verified. Many people including us have done much verification on it. Accuracy on L has been verified to be very good. For most situations, designers are more concerned about the accuracy of the Q-value. Q-value is mainly affected by the loss. There are metallic loss involving field inside the metal, and dielectric loss in the substrates. In an MOM simulator, substrate loss is included automatically in the Green’s functions. As I explained, the field inside a metal trace is modeled precisely in the tube model using the analytical solution of field on the surface of a metallic trace. I believe it should offer the best accuracy as efficiency in MOM EM simulations.

Regarding the stability of triangular cells, it seems to me you are still not convinced that triangular cells can yield the same kind of accuracy as rectangular cells. I understand that it is not easy to convince one person on something when the person considers something is not true. We have included a comparison in the Appendix of the IE3D User’s Manual since more than 10 years ago. We documented a patch antenna using rectangular cells (automatically) and using triangular cells (manually created in order to avoid rectangular cells). It is showing very good agreement. I put together a few PPT slices with the comparison including the involved IE3D geometry files so that those IE3D users can try them out. I compared 3 cases: (1) Mainly rectangular meshing; (2) Mainly triangular meshing; (3) Mainly rectangular and denser meshing. You can see how good they agree. Also, you can see the vector current distribution on the antenna. Even though you see many zip-zags in the triangular meshing, you do not see any problem in the vector current distribution in the results. The vector current distributions are displayed at the center of the meshing edges without any smoothing. You can see how different meshing schemes predict the same kind of vector current distribution both in directions and magnitude. The examples here should demonstrate how triangular meshing and rectangular meshing are equally good in accuracy. Certainly, the triangular meshing will make the simulation much slower and it is why we suggest users to use as many rectangles (and as fewer triangles) as possible in order to improve the simulation efficiency.

Anyway, I think we should go back to the formal topic of this thread.

Best regards.
 

madengr

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Re: Quick question: why does Q of spiral inductor have a pea

I enjoy reading this discussion! I am presently reading Swanson's and Hoer's (sp) EM book so it is interesting comparing the various methods. How about a 45 degree stripline as a benchmark; the two bends would be a problem? By default Sonnet would mesh it as a staircase and perhaps IE3D would default to triangles (I have never used IE3D but will try a demo soon.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian -- You have not stated how you would do a convergence analysis with the tube model. The tube model moves all the volume current to the surface and uses an equivalent surface impedance to get the same DC loss. Using surface current instead of volume current yields a different inductance, the orginal topic of this thread. The change is insignificant for some applications, and significant for others. Telling us that lot's of people have verified the inductance with no details has the same effect as stating that it's accurate (and precisely what do you mean by "accurate"?), we should trust you. Everyone says their software is accurate. I want to know the error. You do not address the inductance error, other than by repeatedly insisting it is accurate. You could address this problem by simply doing a convergence analysis. Why don't you do that? Then you could say the error is +/- X%. End of discussion.

S-parameters are not so sensitive to error. The differences between the S-parameters you plotted actually suggest fairly large error to me. You look at the data and see good agreement. I look at the data and I see the differences.

I still have yet to see what I would call a good current distribution using pure triangle meshing. Your vector plot shows direction and magnitude only at select points. What happens in between those points?
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

In fact, why do I need to convince you that IE3D is accurate? I believe at least 99% of antenna designers agree that the patch antenna modeling using rectangular cells and triangular cells I presented in the PPT in my last poster are quite accurate. I also think many MMIC and RFIC designers know that IE3D can yield accurate results. That is enough. Thanks! Regards.
 

rautio

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Re: Quick question: why does Q of spiral inductor have a pea

Hi Jian – Very nice, 99% is a first class satisfaction rate. You’ve clearly worked hard to earn that. Sorry, I can’t give you a number too, we don’t track that statistic, not sure how I would measure it anyway. Any suggestions are welcome. And, of course, you are under absolutely no obligation of any kind to prove anything to anyone, that is completely up to you.

Just in case anyone else might still be interested, I did a convergence analysis on inductance at low frequency. Some interesting results came up, summarized in the pdf that is in the attached zip file. If you share my interest in this topic, please enjoy! (I will also email this to anyone on request.)
 

jian

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Re: Quick question: why does Q of spiral inductor have a pea

Hi, James:

There are two separate topics:

(1). For my comparison on an antenna using the two different rectangular meshing and one triangular meshing, I see the difference among the 3 models were very small. You mentioned that you see the difference is very big. When you said that, there must be a reason for it. I assume you are very good in doing convergence study using Sonnet's em and you think Sonnet's em is very accurate. When you say that the difference in the simulaiton is very big, it seems to me you must have a way to prove that you can get less numerical error using Sonnet's em on it. Certainly, you can increase the box size to make it closer and closer to the reality for it. Assume it does not break down numerically, you are supposed to get the converged results when you increase the box size and reduce the cell size continously. I am very interested in your comment on it.

(2) Regarding low frequency L, I think I am certainly the right person to comment on it. In fact, I was asked to comment on it when I was offering courses. I may also have written something on it in the IE3D's User's Manunal. I am going to summarize it in the IE3D's User's Manual. I will either put it here or the users can download it from the Zeland's web site (www.zeland.com).

Before, I comment on it, I do want to ask you the following questions:

If you don't know the answer on a specific question, please say "no". If you think some parameter is not critical, please say "it is not critical".

For your study (simulation and experiments):

1. What is the purpose of the study?
2. Is the convergence study for (low and high frequency) circuit designers?
3. What is considered as low frequency?
4. Is DC considered as low frequency?
5. What is the converged value?
6. In your study, what is the box size?
7. What is the conductivity of the bottom plate of the box?
8. What is the conductivity of the top plate of the box?
9. What is the thickness of the bottom plate of the box?
10. What is the thickness of the top plate of the box?
11. What is the conductivity of the side walls of the box? I assume that it is PEC in the simulation because the Green’s functions can only handle PEC. It should not be PEC in measurement because it is impossible to get PEC in reality.
12. What is the thickness of the side walls of the box? I assume you can not provide the thickness information in simulation. In measurement, you may not care about it.

I should write up something on this topic soon. Thanks!

Best regards,
 

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