On 2/1/19 8:57 AM, john@kk9a.com wrote:
I thought that 1000 watts into a perfect 50 ohm load is 220V rms. I would
expect 4:1 SWR to be higher than your calculated 450v. 450v should not
damage N4ZR's antenna system, I would look at the tuner first.
I calculated RMS, not peak. P = E^2/R so E = sqrt(P*R) = sqrt(200 *
1000) = 447.2 Vrms
*1.4 for 630V peak.
That's for 1000W into 200 Ohms.. That probably isn't the voltage though.
It really depends on how long the transmission line is, what the
matching network at the amp is doing, etc. - between the transmission
line and the matching network you could be setting up a resonant circuit
with moderately high Q, and the voltage could climb quite high. See, for
instance, Tesla coils.
These kinds of voltages might not cause "damage" per se, but if it arced
over, it would cause the amp to detect the infinite SWR and shut down.
VSWR = Vhi/Vlo along the line. In a mismatched line, you can think of it
as a forward wave combined with a reflected wave, so where the two are
in phase you get Vhi and where opposite you get Vlo. So Vhi = Vf+Vr and
Vlo = Vf-Vr.
So doing a bit of algebra for VSWR =4 : Vf+Vr = 4*(Vf-Vr)
-> 5 *Vr = 3 * Vf -> Vr = 3/5 Vf.
(that is, the reflection coefficient is 0.6 and a RL of -4.4 dB)
Vf could be the "nominal matched load voltage" (223Vrms, as you
indicated) So Vhi = 1.6*223V = 360V and Vlo = 0.4 * 223V = 90V
But that's a sort of contrived example with Pfwd = 1000 and Prev=360.
John KK9A
Fri Feb 1 11:32:29 EST 2019
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On 2/1/19 6:41 AM, N4ZR wrote:
Last week I was in the 160 CW contest using a jury rigged antenna - my
40M parasitic sloper fed with open wire line from a 4:1 allegedly 5 KW
balun via a long run of Buryflex to the shack. Believe it or not, it
didn't work badly, but...
The swr at the amp's antenna tuner was about 4:1, but when I put 1500
watts on it at the end of the contest the amp would quickly fault
indicating an SWR of 20:1. My guess is that something was flashing
over. However, on 40 meters the antenna continues to operate and take
full power just fine. Its native SWR on that band is under 1.8:1.
So, my question, does higher SWR mean higher voltage peaks on the
feedline? Once the snow stops, I'm going to go looking, but thought any
advice I could get before that would help.
Yes, VSWR = Voltage Standing Wave Ratio = Vpeakhigh/Vpeaklow
so with 4:1, the highest voltage is 4 times the lowest. (neither would
be the voltage you'd get with 50 ohms, BTW)
Also, your tuning network at the amp would have changed the voltage.
Let's, for example, assume that the Z was 200 ohms instead of 50.
TO put 1kW into 200 ohms, the voltage has to be sqrt(1000 * 200) = about
450V.
it might be that you were seeing 12.5 ohms at the amp (also 4:1 to 50
ohms), in which case the current would be high, but the voltage
relatively low. Somewhere else along the line, though, the voltage
would be higher, peaking at the 450V - assuming the line is at least 1/4
wavelength long.
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