[TowerTalk] balun testing

[Steve Hunt]

You must take account of the reactive components. If you have only a 
resistive load, no matter how much you unbalance it you can never 
generate a CM voltage across the balun higher than the differential-mode 
output voltage.

With the reactive components representative of an OCFD you get a 
"voltage multiplication" effect which can generate significantly higher 
voltages. Taking your impedance values:

1kW into the 204.5 Ohm load (8.5+196) would mean a differential mode 
balun output voltage of 452v rms. But with the reactive components 
present, the feedpoint would float upwards to 669v rms.

That higher voltage represents a 2.2 times higher balun power 
dissipation than you could achieve with a completely unbalanced 200 Ohm 
resistive load. More extreme degrees of OCFD offset - for example the 
20%/80% that is sometimes recommended - result in even higher voltages.

All of this is calculable. If you know the unbalanced load impedances, 
you could test using a purely restive load but then make the necessary 
corrections to allow for the unbalanced case.

Steve G3TXQ





On 25/11/2015 19:34, Jim Lux wrote:
>
> I wonder if you even need the reactive component.  What about a 200 
> ohm resistor on one side and a short on the other?
>
> I suppose with reactive components one can get unbalanced circulating 
> currents that are higher..
>
> I happen to have some NEC models here over a wide band. Let's consider 
> a 6 meter long dipole, but with 2 meters on one side and 4 on the other.
>
> the short side is 13-205j (roughly)
> The long side is 130+210j (roughly)
>
>
> Shifting the feed over a bit to get 200 ohms..
> short side 8.5-296j
> long side 196+295j
>
>
>
>
>
>
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