"Dilon Earl" wrote in message
...
On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin"
sabinw@mwci-news wrote:
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?

That would depend on the output impedance of the transmitter. If it is 50
Ohms, all the reflected power would be absorbed by the transmitter. If it is
0 Ohms or its Norton equivalent, it is all reflected, and none is
dissipated.

Tam/WB2TT

On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru"
wrote:


"Dilon Earl" wrote in message
...
On Thu, 17 Jul 2003 09:06:32 -0500, "William E. Sabin"
sabinw@mwci-news wrote:
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?

That would depend on the output impedance of the transmitter. If it is 50
Ohms, all the reflected power would be absorbed by the transmitter. If it is
0 Ohms or its Norton equivalent, it is all reflected, and none is
dissipated.

Tam/WB2TT


Hi Tam,

This is painfully obvious. It is also painfully demonstrative. It
also appears to be painfully avoided in lieu of providing an actual
value (Punchinello seems to wholly ignore his own cries that lacking
numbers renders such whining as ignorance).

I've read for years that the common RF rig is NOT a 50Ohm source, and
absolutely none dare commit themselves to just what value it is (much
less offer their own measure). Being a physical reality, the rig must
present some real value, but vacuous theory seems to bar that
discussion.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru"
wrote:

I've read for years that the common RF rig is NOT a 50Ohm source, and
absolutely none dare commit themselves to just what value it is (much
less offer their own measure). Being a physical reality, the rig must
present some real value, but vacuous theory seems to bar that
discussion.

73's
Richard Clark, KB7QHC

Yeah, seems to be a deep dark secret. If you look at the specs of RF power
transistors, they will give the output impedance vs frequency - BUT you have
to look at the footnote. In virtually all cases what they mean is the
conjugate of the load impedance. It is the jX of the transistor (1/jY), in
parallel with
((VCC-Vsat)**2) /2P.

I have never gotten around to doing this, but I believe the data sheets for
tubes like the 811A and 813 do give the plate resistance, which should make
it possible to calculate the output impedance at the lower frequencies like
160m.

Tam/WB2TT

On Fri, 18 Jul 2003 16:01:43 -0400, "Tarmo Tammaru"
wrote:

Yeah, seems to be a deep dark secret. If you look at the specs of RF power
transistors, they will give the output impedance vs frequency - BUT you have
to look at the footnote. In virtually all cases what they mean is the
conjugate of the load impedance. It is the jX of the transistor (1/jY), in
parallel with
((VCC-Vsat)**2) /2P.

Hi Tam,

Motorola offers quite specific characteristics across frequency.
Reference MRF421, MRF433, MRF454 for examples of dirt ordinary power
transistors found in more than 20 years of transistorized Ham
transmitters. Take their own data, Z transform them through
transformers (not transducers) and you find 50Ohms without any more
sophisticated math than that required of the standard Technology
Certificate of training. Where does it go through after that? A low
pass filter designed for 50Ohms to an antenna jack specified to
deliver full power to a 50Ohm load.

What technical rebuttal do I hear in response to simple engineering
data? "It is impossible to determine the output Z of this source."
For some I can well imagine they do find it difficult....

73's
Richard Clark, KB7QHC

Actually, several people (W8JI among them) have measured the output
impedance of common amateur linear amplifiers by at least a couple of
methods. The most credible measurements show, interestingly, a value
very close to 50 ohms when the amplifier is adjusted for normal operation.

Of course, it doesn't really matter, but people continue to make a big
deal out of it.

Roy Lewallen, W7EL

Tarmo Tammaru wrote:
"Richard Clark" wrote in message
...

On Fri, 18 Jul 2003 09:28:07 -0400, "Tarmo Tammaru"
wrote:

I've read for years that the common RF rig is NOT a 50Ohm source, and
absolutely none dare commit themselves to just what value it is (much
less offer their own measure). Being a physical reality, the rig must
present some real value, but vacuous theory seems to bar that
discussion.

73's
Richard Clark, KB7QHC


Yeah, seems to be a deep dark secret. If you look at the specs of RF power
transistors, they will give the output impedance vs frequency - BUT you have
to look at the footnote. In virtually all cases what they mean is the
conjugate of the load impedance. It is the jX of the transistor (1/jY), in
parallel with
((VCC-Vsat)**2) /2P.

I have never gotten around to doing this, but I believe the data sheets for
tubes like the 811A and 813 do give the plate resistance, which should make
it possible to calculate the output impedance at the lower frequencies like
160m.

Tam/WB2TT