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Please help me wilkinson power splitter/combiner questions - RF Cafe Forums

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Guest
Post subject: Please help me wilkinson power splitter/combiner questions.
Unread postPosted: Wed May 25, 2005 5:29 am

Thingking of 2-way equal wilkinson power splitter/combiner and setting port 1 input(output),port 2 output(input), port 3 output(input)

1. I just know the banlance resistor is 100ohm, but how to get this value by mathematic formula?
2. All these three ports are matched. Looking into port 1, I can understand why it is matched ( port 2 resistor and port 3 resistor at joint are the same 2Z0, and 2Z0//2Z0=Z0). But I can't understand why it is matched looking into port 2 or port 3?
3. Why say the balance resistor provide good isolation? How to compute?
4. If power P enters into port 1, I know the power of port 2 and port 3 are the same, P/2. But if power P enters into port 2, how to compute the power of port 1, port 3, and balance resistor?
5. If port 1 mismatch, how to compute the power consumed by the balance resistor? If port 2 or port 3 mismatch,how to compute the power consumed by the balance resistor?

If possible, Please give me detail information. These questions puzzeled me many days. Many thanks!!!
My email is: sunstar16@163.com


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Guest
Post subject:
Unread postPosted: Wed May 25, 2005 9:44 am

You need to look for EVEN-ODD mode analysis.

You will find this analysis in "Microwave Engineering" from POZAR


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malvinas2
Post subject:
Unread postPosted: Fri May 27, 2005 4:09 am

Or just 'google' around:

http://www.google.de/search?hl=de&q=wilkinson+power+splitter&meta=

gives you plenty of useful links.

BTW: There'll be always power consumed by the resistor, without the resistor the divider wouldn't be matched at all ports. You've to read something about "S-Parameter" (scatter parameter).


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Guest
Post subject:
Unread postPosted: Fri May 27, 2005 9:30 am

Malvinas,

I do not understant why you say there is alway power dissipated in the resistor.

If the system is perfecly balance, I mean same amplitude, same phase at both outputs(inputs), same impedance, then there is no dissipation in the resistor, since both of it's terminals are at the same potential


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malvinas2
Post subject:
Unread postPosted: Fri May 27, 2005 10:27 am

All three ports matched means: Sii=S11 = S22 = S33 = 0

/ 0 S12 S13 \
S =| S21 0 S23 | (3)
\ S31 S32 0 /

no dissipation is expressed by

S_transposed_conjugated * S = E (unity matrix) (4)

e.g.:

|S21|² + |S31|² = 1 (5)
|S12|² + |S32|² = 1 (6)
|S13|² + |S23|² = 1 (7)
S32* S31 = 0 (8)


Symmetry:

Sik = Ski (9)

Let's say S23 = S32 /= 0. according to (8) and (9) S31 = S13 = 0, what can't be because of (5), (6), and (9)


A three-port device can't be matched at all ports and lossless at the same time.

the wilkinson dividider is matched, but not lossless, because his scattering matrix breaks formula (4)


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Rod
Post subject:
Unread postPosted: Fri May 27, 2005 11:42 pm

As a power splitter, there is no power dissipated in resistor .

As a combiner, half the power is dissipated in the resistor.

If you combiner 2 signals in phase and amplitude, no power is dissipated in the resistor.



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