Microstrip transmission line teardrop-shape power divider question

[yefj]

Hello,I have built a microstrip line structure as shown below In photo and article 1.
my lines are er=9.4 h=0.433 line width W=0.015mm spacing between lines 0.065mm and 2 micron thickness.
The line impedance properties is 140 Ohm.
However I need also to build a feeding network for these lines.
I was suggested to use this nature article with the intrested shape.
Howeven i dont see the impedance power dividing logic in this tear drop shape.
I know its supposed to be a transformer of some sort,but there is a huge discontinuety in the width between W1(tear drop shape) and Wt(line width).
What is the impedance logic in this huge transition?
Thanks.
https://www.nature.com/articles/s41598-022-17212-0
https://www.nature.com/articles/s41598-022-17212-0/figures/2

 

[Baluncore]

That is a tapered transmission line impedance transformer.
Your input port #0 has an impedance of Z0
Your output ports #1,2,3,4 have equal impedances Z1,2,3,4
The output ports are connected in parallel to the wide end of the transformer.
The wide end of the transformer will have an impedance of Z1/4.
The width of the track, and droplet, at the input end, will be matched to Z0 over the ground plane.
The width of the taper at the wide end will be the same as a transmission line over ground of Z1/4.
Halfway along the taper, the width of the taper will be for an impedance, being the geometric mean of the input and output.
Z0% = Z0
Z100% = Z1/4
Z50% = √(Z0 * Z1/4) = GM(Z0, Z1/4); the geometric mean.
Z25% = GM(Z0%, Z50%)
Z75% = GM(Z50%, Z100%) ; and so it continues until you have the transformer.
The length of the taper is not critical, but must be over a quarter of a wavelength long at the longest wavelength.

 

[yefj]

Hello Balune core, my output impedance is 140 Ohm
The wide end of the tear drop is Z100%=140/4
few questions:

1.In the article section shown below they use quarter wave type transformer formula for Zt part below.
we have single transformer line between output and Z1 why did they use coeffient 4 in the formula?
In the regular formula its Zt=sqrt(z0z1)

3. what is the logic in putting this quarter wave transformer Lt(Zt) when we already have the tear drop as a transformer?
Thanks.

 

[yefj]

update :
my substrate er=9.4 h=0.43mm w=0.015mm gives each line output is 140 Ohm(I have 4 such lines with 0.065um spacing between lines).
the total tear drop wide side need to be 255um which is 63Ohm.
how do I calculate the transformer inpedance for such case?
Thanks.

 

[Baluncore]

If you design for a single frequency, or for very narrowband, then a quarter wave step transformer can be made from a transmission line, with a GM impedance.

What is the frequency, wavelength and bandwidth of the signal?

how do I calculate the transformer inpedance for such case?

Which transformer?
Maybe.
63R * 4 = 252R, each will become 140R, so Zgm = Sqrt( 252 * 140 ) = 188R.

Use R as the text form of omega.
Write the SI unit "ohm" with all lower case. The man was "Ohm", capitalised.

 

[yefj]

Hello Balune core, so if I have a situation as shown below W1 has impeadnce of Z1 and W2 has ipedance of Z2, then on the connection point between the two sections ,W2 line will feel an impedance of 4*Z1?
and W1 line will feel W2/4? correct?

if I understnad you correctly why does each W2 line will have 4*Z1 impedance connected to it from the left?

 

[Baluncore]

I think you should analyse it as resistors in parallel, by looking at the voltage and the currents. Assume the transition node is matched. The voltage, v, at the transition between w1 and w2, is the same on both sides of the junction. A current of, 4i, that flows in w1 is divided into four lines of, i, in the four w2.
Z1 = v / (4 i ) ;
Z2 = v / i .