I don't have the time today to re-read and study your statements but I think
that I am following you.
However, just a quick thought. Suppose the amp is 100% efficient. (We all
know that it cannot be!) Whether conjugately matched or not, the RF current
flowing through the output of the transmitter is the same as that flowing in
the load.
If you graph the power dissipated in the load at various degrees of
impedance match, the "connect the dots" line will look somewhat like the top
of a sine wave. The peak of the line will be centered on the point that the
real parts of the impedances match and where the reactances cancel, proving
that maximum power transfer does, in fact, occur at resonance--or the
conjugately match condition.
If the same current is flowing in the output of the source as in the load
and the two are equal, do they not act like any voltage divider would. Is
the power not evenly divided across each element?
Billy Dean Ward
>From: Jon Ogden <na9d@speakeasy.net>
>To: Amps Reflector <amps@contesting.com>
>Subject: [AMPS] Conjugate Matching and Efficiency
To: <amps@contesting.com>
>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."
>
>
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