Why doesn’t EM transmit to air for a PCB board?

Hi,

How does EM wave transmit in a PCB board?

Assume that we have a two-terminal component of which one lead is connected to the ground plane (return path), and another to a receiver through a PCB trace.

1. If the lead of both the transmitter and the receiver are on the outer layer of the PCB board so that they are directly exposed to air (assume no solder mask coating), would the EM filed transmit to all directions into the air? After all, this is the solution of Maxwell’s equation in vacuum.
   1. If the EM wave does go to all-directions, then why do we still use a stripline (microstrip, etc.) to guide it?
2. If the transmission line (stipline) is embedded in an inner layer of the board, how does the EM field actually travel? Is it confined to the PCB trace, or is confined to the FR4 dielectric, or could transmit to all directions in the air?
   1. I think it looks likely that the majority of EM wave would not transmit to air due to total reflection between FR4 and air boundary. FR4 has ε_r of 4, so that the incident angle which could cause total reflection is sin^-1(1/4) = 14.47°. Therefore, it might be that only a small fraction of EM energy could FR4-air boundary, either at horizontal edges or from top/bottom layers, to air.
   2. If there the trace is sandwiched by power and ground planes, then due to reflection, no EM wave could penetrate enclosing power/ground layers, so that EM wave only travels in FR4 region enclosed by copper power/ground planes.
   3. Combining (a)(b): which is the actual reason preventing EM wave from radiating into air on a PCB board?
3. However, normally (not considering things like buried resistors) all components are mounted on top/bottom layers, and if assuming no coating then they are directly exposed to the air. For such case FR4-air boundary cannot cause any total reflection to confine signal, and the only reason I could thought of is that EM waves are traveled totally in conductors themselves, not even in FR4. However, most of the calculation in signal propagation time seem to use v=c/sqrt(ε_r). I am confused which speed should I choose for external conductors (without coating).

Could someone who really understands Maxwell equations and waveguide help with the question?


Matt

The question: "How does EM wave transmit in a PCB board?"

Do you mean "efficiently transmit" (as in 100% efficient), or "sufficiently transmit" (%efficeint << 100) in order to meet the definition of 'transmit'?

If you have resonant 'structures' on the board (like 1/4 electrical wavelength resonant traces) you *can* have "efficient radiation" (as in a patch antenna) or more likely incidental radiation from structures which are: L << 1/4 Lamda yield "inefficient radiation" (leakage flux, leakage E-fields and such).

Efficient radiation takes place from structures (usually resonant, although there are a few structures like ridges horn and discone antenna that would seem to defy that requirement) that allow a large amount of 'charge' to oscillate on or near the surface (the surface exposed to air or 'ether' as in the early days the scientists thought) ... Maxwell described the field relationships for EM radiation effect, but I do not think he elaborated on the intensities required for efficient radiation nor the need for certain resonant structures, and I don't recall at this moment whether his laws/equations addressed this either (someone more versed on Maxwell and his equations are welcome to chime in here now and elaborate).

Jim

mmitchell said:

Hi,

How does EM wave transmit in a PCB board?

Assume that we have a two-terminal component of which one lead is connected to the ground plane (return path), and another to a receiver through a PCB trace.

1. If the lead of both the transmitter and the receiver are on the outer layer of the PCB board so that they are directly exposed to air (assume no solder mask coating), would the EM filed transmit to all directions into the air? After all, this is the solution of Maxwell’s equation in vacuum.
   1. If the EM wave does go to all-directions, then why do we still use a stripline (microstrip, etc.) to guide it?
2. If the transmission line (stipline) is embedded in an inner layer of the board, how does the EM field actually travel? Is it confined to the PCB trace, or is confined to the FR4 dielectric, or could transmit to all directions in the air?
   1. I think it looks likely that the majority of EM wave would not transmit to air due to total reflection between FR4 and air boundary. FR4 has ε_r of 4, so that the incident angle which could cause total reflection is sin^-1(1/4) = 14.47°. Therefore, it might be that only a small fraction of EM energy could FR4-air boundary, either at horizontal edges or from top/bottom layers, to air.
   2. If there the trace is sandwiched by power and ground planes, then due to reflection, no EM wave could penetrate enclosing power/ground layers, so that EM wave only travels in FR4 region enclosed by copper power/ground planes.
   3. Combining (a)(b): which is the actual reason preventing EM wave from radiating into air on a PCB board?
3. However, normally (not considering things like buried resistors) all components are mounted on top/bottom layers, and if assuming no coating then they are directly exposed to the air. For such case FR4-air boundary cannot cause any total reflection to confine signal, and the only reason I could thought of is that EM waves are traveled totally in conductors themselves, not even in FR4. However, most of the calculation in signal propagation time seem to use v=c/sqrt(ε_r). I am confused which speed should I choose for external conductors (without coating).

Could someone who really understands Maxwell equations and waveguide help with the question?


