Guest Post subject: How to compute the power comsumed by the balance
resistor Unread postPosted: Tue May 24, 2005 3:05 am I used
Wilkinson 2-way splitter/combiner to combine PA power. The frequency
is 900MHz. The port 2 power is 20W, and port 3 is also 20W, so port
1 is 40W. Normaly, the balance resistor (100ohm) is lossless and can
provide good isolation between port 2 and port 3. But how to choose
the power rating for the balance resistor? And how to compute the power
consumed by resistor for mismatch? Please help me. Thanks.
Top Peter Raynald Post subject: Unread postPosted:
Tue May 24, 2005 10:21 am Offline Captain User avatar
Joined: Tue Sep 07, 2004 3:09 pm Posts: 11 There are two
situations you might want to consider: 1) One of your two power
amplifiers are dead, then your resistor will dissipate 10W, half one
amplifier's power. 2) Your amplifiers become, for an non-identified
reason, out of phase, then the resistor will take a full 40W load.
Top Profile SilencerEx Post subject: Re Unread
postPosted: Tue May 24, 2005 10:50 am Offline Captain
Joined: Tue May 24, 2005 5:41 am Posts: 5 The resistor will only
dissipate reflected waves. So when an amplifier is dead, the full wave
is reflected back. If the power divider has an infinite isolation between
port 2 and 3, the total 20W of reflected power will be dissipated in
the resistor. When both amplifiers are working, it doesn't matter
if they are out of phase or not. The reflected power (caused by impedance
mismatching) of both amplifiers will be dissipated anyway. With a good
amplifier design, the reflected power will be a lot less then 20W.
So when both amplifiers are working correctly, almost no power is
dissipated in the balance resistor. If you want the resistor to coope
with a failing amplifier, you should take one that can handle 20W.
Top Profile Peter Raynald Post subject: Unread
postPosted: Tue May 24, 2005 11:24 am Offline Captain User
avatar Joined: Tue Sep 07, 2004 3:09 pm Posts: 11 The
wilkinson divider, combiner is an in-phase combiner. When both
input signal are in-phase and equal in amplitude, the unbalanced load
of port 1 sees the power and the balanced load (100ohms) sees nothing
since voltages are equal. When both input signals are out of
phase, then there is a virtual ground in the middle of the 100ohms resistor
and also at port one. The virtual ground at port one shows that it becomes
the isolated port and dissipate no power. The virtual ground
at the 100ohms resistor shows it dissipate all power from the system.
Therfore the phase in which both amplifers are combined in does
matter. Also from my understanding, the power dissipated in the
100ohms resistor have nothing to do with reflections unless the output
load of the combiner is bad. Most of the power hiting this resistor
is therefore disbalance power. Top Profile SilencerEx
Post subject: Re Unread postPosted: Tue May 24, 2005 1:25 pm
Offline Captain Joined: Tue May 24, 2005 5:41 am Posts:
5 Ah, I get my error now. I thought it was used to divide the power,
but you're using it to recombine the power. You're right.
Posted 11/12/2012
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