Two-stages class AB amplifier

by Stefanino   Last Updated September 11, 2019 13:25 PM

I'm studying the following two-stages class AB amplifier:

enter image description here

Basically it is a standard two-stages op-amp (i.e. the cascade of a differential pair with active current mirror followed by an inverter with active load) to which the source follower M8-M9 has been added in order to provide the single-ended output signal of the differential pair to the gates of both M5 and M6. I have doubts about some sentences of my book.

1st question

The book says that the gain of the second stage (i.e. the inverter with active load given by M5-M6) is:

enter image description here

Shouldn't the gain be negative? M5 and M6 give raise indeed to an inverter.

2nd question

I think that the book has an error: the gate of M1 should be in-, while the gate of M2 should be in+.

3rd question

Since the book does not provide the expression of the total low frequency gain, I surmise that it is given by the product of the gains of the three stages. Is it correct to assume the gain of the source follower equal to 1 for semplicity?



Answers 2


Q1. Gain is only a positive number, the input amplitude vs the output amplitude, It does not depend on the phase, if it inverts the signal and doubles its amplitude, it has a gain of 2 and a phase of 180 degrees.

Q2. Without simulating I am only assuming, I would take a guess its described like op amp inputs, Increasing the voltage on In+ increases the output voltage, Increasing the voltage on In- decreases the output voltage.

Q3. Voltage follower gain stages have a voltage gain of 1, so yes valid for low frequencies.

Reroute
Reroute
September 11, 2019 12:38 PM

Q2: when voltage at gate M1 is increased then the drain current through M1, M3 and M4 increases that will pull up (increases) the voltage at the gates of M8 and M5. So M5 will conduct less current making the output voltage drop (decrease).

No need to simulate that :-) but if you want, confirm this by simulation.

So yes you're right, the gate of M1 is the inverting input.

Q1: because of that (written above) now the gain becomes positive.

Bimpelrekkie
Bimpelrekkie
September 11, 2019 12:58 PM

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