Derving transfer function

Hi everyone, I am trying to derive the transfer function based on the small signal diagram given below, it is LDO regulator. It is found in this journal
https://ieeexplore.ieee.org/xpl/articleDetails.jsp?tp=&arnumber=6297490&queryText=ldo



I derived manually using KCL based on the diagram and I couldn't express the final form as shown in the attachment, but I tried to equate my coefficient of s^3 and that one in the diagram and they turned out to be the same ! So I assumed that my approach is correct.

Anyhow, is there any more appropriate or more correct way of deriving the transfer function just like those who published their journals ?

Help is appreciated

whompah, all three active units are transconductances - right?
I see that there is a feedback loop and that the output current of the feedback OTA is added to the output current of the first (left-sided) OTA.
Thus, we can apply the classical feedback model resulting in the general equation for the closed-loop transfer function

H(s)= Forward gain (from input to output) / (1-loop gain).

Note that the loop gain (gain of the inner loop) is negative (required for stable operation) and does not involve the first OTA.

LvW said:

whompah, all three active units are transconductances - right?
I see that there is a feedback loop and that the output current of the feedback OTA is added to the output current of the first (left-sided) OTA.
Thus, we can apply the classical feedback model resulting in the general equation for the closed-loop transfer function

H(s)= Forward gain (from input to output) / (1-loop gain).

Note that the loop gain (gain of the inner loop) is negative (required for stable operation) and does not involve the first OTA.

Click to expand...

yes I did that using KCL as well, but I can't get the final expression as shown in the figure, although I expanded my expression and compare with the shown answer, the coefficients are the same. I wonder how these open loop transfer function are derived.

Here is my method:

let Vx be the node between 1/gmc and Cc, Vo1 be the node of R1 and C1.
the current buffer "feedback" is realised by a common gate configuration.

Applying KCL at node Vx:
(Vout - Vx) s Cc = gmc Vx ...1

Applying KCL at node Vo1:
gm1 Vin + gmc Vx = Vo1 (1/R1 + s C1) ...2

Applying KL at node Vout:
gmp1 Vo1 + Vout (1/R02 + s CL) + gmc Vx = 0 ...3

Then I solved these 3 eqns by getting rid of Vx and Vo1, but I couldn't get the final expression as shown in the diagram. Is this a correct way of doing it ?

As far as I can see - all three equations are correct.
But, for comparison with the given formula: What is beta and Adc?

LvW said:

As far as I can see - all three equations are correct.
But, for comparison with the given formula: What is beta and Adc?

Click to expand...

beta is the feedback factor of the LDO from the output back to the input of op-amp (which is 0.5)
Adc is the DC gain, which is (gm1 gmp1 R1 R02)

whompah said:

beta is the feedback factor of the LDO from the output back to the input of op-amp (which is 0.5)
Adc is the DC gain, which is (gm1 gmp1 R1 R02)

Click to expand...

I don`t understand. How can the feedback factor appear in the numerator ?
With beta=0 (no feedback) the whole function is zero?

Following your approach you have to solve a system with 3 equations.
Should work in principle - however, perhaps rather involved.
Instead, following my approach (feedback model and associated equation) you only have to insert the two expressions for the forward gain and the loop gain.
This could be somewhat easier.

LvW said:

I don`t understand. How can the feedback factor appear in the numerator ?
With beta=0 (no feedback) the whole function is zero?

Following your approach you have to solve a system with 3 equations.
Should work in principle - however, perhaps rather involved.
Instead, following my approach (feedback model and associated equation) you only have to insert the two expressions for the forward gain and the loop gain.
This could be somewhat easier.

Click to expand...

The transfer function derived is actually a loop gain of the whole LDO with A(s) times beta
the small signal diagram shown is actually an open loop, in which the loop is already broken, that's why we have to find A(s) = vout(s) / Vin(s)

OK - the only information I have is the diagram as shown in your 1st post.
Based on this diagram I have drawn a block diagram (I hope it is correct).
Perhaps it helps - see pdf attachement.

I forgot to mention that H(forward) has to be calculated without feedback, of course - that means: no feedback current (i,R=0).

On a side note, why is that most such papers describing some complex compensation schemes never derive the equations but it just appears out of nowhere?
I have also faced this problem with some papers regarding advanced compensation techniques. I even tried solving the thing using Matlab's symbolic processing but to no avail.

@whompah
I hope you are able to solve this. Please update this thread with your results since I too am interested in the LDO described in that paper.

Sorry for another error in my pdf attachement.
The loop gain is Hloop=ir/is (with i1=0)