1. Differential Amplifier
amplifies the difference between two input voltages
rejects the average or common mode value of the two voltages
v+, v-, vo are single ended
Diff-mode input: vid=v+ - v-
Com-mode input: vic=(v+ + v- )/2
vo =Avcvic + Avdvid
Avd: Diff voltage gain, Avc: CM voltage gain
Com-mode rejection ratio: Avd/Avc

2. Differential Amplifier
Input common-mode range (ICMR)
vic range over which Avd remains the same
Computed by requiring all MOST affected by vic to be in saturation
Output offset voltage Vos(out)
Voltage at vo when vid = 0
Input offset voltage Vos
vid that is needed to make vo = 0
When we say offset, default is input offset
Vos= Vos(out)/Avd, if Vos(out) is in linear range

3. Differential Input, single-ended output single stage AmplifierVm
N-Channel
vin-
vin+
When vid=0, Vo=Vm.
So, Vos-out= Vm-VoQ,
Vos-in= (Vm-VoQ)/Avd

4. Large Signal Eq. in a N-channel Differential pair
=0.5b1(VGS1-VT)2
=(2ID1/b1)0.5
iD1=0, when iD2=ISS and VGS2=VT+(2ISS/b)0.5

5. Solving for iD1 and iD2
iD1=iD2=ISS/2
VON1=VON2=(ISS/b)0.5

6. INPUT COMMON MODE RANGE
VG1=VG2=ViCM
VSDSAT1=VSDSAT2
=VOD1
VD1=VD3=
VSS+VT3+VOD3
VG1min=VD1-|VT1|
=VSS+VOD3+DVT
VG1max=VDD-
VSD5SAT-|VT1|-VOD

7. Output Range Vomin=Vss+Von4 Vomax=Vicm –|VT2| So what’s the vo range
What’s for the N-ch
circuit.

8. SMALL SIGNAL ANALYSIS AV

9. Common Mode Equivalent Circuit, with perfect match
iC1=VIC/(1/gm rds5)
ro1≈1/gm3
ACM≈
1/ 2rds5gm3
iC1

10. SLEW RATE: the limit of the rate of change of the output voltage
C’Ldvo/dt=i4-i2
ISS
Max |CLdvo/dt|=ISS
ISS
Slew Rate = ISS/C’L
ISS
Output swing: Vosw
GB frequency: fGB
vo(t)=Voswsin(2pfGBt)
Max dvo/dt
=Vosw2pfGB
To avoid slewing: ISS > C’L Vosw2pfGB

11. Parasitic Capacitances
CT: common
mode only
CM: mirror cap
= Cdg1 + Cdb1
+ Cgs3 + Cgs4
+ Cdb3
COUT = output cap
= Cbd4 + Cbd2
+ Cgd2 + CL

12. Cross CM disturbance After feed back, V+ very close to V-. Vcs  0.
Vcs  vic

13. how should the bulk be connected
how should the bulk be connected? Connect to source or connect n-well to VDD p-well to VSS

14. Approximate transfer function AV(s) = AV/(s/p1─1)
Impedances
rout = rsd2 || rds4 = 1 / (gds2 + gds4)
rM = 1/gm3 || rds3 || rds1 ≈ 1/ gm3
Hence the output node is the high impedance node
When vi=0, slowest discharging node is output node with dominant pole p1 = -1/(C’outrout), where C’out = Cout+ Cgd4
Approximate transfer function
AV(s) = AV/(s/p1─1)

15. Gain bandwidth product
Gain AV(0) = gm1 / (gds2 + gds4)
Bandwidth ≈ |p1| ≈ (gds2 + gds4) / C’out
GBW ≈ gm1 / C’out
gm1 = {2*ID1mCoxW1/L1}½
increase gm1  increase GBW
increase W1  increase GBW
But C’out has Cdb2 and Cgd2  W1
Once Cdb2 and Cgd2 become comparable to CL, increasing W1 reduces GBW

16. An approximation of TF: AV(s)=AV(s/z1-1)(s/z2-1)/(s/p1-1)(s/p2-1)
If p1 is dominant, |p1|<<|p2|,|z1|,|z2|; AV(s)≈AV/(s/p1-1)
If p1 is non-dominant, at low frequency, AV(s)≈AV /(s/p1+s/p2-s/z1-s/z2 -1)
1/peq≈ 1/p1+1/p2-1/z1-1/z ≈ 1/p1+1/p2-1/z2 , since |z1| >> |z2|, |p1|, |p2|; ≈ 1/p1, if AV4 is not very large
In either case, BW ≈ p1

17. frequency response AV w -90 PM -180
All in abs val z1 w p1 p2 z2
UGF
-90 PM
-180
Small CL and small gm3: gm1/CL’ > gm3/Cgs3

18. frequency response AV w -90 PM -180 Large CL, gm1/CL < gm3/2Cgs3
All in abs val
UGF w z1 p2 p1 z2
-90 PM
-180
Large CL, gm1/CL < gm3/2Cgs3

19. Observations PM ≈ 90 – tan-1(UGF/z1) If z2 not = p2, UGF < AV*p1
GBW should be at least 2~3 times lower than z1 to ensure good phase margin at UGF
There is conflict between AV and PM
If z2 not = p2, UGF < AV*p1
Design approaches
make z1 high  higher than UGF
make Cgd2 small, gm1 large
make z2 close to p2  better 1st order approx.
make AV4 small
make p1 low  large AV
make gds2 and gds4 small

20. NOISE Model

21. Input equivalent noise source

22. Total output noise current is found as,
Let
Then