Impedance matching

From HvWiki

Impedance matching is a really complicated topic, please hold while we try to figure out a way of explaining it!

Maximum power transfer theorem

Classically, impedance matching is supposed to be done by the maximum power transfer theorem. This consists of determining the Thevenin equivalent impedance [[1]] of the generator, calling this the "output impedance", and making the load impedance the complex conjugate of this.

So if our (AC) generator had a resistance of 50 ohms and 1 ohm of leakage inductance (at the operating frequency) in series, the optimum load would be a 50 ohm resistor in series with a capacitor having a reactance of 1 ohm. This would extract the maximum power possible from the generator.

This assumes that you actually want the maximum possible power, which is not always the case. For instance, the "output impedance" of a wall socket in your home might be a few tenths of an ohm, made up of roughly equal parts resistance and inductive reactance. You really don't want to connect a matched load to that.

It also implies 50% efficiency, since half of the power is lost in the generator's internal resistance.

Impedance matching in amplifiers

This example explains some of the pitfalls of the maximum power transfer approach, using the example of an audio power amplifier. Similar arguments apply to devices such as Class-E RF amplifiers: even though they are designed to operate optimally into a 50 ohm resistive load, they don't have a 50 ohm internal resistance. (They can't, because if they did, they would be at most 50% efficient, and we find that they can be up to 90% efficient in practice.)

The "impedance" of the generator in this case is a fictitious concept that somehow expresses the maximum voltages and currents that its designer would like to see loaded onto his amplifying devices.

...

First, the maximum power transfer theorem. EEs seem to love this for some reason, but we need to drive a stake through its heart today.

The two main problems are that it assumes the resistances to be linear, and that it assumes you actually *want* the maximum power! Let's take the example of a solid-state power amp that uses IRFP460s in the output stage. What is the output impedance really? I can give you several answers.

First of all, there is the small-signal impedance which is relevant to speaker damping. This is a function of the whole amp design, including the negative feedback loop. A typical solid-state amp might have a damping factor of 400 at 1kHz, so the output impedance is 8/400 = 0.02 of an ohm.

Now, what is the impedance that gives us maximum power transfer? Well, maybe it's equal to the impedance we calculated above, 0.02 ohms. Hey, this screwdriver has a resistance of about 0.02 ohms, so I'll just throw it across the speaker terminals. With a 50V B+ I should get 1250 amps out and 15.6kW RMS. Bang! Oh, I guess that was wrong. Let's get some fresh MOSFETs, pick pieces of the old ones out of our foreheads, and try again.

Maybe it has something to do with the actual resistances in the circuit, then. An IRFP460 turned fully on has a Rds(on) of 0.27 of an ohm. And let's say we used a 0.47 ohm emitter resistor. There is a total of 0.74 ohms of actual resistance between the speaker and B+, which might be 50V. So naive application of the MPT theorem suggests an optimum speaker load of 0.74 ohms now, and a resulting power output of 422W RMS.

Of course, trying to get 420W into .7 ohms from two TO-247 packaged transistors in a linear amp will result in instant destruction. If we look at the IRFP460 datasheet, the required peak current of 135A is well over the safe absolute maximum. Bang again! So we wipe MOSFET guts off our safety goggles that we wore this time round, slap a label next to the speaker terminals saying: "4 ohms minimum load", and goodbye maximum power transfer. The situation is similar for high powered tube amps.

Now let's look at the effect of feedback on these two output impedances. The first one, the small signal one that determines the damping factor, is what gets changed. Negative voltage feedback lowers it, and negative current feedback raises it.

In a pentode or beam tetrode amp, it's hundreds of ohms, so negative voltage feedback is mandatory to lower it. However, even with as much feedback as you can apply without going unstable, it's still several ohms for the kinds of circuits in musical instrument use.

In a solid-state amp, it's a few fractions of an ohm even with no feedback, so you may use current feedback to raise it to mimic the above pentode or beam tetrode amp with cheap transformers and modest NFB.

Ultra-linear operation gets down to maybe 8-10 ohms without NFB, and triode lower still, which is why audiophiles who refuse to use negative feedback prefer them to pure pentodes or beam tetrodes. If only the poor fools realised that UL is just another kind of feedback. ;-)

The second impedance we discussed, the minimum load for safe operation, is not affected by any amount of tampering with negative feedback. To change it you need to swap out transformers, add more tubes or power transistors, etc.