[Boatanchors] 813 grid to filament short

[rbethman]

Every power supply I see and have used plugs into the wall.

Therefore it is NOT DC it is AC.  The equipment converts AC to DC.  Even 
100% efficiency can not create more than it consumes.

Voltmeter on 240VAC, Clamp-on ammeter on one leg shows amps.

P = E X I

[ Have I lost you? ]

You can not produce more power than you consume.  Physics dictates this, 
along with a few other specialties.  Even Electrical Engineering states 
the same.

So there is NO possible way to get 1500W from 562.5W.  I don't care what 
you use in terms of fantasy land assumptions of 100% efficiency.

Every transformer mounted on a pad or a pole has an impedance stamped on 
its dataplate.

This dictates loss.

Wikipedia is notorious for erroneous postings/listings.

May just as well quote Facebook.  Try Terman's!

N0DGN



On 10/1/2015 1:52 PM, Michael Clarson wrote:
> Yep. 562.5 watts DC produces 1500w PEP. Defies my common sense filter. 
> I find the following two statements from the Wikipedia page on PEP 
> interesting:
> #1: "Assuming linear, perfectly symmetrical, 100% modulation of a 
> carrier, PEP output of an AM 
> <https://en.wikipedia.org/wiki/Amplitude_modulation> transmitter is 
> four times its carrier 
> <https://en.wikipedia.org/wiki/Carrier_wave> PEP."  There you go -- 
> its 1500 watts PEP by definition!
>
> and, #2: "PEP bears no particular ratio or mathematical relationship 
> to longer-term average power in distorted envelopes, such as a CW 
> <https://en.wikipedia.org/wiki/Continuous_wave> waveform with power 
> overshoot, or with complex amplitude modulated 
> <https://en.wikipedia.org/wiki/Amplitude_modulation> waveforms, such 
> as SSB or AM voice transmissions." which to me means we mortals can't 
> measure it.
>
> On Thu, Oct 1, 2015 at 1:00 PM, rbethman <rbethman at comcast.net 
> <mailto:rbethman at comcast.net>> wrote:
>
>     First issue - power supply draw 562.5Watts - yet output is 1500W?
>
>     So how do you get 1500W?  You have extra power coming from
>     somewhere else?
>
>     Reality is fairly basic.  You never get something for nothing.
>
>     Evert time PEP comes up, all math and reality seems to go away.
>
>     The FCC has NOT figured how to measure PEP in the field.
>
>     None of us has NIST calibrated equipment unless you happen to work
>     there.
>
>     N0DGN
>
>
>     On 10/1/2015 12:09 PM, Michael Clarson 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
>