# Unequal-split Wilkinson power divider simplified

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. 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 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.  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 1.5dB 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). 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