10:1 SWR implies that the impedance is 500 ohms resistive at the maximum point on the line.

E = sqrt(P*R) = sqrt (500w * 500ohms) = 500 Vrms, which is 707 Vpk.

So if the UHF connector happened to be located at a voltage maximum on the line it would see more than its rated 500V peak.

Alan N1AL


On 02/24/2018 05:11 PM, Walter Underwood wrote:
In this case, the lightning protection is on the antenna side of the matching 
unit, so it isn’t at 1:1 SWR or 50 Ohms. At 10:1 VSWR, we would see 500 V peak 
at 500 W.

wunder
K6WRU
Walter Underwood
CM87wj
http://observer.wunderwood.org/ (my blog)

On Feb 24, 2018, at 4:46 PM, Bob McGraw K4TAX <[email protected]> wrote:

Based on the specification of 500 volts peak,    5KW at 50 ohms = 500 volts.  
It would seem the PL-259 is adequate for ham radio applications per 
specification to 0.3 GHZ or 300 MHz without issues. Of course one is dealing 
with 50 ohms, which implies essentially a 1:1 SWR.

73

Bob, K4TAX

On 2/24/2018 4:53 PM, [email protected] wrote:
Really?  An Amphenol UHF connector only has a 500v rating.

John KK9A


Bill Johnson k9yeq wrote

1000v to me could be too low for a 500W amp.  You can for the time being
leave the plug out, just disconnect when you are not using your equipment.
I use a remote disconnect to remove the antennas at my remote grounding
panel.

73,
Bill
K9YEQ

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