Question about feedback stability

In every textbook, the feedback stability is described as one input singal, one output and the feedback signal.
But how to analize the system which has no input singal, such as tieing the output directly with input? For example, a simple LDO, if the input voltage does not change and could be viewed as vitual ground in small signal analysis. If I have no input signal, how could I analyze the stability when some disturbance happens?

What does LDo represent? oscillator?
Oscillator is a good example of feedback stability issue. Stability is not satisfied so that the circuit oscillates. Razavi has some papers about using feedback stability technique to analyze oscillaotr. The key is to suppose there is some noise in the input end and then the normal stability analysis can be done.

The disturbance is your input signal. So you need to find the transfer function from the disturbance to the output and analyze it.

In reality, the LDO has an input: the reference (it may be internal and inaccessible, but it exists). A second input is the feedback input (from your output, possibly with a divider) and then it has the output.
The output (possibly divided down by the divider) is "compared" with the reference. In fact, this operation is done by a differential amplifier that effectively subtracts the feeback from the reference. Generally, this is part of the error amplifier itself. This controls the power stage and ultimately the output voltage. So that is how the output voltage is maintained constant: it is continuously "compared" with the reference, which is a stable voltage. See first diagram.

Now, for the disturbance, consider the reference equal to zero and after some manipulations of the diagram, calculate the transfer function Y(s)/D(s), that is, you consider the input is the disturbance.

Y(s)=D(s)-Y(s)*((Hd(s))*Hea(s)*Hpf(s))

So, Y(s)/D(s)=1/(1+(Hd(s))*Hea(s)*Hpf(s))

Now you analyze this transfer function, which describes the output response to an additive disturbance.

In case you are wondering, the manipulation is based on the fact that the output of the difference block is:

ε(s)=R(s)-Y(s)*Hd(s), so by dividing by Hd(s), we get

ε(s)=(R(s)/Hd)s-Y(s))*Hd(s)

Thanks for you analysis, really great.
but I think you maybe make a wrong deduction from Fig.1 to Fig.2, the transfer block after the error amplifier should be Hd(s) instead of 1/Hd(s).
So according to your analysis, it seems conclution is stability only relates to openloop gain and phase, in other words, 0db phase margin, right?

Yes, that is right, I made a mistake in the diagram, but corrected it now. Thanks!

The stability of the system is analyzed by looking at this new transfer function. It is very similar to the open-loop transfer function of the system. I say only similar because after the manipulations, we no longer took into account the "filter" introduced in series with the reference.

fantaci said:

So according to your analysis, it seems conclution is stability only relates to openloop gain and phase, in other words, 0db phase margin, right?

Click to expand...

i think stability not only relates to 0db phase margin but also relates to -180 degree gain margin. because when the system has zeros, maybe the phase is not degressive.

This seems to be a confusion: 0 deg or 180deg.
Strictly speaking, the stability is assessed when the phase is -180deg.
However, very often in practice, when measuring the open-loop gain, the measurement is done on the feedback block input. Therefore, you include the -180deg phase shift coming from the comparator. Thus, your total phase shift appears 360deg, or 0deg and that is where you assess stability.

But, as I mentioned, this is just another way of seeing things, and many practitioners will refer to the phase margin to mean the difference from 0deg instead of 180deg. The basic stability problem remains unchanged, except for the addition of this fixed 180deg phase shift. It's a matter of convenience.