Features of transmission line impedance calculation in Polar SI9000

Ukhin V.

Delta Design for many engineers is becoming the main tool for developing printed circuit boards. With the introduction of the module SimPCB it has become easier and more convenient to design devices with impedance control. Our team is developing this module and has achieved, as we believe, good results so far. You can read about them in the articles (link), or directly in the program Delta Design (link).

Stop!!! Where is here? Polar SI9000? – the impatient reader will ask. Don't rush, friend, the story is just beginning.

Not long ago, one of the users contacted us with the following problem: it is necessary to design a coplanar transmission line with a wave impedance of 75 Ohm on the RO4003C material with the following parameters: dielectric thickness of 0.203 mm, permittivity of 3.38. Figure 1 shows a fragment of the material specification.

Fig. 1. Fragment of the specification for the material RO4003C

IN SimPCB the following parameters of the coplanar line were obtained (KM-1N) (Fig. 2).

Fig. 2. Calculation result of the KM-1N line — Fig. 2. Result of line calculation **KM-1N**

The conductor width is 3.6 mm, the distance from the polygon to the signal conductor is 0.58 mm.

SimPCB is a very young application, so specialists often recheck the values obtained in it using other software tools. In this case, the engineer recalculated the structure in Polar SI9000 V22.04.00 (line Surface Coplanar Waveguide 1B). The result is shown in Figure 3.

Fig. 3. Result of calculation of the Surface Coplanar Waveguide 1B line — Fig. 3. Result of line calculation **Surface Coplanar Waveguide 1B**

The conductor width is 4.5 mm, the distance from the polygon to the signal conductor is 0.58 mm. The differences in geometry are significant. The difference in the conductor width is almost 1 mm! Who to believe? Polar SI9000 there is authority, but SimPCB just appeared on the market.

Of course, this appeal upset us a little. Did we really make a mistake somewhere?

If parameter W1 and W2 from Polar SI900 substitute in SimPCBthen we get the following impedance value (Fig. 4).

Rice. 4. Line impedance KM-1N — Rice. 4. Line impedance **KM-1N**

The wave resistances in the programs differ from each other by 3.3 Ohm (V SimPCB impedance is 71.8341 ohms, Polar SI9000 – 75.13 Ohm) or almost 4.5 percent. Some readers may say that there is nothing to worry about here, since the difference does not exceed the standard 10 percent. However, this was a fundamental point, since in the process of development SimPCB Our team compared the obtained values, including with Polar SI9000. The error did not exceed 2 percent. Figure 5 shows a fragment of our tests.

Polar SI9000 a serious program trusted by a large number of specialists from all over the world, so we started searching for an error in our solver. Having checked everything several times and found nothing, it was decided to involve an independent expert in the form of the program AnsysFigure 6 shows a fragment of the coplanar transmission line model and the impedance calculated in it.

Fig.6. Transmission line model and impedance value

So, we got three values of wave resistance:

Polar SI9000 – 75.13 Ohm;

SimPCB – 71.83 Ohm;

Ansys – 71.15 Ohm.

From the results obtained, our team concluded that with a high degree of probability, Polar SI9000 The calculation of this structure is carried out incorrectly.

This conclusion was not enough for us and we decided to continue the research, namely, to figure out where exactly it was wrong. Polar SI9000.

Computing module Polar SI9000as in SimPCBis based on the boundary element method. This information is available at websitePolar Instruments. With this method, the area in which the calculation is performed is specified (automatically or manually). Its overall dimensions must exceed the size of the structure (link). For example, for the above coplanar line, the region may look like this (Fig. 7).

Fig. 7. Transmission line and calculation domain

The structure is surrounded by air with a permittivity of one.

So how can we check this structure? Let's imagine that the material with thickness H1 is also air, then for any H1 and the constancy of the remaining parameters of the transmission line, the program should give the same values of wave resistance. Let's check this in SimPCB And Polar SI9000. Figure 8 shows the result of the calculations in SimPCB.

Fig. 8 Calculation result in SimPCB — Fig. 8 Calculation result in **SimPCB**

It can be seen that Z0 does not depend on H1, which is consistent with theory.

Below (Fig. 9) is the result obtained in Polar SI9000.

Fig. 9. Calculation result in Polar SI9000 — Fig. 9. Calculation result in **Polar SI9000**

There is no longer any doubt that in Polar SI9000 when calculating the structure Surface Coplanar Waveguide 1B there is an error, since Z0 depends on H1. At H1=0.103 mm, the impedance is 95.8 Ohm. With increasing dielectric thickness, the wave resistance decreases to 81.6 Ohm. Approximately this impedance value (81.6 Ohm) should be at any value of H1.

