On 11/25/2015 11:34 AM, Jim Lux wrote:
On 11/25/15 11:06 AM, Michael Tope wrote:
On 11/22/2015 11:26 AM, Jim Lux wrote:
On 11/22/15 9:53 AM, Steve Hunt wrote:
Unfortunately that test doesn't subject the balun to the maximum CM
stress it might experience in a typical application - at worst it
subjects the balun to a CM voltage equal to the full differential-mode
voltage at the 200 Ohm point. However in an OCFD, for example, the CM
voltage could easily be as much as four times the differential mode
voltage appearing at the 200 Ohm feedpoint.
The reason is that the impedance looking into the two sides of the
dipole are individually reactive - capacitive on the short side and
inductive on the long side - even though the "composite" impedance at
the feedpoint is purely resistive. And those reactive paths can cause
the feedpoint to float to a very high CM voltage.
Steve G3TXQ
I assume that one could measure "withstand voltage" and "withstand
current".. so I suspect that the question isn't about "breakdown", but
rather "thermal power handling"
So the question is really sort of two parts:
1) what's the loss in the balun (in whatever configuration)
2) Where is that heat generated, and does it get dissipated adequately
And, then, providing way for a user to say "in configuration X
(e.g.OCF dipole) this is the loss".
The symmetric back to back scheme deals with the dissipation, mostly,
I assume from resistive losses in the coax. With symmetry, I'd assume
that the flux in the core is fairly small.
It should be possible to figure out a test fixture which puts a lot of
asymmetry in the system. Whether it's a realistic representation of
an actual antenna probably isn't as important as whether it's a good
way to measure the thermal handling.
What about driving the balanced side of the balun with an unbalanced
input: treat it like a transformer, drive one terminal, ground the
other, load the unbalanced port with something suitable (which
probably isn't 50 ohms).
To take into account the feedpoint imbalance alluded to by G3TXQ, I
suppose you could insert the lumped equivalent series Xc of the short
leg of the antenna (from a NEC simulation) in series with one leg of the
back-to-back balun connection and then put the lumped equivalent Xl of
the long leg of the antenna in series with the other leg of the
back-to-back balun connection. You might need to do some analysis to
determine if the lumped inductor Q contributed significant loss (or
check for inductor heating during power testing), but otherwise I don't
see why this wouldn't be would be a pretty good simulation of the common
mode effects of the unbalanced feed. You might still want to alternately
short either leg of the secondary (i.e. 200 ohm side) of the 2nd balun
to ground in order to force the common mode voltage at secondary of the
1st balun to its respective maximums.
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
The reactances would both be in series with the load resistance, so by
including them in the loop, I think the common-mode voltage at the
output of each balun secondary leg would be highest since in the
worst-case (i.e. grounding one of the reactive elements where it
connects to the 200 ohm load resistor) it would be the load current x
the series impedance. In the worst case the series impedance would be
the load resistance (200 ohms) plus the load reactance (j300 ohms). If
that doesn't burn it up with rated power, then I think you have yourself
a pretty good OCF antenna balun.
73, Mike W4EF
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