# How to design constant impedance tracks on PCB?

I have designed PCB's in the past, but need to draw a constant 50 ohm impedance track on the project I'm planning.

I know the calculations for (for example) a coplanar wave and how to get a straight PCB track around 50 ohms. (I can recommend the free Saturn PCB design software).

But how do I create a track that has 45/90 degree angles or something like this while keeping impedance fairly constant?

I am using Diptrace, which does not support this by itself. The PCB above however is (at least) 25 years old, so I cannot imagine you absolutely need state-of-the-art PCB layout software...

• Are sure that it operates as a coplanar and not as a microstrip line? That means there is no ground layer on the other side of the substrate. I wonder because I would mainly use coplanar when vias are not an option and I think there are vias on the board. With 'operates' I mean the actual transmission line field mode. Microstrip lines often also have ground layers close the the line on the top layer (so they may look kind of a coplanar line from one side), but still operate as microstrip ines. And the answer depends on the transmission line mode. – MagnusO_O Aug 28 '19 at 17:26
• No via's but plenty of through hole, which amounts to the same I guess. Why mainly coplanar when vias are not an option? Clearly, I have no experience here. – Dieter Vansteenwegen ON4DD Aug 29 '19 at 9:14
• What frequency are we talking about here? The requirements for 50 MHz and 50 GHz are very much different. – Phil Frost - W8II Aug 29 '19 at 15:39
• @Dieter Vansteenwegen Coplanar is much more complex (dealding with add. unwanted modes) than microstrip (has some unwanted modes as well, but these are less easily excited). A good reason to go for coplanar is when you have no vias options (no process) to go to a GND on the reverse side, e.g. at on-wafer structures. There you can contact the GND on the sides instead with a coplanar line. – MagnusO_O Aug 29 '19 at 16:23
• @Phil, since this is my first time, I would be happy to get a sort-of working 145/445MHz diplexer or something using one of the atmel mcus with embedded radio in them. So absolutely below 1.5GHz – Dieter Vansteenwegen ON4DD Aug 29 '19 at 16:36

Besides not knowing about the dimensions, substrate relative permittivity $$e_r$$ and backside ground yes/no my guess from the picture is that the transmission line is a microstrip line and not a coplanar line.

Front sides of the microstrip line printed circuit boards often also have ground metal, but that's for shielding or just practical etching reasons and does not impact the transversal electromagnetic wave mode.
However just from the front side these may be mistaken as coplanar lines. I would always check the backside for that reason (still there is no definite prove, but another clue).

Anyway for both transmission lines at bends you get a parallel capacitance to ground due to the fringing fields causes by the sharp inner and outer edges.
That capacitance can to be compensated by adding inductance:

The compensating inductance can be achieved by a narrowed line, e.g. phase cutting the outer edge.

For microstrip lines you can either simulate the compensated bend with e.g. sonnet lite or calculate it with an approximation formula as given by e.g. Microwaves101 Mitered bends:

$$D = W* \sqrt{2}$$ (the diagonal of a "square" miter)

$$X= D* (0.52 + 0.65 e^{-1.35 * (W/H)})$$

$$A = ( X- D/2) * \sqrt{2}$$

W, H being width and height of the microstrip line.

For actual coplanar lines the compensation is more complex as you have to make sure the slot mode is not excited. Here are two links describing such an option:

Coplanar waveguide bend with radial compensation

For these I presume EM simulation is mandatory and you won't get good approximation formulas.

Finally, when there is enough area you can also form a wide radius circle segment to smoothly bend without having to compensate - for both lines.
But that is often prevented due to cost and losses by the additional line length.

• The backside is one big ground plane. Whether that makes it a coplanar or microstrip line, I don't know. I don't have any information on the relative permittivity of the dielectric, but can measure the width of the conductor (3.2mm), gap (2.04mm) and the pcb thickness (1.45mm) figuring it is probably FR4. As a coplanar trace this gives a result of 44 Ohm. Microstrip with the same width and thickness (neglecting the ground plane around it) results in 45 Ohms, so it seems the plane around the trace doesn't matter. – Dieter Vansteenwegen ON4DD Aug 29 '19 at 9:11
• I have no access to the IEEE paper, but will read the other. Also, I'll look at Sonnet Lite. Still leave me to wonder how this was done 25 years ago. – Dieter Vansteenwegen ON4DD Aug 29 '19 at 9:12

At 145/445MHz, things are not so hard. As a rule of thumb, anything under 1/10th of a wavelength doesn't even count as a transmission line. At 445MHz that's about 6.7cm. At 145 MHz, 21cm. Anything shorter than this length and you probably don't even need constant impedance traces.

If you do need such traces (or you just want to do things "right"), you can make mitered bends, like MagnusO_O suggests. But if your design software doesn't support that, you'd probably do just fine making "gentle" bends, with two 45-degree turns. At these low (by today's standards) frequencies you probably won't notice the difference.

• OK, that makes sense. Still, it would be nice to know how it "should" be done. Just because it works without too much effort at these frequencies, doesn't mean it's not worth learning... – Dieter Vansteenwegen ON4DD Aug 29 '19 at 19:15
• Additionally, even though I have things like a vna, I'm still really just beginning to learn how to make proper measurements. So it would be nice to have a pcb that is reasonably "as it should" and start from there – Dieter Vansteenwegen ON4DD Aug 29 '19 at 19:19