Most audio amps will current limit and/or be extremely inefficient if you apply a conjugate match load. As will a nuclear power plant. The Maximum Power Transfer theorem doesn't usefully predict the maximum power you can get from either.
That's because they are nonlinear.
John
You have it exactly backwards. It's Mr. Audiophool who is proposing such nonsense (matching audio circuits using the "maximum power transfer" theorem) and the Fields-Thompson twins are trying to tell everyone he's right when any second-year EE student should know better. Actually, since Thompson has me killfiled, he doesn't know what I've said and is only knee-jerking his twin (obviously without reading that the nut is saying).
The fact is that I've told them that the problem defines a fixed load. The most efficient (best power transfer) driver is zero ohms. ...and amazingly enough, that's what is found in audio gear.
Fields is the one who believes this problem has *anything* to do with "maximum power transfer".
Why don't you read the thread before sticking your head into it? You fell on your ass, too.
Except for intermodulation distortion... my standard test is to play a Mozart woodwind ensemble that has both French Horn and Oboe. If you have significant IM distortion it will glaringly show up :-( ...Jim Thompson
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I love to cook with wine. Sometimes I even put it in the food.
--- Sure.
If they can supply the current, why not?
Here's a cute little table with a 50 ohm source:
. RG RL EL PL .OHMS OHMS VOLTS WATTS . . 50 10 0.33 0.0109 . 50 20 0.57 0.0163 . 50 30 0.75 0.0188 . 50 40 0.89 0.0198 . 50 50 1.00 0.0200
-- Easily! ;)
That's a bit misleading, isn't it? Any amplifier will have current and voltage limits, and SATURATION is what it's called when you explore those limits. Turn the abasolute power level down, and the misbehaving amplifier/load combination becomes linear.
Turn the absolute power level up, and you're exploring nonlinear parts of the operating curve, regardless of the exact nature of the load. Audio amps are intended to pass a test with a dummy (resistor) load, and usually get max-power-before-distortion into a simple resistor of 8 ohms (or 4 ohms, or ... whatever). The max-power-before-distortion is NOT maximum power transfer, though, it's a totally different test criterion. You can do the max-power-transfer test using a millivolt test signal, and no strain on the amplifier's saturation limits.
Oh, he avidly reads what we post, even though he has claimed to have kill-filed us. Makes no sense to me.
John
Go buy a commercial audio amp or receiver. Measure its output impedance. It will typically be below one ohm, usually way below. Now, load it with an equal value resistor. Then see how much power it can deliver.
Or perform the same experiment on an AC outlet in your home.
The maximum power transfer theorem isn't generally useful for nonlinear things like audio power amplifiers.
John
-- _You're_ the one who's fallen the furthest what with all the tall tales of yours about what I said and what I didn't say. You really should go back and read the first two posts I made to this thread and you'll find, since it seems you've talked yourself into believing that they were straight from hell, that in the first what I did was explain to the OP that audio amps are voltage sources with very low impedance outputs and how they act with different loads on them. The second was agreeing with him about his battery circuit and that, indeed, power transfer is maximized in a load when its impedance matches the load's impedance. Simple enough, and I proved it, but your nasty self had to get your knickers all in a bunch because you can't stand to be argued with, especially when you're wrong. Oh, well... Hope you get better. :-)
-- The point isn't how much power the amp can deliver, it's that the load will be dissipating as much power as possible when its impedance equals the source's. Can't you read the tables above and didn't you look at the sim I posted?
Not for a real source, like an audio power amp.
I didn't run the sim. The tables assume a linear, resistive source.
You won't try it? The first part is interesting, measuring the impedance.
That's never stopped you before.
John
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