Why does the feedback mechanism linearize the vi vs vo characteristic curve?

Question

For a unity gain configuration opamp, we arrived at the conclusion that vout = vi * A/(A+1) by using small signal analysis, since A here is the small signal open loop gain. My question is: Why does the formula which we derived by using small signal analysis apply also for large signal? I mean, why is it the case Vo = Vi * A/(A+1). Here Vo and Vi are both large signal voltage. I would explain this based on the fact that feedback can linearize the vi vs vo characteristic curve, so the small signal gain is also large signal gain, since the slope of a straight curve doesn't change according to Vi. But how can linearization actually happen? Or is there any other explanation?

The derivation for vo = vi * A/(A+1) is the following:

A*(vin - vout)=vout

so

vout = A*vin/(1+A)

But my doubt is that A is the open loop gain, which is derived from small signal analysis. How can this derivation apply also for large signal analysis?

Answer 1

Or is there any other explanation?

Forget about op-amps for a couple of minutes and think about a servo control mechanism like this one for example: -

You set a "position demand" with the potentiometer on the left and, the error amplifier "calculates" the voltage difference between "actual_output" (from the position measurement potentiometer) and "position demand". That difference signal drives a motor (a highly non-linear device) to move its shaft in the correct direction and, the "calculated error" reduces accordingly.

As the output from the error amplifier gets smaller, the drive to the motor reduces and, eventually, the error between demand and output position is too low to drive the motor any more. It's an imperfect control system because of motor non-linearities and "real world" problems such as mechanical linkages, stiction and imperfect potentiometers.

If the error amplifier used a significant amount of gain to amplify the error signal then, the final static error reduces and, the output quite accurately matches the input demand. Despite all the real-world non-linearities and imperfections, when you have a decent gain in the error amplifier, you get a decent correlation between input demand and final position.

It's the same formula as the op-amp and has got nothing to do with large or small signal analysis - the gain of the error amplifier drives the system (via negative feedback) to make output equal demand: -

$$V_{\text{output potentiometer}} = V_{\text{demand potentiometer}}\cdot\dfrac{A}{A+1}$$

And, all an op-amp circuit does is bypass the motor and output potentiometer like this: -

Image of control system from this answer.

Answer 2

Question: Why does the formula which we derived by using small signal analysis apply also for large signal?

(1) At first, the input stage of the opamp is a differential amplifier which has a transfer characteristic which mathematically follows a tanh function. For small input signals (millivolt range) this function has a very good linearity.

(2) Due to the high gain properties of the opamp - if operated in its active region (with output voltages still within the supply rail limits), the input signals will never exceed these small voltages (in real practice: µV range).

(3) For all practical amplifier applications, the opamp is used with negative feedback. The advantages are twofold:

(a) Due to the feedback effect, the differential voltage between both opamp terminals will not exceed the above mentioned limitations - even for input signals of several volts. This holds as long as the ouput does not reach the supply limits. In this case, the input differential voltage is Vout/Aol (with Aol very large).

(b) In addition to this effect, negative feedback further improves linearity (as can be observed also in simple transistor amplifier stages). For example, see here: https://www.sciencedirect.com/topics/engineering/feedback-linearization

EDIT: Here is another fine document (distortion as a measure for linearity)

https://www.google.de/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwiMhrCg7Y7tAhUPLewKHVXwD5gQFjADegQIAhAC&url=http%3A%2F%2Frfic.eecs.berkeley.edu%2F142%2Fpdf%2Fmodule15.pdf&usg=AOvVaw1Tg9YfjWvz-Ts-EadSAddh

Answer 3

Write Uout=A(Uin-Uout)+E where E is how much the output differs from the linear gain output. E of course depends in a complex way on the input and output voltages. In addition E isn't unique, it depends on how big linear gain A we assume.

Solve Uout. The formula is Uout=Uin(A/(1+A))+E/(1+A)

Uout approaches Uin if A grows and the amp works in a way that E stays limited.

The question is complex because circuits can behave in complex ways. It's easy to find amps which do not get linearized. For example if Uout stays constant +1 volts we, of course can still write Uout=A(Uin-Uout)+E where E=(1V - A(Uin-1V)) and A is whatever linear gain. But we cannot push E/(1+A) to zero by increasing A.

Answer 4

a larger view of the math is this

• Vout = VIn * G / ( 1 + G * H)

where "G" can be somewhat non-linear IF "H" is a resistive feedback voltage divider.

Audio amplifiers depend on this.

Transformer_feedback RF amplifiers depend on this (using transformer ratios to set the feedback).

Notice the assumption: the H is linear, while the G can be rather NONLINEAR.

Answer 5

Imagine that the open-loop gain of the op-amp is not constant, but rather varies with input voltage A = f(Vin), so it is nonlinear.

For a good op-amp f(Vin) will be a large positive number, regardless of Vin, though it may vary.

Vo = Vi * f(Vin)/(f(Vin) + 1). If f(Vin) >> 1 then Vo ~= Vin.

Similarly, an output offset is reduced by the gain of the op-amp.