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Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views. It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if you would like to post something on RF Cafe's Facebook page, please do.

Below are all of the forum threads, including all the responses to the original posts.

Guest
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

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.

Posted  11/12/2012
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