Remember that the wave resistance is calculated using the formula:

Where L – inductance of the transmission line per unit length, WITH – transmission line capacitance per unit length.

High impedance at H1=0.103 mm indicates that the structure has a primary parameter C smaller than, for example, at H1=1.003 mm. The capacity depends, among other things, on the overall dimensions of the support layer. That is, it can be assumed that the horizontal size of the calculated area of the structure Surface Coplanar Waveguide 1B varies depending on H1, and at low H1 only a small part of the conductors with zero potential enters the region. Below are explanatory figures 10 and 11.

Fig. 10. Calculation domain for small H1

Fig. 11. Computational domain for large H1

As H1 increases, the area of the polygon in the computational domain increases, which leads to an increase in capacitance and a decrease in inductance and impedance. If the theory is correct, then it turns out that for small values of H1 and for large values of W1, W2 or D1, the reference plane for the structure Surface Coplanar Waveguide 1B is represented in the form of a conductor, the width of which can be much smaller than the width of the signal.

It is quite easy to check this using only Polar SI9000since it contains the following transmission line model (Fig. 12).

Fig. 12 Surface Coplanar Strips 1B Transmission Line Model — Fig. 12 Transmission line model **Surface Coplanar Strips 1B**

Let's select the size G1=G2 such that the impedances of the structures Surface Coplanar Waveguide 1B And Surface Coplanar Strips 1B coincided. At the same time, all other parameters are the same in both transmission lines.

So, let H1=0.203 mm, Er1=1, W1=W2=4.5 mm, D1=0.58 mm, T1=0.018 mm. The calculation results are shown in Figures 13 and 14.

Fig. 13 Result of calculation of the Surface Coplanar Waveguide 1B structure — Fig. 13 Result of structure calculation **Surface Coplanar Waveguide 1B**

Fig. 14 Result of calculation of the structure Surface Coplanar Strips 1B — Fig. 14 Result of structure calculation **Surface Coplanar Strips 1B**

Parameter G1=G2=2.85 mm. It turns out that the horizontal size of the calculated area for the line Surface Coplanar Waveguide 1B is approximately 4.5 mm + 2*0.58 mm + 2*2.85 mm = 11.36 mm, and the polygon is represented as two conductors 2.85 mm wide. Let's reduce H1 to 0.103 mm and find the value G1=G2. Figures 15 and 16 show the result.

Fig. 15. Calculation of the Surface Coplanar Waveguide 1B structure — Fig. 15. Calculation of structure **Surface Coplanar Waveguide 1B**

Fig. 16. Calculation of the Surface Coplanar Strips 1B structure — Fig. 16. Calculation of structure **Surface Coplanar Strips 1B**

The width of the polygon became even smaller, since G1=G2=1.1 mm. At the same time, the size of the area is about 7.86 mm.

It is easy to see that increasing the dielectric thickness by 0.1 mm leads to an increase in the calculation area by approximately 1.75 mm. That is, if the dielectric is 0.303 mm thick, then the polygon size will be 4.6 mm on each side. Let's check this by performing the calculation for two structures. The result is shown in Figures 17 and 18

Fig. 17. Calculation of the Surface Coplanar Waveguide 1B structure — Fig. 17. Calculation of structure **Surface Coplanar Waveguide 1B**

Fig. 18. Calculation of the Surface Coplanar Strips 1B structure — Fig. 18. Calculation of structure **Surface Coplanar Strips 1B**

Our guess turned out to be correct. Graphic representation of the line Surface Coplanar Waveguide 1B for certain values of H1, D1, W1 and W2 does not correspond to the structure for which the calculation is performed. The figure assumes that the size of the polygon should exceed the width of the signal line. In practice, however, it turns out that this is not always the case and depends on the geometric dimensions of the cell.

The results of the research were sent to the user with the conclusion that he can safely trust the module to solve this problem. SimPCB.

What conclusions can be drawn?

Developers of computing programs need to at least familiarize users with the methods and principles of calculations, which will increase confidence in the product and allow specialists to avoid mistakes.
IN Polar SI9000 Coplanar structures must be calculated with caution.

Many issues can be resolved using feedback. In the presented case, the specialist received help from the developers SimPCB and confirmation of the quality of the purchased product.