How does a Wilkinson power divider achieve isolation?

Question

Say we have a Wilkinson power divider:

The explanation I've heard about how ports 2 and 3 are isolated is this: say there's a signal generator attached to port 2. This can take two paths to port 3: through the resistor, and through the two transmission lines. With the two transmission lines being a half wavelength, they invert the signal. So the non-inverted signal (through the resistor) and the inverted signal (through the transmission lines) is equal but opposite and so cancels.

That makes sense, but it's not enough to convince me that it actually works. For example, why split in half and not some other ratio which would not result in complete cancellation? And wouldn't port 1 have some significance to the operation?

Answer 1

Microwaves101 provides a good qualitative and intuitive explanation, and you seem to understand the idea. For a more rigorous derivation of the scattering matrix you can read section 7.3 (page 328) of Pozar's Microwave Engineering. In this case he uses what is called Even-Odd mode analysis, which he explains pretty well since it is the first time it is used in the text.

For example, why split in half and not some other ratio which would not result in complete cancellation?

The whole idea of the circuit is that it has the property that ports 2 and 3 are 180 degrees away (through the microstrips) and also 0 degrees away through the resistor. This circuit is the simplest way to achieve this property.

And wouldn't port 1 have some significance to the operation?

Yes. The circuit is only lossless if all ports are matched. If there are unequal voltages at ports 2 and 3 at any time power will be lost in the resistor. Assuming all ports are matches, this means the circuit can divide a signal losslessly (because the V2 and V3 will be equal and in phase) and can combine identical & in phase signals losslessly. It cannot combine independent signals losslessly*. This makes them a good candidate for use in solid state RF power amplifiers.

*It is impossible to construct a three-port lossless reciprocal network that is matched at all ports in general.

Answer 2

I haven't worked out all of the math details, but I think it's going to be possible to show that if power is injected at Port 2, half of it flows through the resistor toward Port 3, and half flows down the transmission line to Port 1.

At Port 1, this power divides further, sending 2/3 of it out Port 1, and the other 1/3 down the other transmission line to Port 3.

At Port 3, the power through the resistor is attenuated by voltage divider action against the net impedance at that port. The power through the transmission line is transformed in impedance and goes through a similar voltage divider action when it arrives at Port 3, but since it has been delayed by 180° it cancels the resistor power.

Again, I haven't written out the full equations yet, but it seems like it should work out.

In any case, this simulation shows the effect rather clearly. Just for variety, I used T networks (rather than your π networks) as transmission line simulators. The component values were selected to have reactances of 70.71 Ω at 435 MHz. There is a deep dip in the power delivered to Port3 at that frequency, and equal amounts of power are being dissipated in R1 and R4.

^{simulate this circuit – Schematic created using CircuitLab}