here to go to our main page on Wilkinson splitters
here to go to our page on N-way splitters
here to go to our page on the Kouzoujian splitter, a great alternative
to the conventional N-way Wilkinson
New for June
2010! This page will provide a basic analysis of planar, three-way
Wilkinson splitters in 50 ohm system impedance. This is splitter
is imperfect, because it is missing an isolation resistor between
the two outer ports, however, this is what makes it easy to lay
out. We divided the analysis into three "types" which
are described below, and ranked according to bandwidth potential.
Type 1 splitter
The Type 1 splitter is the simplest
network possible. The three arms each employ a single quarter-wave
impedance transformer. If you were to impedance match port 1 at
center frequency, the transformers would all be 86.6 ohms (transforms
each 50 ohm leg to 150 ohms, and three 150 ohms in parallel is 50
ohms). Sorry about the crummy schematic, you'd think by now Agilent
would provide a means for graphic capture beyond the usual copy-and-past
into Powerpoint, then shrink image with PaintShop...
The variables R1 and Z1 were
allowed to optimize to provide the maximum "-15 dB bandwidth",
in other words, it was required that reflection coefficients S11,
S22 and S33 and isolation terms S23 and S24 were less than -15 dB.
Note that by symmetry, it is not necessary to optimize (or monitor)
S44, and S43.
The choice of -15 dB as the goal
is arbitrary. However, in the planar three-way Wilkinson, you can't
expect perfect isolation of the split ports, and without perfect
isolation, you won't get perfect return losses for S22, S33 and
The transmission line impedance
values were restricted to no more than 100 ohms (in all three types),
to make the circuit realizable in microstrip. It might be better
to restrict to a lower value depending on what substrate is chosen,
and yield considerations. The subject for another day!
For Type 1, Z1 was found to be
91.4 ohms and R1 was 83.6 ohms, resulting in -15 dB bandwidth of
Below is the transmission coefficient,
showing the near-ideal value of 4.78 dB
[-10*log(1/3)]. All three transmission coefficients are equal in
this simplified analysis, in practice the outer branches will behave
slightly differently from the center branch.
Here are the return losses and
isolation. Note that all parameters are somewhat compromised, as
expected. This is the price that is paid for a planar design with
only two isolation resistors.
In Type 2 we added another transformer
section, in this case a single transformer at the input, and component
values were reoptimized for maximum -15 dB bandwidth.
The optimized results are shown
below. We didn't bother to plot the transmission coefficient, it
is similar to that of Type 1. Here, Z1 was found to be 41.9 ohms,
Z2 was 91.4 ohms, and R1 was 76.9 ohms, resulting in -15 dB bandwidth
In Type 3, the second transformer
section was moved to the output ports, and component values were reoptimized
for maximum -15 dB bandwidth.
The optimized results are shown
below. Here, Z1 was found to be 99.3 ohms, Z2 was 54.7 ohms, and
R1 was 98.6 ohms, resulting in -15 dB bandwidth of 53%.
In conclusion, Type 2 offers
the best bandwidth of the three splitters we examined. Note that
if you needed more bandwidth, you could add more sections and a
second (or third) set of dump resistors. Planar dividers have been
made that span 2-18 GHz in this fashion, and people have been happy
to spend $1000 on them!
This example is of a two-stage
three-way Wilkinson with two sets of isolation resistors. It was
EM-analyzed by some students from India. Bravo!
Below is the linear schematic.
We accidentally cropped out the microstrip parameters. The microstrip
substrate has ER 4.4, and is 1.5 millimeters in height. We left
the conductivity as a PEC (perfect electrical conductor).
And below is the optimization
goals and constraints we put on the network. Here we didn't demand
15 dB for the isolation and return losses. We asked for 20 dB, but
didn't try to calculate the exact bandwidth that it was capable
of, over 65% bandwidth.
Here is the result after optimization
(the final values are given in on the schematic). We put a constraint
on Z2, it was not allowed to exceed 100 ohms (it is often difficult
to realize very high impedances). The parameter that has the least
20-dB bandwidth is S33, which is the return loss of the center leg.
Whoops, looks like we also constrained R2 to 150 ohms, there was
no reason to put any constraints on the isolation resistors' values.
What happens when you don't have
any resistors in the network? In the schematic below the resistors
have all been deactivated.
The input match (S11) looks the
same, while the output matches and isolations have been severely
degraded. This is obvious if you study even
and odd mode analysis.
Here is the EMed layout of the
divider. You can see how the tool divided the network into tiny
triangles, upon which Maxwell's
Equations are imposed. The resistors (drawn and multi-colored
ovals) are not part of the EM analysis, so we don't expect much
isolation. The Line lengths in the layout have been folded to achieve
approximately 44 mm length to agree with the linear schematic. Why
are their elbows in the upper and lower paths? The electrical length
through an elbow is some non-zero value. By putting equal number
of elbows in each path, you can compensate for them all at once
by trimming the three legs' high impedance sections simultaneously.
Note that in this layout there are three elbows in the upper and
lower paths, and four in the center one, so we slightly violated
And here are some of the parameters,
analyzed in an EM tool. Here's the reflection coefficient, which
looks similar to the linear analysis but is a more accurate representation
of what the circuit will do. Note that it has less return loss at
the center of the band than we wanted. Welcome to the real world!
And below are the three transmission
coefficients, which are remarkable similar. Good job!
A postulated rule
One of the most basic rules of
network theory is for a three-port circuit, you cannot have
all ports simultaneously matched if it is lossless (no resistors).
We're not going to step out on a limb and say that the same is always
true for a four-port (it's not, one example of a matched lossless
four-port is a rat-race). But for "related reasons" that
we are to lazy to prove, let's postulate that an "N-port Wilkinson-like
power divider without resistors" can never be impedance matched
at the output ports.