Question regarding the electrical signal traveling in circuits.

[yungman]

This is actually a continuation of another thread here:

https://www.physicsforums.com/showthread.php?t=618868

I don't want to hijack that thread. But that thread make me think about how the signal travel.

Let's consider a pcb trace on top of a ground plane...which is microstrip. EM wave travel in the dielectric guided by the microstrip structure. If you probe on top of the trace, you get a signal. You solder a connection on top of the trace, you get the signal.

My question is what is the path from the EM wave traveling between the bottom of the trace and the ground plane, to the top of the trace ( the other side). Remember from post #7 of Studiot, the velocity of the EM wave in good conductor ( the trace) is much slower.

v = \sqrt {\frac{{2\omega }}{{\mu \sigma }}}

1) So if the EM wave travel from the bottom of the trace to the top through the metal, not only should the EM wave get attenuated, but also delayed.

2) Say if the EM wave travel from the surface from the bottom, to the side and then to the top, it still have delay. Particular if you think of a case where the driving circuit is on one side of a large thick metal plate, you can connect the return of the driver on one side of the plate, the input of the next stage on the other side of the plate. You don't get extra delay. If it is true the EM wave travel on the surface, the wave has to travel a long distance if the plate is big.

At high frequency, the wave length is very short, even a trace width will cause a change in phase...delay if you want to call it.

I don't know what to think in this case.

 

[Antiphon]

Y,

1) The signal you measure at the top of the trace represents the voltage at that point along the top or bottom of the trace. It's the same voltage as on the bottom but the signal didn't go *through* the strip. The voltage at the top and bottom of the trace is essentially exactly the same even though the higher field magnitude is on the bottom side of the strip.

2) The delay would be large if it had to wrap around a large plate. But it's not seen this way because the excitation is usually balanced and begins at the generator far from the plate.

Example: you take a pulse generator a mile away with a transmission line. You come to a large metal plate where one wire goes to either side of the plate and then connects to the center of the plate.

In this case the signal enters from the generator in a symmetrical way and goes to either side of the plate. But does the current go through the plate? No, but its unclear how.

But suppose you put the generator on one side of the plate right where the wire attaches to it and put a matched load a mile away on the end of the transmission line after it has cleared the plate by a mile. What does the signal look like?

you would send a pulse down the one wire only from the generator with the plate itself as a return like micro strip. The wire on the other side of the plate has no current. Once the plate edge is reached, the fields will 1) wrap around the edge of the plate and start heading down the micro strip on the backside and 2) get launched down the transmission line toward the mile-away load. A portion will also be reflected back toward the generator. So you will have three separate waves; one on each side of the plate and a third down the transmission line.

There is never any current flowing through the bulk of the plate.

 

[yungman]

Y,

1) The signal you measure at the top of the trace represents the voltage at that point along the top or bottom of the trace. It's the same voltage as on the bottom but the signal didn't go *through* the strip. The voltage at the top and bottom of the trace is essentially exactly the same even though the higher field magnitude is on the bottom side of the strip.

Why the voltage is the same on both side. In electrostatic, there is no potential inside the conductor. But why does that hold true for RF wave?

2) The delay would be large if it had to wrap around a large plate. But it's not seen this way because the excitation is usually balanced and begins at the generator far from the plate.

Example: you take a pulse generator a mile away with a transmission line. You come to a large metal plate where one wire goes to either side of the plate and then connects to the center of the plate.
I don't quite get this. Do you mean there is two wire from the generator, one connects to one side at the edge of the plate, the other wire connects at the edge on the opposite side of the plate? Then what is the center?
In this case the signal enters from the generator in a symmetrical way and goes to either side of the plate. But does the current go through the plate? No, but its unclear how.

But suppose you put the generator on one side of the plate right where the wire attaches to it and put a matched load a mile away on the end of the transmission line after it has cleared the plate by a mile. What does the signal look like?
I don't get the picture on this one. Do you mean the generator right on one edge of the plate, then the opposite edge has a mile long tx line?
you would send a pulse down the one wire only from the generator with the plate itself as a return like micro strip. The wire on the other side of the plate has no current. Once the plate edge is reached, the fields will 1) wrap around the edge of the plate and start heading down the micro strip on the backside and 2) get launched down the transmission line toward the mile-away load. A portion will also be reflected back toward the generator. So you will have three separate waves; one on each side of the plate and a third down the transmission line.

There is never any current flowing through the bulk of the plate.

Thanks for your reply, I still don't quite get this, how does the signal go through the metal plate/trace from the bottom surface to the top surface.

 

[sophiecentaur]

Thanks for your reply, I still don't quite get this, how does the signal go through the metal plate/trace from the bottom surface to the top surface.

It doesn't "go through". It travels as fields between the conductors and may taker the long route from one side of the conductor to the other. In the case of a 'good' coax, the signal goes all the way to the end termination and then may travel along the outside, back to the point of interest if the termination is not perfect.
That's why a TV antenna works. There is no 'pickup' on the down-lead - just on the antenna itself. (Assuming your feeder is not rubbish and 'leaky' due to gaps in the braid).

