circuit analysiscircuit-designcmosfeedbackintegrated-circuit
From this online lecture (page 190-20), I see that the negative feedback (Rs in common source degeneration) helps increases linearity range.
I am wondering if anybody know an explanation for this intuitively or mathematically.
Another question is what is the difference between the left and right hand pictures in the image below?
For small signal, they are exactly the same.
Thank you.
"So why do we only assume negative feedback takes place"
It is not correct that "we assume negative feedback" only. Who say this?
The shown two-opamp circuit (introduced by A. Antoniou) can be interpreted as a combination of two "Negative Impedance Converter (NIC)" circuits. There are two basic NIC types (current-inversion - INIC, and voltage-inversion - VNIC) and both exhibit a negative input impedance. However, a stable combination results if we replace the grounded output resistor of the first NIC unit (INIC) with the negative input impedance of the second NIC unit (VNIC).
This forms already a circuit called "Generalized Impedance Converter GIC". However, this form has some disadvantages - and therefore a modification is used which is known as Antonious GIC circuit (as shown in the question). It is easy to show that this alternative has the best properties of all possible modifications - as far as influence of real opamp parameters is concerned. This form can be derived from the simple NIC combination exchanging some opamp input nodes which are (for ideal opamps) at the same potential.
This GIC is extensively used in active filter realizations as an "active inductor" and/or as a "Frequency-Dependent Negative Resistor (FDNR)".
The bias current is set by the current source Ibias, it is mirrored to the active load Q2 and the current through Q1 has to be about equal to the current delivered by Q2.
This is not done by some sort of bias network but through application of feedback. Imagine this stage as the second stage of an OpAmp as shown below.
The CS Stage consists of M3 and M8, biased by M1.
In order to use this Opamp feedback is required. For example OUT could be fed back to IN-. The first stage of the OpAmp will take care of the biasing. The voltage at the gate of M8 will be such, that OUT is almost equal to IN+.
Best Answer
Suppose you have a nonlinear transfer function \$\frac{v_O}{v_{IN}} = a(v_{IN})\$
If you add negative feedback 1/m such that:
\$ \frac{v_O}{v_{IN}} = a(v_{IN})(v_{IN} - \frac{ v_O}{m})\$
so
\$ \frac{v_O}{v_{IN}} = \frac {a(v_{IN})}{1+\frac{a(v_{IN})}{m}}\$
For \$\frac{a(v_{IN})}{m} \gg 1, \frac{v_O}{v_{IN}} \approx m \$ (linear gain of m)
For \$\frac{a(v_{IN})}{m} \ll 1, \frac{v_O}{v_{IN}} \approx a(v_{IN}) \$ (nonlinear with negligible feedback)
As you can see, the closed loop gain must be reduced significantly from the open-loop gain: \$m \ll a(v_{IN}) \$ for there to be a large reduction in nonlinearity.