operational-amplifiervoltage-regulator
I was watching a video about power supply design and it was mentioned that an opamp being used as a regulator like this:
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?
Yes, the divider definitely will affect the gain calculation, because its equivalent source resistance (Thévenin resistance) of 6.23 kΩ is in series with one of your 10 kΩ gain-setting resistors. The overall gain will be -0.2325, rather than the -0.375 that you're looking for.
If you want an overall gain of -3/8, it would be simpler to forget about the voltage divider and just set up your opamp with 20 kΩ input and 7.5 kΩ feedback resistors.
Also, be sure to use an opamp whose common-mode input range includes ground.
First of all, a 9V battery will not output 2Amp. This alkaline battery http://data.energizer.com/PDFs/522.pdf can give you 500mA but even then the rated capacity is halved.
Next, your "Zener circuit" will start conducting above 17V (it starts conducting when Q2 base is 0.7V above ground, so when current through voltage divider reaches 0.7V/4.7kOhm=0.149mA ; this happens when input voltage is 114.7kOhm * 0.149mA = 17V). As-is it is equivalent to the voltage divider alone ; your + input is therefore about 0.2V below 9V. So the op-amp will be fully-on all the time.
Finally, did you use this particular amplifier? That's not an op-amp (gain is 20x) and the output is referenced to Vcc/2 (here to 4.5V). So, when inputs are equal the output voltage for your circuit is 3.8V.
I'd suggest you replace it with a proper op-amp, and choose it so that the output voltage can swing to at least 1V above output (that's 3V below rail) while delivering significant current. Also mind the input voltage range...
Or, if you're not after high precision (as your use of a non-temperature-compensated reference suggests) why not use directly your zener "circuit" and a pass transistor? Something like this :
simulate this circuit – Schematic created using CircuitLab
This works because the zener steals the current from the transistor base as soon as the voltage is above the zener threshold, which is 0.7V above the output (because of the base-emitter voltage drop of the transistor).
Obviously for 2A you also need a beefier pass transistor, something like TIP41 instead of poor little BC547... It will have to dissipate max 2A*4V, that's 8W of power!! Then lower R3 to 47 Ohm.
Best Answer
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: -
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.