Loop-gain calculation

[akbarza]

hi,
i think this pic belongs to Allen cmos analog circuit design book, i find it in an internet file.
i want to know how can i calculate loop-gain in this pic?
thanks.

 

[LvW]

Are you familiar with the loop gain definition?
Do you know how you would measure/simulate the loop gain function?

 

[akbarza]

thanks
i read a file about loop-gain and also i know from control lesson that A(CLOSE LOOP)= A(openloop)/[1+B.A(openloop)].
and the multiplication of B.A(OPENLOOP) is called loop_gain

 

[LvW]

So - the question remains: What is "B" in your circuit?
What is the name of the quantity "B"?

 

[FvM]

So - the question remains: What is "B" in your circuit?

Does the shown circuit represent the complete feedback amplifier? If so, the topology is different from the basic "loop gain A, feedback factor B" topology by an additional forward gain factor C. You get a closed loop gain expression like


G = C*A/(1 + A*B)
signs are a matter of convention, assuming positive A and B you get
B = R1/(R1 + R2)
C = -R2/(R1+ R2)

- - - Updated - - -

To be done: derivation of open loop amplifier gain A

 

[akbarza]

hi
i thinked befor that for obtaining loop gain , we had to multiply B(feedback factor) and A_O(open loop gain). BUT I KNOW IT IS NOT RIGHT.
Thanks for answers

 

[FvM]

In fact it is right.

 

[LvW]

hi
i thinked befor that for obtaining loop gain , we had to multiply B(feedback factor) and A_O(open loop gain). BUT I KNOW IT IS NOT RIGHT.
Thanks for answers

As mentioned by FvM: Lopp Gain=A*B.

Do you know how to find the open-loop gain A?

 

[Easy peasy]

Respectfully - there is no loop in the ckt.

 

[LvW]

Respectfully - there is no loop in the ckt.

So - you do not see any connection between output and input?
I must admit, I see such a feedback path.
Without it, the circuit would not work because of a missing DC biasing point.

 

[FvM]

The circuit has internal feedback, one of the possible readings is that its loop gain should be calculated, see analysis in post #5.

 

[LvW]

The circuit has internal feedback, one of the possible readings is that its loop gain should be calculated, see analysis in post #5.

Yes - the circuit exhibits exactly the same feedback method as can be found in the inverting opamp configuration.

 

[Easy peasy]

is there is some marginal internal feedback? - nope. Loop gain refers to gain around the complete loop, there is no ckt shown to complete the loop.

what you are seeking is the forward transfer function of input to output with specified load.

 

[FvM]

Consider a setup without additional external feedback, e.g. Vin connected to a voltage source. Then the internal feedback through R2 (feedback factor R1/(R1 +R2) ) is the only feedback in this circuit. It's not "marginal" but - among other parameters - determining the circuit gain. A loop gain can be calculated.

 

[LvW]

is there is some marginal internal feedback? - nope. Loop gain refers to gain around the complete loop, there is no ckt shown to complete the loop.
what you are seeking is the forward transfer function of input to output with specified load.

The two transistors can be used as a linear amplifier.
The same applies to an opamp.
However, the problem in both cases is that - without negative feedback - there will be no usable DC bias point in the middle of the quasi-linear region.
(I know - the reason for the missing bias point may be not the same...but this is not of any importance with respect to the subject under discussion).

Therefore, we use two resistors (forming a simple voltage divider) for feeding back a part of the developped output voltage to the inverting input.
However, it is to be mentioned that - in the circuit under discussion - there must be 100% feedback for DC (CMOS inverter with Dcin=DCout).
Therefore, an additional capacitor at the input will be necessary. This results in a feedback factor k=1 for DC (no DC current through R2) and k=R1/(R1+R2) for AC.
For calculating the gain of the complete circuit, we are using the classical formula for the closed-loop gain of an amplifier with feedback (given in post#5): G=C*A/(1+A*B).

Easy peasy - anythig wrong ?

 

[Easy peasy]

R2 provides degeneraton - a loop requires connection to the actual input ...

 

[FvM]

a loop requires connection to the actual input

Why particularly the "actual" input? A loop is created by feeding back a signal from amplifier output to input. In the given circuit, the drain node can be recognized as output and the gate node as input.

 

[LvW]

R2 provides degeneraton - a loop requires connection to the actual input ...

What is an "actual input"? What is your definition? The CMOS inverter has only one single input node - the common node for both gates.
Is there any doubt that R2 is connected between output and input?
As an example, consider the non-inverting opamp configuration.
The feedback path is connected to the inverting input - and the "actual" input (do you mean: signal input?) is the non-inv. opamp terminal.
The circuit under discussion shows the classical negative feedback scheme.
(by the way - "degeneration" is nothing else than negative feedback; see the "degeneration resistor" RE in the emitter path of a common emitter gain stage).

 

[CataM]

To be done: derivation of open loop amplifier gain A

Here it is. Negative feedback looks fancier than your negative feed-forward ("C").

 

[mtwieg]

I had a professor in undergrad who absolutely insisted that simple circuits like this be analyzed in terms of shunt/series feedback systems. What fun we had, spending half an hour analyzing calculating the gain of an inverting opamp amplifier, which could be solved in seconds using KVL and KCL.

For more info on this horribly inefficient and outdated method of analysis, see: **broken link removed**