More on Q Sections

In my previous post about creating a phasing harness for stacking two WA5VJB yagis I mentioned the use of Q sections (that is, 1/4 wave length sections of transmission line) to transform the impedance to 50 Ohm – from down at 25 Ohm. What I'd like to do now is go over the how and why this works, and then provide the formula used to calculate what Q section you might need if you wish to setup an impedance transformer.

My reference for this is Chapter 24 (Transmission Lines) of the ARRL Antenna Book. Of specific interest is pg 24-13, Special Cases.

They key elements of theory that Q sections seem to rely on is that of Phase and Characteristic Impedance of transmission lines. Firstly, we know that transmission lines have a Characteristic Impedance (ie. RG-58 has 50 Ohm and RG-6 has 75 Ohm). The next piece of important information then is that in an AC current, the current (or voltage) alternates from 0 to a positive peak, then back to zero and to a negative peak, and finally finishing at zero again. This alternating of a current is it's phasing.

Now due to the relationship between current and resistance, as the variance in phase occurs so does characteristic impedance. The characteristic impedance will vary from minimum impedance (so when the current is at 0 – at 0 degrees and 180 degrees) up to maximum impedance (when the current is at peak – at 90 degrees and 270 degrees).

If we list this out in full we get:
0 current – 0 degrees – 0 wave length – minimum impedance
peak positive – 90 degrees – 1/4 wave length – maximum impedance
0 current – 180 degrees – 1/2 wave length – minimum impedance
peak negative – 270 degrees – 3/4 wave length – maximum impedance
0 current – 360 degrees – 1 wave length – minimum impedance

Now in relation to the Q section, we want maximum impedance so that it will have the full effect when utilised in a parallel configuration as needed for a phasing harness. So we can see above that we need it to be at peak current and this occurs when it is at either 90 degrees or 270 degrees of the phase. In turn we can see this only occurs at 1/4 or 3/4 wavelengths which lead us back to the notion of: A Q section being an "odd multiple" 1/4 wave length of transmission line.

The ARRL Antenna Book puts this as:
"[..] such a line will have a purely resistive input impedance when the termination is a pure resistance."

Above then, you have a basic outline of why a Q-section works. But how do you calculate what transmission line characteristic impedance you need to work with to setup an impedance transformer for your antenna?

The basic formula is:

Zi = Zo^2 / Zl

Zi = input impedance
Zo = output impedance
Zl = load impedance

This can be re-arranged such that:

Zo = sqrt(Zi x Zl)

Now if I apply that formula to the case where I want to stack two yagis with an impedance of 50, I first use Ohm's Law to figure out that each individual yagi needs to present an impedance of 100 Ohms. (Because putting 100 Ohms in parallel will provide an impedance of 50 Ohms – the target.) This means I have the following inputs:

Zi = 100 – basically the impedance presented at the input to each yagi (as stated above we need 100)
Zl = 50 – the impedance load as presented by the antenna

With these values, Zo will then be the value for the characteristic impedance of the transmission cable we use for our input transformer. So the algorithm will be:

Zo = sqrt(100 x 50)
Zo = sqrt(5000)
Zo = 70.71 Ohms

So there we have it, I will need a transmission line with an impedance of 70.71 Ohms. Now we know there isn't a transmission line readily available with that impedance, so we go with the closest match – 75 Ohm. We then cut that to an odd multiple of 1/4wl (keeping in mind it needs to be an _electrical length_) and we'll have our impedance transformer to present a load of ~100 Ohms when connected to our 50 Ohm antenna.

Hope that helps. It helped me once that all came together, so hopefully I've portrayed it correctly above – I'm actually off home sick today so my head might be a bit fuzzy.

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