What is the advantage of using transistors on the power rails of an opamp over more typical boost circuits?

I have seen a circuit topology called the "Widlar Boost" made famous by Bob Widlar.

The power rails on the opamp have series resistors and the bases of the boost transistors are attached between the resistor and the opamp. Like this: Widlar Boost

There are other variants where the power to the opamp's rails go through the transistors emitter/collector or drain/source. Like this: enter image description here

What would these circuits do better than what a more common approach could (like transistor(s) on the output of the opamp to boost current capability)?

Edit: I am pretty certain I understand how these circuits boost the output: voltage drop across the series resistors (which is directly related to output current) affects the transistors in such a way that they provide the bulk of the current to the load.

My question is why would we want to do it this way? I am having trouble understanding how this could be better than hanging one or more transistors off the output of the opamp.

It seems to me that causing the voltage at the opamp's rails to fluctuate like this would only serve to make noise or distortion worse.

The second circuit I posted is the more mysterious one to me, even though it shows the entire circuit. It strikes me as needlessly complicated for what it is trying to accomplish, so I would like to learn what makes this a good solution (i.e. what problem is it trying to solve that a more basic circuit doesn't?).

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    \$\begingroup\$ That second schematic is particularly interesting to me, because it has what look like common-base amplifiers between the op-amp itself and the power rails. \$\endgroup\$ – Hearth Apr 27 '18 at 17:58
  • \$\begingroup\$ Well, I posted this answer a while back: Lighting either of two LEDs, or none. And this question: two-quadrant constant current pin-driver circuit. But I'd like to first see your analysis of these ideas before commenting. I'm probably not the best person to write a survey on this, but your examples seem easy enough to follow. \$\endgroup\$ – jonk Apr 27 '18 at 18:04
  • \$\begingroup\$ By typical boost circuits do you mean triple complementary Darlington Emitter followers which have much lower output impedance than these current sources with voltage feedback \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 27 '18 at 18:11
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    \$\begingroup\$ @Tony Stewart: Yes, not necessarily a triple darlington, but yes. I noticed you specifically mentioned how these would have a lower output impedance and identified these circuits as current sources with voltage feedback...are you giving me hints? \$\endgroup\$ – evildemonic Apr 27 '18 at 19:39
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    \$\begingroup\$ @evildemonic So, you know that there are two outputs with the long-tailed pair. Both collectors can be (aren't always) used. So 2 inputs, 2 outputs. Right? Well, stupid opamps only give you one output that is all decked out for your use. But they do have a lot of really nicely designed stuff in there. One way to get those two outputs back is to just drive the output into the ground through a resistor and instead grab up those the two rails as your outputs. Of course, they are current signals. But that's often fine. Does that make sense? \$\endgroup\$ – jonk Apr 27 '18 at 20:53

Your second schematic allows Vcc-Vee to be higher than the opamp can handle. The two small transistors keep its vcc/vee dc biased at +-19V which is as high as it can handle. The darlingtons use the +-38 V

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  • \$\begingroup\$ +10 if I could. This is the main purpose, to de-couple the power stage from limitations of OPA. However, I am wondering how this arrangement would keep the control over the bias current across the push-pull stage, say with temperature drift... \$\endgroup\$ – Ale..chenski Apr 27 '18 at 21:16
  • \$\begingroup\$ The 680 ohm resistors are probably not low enough. The darlingtons are supposed to be biased off. This circuit can't control any quiescent current so it will have some crossover distortion. \$\endgroup\$ – τεκ Apr 27 '18 at 22:16
  • \$\begingroup\$ One of my earlier thoughts was that maybe this was a creative attempt to eliminate crossover distortion, but as you say, this is not the case. I understand this is to have output capabilities beyond that of the opamp, but why is doing it with the rails better than doing it off the output? \$\endgroup\$ – evildemonic Apr 27 '18 at 22:18
  • \$\begingroup\$ This circuit can't control any quiescent current so it will have some crossover distortion.. Right. Or it will instantly fry out the expensive output pair of transistors (which did frequently occur in my practice when I was much-much younger and stupid). So, a much more complicated circuitry is required, to possible disappointment of OP. Sorry. \$\endgroup\$ – Ale..chenski Apr 27 '18 at 22:41

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