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I am creating a precision 3A current sink which functions when connected to 12+ volts. I am using the following circuit:

The voltage reference is created by a 0.4V precision reference chip (LT6650CS5) and divided down to 0.3V.

For some reason the output of my Op-Amp is oscillating instead of settling which is causing the output of the FET to oscillate.

Op-Amp output (vout1) Op-Amp output (vout1)

Any ideas?

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    \$\begingroup\$ Add a small capacitor(~100pF range) right between your output and negative opamp input. Go further, connect resistor in kOhm range between vfb1 and negative terminal so that capacitor has some impedance to work against \$\endgroup\$ – Bip Jun 26 '17 at 19:02
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    \$\begingroup\$ I'm not real good at this, but I think the gate capacitance is too high for the opamp. The gate capacitance (if I'm reading the datasheet correctly) is 1600 pF. The opamp datasheet says the maximum capactitive load is 1nF (1000pF.). That coukd be enough to make the opamp unstable. \$\endgroup\$ – JRE Jun 26 '17 at 19:08
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Do this to isolate the heavy load of the gate from the AC feedback path of the op-amp. The DC feedback path remains accurate.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The missing bit: MOSFET gate is a capacitor Cgs. Load at the opamp's output is Cgs in series wien R3, but R3 is too small to matter. Problem with capacitive loads is they add a phase shift in the feedback loop, which reduces phase margin and means Nyquist's criterion for stability will not be met, and the opamp will oscillate. Spehro's circuit decouples the opamps' feedback loop from the capacitive load and thus makes the opamp stable. \$\endgroup\$ – peufeu Jun 26 '17 at 21:36
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You will have to add a small series resistance to the output of the op amp and a small feedback cap to allow the op amp to drive the capacitive load of the MOSFET gate.

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    \$\begingroup\$ You were downvoted (not by me) because you were fundamentally wrong. The desired loop gain is 1, which the direct feedback provides, and the desired error is very small - again the direct feedback produces it. The comments and Spehro's answer provide the proper approach. \$\endgroup\$ – WhatRoughBeast Jun 26 '17 at 21:23

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