1. VDD
VDD
Vo<VG6+|Vt6|
=VDD - |Vtp| - 2Veff M7 M5 M8 M6 vo
Vo>Vbb-Vtn
>Vicm-Vtn+Veff M3 M4 CL
Vbb M1 M2
vin+
vin-

2. VDD
VDD
Vo<VG6+|Vt6|
=VDD - |Vtp| - 2Veff M7 M5 M8 M6 vo
Vo>Vbb-Vtn
>Vicm-Vtn+Veff M3 M4 CL
Vbb M1 M2
vin+
vin-

3. VDD
VDD M7 M5
Make I1R=Veff
or 1.5Veff M8
Vo < VDD - 2Veff M6 vo M3 M4 CL
Vbb
Vo > Vicm-Vtn+Veff M1 M2
vin+
vin-

4. M3 M1
vin+ Vo Vo

5. VDD
VDD Vo Vo
Folded cascode

6. VDD
Vin CL
Vbb
flip up-down for
I sources
VDD M1
Vin CL M2
Vbb

7. VDD
Vin+ CL
Vin-
Vb3
folded cascode amp
Vb2
Vb1
Vb4
Vb5

8. Output swing range and input common mode range are now decoupled.
OSR: 2Veff to VDD-2Veff
total: VDD-4Veff
ICMR: 2Veff+Vt1 to VDD-Veff +Vt1
total: VDD-3Veff
But, if Vic cannot exceed VDD,
ICMR: 2Veff+Vt1 to VDD
VDD-2Veff-Vt1

9. Miller equivalent circuit
I2 =Y(V2-V1)
=Y(V2+V2/A)
=(1+1/A)YV2 
Y2=(1+1/A)Y Y
Chapter 4 Figure 15
I1=I=Y(V1-V2)
=Y(V1+AV2)
=(1+A)YV1
Y1=(1+A)Y

10. The above is proven by assuming constant A.
But in reality, A depends on Y and connection.
Correct use of Miller is quite tricky!

11. v2 = -Av1 v2
Agm1R2
This is for DC.

12. But this is missing the positive zero: z1 = +gm1/Cgd1
v1/vin=1/(1+sRs(Cgs1+(1+gm1R2 )Cgd1))
vo/v1= -gm1/(G2 + s(C2 +Cgd1))
Multiply: vo/vin= -gm1/{(G2 + s(C2 +Cgd1))(1+sRs(Cgs1+(1+gm1R2 )Cgd1))}
But this is missing the positive zero: z1 = +gm1/Cgd1
 Cgd1
A better approach: first set Rs=0, find TF from gate to vo, as we did before;
then use Miller effect, find TF from vin to v1 (left circuit);
then multiply together to get TF from vin to vo.
 vo/vin = (sCgd1 -gm1)/{(G2 + s(C2 +Cgd1))(1+sRs(Cgs1+(1+gm1R2 )Cgd1))}

13. g'm=gm1+gs11.2gm1
KCL at vo:
-g’mvs+(vo-vs)gds+vo(GL+sC2)=0
vo/vs=(g’m+gds)/(gds+GL+sC2) ’

14. Zero value time constant:
If a TF=n(s)/d(s) has only real poles
d(s)=(1+t1s)(1+t2s)(1+t3s)…
=1+(t1+t2+t3+…)s + (t1t2+t2t3+…)s2+…
For low frequency, s small, we have
d(s)  1+(t1+t2+t3+…)s
Hence the lowest freq pole is approximately
p1  -1/(t1+t2+t3+…)
This only good for low freq, and real poles.

15. Application:
Perform series/parallel simplification
For each remaining capacitor, find resistance it sees by: Short V-source, open I- source, open other caps
teff = R1C1+R2C2+…

16. C2 is CL||Cdb1||Cdb2
R2 is rds1||rds2
Cgs1 sees Rs, C2 sees R2
For Cgd1, more complex.

17. Resistance seen by Cgd1 can be found by applying a test v3 across Cgd1, finding the resulting current i3, and using R3=v3/i3
R3 = R2 + (1+gm1R2)*Rs
teff=RsCgs1+R2C2+R2Cgd1+ (1+gm1R2)*RsCgd1