 

[marcusl]

You can't pretend that an EM wave travels just between the conductors and that you can ignore the rest of the problem. If you solve for the potentials and fields everywhere, you will find that the top of the microstrip trace has a potential and even carries some current (more near the edges, less in the center). Thinking of a wave trying to propagate through the metal is not a correct model. If you must use an analogy, then think of the situation as reminiscent of eddy currents--at low frequency there is current throughout the top conductor but as f increases, the current crowds towards the underside.

 

[yungman]

It doesn't "go through". It travels as fields between the conductors and may taker the long route from one side of the conductor to the other. In the case of a 'good' coax, the signal goes all the way to the end termination and then may travel along the outside, back to the point of interest if the termination is not perfect.
That's why a TV antenna works. There is no 'pickup' on the down-lead - just on the antenna itself. (Assuming your feeder is not rubbish and 'leaky' due to gaps in the braid).

So if the signal don't go through the metal strip, then it has to travel on the surface from the bottom, around the edge and on the top surface to the "pickup" point? I draw a diagram of a section of microstrip where the signal is pick up at point A. I showed the EM wave between the bottom of the microstrip and the ground plane in pink. The path from the bottom to point A in green. So this is what you mean?

153740[/ATTACH]"]

 

[yungman]

You can't pretend that an EM wave travels just between the conductors and that you can ignore the rest of the problem. If you solve for the potentials and fields everywhere, you will find that the top of the microstrip trace has a potential and even carries some current (more near the edges, less in the center). Thinking of a wave trying to propagate through the metal is not a correct model. If you must use an analogy, then think of the situation as reminiscent of eddy currents--at low frequency there is current throughout the top conductor but as f increases, the current crowds towards the underside.

I agree, there must be more to it than just EM wave sandwiched between the line and the ground plane. But what? I don't believe it is through the metal of the microstrip. What I really want is the explanation in terms of Maxwell's equations.

All the eddy current etc. are a simplified way of explain EM theory, it is the EM wave that is really doing the transmitting. My question is what is the mechanism of transmitting from the bottom to the top. Even if you want to explain in terms of potential or current, it is still a phasor, it is not instantaneously travel from the bottom to the top. What is the path of the potential or the current travelling?

 

[Studiot]

Sophie Centaur is correct, but his answer is not complete.

It's too late tonight for a long answer so here is a short one till I have more time.

If the wave travels round the plate it re -enters the plate everywhere it is in contact. However it doesn't get very far due to skin depth (I will post the formula).
At each point of entry it has traveled at the dielectric velocity so is not delayed and the potential it sets up locally is also not delayed.
This is what you are measuring.

Does this help?

 

[yungman]

Sophie Centaur is correct, but his answer is not complete.

It's too late tonight for a long answer so here is a short one till I have more time.

If the wave travels round the plate it re -enters the plate everywhere it is in contact. However it doesn't get very far due to skin depth (I will post the formula).
At each point of entry it has traveled at the dielectric velocity so is not delayed and the potential it sets up locally is also not delayed.
This is what you are measuring.

Does this help?

I would really like to see the formulas. EM is not very intuitive, the only way is to look at the formulas and get the feel of it. I don't think you can explain in simple common sense terms. Particular at microwave frequency where λ is comparable to the dimension of the structure, I don't think you can use simple equivalent current and voltage to explain away. Everything has to be in phasor form.

As I drawn, even the distance is short and velocity is calculated using the dielectric, still it takes time. I was going to post a diagram from a design of matching network of a RF power amp. It is using a low impedance microstrip to feed into a power transistor with the capacitor at the input ( the termination of the microstrip). As you know the input impedance of power transistor is usually very low. To transform to low impedance, you use very low impedance tx line. I used something like 25Ω quarter wave which is easily over 100mils wide. When I designed for 2.4GHz using FR4, the λ/4 is only about 0.6", you don't have much room to play with the extra distance.

 

[sophiecentaur]

This link has a good diagram of the fields on an idealised piece of microstrip.

 

[Studiot]

Sorry it's taken a while to come back.

The thickness of a typical pcb land is 25 - 50 micrometres.

The skin depth at 2.4Ghz in copper is about 1.5 micrometres, beyond which the signal is effectively zero.

So at this frequency the signal does not travel through the land from bottom to top.

skin depth, d = .075/√f for copper.

If you want some text details, PM. There is about 10 - 12 pages.

Edit : Emboldened figure.I thought I changed this last night, don't know why it didn't 'take' then. There was a typo between . and /

 

[yungman]

Thanks for both your reply. I was looking at the diagram in the link Sophiecentaur provided, notice the B is circling around the trace? From

\int_C \vec B \cdot d\vec l'= \mu I

This get me thinking about the I being surface current, is also exist on the top of the trace as well as on the underside. This will be a sure way to communicate the current to the top. But still my next question is how long it take for the B to communicate from the bottom ( where the EM wave travel) to the top. Is there a propagation delay in the B field?

