Discussion Forum

Any thoughts are welcome on this please.

Thanks

An electromagnetic wave will not induce a voltage on or across a perfect conductor, because the tangential component of E on a perfectly conducting surface is always zero. That said, such a wave will still induce a non-zero surface current density. If your surface has a finite (not infinite) conductivity, then a line-integral along (not perpendicular to) the electric field from one point to another, will produce a quantity with units of volts, and this may be non-zero.

Hi thanks.

The problem is that I am trying to couple an em-wave (normal incidence-planner) with an patch antenna (flat metal surface) and to calculate the potential produced on the antenna. The antenna has to be a PEC since without that I won't have any reception. But like you said if its a PEC then there won't be a potential on it.

Do you think theres a way around it. can instead of defining the antenna patch as a PEC can i define it as having finite conductivity (a high value but finite) in its material properties. ??? And even if this works how will I know the value of the potential of the entire patch as a line integral will give me the potential difference between the edge end-points.

Thanks once again

Well now, that is a different question entirely. First of all, your patch antenna must not be a single, disconnected, patch, floating in empty space. Such a surface will possess no unambiguous potential. Rather, to actually be a patch antenna, and not just a meaningless piece of metal, it has to be connected to something, or at least be near something, so that you can at least define and understand some specific point on the patch as the point of interest to you, and then you need to be able to compare that point to another specific one somewhere else (generally not on the patch itself). For example, your patch might be connected to the center conductor of a coax line, or simply to a trace on a pc board. Ultimately, there will then exist two points *between which* you might be interested in observing or calculating a voltage. For example, such a point pair could be: (1) between the inner and outer conductor of a coax, or (2) between the pc board trace and the ground plane. It is only between two well-defined points that you should seek to compute a voltage difference. And one way to find that voltage difference is to integrate, from one point to the other, the component of E that is parallel to a path between those two points.

Hi Sir,

I appologize for not being more forth coming. If you dont mind to please bear with me one more time ill explain the complete idea and if you could provide me with some more comments ill be most greatfull.

The idea that I am trying to simulate is to have an patch antenna (modelled currently as a 2D PEC surface) (x-y plane) on top of a 3D substrate. directly under the antenna is a layer of PZT (piezoelectric material) 1microns thick. below that is a 10 microns thick layer of silicon. The idea is to capture incoming (normal incidence x-plane polarized) em-waves by the patch antenna and to try and calculate the resulting potential on the antenna patch which also doubles as an actuating electrode for the piezoelectric layer beneath it. And hence excite this layer to deform. The 2D surface of the antenna is modelled as a PEC and the bottom surface of the silicon is also a PEC which is to serve as the ground electrode of the patch antenna.

This I believe is my simple but complicated problem which I need to solve. Any ideas on this.

I understand that when a patch antenna transmits the fringing electric fields line up and that causes the radiation. So when it receives the fringing fields also must line up. And I can see that happening thorugh the arrow plots of the electric field in the simulation (post processing) however I cant see any displacement of the PZT. and I understand through your comments as to why that may not be happening.

But the strange thing is that in the derived values section of the results if I line integrate any edge of the patch wrt the normE I get some real value for the potential in V. Why do you think that is happening since I completely agree with your comments on why it should be zero. One reason I think is that the E-field due to internal reflections within the air domain does not maintain being x-plane polarized and does develop other components which may lead to a finite line integral value on any given edge. But still. My air domain is about 300 microns by 300 microns while my substrate geometry is about 100microns by 200 microns. Do you think increasing the air domain might help.

Thanks for your responses please, I am most grateful.

You said "if I line integrate any edge of the patch wrt the normE I get some real value for the potential in V" and "I completely agree with your comments on why it should be zero."

Your comments suggest that you have some misunderstandings of the physics here.

