# How to Compute the Power Consumed by the Balance Resistor - RF Cafe Forums

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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.

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Peter Raynald
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Unread postPosted: Tue May 24, 2005 10:21 am
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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.

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SilencerEx
Post subject: Re
Unread postPosted: Tue May 24, 2005 10:50 am
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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.

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Peter Raynald
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Unread postPosted: Tue May 24, 2005 11:24 am
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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.

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SilencerEx
Post subject: Re
Unread postPosted: Tue May 24, 2005 1:25 pm
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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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