5
\$\begingroup\$

I have a question. Imagine that I need to have a 50 ohms impedance in a PCB trace. For simple cases it's just applying the formula found all over the internet or use one of the online calculators

While looking for this information online I came across an article where a guy created a low cost electric field probe and a low cost magnetic field probe and then proceeded to perform tests on different configurations of microstrips on a dual layer board. His goal was just that, measure magnetic field and electric field with a near-field probe, with a low cost set up, so that small companies can perform their own tests before spending lots of money certifying the end product.

The configurations were as follows:

  1. L-shape: One side all ground, not corroded. The other side just had a line trace in the shape of an L.
  2. H-shape: One side all ground, not corroded. The other side had traces forming an H, and the middle of the H had a line going to the edge of the board (the other traces, parallel with this one didn't go to the edge of the board).
  3. Radius-shape: This one was one sided.ç It was connected to the edge of the board on one side, it made a loop and the end connector was just beside the entry point.

It was necessary for the measures that the impedance on those lines were 50ohms. I honestly can't see how he calculated the impedance for those odd shapes, and I can't find any material on the web concerning how he was able to assure 50ohms on those lines.

Can someone point me in the right direction?

EDIT:

Did he just add termination resistors in specific places? Maybe in the H-shaped one he simply added 4 200 ohms resistors one each of the "legs" of the H? Does this guarantee 50 ohms? Then how did he do it to the radius-shaped one?

EDIT2:

Original article: Analysis of Electromagnetic Emission from PCB by using a Near-Field Probe

\$\endgroup\$
  • \$\begingroup\$ Can you provide a link to the article you're referring to? From what you wrote, I'm not clear on what you're describing with the "L", "H", and "radius" shapes. \$\endgroup\$ – The Photon Oct 6 '13 at 23:59
  • \$\begingroup\$ Also, why did you specify these shapes are "not corroded".? Is there some reason we would normally expect to use corroded materials to build these? \$\endgroup\$ – The Photon Oct 7 '13 at 0:03
  • \$\begingroup\$ I was just trying to be as specific as possible. I just put the information out. Maybe someone thinks it's necessary information \$\endgroup\$ – morcillo Oct 7 '13 at 12:58
  • \$\begingroup\$ To view the article you must be allowed to read IEEE articles. As it is a payed site, I thought it would be better if I explained it. \$\endgroup\$ – morcillo Oct 7 '13 at 13:00
  • 1
    \$\begingroup\$ Some of us do have access to IEEE articles and it would help us to see the original article. For everyone else, it might be more clear if you quote the key parts of the article (just enough to explain your question). \$\endgroup\$ – The Photon Oct 7 '13 at 15:55
5
\$\begingroup\$

When we talk about the "impedance" of a pcb trace, we are talking about the characteristic impedance of a uniform transmission line.

enter image description here

The characteristic impedance depends mainly on H, W, and \$\epsilon_r\$ in the figure above. To get a 50 Ohm characteristic impedance, you just have to define your trace width in the proper proportion to the H of your dielectric (with some small adjustments for Tmet and dielectric and conductor loss terms). As other answers have said, there's numerous online calculators and high priced tools available to help you do that.

The characteristic impedance doesn't depend on any features that change along the "L" direction. For example the different patterns in your example ("H", "L", and so on) don't change the characteristic impedance of the traces that make up the patterns. However, bends, curves, or branches in the lines do create discontinuities and regions where the characteristic impedance doesn't accurately model the transmission line behavior, as other answers have discussed.

Now, about that paper

For the record, the paper you referred to is

P. Sujintanarat, et al., "Analysis of Electromagnetic Emission from PCB by using a Near-Field Probe", International Symposium on Communications and Information Technologies (ISCIT), 2006.

It appears that they are just using the different patterns of microstrip traces to produce fields that they can measure using the probe that they developed and which they are reporting on. It is not obvious that they had to use 50 Ohm microstrip, but it may have been convenient to do so because that would minimize back reflection where their coax line feeds the the test patch and maximize the signals seen on the samples.

They would have seen very similar results if they had used 40-Ohm traces, 60-Ohm traces, or even 100-Ohm traces. However, they had to pick some uniform characteristic impedance, and 50 Ohms is convenient, and less likely to confuse the pcb shop that built their test patches.

Incidentally, I was somewhat disappointed in the paper, because they started out saying that their goal is to have a low-cost test method that helps to predict the results of radiated emissions tests for EMC. However, they only compare their measurements to simulations; it doesn't appear that they ever compared their probe results with actual EMC measurements, so there's no demonstration that they accomplished what they set out to do.

Did he just add termination resistors in specific places?

It doesn't appear that they terminated the traces on their test patches, but they didn't need to for their purposes.

The signal comes in from the coax feed. It reflects off the various feature and the unterminated ends of the transmission lines. And it creates a field whether they terminate the traces or not.

Since they're only interested in whether they can measure that field, it's not important to their results to correctly terminate the transmission lines.

\$\endgroup\$
  • \$\begingroup\$ Incredible answer. Thank you so much, this helped me a lot since I'm just starting into this field. \$\endgroup\$ – morcillo Oct 9 '13 at 11:26
3
\$\begingroup\$

Using formulas to calculate strip/mictrostrip impedance's used to be good enough, but with modern PCB technology, this does NOT work well. On a typical high speed multilayer board you may have something around 100um trace width and 100um dielectric thickness. This puts you outside of the range where the approximations in any of the formulas I have seen works well.

The tool of choice is a so called field solver. You find those embedded in tools like Hyperlynx, SigXplorer, ADS etc. Costly tools. The only free tool I have found is called TNT and is developed by an American university.

If you want to look at more complex structures (not sure why?) you would need to look for a 3D tool like Ansoft HFFS or similar.

As for measurement (if that is what you are after?) the tool of choice is called a TDR. Also a quite costly tool. TDR's are available from companies like Agilent, Tektronix, LeCroy etc. (Look for "impedance test trace measurement" on YouTube to see how that works).

\$\endgroup\$
  • \$\begingroup\$ @RolfOstergard About the more complex structures, the only reason I want to know how to calculate it by hand is because that guy did it. He only had a multilayer board and the only thing he did was to put the microstrips there just to measure. I got really curious as to how he did it, since he didn't use any tool for calculating the impedance \$\endgroup\$ – morcillo Oct 6 '13 at 14:44
  • \$\begingroup\$ Are you looking at lower frequencies where this can be considered a lumped element? In that case you may be able to find L and C for the complete structures by hand. Divide the two and take the root to find impedance. \$\endgroup\$ – Rolf Ostergaard Oct 6 '13 at 18:44
  • \$\begingroup\$ Ok I'll try to do that. The problem is that this article was poorly written. There were many things missing: equations, precise photos, etc. And alos the english in it was horrible. I'm not saying that my english is that good, but it's certainly better than that. Maybe I'm so obsessed with knowing that because it was missing so much info. \$\endgroup\$ – morcillo Oct 6 '13 at 20:55
  • \$\begingroup\$ @RolfOstergaard, There are still plenty of applications where calculators can be used (and also there are better and worse calculators out there). OP's application probably doesn't involve 0.1 mm dimensions since (AFAICT) he's asking about something that wouldn't require a multilayer stack-up. \$\endgroup\$ – The Photon Oct 7 '13 at 15:54
  • \$\begingroup\$ You might also mention Polar among the simulation tools...It seems to be the de facto standard for doing this work at board houses, where they just want to calculate geometry vs characteristic impedance and they don't need any of the other features of tools like HyperLynx, ADS, etc. \$\endgroup\$ – The Photon Oct 7 '13 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.