Heat Cantenna

[Stephen Tetorka]

Other factor that diode is not linear, so I'm told.

Steve
WA2TAK


Glen Zook <gzook at YAHOO.COM> wrote:

>Remembering my Algebra!
>
>P = VI
>
>P = (I^2)(R)
>
>V = IR
>
>I = V/R
>
>P = I x I x R
>
>P = (V/R)(V/R)(R)
>
>Therefore P = (V)(V)/R  = (V^2)/(R)
>
>Or, you square the voltage and then divide by the
>resistance of the Heath Cantenna.  Now, this should be
>50 ohms, but the resistors do drift and you should
>measure the exact resistance with a suitable meter.
>
>If I am not doing my Algebra correctly, someone please
>tell me.
>
>Glen, K9STH
>
>
>--- James Sutherin <JSutherin at FBAINC.COM> wrote:
>
>Can anyone pass on to me what the DC output voltage to
>power is for the Cantenna dummy load?
>
>=====
>Glen, K9STH
>
>Web sites
>
>http://home.comcast.net/~k9sth
>http://home.comcast.net/~zcomco
>
>
>
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