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
Top 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 Top malvinas2 Post subject:
Unread postPosted: Fri May 27, 2005 4:09 am Or just 'google'
around: https://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). Top 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 Top 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)
Top 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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