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My educated guess: It's the terrible power supply rejection ratio of this circuit. It's got a maximum voltage amplification of 1000. Your input voltage is fixed, yet the 1.65V reference voltage depends on the power supply. Therefore it amplifies changes in the power supply voltage by a factor of 1000, so that a 3.3mV change in VCC already drives the ...


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Many op amps have a maximum capacitive load on their output before they become unstable; you're giving this one a 10 nF load, since the integrating capacitor is connected between the output and the virtual ground. Check section 4.3 (page 13) of the datasheet. Microchip recommends putting a small resistor in series with the output for stability if the ...


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If you'll look at the examples/Educational/UniversalOpamp2.asc you'll see that there are 4 levels for the UniversalOpamp2: level1 is a basic VCCS and an R||C output level2 is a single pole opamp level3a is a two pole opamp and programmable phase margin level3b has an additional dominant pole and delay You used level2, so a simple opamp with a single pole, ...


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For the regulator to regulate accurately to a voltage of Vref the regulator needs to have a very high dc open loop gain. Resistive compensation will reduce the open loop gain at all frequencies including dc which will reduce regulation accuracy. Capacititive regulation (CC) provides a frequency dependent roll off of the open loop gain. So the open loop gain ...


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Since \$|n-n_0 +1|>|n-n_0|\$, since all \$n-n_0\$ are non-negative, the statement of all the inequality signs is accurate. Lugging around the extra \$+1\$ doesn't help!


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I learned to use Spectre (and other simulators) when there was no STB type source. So what I used (and still do) is a lowpass filter (RC) and a VCVS. Here I use that to keep the feedback loop working at DC but at AC the loop is open as the RC filter prevents an AC signal from getting through. To make a Bodeplot of the AC transfer, I add a voltage source with ...


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This is not an answer, but extended comment. The target readers of this post are visitors that come here through searches or links. There are a few questions on electronics SE about frequency compensation techniques for opamp circuits with capacitive load. The questions have received answers of different quality, and one answer even insists on being "...


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You are right to think that the closed-loop inductive output impedance is not responsible for the potential instability. The stability of the loop is determined by the loop gain, which does not contain the inductive element that the opamp looks like after closing the loop. Instead one must look at the open-loop output impedance. As a final note, the open-...


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I reckon you're missing the fact that whatever's connected to the output of an opamp, actually does form part of the feedback loop. It's not obvious from the classic opamp follower diagram, because that diagram does not show such an influence "inserted" into the loop itself: simulate this circuit – Schematic created using CircuitLab However, ...


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