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

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

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

4. 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))}

5. 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.

6. Application:
Perform series/parallel simplification
For each remaining capacitor Ci, find resistance Ri it sees by: short V-source, open I-source, open other caps, and compute R
teff = R1C1+R2C2+…

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

8. 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

9. For a differential amplifer: vin+=vic+vid/2, vin-=vic-vid/2
As vid change, Vs=const
Vs is virtual ground.
left half = -(right half)
Can use half circuit: M2
vin- CL M8 M1
vin+ M7
Vyy
vo+
vo- Vs
VDD
VDD
VDD
Vyy M7
R,C
vo+
vo+ CL CL M1 M1
Can further use quarter circuit for analysis with suitable parasitic R, C:

10. VDD
VDD
Telescopic
cascode
amplifier
Quarter circuit: M7 M5
Vyy
Vxx M8 M6
VDD M3 M1
Vbb
Vin CL Rb Vo
vo-
vo+ CL M3 M4
Vbb CL M1 M2
vin+
vin- Vs

11. VDD
VDD
VDD Vo
Vb3
folded cascode amp
Vb2
Vb2
Vb3
Vin+
Vin- CL
Vb4
Vb1
Vb5
Quarter circuit

12. Both have the same small signal model
VDD M3 M1
Vbb
Vin CL Rb Vo
VDD Vo
Vb3
Both have the same small signal model

13. a a >1 First take Rs = 0, find TF from vg1 to vo
then from TF from vin to vg1,
Finally, multiply the two together.
Include gs2
Include Cgs2,
Cdb1, Csb2
Include ro of I-source if folded a
a >1
vg1/vin =1/(1+sRsaCgs1)

14. To find TF from vg1 to vo:
Write KCL at vs,
KCL at vo,
Two equations, eliminate vs, and solve for vo in terms of vg1
 vo/vg1

15. ro  rds1*Av2 || RL
Av(0)  - gm1ro
p1 = - w-3dB = -1/roCout
GBW = gm1/Cout
z1 = +gm1/Cgd1
p2  -1/{(rds1|| rincg)Cs2}
p3  -1/RsaCgs1

16. Design steps:
SR, CL  IQ
OSR  Veff ranges
Ibias or Iref  size current source
GBW, CL  gm
IQ, gm  (W/L)1, round to nice numbers
Feedback to generate VinQ
Check Veff, if too large, increase W1, (this leads to over design of gm and GBW)
Check Av or ro, if not sufficient, use larger L, and adjust W accordingly to keep the same W/L
For cascode structures, modify the steps a little.