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Middlebrook's Method VS Tian's Method VS Middlebrook's General Feedback Theorem

Dominik Przyborowski

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Here is nice paper comparing various stability methods, including Middlebrook and Tian.
 

    promach

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promach

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Middlebrook's method makes the assumption that there is no forward-transmission through the feedback network. Tian's Method is more general because it doesn't make this assumption.
Someone told me the above which I am bit confused. Do you have any comment ?

Is this related to impedance looking into the negative input of the opamp is much larger than the impedance looking into the feedback network from the input of opamp ?

Here is the loop gain test circuit for Middlebrook method. Why use Vt= AC 1 on the feedback path ?

How to derive this more accurate formula for loop gain (equation 3.64) ?





 

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> Middlebrook's original method requires that the injected loop test signal be inserted between a low impedance point
> (usually the output of the power supply) and a high impedance point (usually the feedback amplifier network).
> There is no backwards loading with such a insertion point.
But why use "V2 = AC 1" on the feedback path ?
Why no backwards loading ?
Why must the test signal located between low impedance point and high impedance point ?

> Tian's method allows for backwards loading so that the signal insertion point may be anywhere along the loop.
> To do this, it requires two measurements and some simple calculations to combine them.
How does Tian's method accounts for reverse feedback ? Is backwards loading equivalent to reverse feedback ?
What does it mean by the inserted probe elements result in smaller, sparser circuit matrix ?

> The General Feedback Theorem allows for multiple influential feedback loops all of which may have backwards loading.
Why GFT allows multiple feedback loops while Tian's Method and Middlebrook's method do not ?
 

promach

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> Middlebrook's original method requires that the injected loop test signal be inserted between a low impedance point
> (usually the output of the power supply) and a high impedance point (usually the feedback amplifier network).
> There is no backwards loading with such a insertion point.

This quote is not from my webpage. I don’t think that it’s correct either. In his original paper from 1975, Middlebrook did take impedances into account, but not backwards transmission. The conditions given above are for voltage loop gain.

But why use "V2 = AC 1" on the feedback path ?

For the voltage loop gain analysis, you can use any AC value, because you are calculating a ratio of two voltages and AC analysis is linear.

Why no backwards loading ?

Why must the test signal located between low impedance point and high impedance point ?

To understand the conditions for voltage loop gain, please read https://www.omicron-lab.com/fileadmin/assets/Bode_100/Documents/Bode_Info_LoopGain_V1_1.pdf (which is also linked from my webpage).

> Tian's method allows for backwards loading so that the signal insertion point may be anywhere along the loop.
> To do this, it requires two measurements and some simple calculations to combine them.

How does Tian's method accounts for reverse feedback ? Is backwards loading equivalent to reverse feedback ?

Tian’s formula is symmetrical. He basically adds forward and reverse loop gain in his formula as you can see in his article.

What does it mean by the inserted probe elements result in smaller, sparser circuit matrix ?

You get a smaller matrix because you don’t need two copies of the circuit as in the original LoopGain.asc example.

> The General Feedback Theorem allows for multiple influential feedback loops all of which may have backwards loading.

Why GFT allows multiple feedback loops while Tian's Method and Middlebrook's method do not ?

The basic version of the GFT also allows for only one feedback loop. However, it can be extended for multiple loops, see https://web.archive.org/web/2018100...l-network-theorem/gnt-integration-in-virtuoso
The above quote is reply from Frank Wiedmann, the author of the loop gain article


Now, how to derive equations (17) and (19) below ?


Middlebrook's Method_Proof_1.png

Middlebrook's Method_Proof_2.png
 

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promach

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Equation (17): Looking at Figure 13a and calling the current through Z2 (from + to –) i, we have (with vx=v being the voltage over Z1, incorrectly also labelled vz in the figure):

vy = Z2 * i
vx = Z1 * (i – Gm * vx)
Solve the first equation for i, insert it into the second one and solve for vy/vx.

Equation (19): Looking at Figure 13b, we have:
v = Z1 * ix
v = Z2 * (iy – Gm * v)
Insert the first equation into the second one (twice) and solve for iy/ix.
 

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