> One of the problems with darlingtons is transient intermodulation distortion
I don't see how that statement can be generally valid.
> we still have a problem with noise figure, right?
Where? The darlington emitter follower comes after a moderately high gain stage, that is moreover much noisier than the worst (modern) BJT. FET noise is (say) 1uV over the audio band, stage gain is like 30, so it outputs 30uV. Qa noise is maybe 1uV, Qb noise is more like 0.5uV. So using thumb-math, output noise is 30+1+0.5 or 31.5uV, only 1.5uV from the BJTs. As bcarso says, you are supposed to do a thing with rooted squares instead of simple addition; the long answer is still a bit over 30uV and almost all from the FET.
> So what if we changed Q2 to a Sziklai pair?
Scott spent a lot of time trying variations and listening to the sound. We can argue theory until the Corgis come home. But if we want sound, a well listened-to project beats all theory. Or at the least, you should try it Scott's way before wandering into your own variations and listening fugue.
> our output impedance goes up.
Does it? Given a relatively low-Z source (the drain resistor seems low enough), 30mA current, a straight Darlington with intermediate resistor comes out around 1Ω. The Sziklai depends on first stage current and second stage current gain; taking 1mA for the first and 30 for the second number, we again come to around 1Ω.
> Does it really matter, or is it purely academics in this case?
Hardly worth academics.
> the overall current gain that comes from a Darlington, provided that's necessary. This would make it a good candidate for driving a long cable, right?
No. In a reasonably synergistic design, current gain is just the number of transistors. The Darlington tries to fool you, being two transistors that look like one. Darlingtons ARE very popular in audio. But they have problems too. Maybe the biggest problem: it is "too simple". It seems like one transistor but in many ways it MUST be understood as two. A single transistor's current gain drops 6dB/octave. A Darlington ends up with 12dB/oct over a large range. This does strange things, like turning capacitive loads into inductive inputs (I think).
Get exact specs on your input and output. Say 1V into 10K in, 10V into 10Ω output. Input current is 0.1mA, output current is 1A, we need current gain of 10,000. This certainly needs more than one BJT. Two selected BJTs would do it, except leakage stability usually suggests working current gain much less than device current gain. At working current gain of 30 per BJT, we need 3 devices. These days we will probably run overall feedback. The feedback loop can't hog load current, and the feedback input (not defined by external box-level specs) may be lower impedance than the external input. So we may need more gain and more devices to close the NFB loop well. We see that loudspeaker amps rarely have less than 4 stages of current gain, and often more.
How much current do you think a cable sucks? 100 feet of cable is around 3,000pFd, which is around 300Ω at the top of the audio band. (Now you see why audio line impedances are generally 100-600Ω: even "unloaded" they run about that much in the treble.) If you can drive 300Ω, you can probably fiddle it to drive 100 feet of cable. You have the power. You may have stability issues but more power may not be the way to fix that (though it can be).
If you go to very long cables: 10,000 feet of fat wire costs a fortune. So we buy skinny wire. Now the resistance of the wire may be hundreds of ohms. The last third of the wire may be 100,000pFd or just a few ohms at the top of the audio band, but we see it through hundreds of ohms of plain copper. Also any conductor has inductcance, which leads to a nice sophmore-year EE course in general fields and lines. But skipping ahead to the real world of audio: a very long line has a Characteristic Impedance, for almost any kind of cable-pair it is 100Ω, and if we put 100Ω on both ends then we don't see any capacitance at all, just resistance. Or in the modern world, we just source lines with a 60Ω resistor, load them with much-higher-than 100Ω assure that a shorted line will not burn-up the amp, and everything is good.
featherpillow, do you have an image-cropping tool? Maybe I just hate to see a small schematic on a big sheet because I remember beating reeds into papyrus for my blueprints....
