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

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

3. D1 connects to G2, two stages
VDD
VDD
VDD
VDD
two stage
CS amplifier
CS amplifier with a
source follower buffer

4. VDD
VDD
VDD Vx Vx
Same as above,
only T2 is pMOS
Connecting S1 to D2
makes ro really small
buffer or output stage

5. VDD
Vbp
Vbn
vin vo

6. 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

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

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

9. 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))}

10. 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) ’

11. 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.

12. 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+…

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

14. 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

15. 4.2.6
If vin is v-source,
Small signal model
Should be: ~

16. VDD
VDD M7 M5
Vyy
Vxx M8 M6
vo-
vo+ CL M3 M4
Vbb CL M1 M2
vin+
vin- Vs

17. VDD
VDD
VDD Rb M2
Vbb CL M1
Vin Vo Vo
Both have the same small signal model