RF cable model

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New for November 2020! Here is a simple model that takes into account the loss of coax, the loss of connectors, and their VSWR interactions. We will show you how to specify a 40 GHz, 24-inch cable for laboratory use. Note that for this exercise we did not consider the various options for dielectric, we just assumed DK of 2.2, consistent with PTFE.

A bit of nomenclature: when we refer to "coax" we are talking about just the cable material. When we say "cable" we mean the assembly of the coax with connectors on either end.

In the simplest form of analysis, the composite VSWR created from cascading two components would be found by multiplying them together. Therefore, two connectors with 1.3:1 mismatch would have 1.69:1 worst-case mismatch. But in cable design, the composite VSWR will be less than the worst case, as you will see.

Connector model

Most (but not all) connector data sheets include VSWR and loss specs over frequency. The loss is often given as a function of SQRT of frequency, as it is mostly metal (skin depth) loss. In our model, we use the _FREQ variable, which Microwave Office sets to the frequency in Hertz. We converted it to GHz, because, why not? The loss coefficient of 0.04 is in units of dB/SQRT(GHz). If you have a calculator, you can see that at 40 GHz the cable should have 0.253 dB loss (0.04 x SQRT(40). If you made that calculation in your head, please send us your resume.


Below is a plot of connector loss. Recall that it should have 0.253 dB loss at 40 GHz... But what's this? it shows 0.3265 dB? It's called mismatch loss.  It's not really real, you can get some or all of it back when you cascade components. Or make it worse! 

Now let's relook at the loss, with the inductance set to zero...

Voila, connector loss is indeed 0.253 dB.

Coax model

Now, let's make a coax model using the TLINP element. Cable vendors often specify attenuation in dB/100 foot at discrete frequencies, as if anyone would want a 100-foot cable.  We are interested in 40 GHz so we grabbed that point and converted it to dB/foot. We converted it to "real" units of dB/meter, which TLINP will scale as SQRT frequency. Oops we gave away the manufacturer in the element ID....

Not that coax cannot be assumed to be perfect 50 ohms, it will have a manufacturing tolerance on dimensions. This is taken into account by the VSWR variable, which we converted to its effect on Z0.


Below is the attenuation curve, at perfect 50 ohms.

Full cable model

Now let's put the cable model together. We used two of the connectors, plus the coax model, and we specified the coax to have 1.15:1 VSWR. We decided to make our cable 24 inches long (not including the connectors), for use in a laboratory environment, such as for a VNA.

And finally, here is the prediction of our full cable model. The VSWR is 1.4:1 or less, even though the two connectors put together would result in 1.69:1 worst-case.  This is because the loss of the cable pads down the connector mismatches. Note that if you added the losses of the connectors (2.53dB x 2) and the cable (2.73 dB/ft x 2 feet) you would predicted 5.966 dB and have slightly underestimated the maximum loss.

If we were going to generate a spec for this cable, we would cut the manufacturer some slack and allow it to have 6.5 dB loss at 40 GHz, and 1.5:1 maximum VSWR. There will be nothing perfect about the solder joints or the crimp joints holding the cable together.

What can you learn from this exercise? For one, longer cables will have lower VSWR than shorter cables, because the loss of the cables pads down the VSWR interaction of the connectors. 



Author : Unknown Editor