[TowerTalk] Voltage and SWR - an ignorant question

jimlux jimlux at earthlink.net
Fri Feb 1 13:20:37 EST 2019


On 2/1/19 8:57 AM, john at 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
> Previous message (by thread): [TowerTalk] Voltage and SWR - an ignorant
> question
> Messages sorted by: [ date ] [ thread ] [ subject ] [ author ]
> 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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