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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