Unequal-split Wilkinson power divider simplified

Click here to read about the "normal" unequal-split Wilkinson

Click here to read about the "rest of the story" of the unequal split Wilkinson

Click here to go to our unequal split power divider

Updated for December 2023: thanks to Parker, we now have equations for determining line impedances based on split ratio, which is way more convenient than our original solution when you pick Z1. also, we placed a Microwave Office project on this topic in the download area.

New for July 2023. Are you frustrated with having to accommodate output transformers on an unequal-split Wilkinson power divider because they take up room on your layout?  What if you did away with them?

Three things happen:

  1. The line impedances required for a given power split get a lot harder to achieve.
  2. Power starts to burn up in the isolation resistor, but not much...
  3. You won't get perfect isolation and return losses at center frequency, but you never do in real life so this is a non-issue.

But it turns out, in many instances this is a good compromise to look at.

Mitered Bends After looking at this topic we came up with a new Rule of Thumb #128... regarding unequal-split Wilkinson power dividers, for power splits up to 1.2:1, you don't need the impedance transformers on the split ports!

Let's start by looking at the schematic, in Microwave Office.  Note that we put a 20 dBm source on port 1, or 100 mW.  There's a reason for this...

Now, let's look at equations to satisfy an unequal split with S11 matched to Z0, using Z1 as the independent variable. ZIN1 is the impedance looking down the Z1 line by itself, and ZIN2 is the impedance looking down the Z2 line.  The ratio of the square-root of these impedances is the split ratio.  When you put them in parallel you need to see Z0.  It's all pretty simple algebra, but we counted four crumpled-up sheets in the trash before we came up with the correct answer.  If you don't stay on top of your math skills you will find they degrade with age. If someone wants to be a hero, he/she/they could solve the equations so that the split ratio is the independent variable for us and we'll post the result.  Hold that thought, the answer is shown below thanks to an anonymous contributor! While you're at it, you can calculate return losses and resistor dissipation...

For a design example, we picked Z1 to be 85 ohms, which we consider the maximum practical impedance for a variety of media.  This makes Z2=62 ohms, and the split ratio approximately 1.4 dB.

Before we look at RF performance, let's look at the impedances plotted against split ratio.  Captain obvious wished to point out that at 0 dB but lines are 70.7 ohms. the ordinary Wilkinson power divider. If you are looking for 1.5 dB split ratio or more, this network is not going to work because the required impedance for Z1 exceeds 85 ohms.

Now we boot up Microwave Office. Here are transmission coefficients at Z1=85 ohms and Z2=62 ohms. Indeed, you get 1.4 dB split ratio. 

Here are the return losses. None of them, including S11, are actually perfect at center frequency.  But they are certainly good enough for most designs of moderate bandwidth.

Here's isolation.  It is ~12 dB minimum from 5 to 15 GHz.  Some antenna feeds don't even care about isolation, in those cases you can eliminate the resistor from the design.

As for bandwidth, bandwidth is in the eye of the beholder.  We recommend this for 100% BW or less (5 to 15 GHz when centered at 10 GHz).

Now let's see what is dissipated in the resistor. Recall that we injected 100 mW into port 1.  Therefore, if we plot resistor dissipation in milliwatts, it will be the same as percent.  Brilliant!  In Microwave Office you have to use the "power" measurement type (see below), then select the component or node you are looking at. Harmonic balance is used to evaluate the dissipations, even though this is a linear network it adds a lot of annoying computation time.  If you didn't rent the HB option, you are on your own for calculating resistor dissipation.  Again, it's just an algebra problem... Put one volt on Port 1, then calculate the voltage between ports 2 and 3.

Here is resistor dissipation, just 0.6 percent.  Looked at in dB, you are throwing away -0.26 dB. For splits closer to 1:1. the effect is even less. In case you were wondering how we came up with -0.26 dB, it is 10*log10(1-0.006).

Update December 2023: thanks to Parker, we now have equations for determining line impedances based on split ratio. We show them in two different formats for the first time in the history of the internet, right here,  Thanks!

First, based on voltage split ratio, the equation for Z1 and Z2 in text form are

Z1=SQRT(Z0^2*(SplitRatio^4+1))

Z2=(Z1*Z0)/SQRT((Z1-Z0)*(Z1+Z0))

Z2 calculation is the same as we showed you in the orginal set of equations:

Here is how it looks in our Microwave Office file (available in the download area).

Here's second equation set, based on split ratio in dB.  This is arguably the most useful form of the equations!

In text form, Z1 is solved as:

Z1=SQRT(Z0^2*(10^SplitRatiodB/5)+1))

It's kind of cool how the split ratio in dB has to be divided by 5.  Trust us, it's correct.

Here is the equation block from our MWO file.

These equations are provided in a MSTLAB snippet below, if that's your thing:

%Unequal split calculator for ratios under 1.2:1

clear; clc;

%Using linear ratio

disp('Linear ratio results')

Rlin = 1.173; %ratio on microwaves 101 site

Zo = 50;

Z1 = sqrt(Zo^2*(Rlin^4+1))

Z2 = Z1*Zo/sqrt((Z1-Zo)*(Z1+Zo))

Zin1 = Z1^2/Zo

Zin2 = Z2^2/Zo

Zin = Zin1*Zin2/(Zin1+Zin2)

%Using dB ratio

disp('dB ratio results')

RdB = 1.382;

Z1 = sqrt(Zo^2*(10^(RdB/5)+1))

Z2 = Z1*Zo/sqrt((Z1-Zo)*(Z1+Zo))

Zin1 = Z1^2/Zo

Zin2 = Z2^2/Zo

Zin = Zin1*Zin2/(Zin1+Zin2)

Now, we will show you something that is slightly disturbing but not a big deal so don't panic and change careers or become a program manager or sales rep  You can calculate the split ratio from the S-parameters easily in a Microwave Office equation block, shown below.  The intended split ratio is only valid at the exact center frequency of course.

We tried out the equations where split ratio is specified in dB and entered exactly 1.5 dB. You'd think it would come out exactly, but it doesn't... the S-parameters show 1.489 dB instead of 1.500. Cue the Twilight Zone theme... If anyone can tell us why that is happening, please enlighten us and while you are at it, feel free to rewrite the equations so they come out exact!

 

 

 

 

Questions, comments or corrections, shoot us an email.  Just about any address at M101 should work, try:

themostannoyingperson@microwaves101.com

for example...

 

Author : Unknown Editor