Amps
[Top] [All Lists]

[AMPS] Conjugate Matching and Efficiency

To: <amps@contesting.com>
Subject: [AMPS] Conjugate Matching and Efficiency
From: na9d@speakeasy.net (Jon Ogden)
Date: Thu, 17 May 2001 00:28:56 -0500

After noodling all of the ideas tossed back and forth regarding conjugate
matching and efficiency, I have come up with the following simplifications
to try to clarify the idea.  If there are any errors in what I am putting
across, I'd be happy to have them pointed out.

For simplicity let's assume we have a 100 Watt transmitter that is 100%
efficient.  Obviously, this isn't possible, but for the sake of a model,
let's assume that.  Therefore, for 100 Watts output, the radio consumes 100
Watts of power.

Now, let's hook up a power meter on the output of this radio and a 50 Ohm
load.  Let's also assume that the output impedance of the transmitter is
50+j0 Ohms and the load is also 50+j0 Ohms.

Now key the transmitter.  How much power will be delivered to the load?
Well, I believe that it will be 100 Watts.  The radio puts out 100 Watts,
there is NO mismatch at either the power meter (ideal) or the load.  Maximum
power is transferred to the load.

Now for those who believe that a conjugate match means 50% efficiency, then
that means that my power out would only be 50 Watts.  To me, this does not
make sense.  Also, it would follow then that having a load that is something
other than a perfect 50 Ohms in this case would mean more efficiency.  How
can my system be more efficient when some of the transmitted power is being
reflected from the load back towards the transmitter?  If I am 100%
efficient and I have a 2:1 mismatch, 10% of my power is being reflected back
to me.  Therefore, the power actually delivered to the load is 90 Watts.
Certainly that's not more efficient than the case of a conjugate match!

Now, let's add an amplifier in line.  We know nothing about this amplifier
other than the fact that it is perfectly terminated on each port in 50 + j0
Ohms.  It has a gain of 10 dB.  Therefore for 100 W in, it will put out 1000
Watts.  We know nothing about the efficiency.  How much power will be
delivered to our ideal load?  1000 Watts.  100 W in + 10 dB of gain = 1000
Watts.

We could play with these numbers all day long.  If we make the transmitter
50% efficient, it will still put out 100 Watts of power at its output.
Sure, it will dissipate 200 Watts, but that loss of efficiency has more to
do with the real world fact that there is no such thing as "lossless."

The fact of the efficiency of an amplifier does not matter.  When analyzing
the system it's an impedance either on the input or the output.  It's a
black box.

I agree with Tom Rauch that a conjugate match tells us nothing about the
efficiency of the amplifier but only about the amount of power transfer.  If
maximum power is being transferred to the load under a state of conjugate
match, this means that in simple terms, more power is being drained from the
PA tank circuit than at any other state of match.  If this case means a
maximum of 50% efficiency, then how can you have an amplifier designed to
operate into a 50 Ohm load (50 Ohm load line) with an efficiency greater
than 50%?

Thinking this out some more, when I tune up my amplifier, I am tuning for a
plate current dip as one of the parameters in my match.  If I am tuned for
maximum output power, I believe that I am conjugately matched at the output
of the PA tank circuit.  If my load is 50+j10, then looking into the output
of the PA, I would see an impedance of 50-j10.   My efficiency is
predominantly based on the power I put out divided by the power I consume in
the amplifier (approx Vplate * I plate).  Since the dipping the plate
reduces my plate current, it therefore decreases my total consumed power.
By dipping the plate, my output power as measured on a power meter also
increases (which would indicate maximum power transfer).  This totally
increases my efficiency.

How is it therefore, more efficient to operate at some other point of plate
current out output power?  If I tune to increase my plate current, I also
decrease my output power, not increase it.  Therefore, my efficiency drops.
The 50% efficiency argument doesn't make sense.

I do think that people have taken models beyond what they are supposed to
say, as Tom mentions.

73,

Jon
NA9D

-------------------------------------
Jon Ogden
NA9D (ex: KE9NA)

Member:  ARRL, AMSAT, DXCC, NRA

http://www.qsl.net/ke9na

"A life lived in fear is a life half lived."


--
FAQ on WWW:               http://www.contesting.com/FAQ/amps
Submissions:              amps@contesting.com
Administrative requests:  amps-REQUEST@contesting.com
Problems:                 owner-amps@contesting.com


<Prev in Thread] Current Thread [Next in Thread>