Matt

Click to expand...

All of your questions can likely be answered if you look at the introduction to microstrip transmission lines in a good microwave engineering textbook (recommend Microwave Engineering by Pozar).

When you look at a structure like microstrip, it's primary frequency range of operation is operated in what's called a quasi-TEM mode. You need to operate at a frequency thats sufficiently low for modes greater than the dominant mode (TE010, IIRC). You are correct, when you have microstrip, the E and H fields travel in all directions (into the PCB and out into space). If you look at a good microwave engineering textbook, you'll see that microstrip transmission lines have a calculated ε_effective, which is based on ε_r for the PCB dielectric, but is lower than ε_r. This takes into account the air and board dielectric for the quasi-TEM mode and gives the user an effective permittivity for a homogenous dielectric that surrounds the microstrip.

**broken link removed**

Electric field lines originate on one electric conductor, traverse a dielectric medium, and terminate on a lower potential (ground) conductor. That's how everything works.... EM fields are not contained in the metal, they must exist in a dielectric; that's fundamental.

Stripline operates similar to microstrip. The majority of the E-field is contained in the PCB between the two conductors. Some energy can radiate up and out of the PCB, into the air. However, that's a small amount due to the long path length and reflections at the interfaces. If you cover the stripline with power/ground planes, even less energy gets radiated into space, but some can still radiate "across" the board and come out the sides. To reduce this, PCBs are often edge-plated, or via fences are used to create a "wall" that the E-fields can terminate on, rather than create a loop and radiate into space.

Without getting to involved in maths and theory, the following articles by Henry Ott were designed for people like me, to explain how the waves can propagate into the wider world... DIPLOLES:shock:
http://www.hottconsultants.com/pdf_files/dipoles-1.pdf
http://www.hottconsultants.com/pdf_files/dipoles-2.PDF
http://www.hottconsultants.com/pdf_files/dipoles-3.PDF

I came up with a plausible explanation for the use of conductors (microstrip, stripline) to transmit signal:
consider lead A on transmitter and lead B on receiver and conductor AB connecting them. AB would have a certain cross-section area, and assume it to be uniform even at the soldering joints with A and B. If I imagine myself standing at A and looks out, my signal can travel at all directions and the strength decreases with 1/R^2. However, a certain portion of the all-directions wrapped by the solder joint with conductor, say, perhaps 20% of all directions. Then, in 80% directions energy would still be radiated in spherical directions, but for the 20% directions wrapped by solder joint, EM wave will be contained to conductor AB and propagate along it without decreasing in strength due to reflection at the conductor reflection.

Therefore, if we consider the signal strength at length lead B, it would still be the original strength at the 20% directions covered at the solder joint at lead A; but for other space points equally distant from lead A just as lead B, the EM strength would be r^2/AB^2 smaller, where r denote the radius of the very small sphere at the solder joint connection from the idealized concentrated point signal source. The simple idea could be summarized that: conductor AB contains EM wave with conductor boundary reflection, and by this signal strength is preserved. This is why a conductor is necessary, otherwise signal at B would be very weak, and due to the presence of other component leads on the board, say, transmitter leads C, D, E, ..., Z from other components, it would be impossible for B to differentiate from which lead the EM it sensed was coming from.

Is this explanation valid?

enjunear said:

EM fields are not contained in the metal, they must exist in a dielectric; that's fundamental.

Click to expand...

I would therefore not agree with the description that EM fields are not contained in the metal. At least for the case of conductor AB above, the EM wave traveling through it is contained within. The fact that at internal points of conductor AB there would be no E field would be due to the cancellation between the induced E field of surface charges and the original E field. The vector sum is zero, but the original EM wave still exists and travels along conductor AB.

enjunear said:

Electric field lines originate on one electric conductor, traverse a dielectric medium, and terminate on a lower potential (ground) conductor. That's how everything works.... EM fields are not contained in the metal, they must exist in a dielectric; that's fundamental.

Click to expand...

It seems like your are describing stripline? For bare uncoated copper microstrip, how can EM field still be connected in the dielectric?

enjunear said:

Stripline operates similar to microstrip. The majority of the E-field is contained in the PCB between the two conductors. Some energy can radiate up and out of the PCB, into the air. However, that's a small amount due to the long path length and reflections at the interfaces. If you cover the stripline with power/ground planes, even less energy gets radiated into space, but some can still radiate "across" the board and come out the sides. To reduce this, PCBs are often edge-plated, or via fences are used to create a "wall" that the E-fields can terminate on, rather than create a loop and radiate into space.

Click to expand...

I could understand how "edge-plated" could be used to reflect radiation; but how could "via fence wall" achieve that? Does that require a large number of vias arranged that for each D(diameter of via) length there is one via, so these vias are connected together and that their plating would then be equivalent to a fully-plated edge? But in this case there are no separation between adjacent vias and all of them are effectively connected together, and I could not understand why there is any motivation of using it rather than directly plating the edges which would be much easier and leaves a flat edge rather than the strange concatenation of half-circles edge formed by connecting vias.


