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Click here to go to a page on mismatched terminations (new for October 2020)
Click here to go to a page on phase-variable mismatched terminations (new for November 2020)
New for November 2020. This may seem like an esoteric issue, but how much power is absorbed in a mismatched load? The true answer is that unless your power amplifier is isolated from the mismatch using a ferrite isolator, you might never know. This is because the mismatch will load-pull the amplifier, it is not a linear effect. Chances are you won't have a model of the amplifier that would predict the load-pull effect.
But, what if we consider that the load-pull effect is similar to mismatch loss? Then the problem becomes simple. We further postulate that mismatch loss would be less than load-pull power reduction, so this means our analysis will be worse-than-worst-case, which is a good way to approach any problem. If you have any data to refute this, please kick it in!
We made a spreadsheet where the VSWR of the load is the independence variable, then subtracted the mismatch loss term from it and put the answer in percent. For perfect match, 100% of the power is dissipated. At 2:1 mismatch, 89%, at 3:1, 75%, etc.
Now, suppose you were testing a 100W amplifier into a 5:1 mismatch. Would you have to specify the 5:1 mismatch to absorb 100W? the plot reveals that 56% of power will be absorbed, so you can safely get away with a mismatched load that can handle 60 watts.
VSWR effect on absorbed power
Now look on this page on mismatched loads. Supposed you create a 5:1 mismatch using five 10W loads. You should be safe to expose this to a 90W amplifier without smoking up the lab.