1. M2 M1
Vbb
Vin CL M4 M3
Vyy
Vxx
VDD
VDD
Vo<Vxx+|Vt3|
flip up-down
for source Rb M2
Vbb CL M1
Vin
connecting D1 to S2
cascoding
Vo> Vbb-Vt2
ro = rop||ron, Av=-gm1*ro

2. VDD
VDD
flip left-right
to get this
differential
telescopic
cascoded
amplifier M7 M5
Vyy
Vxx
Vo<Vxx+|Vt3| M8 M6
vo-
vo+ CL M3 M4
Vbb CL M1 M2
vin+
vin- Vs
Vo> Vbb-Vt2
Tail current to change
gnd to virtual gnd

3. Let vin+ = Vic + ½ vid, vin- = Vic - ½ vid
The cross source: Vs = Vic – VGS1Q.
Left half circuit same as two slides back, but with vin = ½ vid.
So vo- = - gm1ro-(½ vid), where ro- = ro5,7||ro1,3.
Similarly, vo+ = - gm1ro+(-½ vid), where ro+ = ro6,8||ro2,4.
Hence: vod = gm1rovid
Av = vod/vid = gm1ro
where ro = ro- = ro+

4. ICMR M3 M4
Vbb
What is Vicmax?
VD1 = Vbb-Vt3-Veff3
For M1 in saturation, need VD1>VG1Q–Vt1 =Vic–Vt1. So, Vic<Vbb-Vt3-Veff3+Vt1
Vicmax=Vbb-Veff3
What is Vicmin?
Vs>Veff9 for M9 in saturation.
Vic>Veff9+Veff1+Vt1
Vicmin=Veff9+Veff1+Vt1 M1 M2
vin+
vin- Is M1 M2
vin+
vin- M9

5. VDD
VDD
Diode connection
to turn diff out
to single-ended
output  M7 M5 M8
Vo<VG6+|Vt6| M6 vo
Same ICMR.
But Vo range is different.
VG6=VDD-2Vtp-2Veff M3 M4 CL
Vbb M1 M2
vin+
vin-
Vo>Vbb-Vt2
Same Av, ro

6. VDD
VDD Vo Vo
Folded cascode

7. VDD
Vin CL
Vbb
flip up-down for
I sources
VDD M1
Vin CL M2
Vbb
connecting n-D to p-S

8. VDD
VDD
folded cascode amp
Vbb
Vin+
Vin- CL

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

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

11. Frequency dependent analysis
By including the effects of capacitors and by using s-domain impedances

12. MOST model: saturation
Chapter 4 Figure 11
Cgs  2/3 CoxWL, Cgd = CoxWLov= Cgd0W
Cdb = CjAD+CjswPD, Csb tricky, mostly short

13. Non-symmetric operation  Csb > Cdb
Chapter 4 Figure 12
Non-symmetric operation  Csb > Cdb
If S-B together, Csb is shorted. Easy for PMOS, but for NMOS, true only if S=gnd.

14. 1:M
Voutmax
Vin
I1Q=I2Q=
Pcs=
Ptot=
Voutmin

15. For Zo, set vin=0; Yo=gds1+gds2+s(C2+Cgd1)
KCL at vout: gm1vin-sCgd1 +Yovout=0
Av(s) = (sCgd1-gm1)/Yo
C2=CL+Cdb1
+Cdb2
1/R2=
gds1+gds2

16. Av(0) =
p1 =
z1 =
BW = |p1| =
GBW = Av(0)* BW=
SR = max dvout/dt =