on 5/17/01 11:01 AM, Billy Ward at billydeanward@hotmail.com wrote:
> 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?
Hi Billy,
I've been trying long and hard to come up with a "scientific" answer to your
point and have been consulting with others as well as noodling it around in
my head.
Let's see how this works.
I think one of the problems folks have when talking about source impedance
is that they think of it in terms of a lossy or dissipative resistance. A
50 Volt, 50 Ohm source would have a lossy 50 Ohm resistor in series with its
output. And so the statement you made above would follow.
However, I don't know if you can treat the source impedance as a lossy
impedance in this case. "How can you do that?", you ask. Well, let's look
at another RF type component. Coax cable. What is its impedance? Well,
generally it is 50 Ohms. But is it dissipative (other than it's
characteristic loss)? No, it is not. In most models it is an ideal
component. It is a transmission line with a particular characteristic
impedance. Yet if you put your Ohm meter across it through it or whatever,
you won't get a measurement of 50 Ohms.
A tank circuit has some transmission line characteristics as well. It has
delay characteristics, series L, shunt C, etc. And as one person I talked
to say, it may even have a characteristic impedance although we might not be
sure what that is. Put your Ohm meter across the output of your amp or
transceiver, do you get 50 Ohms? No, likely you'll see zero Ohms or an open
circuit.
Let's look at this concept of the 50 Ohm transmission line some more. If we
assume that impedance is impedance is impedance and must be taken into
account, then we have to look at the following of what is in our circuit.
We have a voltage source with a series resistance of 50 Ohms. Then we have
another 50 Ohm block of transmission line. Then we have a 50 Ohm load.
Since these are all equal, each element has the same voltage drop across it.
All currents are equal so each element absorbs the same power. Therefore,
my efficiency is only 33%!
"But, you can't say that!", will certainly be said. Of course not. It
doesn't make sense. But it is almost exactly what you said with the
exception of adding the impedance of the transmission line.
The concept of assigning the source impedance a dissipative resistance is
assigning an incorrect model. And since we are talking about models, if a
tank circuit is made up of lossless, ideal elements, then where does this
50% of the power, that you and other allege gets lost, go?
Now, let's get a little more mathematical here.
The model that we typically use for sources is Thevenin. That is a voltage
source with a series impedance. But as others have pointed out, we forget
about another model, the Norton model. The Norton model is a current source
with parallel source impedance. Now I went to engineering school and we
were always told that you could interchange the two models. They are
equivalent. Therefore, if a model is proper, the mathematical analysis that
works in one will work in the other. Does anyone want to disagree with the
concept that the Thevenin and Norton models are interchangeable?
So let's take a 50 Volt Thevenin model with a 50 Ohm resistor as source
impedeance. What's the Norton equivalent? It is a 1 Amp current source
with a 50 Ohm shunt resistor. Now let's place a 50 Ohm load across each of
these and do some math. In each, we get 25 volts across each resistor and
in each case we develop 25^2/50 or 50 watts. The internal resistor is also
dissipating 50 Watts in each case. OK, so everything is holding up. This
"sanity check" seems to show that we have valid models for maximum power
transfer as well as dissipation.
But let's take things further. What if we use a 10 Ohm load resistor
instead of a 50 Ohm load?
In the Thevenin circuit, we will have an output voltage of 50*10/60 or 8.333
volts. The power dissipation in that resistor is 6.9444 Watts. In the
Norton circuit, our one amp source is now driving 50 Ohms in parallel with
10 Ohms which ends up being 8.3333 Ohms. Clearly, a one amp source across
an 8.333 Ohm resistor develops 8.333 volts into that load. Again, our 10
Ohm resistor with 8.333 volts across it dissipates 6.9444 Watts.
So, hey, it works! We can swap the two models back and forth and get
accurate analysis of voltages, currents and power dissipated in the load and
source. Cool! But now at this point, we might be misled into thinking that
the models will therefore work for efficiency calculations. Let's see if
that's valid.
In our Thevenin model, the internal resistor drops what the load didn't.
That would amount to 50-8.3333 or 42.6666 volts. The 50 Ohm resistor with
that voltage across it dissipates 36.4 Watts. If we proceed with an
efficiency analysis, we can calculate efficiency as Pload/(Pload + Pdiss),
where Pdiss is the power dissipated in the source. This in our example will
be 6.9444/(6.9444+36.4) = 16.02%. We say, no big deal. We aren't matched,
efficiency should stink. So we think we are doing OK.
Now let's switch to the Norton model. If you remember in the Norton model,
we had 8.333 volts across both the 50 Ohm resistor and the 10 Ohm resistor
in parallel. So the Pdiss in the 50 Ohm resistor with 8.333 volts across it
is 1.388 Watts. The load resistor, meanwhile is dissipating 6.944 Watts.
Our efficiency therefore is 6.944/(6.944 + 1.388) = 83.4%
STOP! HOLD THE PHONE! WHOA! Our model just failed! Clearly, the Thevenin
and Norton models are NOT equivalent for efficiency calculations, so it
would be ludicrous for us to expect either one to be correct for efficiency
calculations.
As has been pointed out, the models must be used within their proper
context. While the models are valid for calculating voltages and currents,
the are NOT valid for calculating efficiencies. If the mathematical model
is to be used and be used accurately, it must be valid in all cases.
Conjugate matching impedances therefore, tells you NOTHING about the
efficiency of the source or the system. It is quite possible that Zout and
efficiency have absolutely nothing to do with each other.
There, I am done. Many thanks to those who have discussed this with me off
line and given me some good positive ideas and examples.
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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