Two stage opamp dumb question

Hi everyone,

I have a question regarding a two stage opamp in which I can't visualise/imagine in an intuitive way but I see everybody saying the same thing leading me to wonder if those people know really why.

What I normally hear people saying, for example, regading a two stage opamp with the miller cap for compensation is something like this: because this is a high gain node the voltage here will not vary much, the amplifier with try to maintain this voltage always constant. Say at the gate of the second stage MOSFET.

Question: Is this really true, this affirmation? I mean, you can see that it is true through simulation, but why? Can this be explained wirh words or with equations? OK, this is providing that the opamp is in negative feedback - I guess.

Regards.

Hi,

I don´t see your circuit, nor your simulation, therefore just a general answer:

Assume you have a gain of 100,000.
so for an output voltage of:
* 0V you theoretically need 0V/gain = 0V / 100,000 = 0V input voltage difference
* 1V you theoretically need 1V/gain = 1V / 100,000 = 0.00001V = 10uV input voltage difference

The difference of just 10uV usually is hard to measure, because it will be negligible with respect to
* offset voltage
* offset voltage drift
* noise

Klaus

Hi, thanks for the reply.

The circuit is this one:



The point where A is shown is the point on the circuit that I was referring to not changing much because of the high gain stage. That's the argument that I sometimes hear/read.

Why?

CAMALEAO said:

... because this is a high gain node the voltage here will not vary much, the amplifier with try to maintain this voltage always constant.

Click to expand...

View attachment 140126
The point where A is shown is the point on the circuit that I was referring to not changing much because of the high gain stage.

Click to expand...

A node where the voltage doesn't change much is a low impedance node. The A node in your OTA schematic above is a low impedance node due to the Miller effect: its original impedance Z is reduced by the Miller feedback to Zin = Z/(1+Av) , where -Av is the gain of the following stage, s. e.g. this Wiki explanation.