# nanoVNA – measure Transmission Loss – example 5

This article is demonstration of measurement of Transmission Loss in a section of two wire transmission line embedded in a common mode choke. The scenario is based on an online article  MEASURING DM ATTENUATION of YOUR CMC USING THE NANOVNA AND NANOVNA SAVER.

The reference article publishes measured attenuation or loss being -1.45dB @ 28.4MHz. Of course, the -ve value hints that the author is lost in hamdom where all losses MUST be -ve dB..

The meaning of loss in a generic sense (ie without further qualification) is $$loss=\frac{Power_{in}}{Power_{out}}$$ and can be expressed in dB as $$loss_{dB}=10 log_{10}(loss)$$.

Some might interpret the result to imply that $$(1-10^{\frac{-loss}{10}})*100=28 \%$$ of input power is converted to heat in the choke.

The result given (and corrected) as 1.45dB was taken simply from the nanoVNA $$|s21|$$ result, and so it is actually InsertionLoss, not simply Loss.

What is the difference?

Above is a Simsmith model of a similar scenario. I have calculated |s21| and |s11|, and $$TL=10 log_{10}\frac{Power_{in}}{Power_{out}}$$. PVC insulation was assumed.

The type of source specified would result in 1W or 0dBW in a 50+j0Ω load, this is the AvailablePower.

So, firstly the power in the load is -1.39dBW, that is 1.39dB less than the AvailablePower from the source (ie into a matched load), so the InsertionLoss=1.39dB.

In fact the input power to T1 is not 1W or 0dBW, it is -1.28dBW. So, we can find the value of Loss (or TransmissionLoss for clarity) which is the ratio $$\frac{Power_{in}}{Power_{out}}$$ by subtracting the dBW figures to obtain $$Loss=Power_{dBWin}-Power_{dBWout}=-1.29 – -1.39=0.10dB$$. This is affected by rounding error, the more exact calculation given at TL of 0.09561dB is more correct.

So, how much input power is converted to heat?  Easy, $$(1-10^{\frac{-TL}{10}})*100=2.1 \%$$ of PowerIn.

## How do we apply this to a ham transmitter scenario?

The first thing to note is that the VNA is intended to be:

• a near perfect 50+j0Ω Thevinin source; and
• a near perfect 50+j0Ω load.

Neither of those apply to most ham transmitter scenarios, especially one that uses 1m or so of twisted #14 insulated wires as in the reference article.

Such a choke might well be used with an ATU, and if we assume that it was adjacent to the choke, a good ATU would deliver most of the rated transmitter to the choke, ie for a 100W transmitter you would expect that PowerIn is almost 100W.

Now the TransmissionLoss above was calculated under a specific standing wave scenario, Zload=50+j0Ω. The actual TransmissionLoss will depend on the distribution of voltage and current on the choke’s two wire line section, so the figure calculated for the VNA scenario doesn’t apply and is not simply extended to a real transmitter scenario.

## Conclusions

The Transmission Loss of a line section may not be directly given by any measures displayed by a VNA, it may take some interpretation and some accounting for elements that can be measured.

The discovered TransmissionLoss of the line section in a common mode choke applies to the exact standing wave scenario and is not simply applied to other (practical) scenarios.