Matt

I have problems to understand the motivation of your question. The title "Why doesn’t EM transmit to air for a PCB board?" seems to involve a wrong assumption, otherwise life would be much easier related to radiated interference emission.

The EM phenomena effective on a PCB can be analyzed in detail, in fact none of them is specific to PCB, they all refer to more general EM theory and can be observed in other situations as well.

When looking at real PCB behaviour, you shouldn't get confused by the simplifications that are used in the analysis if basic transmission line structures. E.g. the micro strip model is assuming an infinite passive ground plane. But a real PCB has one or more planes of finite size, which act as resonators itself, so there will be possibly an electric field parallel to the ground plane. Multiple ground planes don't have perfect interconnection, so there will be a potential difference between the planes.

Planar transmission lines on the outer layers are generally dispersive (respectively radiating energy). But the radiated share is of course much smaller than 20%... I fear however, that you need to dive considerably deeper into EM theory for a quantitive analysis, than it's sketched in your post.

You are able to get a realistic view of the complete PCB's EM behaviour using a 3D solver, if you really need to know. Or apply design rules for "low radiation" design and verify their success in an EMC measurement.

P.S.: To get started, you should be able to find an analysis of the radiation properties of micro strip transmission lines in antenna text books, e.g. Balanis.

mmitchell said:

Does that require a large number of vias arranged that for each D(diameter of via) length there is one via, so these vias are connected together and that their plating would then be equivalent to a fully-plated edge? But in this case there are no separation between adjacent vias and all of them are effectively connected together,

Click to expand...

A via fence is created using standard-width vias (depends on your substrate), that are separated by no greater than λ/10 spacing at your maximum frequency of interest. λ/10 is a routinely used starting point for EMI/EMC mitigation strategies. In above-resonant structures, it gives significant isolation to lower frequency signals trying to couple/radiate along or through said structure (gap, edge, slot, line, etc).

enjunear,

I got it, thanks for this explanation.

Bob

mmitchell said:

It seems like your are describing stripline? For bare uncoated copper microstrip, how can EM field still be connected in the dielectric?

Click to expand...

I was describing ALL field paths in a transmission line structure. The dielectric can be air or the PCB; EM fields will exist in both regions. In microstrip (or stripline) the E-field lines originate from all sides of the high-potential conductor, and travel either through freespace or the PCB dielectric, to reach the lower-potential (ground) conductor.

As you can see from the density of the field lines in the image below, the majority of the E field is contained in the dielectric, but some go through space for microstrip.


For stripline, you'll get something closer to (c) in the next image.

Caption: Common, multiconductor, planar transmission lines and their TEM mode E-field distributions: (a) parallel-plate, (b) microstrip, (c) stripline, (d) coplanar waveguide, and (e) slot line.

from 3.5: Planar Transmission Lines

enjunear,

Thanks for this illustration. I would refer to the Holzman book.

Bob

---------- Post added at 01:52 ---------- Previous post was at 01:45 ----------

FvM said:

The title "Why doesn’t EM transmit to air for a PCB board?" seems to involve a wrong assumption, otherwise life would be much easier related to radiated interference emission.

Click to expand...

I think I was wrong here.

FvM said:

When looking at real PCB behaviour, you shouldn't get confused by the simplifications that are used in the analysis if basic transmission line structures. E.g. the micro strip model is assuming an infinite passive ground plane. But a real PCB has one or more planes of finite size, which act as resonators itself, so there will be possibly an electric field parallel to the ground plane. Multiple ground planes don't have perfect interconnection, so there will be a potential difference between the planes.

Planar transmission lines on the outer layers are generally dispersive (respectively radiating energy). But the radiated share is of course much smaller than 20%... I fear however, that you need to dive considerably deeper into EM theory for a quantitive analysis, than it's sketched in your post.

Click to expand...

I do need some more quantitative analysis.

FvM said:

P.S.: To get started, you should be able to find an analysis of the radiation properties of micro strip transmission lines in antenna text books, e.g. Balanis.

Click to expand...

[/QUOTE]
I will refer to the book.


Thanks,
Bob

---------- Post added at 01:53 ---------- Previous post was at 01:52 ----------

Jim,

I need to read more in books before understanding your comments.

Bob

---------- Post added at 01:55 ---------- Previous post was at 01:53 ----------

marce,

Thanks. I am reading them.

Bob

Ralph Morrison, "The fields of electronics" anther good reference, that gives the basic physic behind signal transmission. Traces on a PCB are effectively wave guides at high speed, it is the wave that travels at some velocity near the speed of light, depending on the dialectric, electrons being generaly lazy drift along at about 84mm an hour.

marce,

I see it, thanks for the suggestion.

Bob