Question about isolation resistor in wilkinson power divider

[alard]

isolation resistor

hello,

I have a question regarding this article 'A General Design Formula of multi-section Power Divider Based on Singly Terminated Filter Design Theory' on wilkinson power divider design.

Let's take a 4-section power divider just for verifying the formulas.First, I got the impedance of optimum transformer (120% bandwidth ,0.01 ripple). All values have been normalized.

Z=[1.1157 1.2957 1.5435 1.7926];
then
Y=[0.8963 0.77178 0.64788 0.55785];
On the other hand, i calculated the prototype element values for the singly teminated chebyshev lowpass filter, also at the ripple level of 0.01
g=[0.3564 0.8945 1.1547 1.0421] (should be right, already verified)

then according to the formulas in the paper, I got the G values
G=[1.4028 0.0913 3.3695 0.075] which correspond to my final result
R=[0.7129 10.9530 0.2968 13.3419]

However the resistor value that COHN provided in his paper(BROADBAND THREE-PORT TEM-MODE HYBRIDS) is
R=[9.6432 5.8326 3.4524 2.0633]

There must be some misunderstanding or wrong calculation in my above steps. Really makes me confused now. Could someone take some time and help me to check it? Thanks in advance.

Best Regards,
kevin

 

[biff44]

wilkinson broadband resistors

I am not sure of the question. Those values look within the ballpark to me. Did you simulate it to see.

You DO know that those formulas are "normalized" to a Z0=1 ohm measurement system. You have to convert the equation's answers to figure out the corresponding values for a Z0=50 ohm system. For example, in a 50 ohm system:

If R1=1, R1'=1*50=50 ohm
If characteristic impedance Z1=0.5, Z1'=0.5*50=25 ohm
If characterisctic admittance Y1=2, then Y1'=2/50=0.4 mho

For filter designs with lumped reactance, usually you divide the capacitance by Z0, and multiply the inductance by Z0.

 

[alard]

wilkinson chebyshev broadband

Hi biff44,

At first I also thought the way in ' 'A General Design Formula of .....' may give me another solution for the isolation resistors other than the one given by COHN. However, I got a very bad isolation level when simulate the resistors (of course, with the denormalized values) in the circuit.

The reason I use the normalized values is that I want to compare the result with the 'proven' one in COHN's paper afterwards.

BTW, I am not designing filters now. Just in this paper, the auther caculated the resistors based on the singly terminated filter design. The paper you can find it in the following link


Best Regards,
kevin

 

[biff44]

power calculation wilkinson

I am not sure how a single terminated design works for a power divider!

A singly terminated prototype is useful when you design a diplexer filter, like a lowpass/highpass combination that is fed from a comon junction. The lpf output, and the HPF output are loaded in 50 ohms, but the common junction has an Admittance (assuming a parallel connection at the junction) of:

LPF:

g=1 for f<fc, and g=0 for f>fc
y= Y(f)

HPF:
g=0 for f<fc, and g=1 for f>fc
y=-Y(f)

If you get my drift.

 

[hxjicn]

Re: Question about isolation resistor in wilkinson power div

R =

5.63399643942376 2.79204551958272 2.34187961232433 3.39763259896539