First of all, integrating normE along the edge of a patch doesn't seem useful to me. Yes, if you do so, you will obtain a quantity that has units of volts. That is simply because E [V/m] times distance [m] yields V. Well, you might as well be saying that torque = energy because they have the same units (both are force * distance). The only meaningful voltages you will get here by integrating E along a path is if you integrate the scalar product of the vector E with the vector path element, i.e., if you integrate the component of E that is parallel to a path, along that path, between two points, to find the potential difference there. Integrating normE anywhere on/around your patch edges is not doing that, but you are generating numbers that happen to have units of volts.

Second, I'm not saying that such integrals would or should yield zero. Rather, I am saying that integrals of the tangential component of E along any path on a PEC surface will yield zero. The norm of E includes the component of E perpendicular to the surface. That's very different. There are circumstances in which one might want to integrate that (such as for computing charge density) but that is not the same as computing potential.

Finally, unless it is electrically very small, your patch antenna does not even have a single-value potential to it. It is an antenna! Antennas have distributions of currents and charges on them. You can compute those. And you can compute the fields on and around the antenna. And you can compute power densities, power flow, etc. And you can even compute voltages and currents at the feed of such an antenna, if it has one.
But an antenna problem is not an electrostatics problem. A patch antenna is not simply a capacitor, at least not if it is being operated at the frequencies for which it behaves like an antenna.

Thank you very much sir, this does help me improve my concepts about this physics. As you may have guessed I haven't been much experienced with it.

From what I have read the patch antenna does have a voltage distribution, as you mention. So for a half wave patch the voltage looking at the antenna in the cross section x-z plane (with the antenna at the top the substrate below the antenna and the ground at the bottom) the length of the antenna being from x = 0 to x = L and z perpendicular to x, the voltage is positive maximum on x = 0 it decreases to zero in the middle of the antenna and then increases in the negative direction to a negative maximum at x = L ( like a cos curve between 0 and 180 degrees) while the current is zero at x = 0 and L and maximum in the middle (like a sin curve between 0 and 180 degrees). This can be a typical fundamental mode of operation of the patch. If you agree with this sir do you think that the voltage which has a different value at each point along the length of the antenna (still looking at the x-z plane) can be applied (if we can calculate it somehow) to the piezoelectric layer underneath the antenna ??? since the voltage is varying it will cause a varying strain in the pzt material and deform it.

My second question is that can we calculate this varying voltage along the length of the antenna in COMSOL ???

I really appreciate your comments.

Thanks again.

Hello,

After these years, I think I am exactly experiencing the same situation. Please let me know, how you managed to compute the induced voltage between the antenna.

One thing, I would like to know, the electric field distribution that you are going to integrate is in frequency domain or in time domain?

Thanks,
Rishad

At that time I couldn't even work out a patch antenna. The problem is that when you make your patch dont make it a PEC make it a metal like gold or something. You have your patch antenna and below it is your substrate then below it is your ground. Make sure you put a LUMPED PORT between the patch and the ground the potential will come out of the lumped port by itself. No integration or anything required. This lumped port is what I was missing. you can make the patch any thickness or any metal and it will work as long as the lumped port is there. Look up these ports for more info on them.

I am working with Optical antennas (in THz ranges). I got a bowtie antenna in my design. Now, same as like your case, I need to compute the voltage that is induced in the antenna between two points. This is my main objective.

-I have the e-field distribution in around the bowtie antenna, in frequency domain. I thought, if I can take the time dependent E-field distribution of the structure which will be non-sinusoidal in nature.

-Then If I take the average of the time -dependent E-field distribution which will result in a non-zero value.
- Then taking the line integral of the E-field along the path will lead to a non-zero voltage value.

Please let me know, what you think about this method. However,I don't know whether I can use port in THz ranges.

Thanks a lot Mateen,

I am not sure of the time domain method, seems plausable. But if you want to know the potential between two points in comsol just put a lumped port between them. That is why they have been created in COMSOL.

Thanks. I will try to work it out. I might ask for your help again. Thanks a lot.

-RA