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So as far as I understand, the basic idea of a linear regulator is to use a series-pass element along with an opamp with negative feedback to stabilize the output. This is what I usually see as an example:

schematic

simulate this circuit – Schematic created using CircuitLab

Now, to use a MOSFET in place of the BJT, can I just do a direct replacement like this:

schematic

simulate this circuit

Or should the gate of the MOSFET be driven by another BJT like below?

schematic

simulate this circuit

Or does it not make any difference? Is one more stable than the other?

Also I remember reading somewhere on the internet that in order for the opamp to regulate correctly, a small current should flow out of the output through the negative feedback loop, but a MOSFET gate doesn't allow any current in, which is also confusing.

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  • \$\begingroup\$ Remember that a MOSFET has a minimum gate-source ON voltage, thus your op-amp and Vin should be able to source the voltage needed to saturate the MOSFET ON. For standard MOSFET's this is about +10 volts. Thus if Vin is => 12 volts it should work just fine. \$\endgroup\$
    – user105652
    Jun 7, 2020 at 4:57
  • \$\begingroup\$ @Sparky256 you mean just the MOSFET by itself, right? Do I not need a BJT to drive the gate? But yes, the input voltage is greater than 12. \$\endgroup\$
    – darksky
    Jun 7, 2020 at 5:01
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    \$\begingroup\$ Don't think the third one will work. The BJT will dump charge into the MOSFET, but there's no discharge path. Gate leakage (if any) likely won't be enough to discharge it either. So this will end up just turning on the MOSFET and no proper regulation. regarding your comment on the current feedback: maybe add an RC network between the output and negative terminal? In general that configuration can help improve phase margin. \$\endgroup\$ Jun 7, 2020 at 5:01
  • \$\begingroup\$ @MatthewDiNardo I see. Can you please show me what that RC network looks like? \$\endgroup\$
    – darksky
    Jun 7, 2020 at 5:03
  • \$\begingroup\$ parallel resistor and capacitor \$\endgroup\$ Jun 7, 2020 at 5:04

2 Answers 2

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It is quite possible that none will work effectively without a compensation network applied directly to the op-amp. Whether it's an op-amp supplying a load directly or via a buffer transistor, adding bulk capacitance to the voltage output will shift the phase margin towards the point of instability: -

enter image description here

The object of the above modification is to reduce the gain of the op-amp at higher frequencies so that the excessive phase shift brought about by the bulk output capacitance won't act on a part of the spectrum where the loop-gain is greater than unity. You might have noticed that some regulators do state that they have a maximum capacitance that can be added to their output - this is the same issue - they have internal compensation that counters "so-much" bulk capacitance but no-more.

Regards driving the MOSFET, because it is a source follower, it's gate-source capacitance won't be fully seen at the op-amp output and it might be OK stability wise but the bulk output capacitance is still present and is still a problem (see above). There's no real reason to use circuit 3 given what I've said above and, if you did you would need to add an emitter resistor to ground.

Simulators are a good friend for fixing and uncovering the problems mentioned above.

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On the face of it your first circuit will work .Your second circuit could work if there was a pulldown resister on the gate of the mosfet .Now if the mosfet is big it will load up the opamp with lots of capacitance affecting opamp stability .You will then have to work on the control loop.Also remember that there is no current limiting so if you do not add this the fet could blow up if the load was a short circuit.

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