Pete, N4ZR wrote:
My shunt-fed tower requires an Omega
match in order to get a good
> feedpoint SWR. My current Omega
match has 3000-volt air variables in
> both positions. I
understand the calculation for the voltage across a
> single
series capacitor, but not how to calculate it for this matching
>
system. I have never run over 200 watts to this antenna, and am
>
wondering what will happen with QRO. Can someone explain the calculation?
>
> 73, Pete N4ZR
Since I have seen no
other answer on the reflector, I will give it a shot and let others
comment on what I think is the answer.
First let us assume that one
is running 1500 watts at 50 ohms on 1830 KHz
Since Power = I squared
R or I x I x R, I squared = P/R
the square root of (1500/50) is 5.48
amps.
I will use 5.5 amps for the rest of this.
For the Shunt
Capacitor:
The RMS voltage V = I x R, so Vrms = 5.5 x 50 = 275
volts.
The peak voltage (assuming a sine wave) is about 1.42 x Vrms =
390.5 volts peak.
This has to be adjusted upwards for VSWR, but with
a VSWR of 2.5 or less a 1000 volt capacitor should be sufficient for the
shunt capacitor between the feed point and ground.
For the
series capacitor from the feed point to the bottom of the omega match wire
or tube,
The voltage across the series capacitor depends on the
reactance (Xc) times the current.
Using 200 pfd as the value of the
series capacitor,
Xc = -435j ohms. from the standard formula,
Xc = -1/(2 x Pi x F x C)
where F is frequency in cycles or Hertz and
C = capacitance in farads
The voltage across the series
capacitor is then Vcs = 5.5 amps x 435 ohms =
2391 volts rms and 3395
peak volts, at a VSWR of 1.0
Multiply the peak value by
the VSWR to get a recommended minimum voltage rating
for the series
capacitor. In this case I would use a 10 KV vacuum capacitor, since
a 2:1 VSWR gives almost 6800 volts across the capacitor.
I hope
this helps.
73 to all
George K8GG
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