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I was watching a video about power supply design and it was mentioned that an opamp being used as a regulator like this:

schematic

simulate this circuit – Schematic created using CircuitLab

in practice, isn't an inherently stable configuration - "it's very tricky to get it stable over a range of output loads and capacitances"

Why is this? What happens in this circuit to make it unstable?

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I don't really see the big deal about this. Most voltage regulators can be made unstable by one means or another. A perfect op-amp and a perfect darlington might be unconditionally stable (in all circumstances). On the other hand, an imperfect op-amp and imperfect darlington might be just fine for such a device as the 7805 voltage regulator. Take a look at the innards of the uA7805: -

enter image description here

It looks complex but it has a darlington transistor output and a makeshift op-amp driving it. There is feedback to the inverting input of the makeshift op-amp and the non-inverting input receives a voltage from a reference circuit (far left).

The darlington is acting as a voltage follower, providing no gain at its emitter - why should this configuration be any less stable than an op-amp (unity gain configuration) with a load capacitor? Those types of circuits can be stable - it all depends on the op-amp of course.

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  • \$\begingroup\$ I think the question had to do with stability, not the novelty of the voltage regulator idea. \$\endgroup\$
    – Joe Hass
    Commented Apr 18, 2014 at 10:50
  • \$\begingroup\$ @Joe - my point is that you can't just draw an op-amp and transistor and expect to analyse/predict whether it will be stable or not. I'll add to my downvoted answer. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2014 at 10:51
  • \$\begingroup\$ @Andy_aka Well I think you can make a pretty good guess why there might be a statement that it's hard to get stable when there's a lag in the feedback loop. But your 7800 point does raise the question of what conditions might make a 7800 unstable. \$\endgroup\$
    – gwideman
    Commented Apr 18, 2014 at 11:03
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    \$\begingroup\$ @gwideman The thing is, an "artist's impression" of an op-amp and darlington is just meaningless - I've seen the same diagram (virtually) used to explain how a voltage regulator works. If the op-amp type (real world) had enough phase margin, it wouldn't necessarily be unstable. It might be close to instability or it might be perfectly fine. \$\endgroup\$
    – Andy aka
    Commented Apr 18, 2014 at 11:12
  • \$\begingroup\$ @Andy_aka Agreed. \$\endgroup\$
    – gwideman
    Commented Apr 18, 2014 at 11:27
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If there was no C1, only a resistive load, then you have negative feedback around the loop = stable. With C1 in the picture, feedback is delayed, which, at some frequency, looks like positive feedback... leading to oscillation.

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  • \$\begingroup\$ Yes - we have the same situation if an opamp hs a pure capacitve load (danger of instability due to a new pole). However, because such a capacitor has several advantages (noise filtering) it should have an appropriate small seires resistor (0.5 to several ohms). \$\endgroup\$
    – LvW
    Commented Apr 18, 2014 at 14:41

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