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Old March 23rd 07, 05:21 PM posted to rec.radio.amateur.antenna
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Default Revisiting the Power Explanation

On Mar 22, 10:43 pm, Richard Clark wrote:
On 22 Mar 2007 20:08:59 -0700, "Dr. Honeydew"
wrote:

What sort of output impedance does this generator have?


With the "accurate RF power meter" determination of 1W, it appears to
be a precision source (of 1W). The output impedance IS the 50 Ohm
resistance you applied to it. This loading at the output terminals is
a conventional usage of a precision source.

73's
Richard Clark, KB7QHC



Thank you, Mr. Clark. But I'm having trouble reconciling what you
wrote with what Mr. Harrison posted about the output impedance being
equal to the conjugate of the load which dissipates the most power.
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so
we don't see how your answer and Mr. Harrison's posting can both be
correct.

From the labs,

Dr. Honeydew

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Old March 23rd 07, 06:15 PM posted to rec.radio.amateur.antenna
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Default Revisiting the Power Explanation

Dr. Honeydew wrote:
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so
we don't see how your answer and Mr. Harrison's posting can both be
correct.


It appears that the internal resistance of the
Thevenin equivalent circuit was chosen to be
~1 ohm. Maximum power transfer of ~13 watts
should occur when the load is ~1 ohm and the
current is ~3.6 amps with ~3.6 volts across
the 1 ohm load and across the ~1 ohm generator
impedance. How one gets 13 watts out of a source
rated for one watt is a non-linear physical
design problem, not a linear theory problem.
--
73, Cecil http://www.w5dxp.com
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Old March 24th 07, 12:21 AM posted to rec.radio.amateur.antenna
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Default Revisiting the Power Explanation

On 23 Mar 2007 09:21:07 -0700, "Dr. Honeydew"
wrote:

Thank you, Mr. Clark. But I'm having trouble reconciling what you
wrote with what Mr. Harrison posted about the output impedance being
equal to the conjugate of the load which dissipates the most power.
Clearly the 40 ohm load dissipates more power than the 50 ohm load, so
we don't see how your answer and Mr. Harrison's posting can both be
correct.


It is a problem of apples and oranges. Your source is not
representative of either a commercial transmitter, nor a amateur
transmitter. It is a precision source. It has no need to exhibit
impedance matching, and is thus not designed to perform that purpose.
It is happy to run quite inefficiently as efficiency is a discardable
quality. Rather, it is incumbent upon you to enforce matching by
careful load selection. Otherwise, its utility is wasted and makes it
a rather expensive choice for a general purpose signal generator.

73's
Richard Clark, KB7QHC
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