Why is the opamp output a voltage?

[AlexCaledin]

It's not really "charging" that capacitor. That is likely the "dominant pole" . . .

- well, at least, when that pole is actually dominant - that is, when the Phase is approximately 90 degrees - the amplifier works practically according to "charging" the capacitor with high frequency (from several kHz to several MHz) current. Of course, at low frequencies a 30 pF capacitor can hardly define anything and the output is more like current (amplified by Q15, Q19 and Q14/Q20).

 

[berkeman]

well, at least, when that pole is actually dominant

Sorry, I'm not following. Do you routinely add a different dominant pole in your feedback opamp circuits? If you do, how do you manage to maintain a reasonable phase margin and ensure stability?

EDIT/ADD -- And can you list an opamp part number where you actually are required to add your own dominant pole to keep your feedback circuit stable?

 

[Tom.G]

EDIT/ADD -- And can you list an opamp part number where you actually are required to add your own dominant pole to keep your feedback circuit stable?

LM108
http://www.mit.edu/~6.301/LM108.pdf

 

[AlexCaledin]

Do you routinely add a different dominant pole in your feedback opamp circuits?

- sorry! What I tried to say was simply "when the process of amplification - not only stability - is actually determined by that pole", that is, when the frequency of the amplified signal is such as to make the open-loop phase shift be about 90 degrees.

No, I routinely choose opamps that are stable with simplest feedbacks. (I am thinking how to eliminate the crossover distortion in an audio amplifier, using actually two quite simple amplifiers working together, one of them forcing another, via a little transformer, to have non-distorted output current - it seems very promising when simulated. To find the principles of connecting two systems together, I have to keep each of them simple.)