To clarify, I am not talking about the propagation of the EM wave. I am referring to the B line travel from the bottom of the trace to the top. ie: Let's call propagation of EM wave in z direction and \vec B =\hat \phi B_{\phi}. My question is how long it takes for B to go from \phi = 0 to 180 deg. I am not even sure I can look at it this way, bottom line, I am confused!

 

[Studiot]

I do believe that the diagram in Sophiecentaur's link is a section across the microstrip and ground plane, with the signal traveling into the plane of the paper.

The electric field is shown as the blue lines. The magnetic field will be closed contours around the strip, like the red ones marked equi potential lines are.

 

[yungman]

That's the typical field pattern of a microstrip. I understand microstrip is not perfect where the velocity at the top and the bottom is slightly different.

But I want to find out how the signal goes from the EM wave to the top of the trace. So far, I can see the closed loop magnetic field lines induce current on the top side of the trace. But still my question is whether it is instantaneous or still having a delay.

 

[sophiecentaur]

That's the typical field pattern of a microstrip. I understand microstrip is not perfect where the velocity at the top and the bottom is slightly different.

But I want to find out how the signal goes from the EM wave to the top of the trace. So far, I can see the closed loop magnetic field lines induce current on the top side of the trace. But still my question is whether it is instantaneous or still having a delay.

How can there not be a delay? EM effects always involve a delay with distance. This delay will manifest itself as a small phase lag of the signal on the top wrt the signal on the bottom. Considering the speed of light (ok, a bit less) and the dimensions involved, this phase will be small for practical signal frequencies.

 

[Studiot]

Delay?

I'm sorry I don't understand the question in relation to my comment.

Sophiecentaur, please confirm or deny my understanding of the diagram you linked to.

If what I am saying is correct and the signal is traveling ar right angles to the diagram it is not localised and so does not travel from one point on the section 'around the rectangle' - in effect it is already there when it arrives at the plane of the paper.

 

[yungman]

How can there not be a delay? EM effects always involve a delay with distance. This delay will manifest itself as a small phase lag of the signal on the top wrt the signal on the bottom. Considering the speed of light (ok, a bit less) and the dimensions involved, this phase will be small for practical signal frequencies.

I am referring to my post #12. I am not talking about the EM wave propagation delay traveling down the stripline. I am talking about each of the B closed circle. The EM wave travel between the bottom of the trace and the ground plane. The B component of the EM wave is a closed loop around the trace as in your diagram. Say when the signal first arrive at a point on the stripline, the B just started. My question is whether there's a delay of the B field from \phi=0 to \phi = 180 deg. This is not EM propagation. Please refer to post # 12.

From the B circling the trace, surface current exist on both the top and the bottom surface of the trace. Ultimately, my question is whether the top current and the bottom current are exactly the same phase or whether the top current has a delay from the bottom current.

 

[Studiot]

You need to think in 3D for this.

 

[yungman]

I know this is a tedious question. Everyone know you can pick off a signal by solder on the top of the trace. I wonder anyone stop and think about how and why that happen!

 

[Studiot]

I don't want to comment further until Sophiecentaur has confirmed my understanding of his picture, in case I am barking up the wrong tree.

 

[sophiecentaur]

I don't want to comment further until Sophiecentaur has confirmed my understanding of his picture, in case I am barking up the wrong tree.

I don't see the problem here. Where the wave is launched, there will be a delay (phase lag) in the fields on top. Without a dielectric to confuse things, the field at a point on top of the line will be the result of waves reaching it from under the line and further back down the line. At a large distance from the source, the path lengths will become equal to all points at a given distance from the source so the phases would be equal, I suppose. With a dielectric then you could get a phase tilt forward because the speed on top is nearer c than the speed below. But this tilt will settle down to a constant value as happens with an mf ground wave over a uniform ground plane.

@yungman
If you were using wide microstrip then I might suggest that the impedance presented by your soldered connection 'in the middle' of the line might be very different from the impedance if you terminated directly between trace and earth. But where there is a discontinuity, the situation would actually be different and the E fields and currents would be disturbed in any case. Your 'pick up point', soldered to the top of the line is only half of the connection; there needs also to be an Earth connection if you want to take a signal from the line. Everything would be upset.

 

[Studiot]

I don't see the problem here.

My comments have nothing to do with launching a wave and everything to do with the geometry of the situation.

I am suggesting that the picture you posted is a cross section through an already fully developed (wave? signal?) (travelling?) at right angles to the plane of the section.

 

[sophiecentaur]

Yes. It shows the 'static' situation. Isn't that clear in the text? I'm afraid I recognised it immediately and didn't search for the detailed description. It shows that the field is stronger between the plates - not surprisingly.

Interestingly, when discussing what happens when you pick a signal off the line, it is the potential that makes the difference and not the fields. This is an old chestnut for ordinary circuits too.