1. vs vb cc VDD M9 M12 M11 Iref M1 M2 vo vin- vin+= voQ = vo CL M3 M4 M13

2. DC gain: Av(0)={gm1*(ro6,8||ro2,4)} * {gm10*(ro10||ro11)
=Av1(0) * Av2(0)
With Miller simplification:
p1 = -1/{(Av2*Cc + Cgs10 +…)*(ro6,8||ro2,4)}
 -1/{Av2*Cc*(ro6,8||ro2,4)}
p2  - gm10/{CL+Cc+Cdb10+Cdb11}
BW = |p1|
GBW = Av(0)*BW  gm1/Cc

3. But Miller simplification gets rid of bridge cap, hence hides zero.
Going back to schematics, examine vo1  io  vo
io = vo1*(sCc – gm10)
vo = io/(sCLtot + go) =vo1*(sCc – gm10)/(sCLtot + go)
 z1  +gm10/Cc

4. wp1
gm10/CL
A1(s)A2(s)
A1(s)
wp2
wz1
A2(s) w
0dB
(s/z1 - 1)
A(s)A(0)
(s/p1 - 1)(s/p2 - 1)
UGF
GBW

5. Phase Margin:
PM = 180 +(A(s))= 180 +A(s)
= 180 +(s/z1-1) - (s/p1-1) - (s/p2-1)
= 180 – atan(GBW/z1) – 90 – atan(GBW/|p2|)
= 90 – atan(gm1/gm10) – atan(GBW/|p2|) 
Make GBW < 0.5 |p2|
Make gm1 < 0.1 gm10

6. In the previous, we assumed  to be frequency-independent.
How can we do that?
Examine 3 cases:
Buffer connection
Resistive feedback
Capacitive feedback

7. vs vb cc =1 VDD VDD VDD M9 M12 M11 Iref M1 M2 vo vin+= voQ Vin- = voCL vb M3 M4 vo cc R _ + M5 M6
M10 M7 M8
vin2
=1

8. vs vb cc VDD VDD VDD M9 M12 M11 Iref M1 M2 vin- vo vin+= voQ CL M3 M4

9. vs vb cc VDD VDD VDD M9 M12 M11 vo Iref vin- M1 M2 vin+= voQ CL Rf M3Ri cc R M5 M6
M10 M7 M8
vin2

10. This is s-dependent.
To create an s-independent b an, place a Cf in parallel with Rf, with Cf*Rf = Cin-*Ri.
Then,
is real.

11. vs vb cc VDD VDD VDD M9 M12 M11 Iref M1 M2 vo vin- vin+= voQ CL M3 M4

12. vs vb cc VDD VDD VDD M9 M12 M11 vo Iref vin- M1 M2 vin+= voQ CL Cf M3Ci cc R M5 M6
M10 M7 M8
vin2