compensation des poles dominants
Since nobody has answered, I'll give some background information.
The loop gain (return ratio) of an amplifier can be written as
Aβ(s) = Aβ(0)N(s)/D(s), where Aβ(0) is the DC loop gain.
Now the closed-loop gain of an amplifier is given by
A_cl = A_ideal • -Aβ(s)/(1-Aβ(s)), where A_ideal = 1/β.
Substituting the above in A_cl gives
A_cl = A_ideal • Aβ(0)N(s)/[D(s) - Aβ(0)N(s)]
So the poles of the closed loop gain (system poles) is given by the characteristic equation CP(s) = D(s) - Aβ(0)N(s) = 0.
To simplify loop gain calculations, we'll neglect C_µ (C_gd for FETs) in the hybrid pi-model. This is equivalent to a case with n poles and no finite zeros, that is, N(s) = 1 and
D(s) = (s-p1)(s-p2)...(s-pn).
Substituting this for the characteristic equation gives (after some manipulations)
s^n - s^(n-1)Σpi + ... + [1-Aβ(0)](s-p1)...(s-pn) = 0, where Σpi = p1+p2 + ... pn. Compensating an amplifier means that we want to move the system poles in a certain position, that is, the characteristic equation should be
CP2(s) = (s-ps1)(s-ps2)...(s-psn) = s^n - s^(n-1)s^(n-1)Σpsi - ... = 0, where psi are the wanted system poles.
We can now compare CP(s) with CP2(s) and to compensate the amplifier we want Σpi = Σpsi. According to my text, it is not always possible to move all poles such that Σpi = Σpsi. It only works if Σpi ≥ Σpsi, that is if the sum of the loop poles is larger or equal than the sum of the wanted system poles. So, why is this true? I guess one should be able to understand that from the above? I should also mention that I have little knowledge of analog electronics, I just started learning frequency compensation.