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Maximum
power transfer theorem
Updated June 22,
2005
New for June 2005!
The importance of matching a
load to a source for maximum power transfer is extremely important
in microwaves, as well as all manner of lower frequency stuff such
as stereo sound systems, electrical generating plants, solar cells
and hybrid electric cars. It is very simple to prove using Ohm's
Law, and that's what we'll do here.
The first engineer to understand
the importance of source and load matching was American Joseph Henry.
Check out his portrait in our Microwave
Hall of Fame!
A circuit diagram of a source
and load is shown below. Feel free to think of everything as direct
current because that is the simplest case (RF follows ohm's law,
so the theory is valid at any frequency). The source has a series
resistance (often called the Norton or Thevenin equivalent resistance)
built into it. In microwave engineering, the generator's resistance
is the same as the characteristic impedance of the transmission
media, usually 50 ohms, and it is usually called "Z0"
(sometimes ZC for "characteristic impedance".
The load resistance is called ZL.

Notice that the
generator voltage is "2V" in the schematic. Some people
are wondering, "if I buy a 1.5 volt D-cell battery, does that
mean that it actually can put out 3 volts?" No way, Jose! What
we are trying to show here is that under maximum power transfer
conditions, only half of the generator voltage makes its
way to the load... so in the case of a D-cell, only 0.75 volts would
be available at maximum power transfer. Then you ask, "why
doesn't maximum power transfer happen when I put a short circuit
on the output, I know that the battery will discharge quickly under
that condition?" The problem here is that total dissipation
will be maximum with a short circuit load, but no power will
be transferred out of the battery. All of the power dissipation
is internal to the battery, which is why it is getting too hot to
hold....
Using Ohm's law
we can solve for the transferred power. First note that Z0
and ZL form a voltage divider. Then recall that power
is equal to voltage squared divided by resistance:

Now let's look at output voltage,
current and as a function of load resistance. For convenience we
will normalize the load resistance to the characteristic impedance
and let them both equal 1; for ZL=twice Z0,
the normalized load resistance is 2.0 for example. We further normalize
the generator voltage 2V to 2 volts.

The equation for power, plotted
in purple, has a funny shape with a clear-cut maximum. For homework
you can use calculus to find the maximum, but trust us (and the
plot), it occurs at exactly (ZL/ZC)=1!
We can see from
the plot, the maximum output current occurs into a normalized impedance
of zero (a short circuit). Here we get a DC current of 2, which
is because the generator has a voltage of 2. Looking at an open
circuit (off to the right of the plot) we can see that the load
voltage asymptotically approached 2 volts. By multiplying the current
and voltage, power is obtained.
Under open-circuit conditions
(ZL=infinity) the output voltage is "2V". Under
short-circuit conditions, a current of 2V/Z0 would result.
Going back to our D-battery analogy,
how can you find the source impedance of a battery? You could put
a variable load on it, and when the load voltage hits 0.75 volts,
it would have the same value as the generator impedance. A faster
way is to measure the short circuit current (quickly!), then divide
the open circuit voltage (1.5 volts) by the short circuit current
(perhaps three amps) and you will arrive at the generator impedance
of maybe 0.5 ohms.
Efficiency versus load impedance
Now lets look at efficiency versus
load impedance. This time we'll plot a wide range of load impedances
using a log scale. The log scale chart is interesting from the point
of view of the symmetries involved. In addition to the maximum power
point at RL/RG=1 providing 50% efficiency, the point at which RL/RG=3
provides exactly 75% efficiency. One lesson you could take away
from this is that for highest efficiency, the load impedance must
be much greater than the generator impedance even though maximum
power transfer will not take place under this condition. In
the two extreme condictions, a short circuit is 0% efficient, and
open circuit is 100% efficient, even though in both cases, no power
is transferred to the load.

By analogy, for a current source
the efficiency increases with GL/GG. In this case a diminishing
load resistance leads to higher efficiency. The two types of sources
have opposite behaviors as far as efficiency relative to load resistance
is concerned.
Whet does this say about amplifiers?
Regarding power
amplifier efficiency, varying the slope of the load line in
Class A does not have such a predictable effect on efficiency. For
amplifiers in or near compression power and efficiency are complex
functions of many factors including the outermost excursions of
the dynamic loadline near the pinch-off area and near the Imax boundary.
Too much of a topic to squeeze onto this one page!
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