Unequalsplit
Wilkinsons  the rest of the story
Updated July
27, 2012
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here to go to our main page on Wilkinson power splitters
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Originally posted November
2008. This topic is so novel and important to microwave engineers
that we actually considered whether it should be first published
in the IEEE transactions. But that would
delay the release of the material by six or eight months, to a far
more limited audience than we have, what were we thinking? The world
has been waiting 43 years for this topic to be further explored,
and that's long enough. Hopefully the next IEEE author that discusses
unequalsplit Wilkinsons will reference our material, or they will
hear from us on this page. So when the time comes, be sure to credit
Editor, T., and Kouzoujian, J., affiliated with Microwaves101 and
Matrix Test Equipment respectively,
for developing the following material...
The original unequalsplit Wilkinson
article is entitled Split Tee Power Divider, by Parad and
Moynihan, published in IEEE Transactions on Microwave Theory and
Techniques in January, 1965. This article has been referenced in
many subsequent works, including Harlan Howe's Stripline Circuit
Design and Pozar's Microwave Engineering. In the article,
Parad and Moynihan develop a singular set of equations for two unequal
splitters. The simpler one is shown below (the second one adds an
input transformer). For now, we will reanalyze the simpler one here.
We created our own
nomenclature for the problem, which is provided in the figure below:
Referring to our figure, the
article states that the impedances ZC' and ZD' are equal to:
These equations are based on
an assumption that is not fully explained in the paper. If anyone
can explain it to us, you will win a Microwaves101
pocketknife! The four line impedances can then be uniquely solved
to agree with the solution that Parad and Moynihan provided. We
rearranged their equations slightly but ours give the same result:
Update June 2011: Igor offers the following explanation for how Parad and Moynihan came to this singular solution:
The ZC' and ZD' are set such that their geometric mean is equal to Z0.
ZC'=(P3/P2)^0.5*Z0
ZD'=(P3/P2)^0.5*Z0
(ZC'*ZD')^0.5=Z0
This minimizes the impedance transformation ratios for a given power split and thus maximizes the bandwidth.
Thanks for that! The knife is in the mail...
Note: there is some confusion
on this page about the definition of the power split. The way Parad
and Moynihan defined it, the ratio of power delivered to port 2
to power delivered to port 3 is 1/K^2. Later, we might define this
differently, so pay attention... we'll try to clean this all up
one day!
The above equations are used
in the Microwaves101 unequalsplit
Wilkinson online calculator, and have been used by thousands
of engineers for over 40 years.
Taking a fresh look at the problem,
we have found just three equations that must be satisfied for the
four unknown impedances, which reveals that there is a degree of
freedom that has been ignored for the past 43 years. The solution
that Parad and Moynihan chose provides good bandwidth, but bandwidth
could perhaps be traded for more realizable impedances, and design
engineers should make that choice themselves.
The equations that are incontrovertible are based on three criteria
that must be met. The impedances of the two arms have to combine
in parallel to provide the system impedance Z0. In our figure, these
impedances are denoted ZA' and ZB'. Then:
(Equation 1)
Because the voltage at the input
node is the same looking into each arm, the power split is proportional
to the current split, which will be inversely proportional to the
impedances ZA' and ZB' looking into the arms:
(Equation 2A)
Therefore:
(Equation 2B)
Substituting (2B) into (1) and
solving for ZA' as well as ZB' we get:
(Equation 3)
Noting that ZA' and ZB' are functions
of the two stage transformers ZA cascaded with ZC, and ZB cascaded
with ZD, we see that:
(Equation 4A)
Which can be rearranged to:
(Equation 4B)
The voltage across the isolation
resistor has to be zero, when power is applied from the common port.
In order for this to happen, the impedance ratio must be maintained
through the first legs of the splitter (ZA and ZB). Therefore:
(Equation 5)
From (4B) and (5) we now have
three equations and four unknowns. We chose ZA as the independent
variable and solved for ZB, ZC and ZD:
(Equation 6)
Finally, we need to calculate
the load resistor. This can be done easily by noting that in the
odd mode, power incident into port 2 must see impedance matched
to Z0 and power incident into port 3 must also see impedance matched
to Z0 (see figure).
Therefore:
(Equation 7)
With (6) and (7) our solution
to the unequal split Wilkinson is complete. Now we can look at the
impedance values when we choose different values of ZA. Let's look
at a divider that splits two thirds energy to port 2, and one third
to port 3 (P3/P2=0.5, see table below). Parad and Moynihan's results
are ZA=51.5 ohms (shown in bold). Here the impedance for ZB is very
high (103 ohms). By moving to ZA=40 ohms, the impedance range of
the four lines is now 32.7 to 80 ohms, which may be more realizable
in microstrip applications.
What does this do to the frequency
response? Below is a model of the splitter in ADS. We used this
to look at the effects of varying ZA but maintaining a fixed power
split. The equations in the "VAR" block are valid versions
of the solution we have just provided. In this case, "ratio"
is the ratio of power at port 3 to power at port 2, so "ratio=0.5"
means that port 2 is favored with twice the power as port 3.
Let's look at the response as ZA is varied. The center column of
the figure below represents Parad's and Moynihan's solution, for
a power divider where P3/P2 is 0.5 (ZA=51.5 ohms). Beside it are
plots of the same power split but with PA=40 ohms, and PA=60 ohms.
The markers shows that the power split is maintained. Click on an
image to zoom in on it.
Isolation

Isolation

Isolation

Match

Match

Match

Transmission

Transmission

Transmission

ZA=40
ohms

ZA=51.5
ohms

ZA=60
ohms

Before we get much further, we
want to point out one thing. For equations for solving the line
impedances of an unequal splitter, you'd expect that for P3/P2 =x
and P3/P2=1/x, you should have the same solution, only "flipped",
right? If you use Parad and Moynihan's equations, this is what happens.
For other solutions, this is something to watch out for (including
the one below).
Maximally flat transformer applied
to unequalsplit Wilkinson
Incorporating the maximallyflat
impedance transformation to determine the section impedances provide
for superior input match compared to Parad and Moynihan's solution.
The exact solution to maximallyflat
impedance transformer is not easy to solve, except in the case
where the transformer only has two sections.
Unfortunately the dog ate this
part of our homework, but when it is completed the unequal/max flat
divider design will appear on this
page. Stay tuned!
