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

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

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

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

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

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

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

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

vs vb cc VDD VDD VDD M9 M12 M11 vo Iref vin- M1 M2 vin+= voQ CL Rf M3 Ri cc R M5 M6 M10 M7 M8 vin2

  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.  

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

vs vb cc VDD VDD VDD M9 M12 M11 vo Iref vin- M1 M2 vin+= voQ CL Cf M3 Ci cc R M5 M6 M10 M7 M8 vin2