[Boatanchors] 813 grid to filament short

[W2HX]

Well you are correct. But many of us will set up into a dummy load. You won't get any more power into a non 50J0 antenna than you will into the DL. So tune up into the DL and you'll be fine with the antenna.

 
73 Eugene W2HX
 


-----Original Message-----
From: boatanchors-bounces at mailman.qth.net [mailto:boatanchors-bounces at mailman.qth.net] On Behalf Of Rob Atkinson
Sent: Thursday, October 01, 2015 1:00 PM
To: Michael Clarson
Cc: Boat Anchors List
Subject: Re: [Boatanchors] 813 grid to filament short

Forgot to mention another reason why this notion of precision measurement of power is ludicrous--who delivers transmitter power into a 50J0 load when operating.

Run your rig, whatever carrier it puts out--fully modulate it and forget it.  Leave the illusion of "375 watts" power nonsense to phony operators who think they can measure it.  I usually tell folks I run
300 watts give or take 25 to 50.  That's into a Bird dummy load with a thermocouple RF amp meter and Bird 43 and I interpolate.

73
Rob
K5UJ

On Thu, Oct 1, 2015 at 11:09 AM, Michael Clarson <wv2zow at gmail.com> wrote:
> On the PEP issue, here is a brain exercise. I know there is a 
> mathematical basis for PEP, and, that mathematically, 375 watts AM is 
> 1500 watts PEP. So, lets assume (to simplify the math) that our transmitters are 100% efficient.
> On CW, if we output 1500 watts, PEP is 1500 watts, and our 100% 
> efficient amplifier draws 1500 watts from the power supply. Makes 
> sense. Now, if we go to AM, and we put out 375 watts 100 % modulated, 
> the RF puts out 375 watts, and the modulator adds 187.5 watts. Total 
> draw from the power supply is
> 562.5 watts, yet produces 1500 watts PEP. The 1500 W PEP on CW 
> requires 1500 watt dissipation in a dummy load, yet the AM 1500 watt 
> PEP only requires a
> 562.5 watt load. Perhaps a single PEP value is not he best way to 
> specify transmitter power. --73, Mike, WV2ZOW
>
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