1. Voltage doubler for gate overdrive

2. OUTPUT AMPLIFIERS Requirements Types of Output Stages
Provide sufficient output power in the form of voltage or current.
Avoid signal distortion for large signal swings.
Be power efficient.
Provide protection from abnormal conditions.
Types of Output Stages
Class A amplifier.
Source follower.
Push-Pull amplifier ( inverting and follower).
Negative feedback (OP amp and resistive).

3. Power efficiency It is most power efficient at maximum signal level
Let VSS= ─VDD, Vin is sinusoidal such that Vout reaches Voutmax
PRL = ½ (Voutmax)2/RL
Psupply>=average((VDD or VSS)*IRL) =VDD*average(Voutmax*sin()/RL) =2*VDD*Voutmax/RL/p
Power efficicy = PRL/Psupply<=p/4 (78%)

4. < 25% CLASS A AMPLIFIER Constant bias current Iout max = IQ PRL
Power effic =
Psupply
0.5voutmaxIQ
< 25% =
IQ(VDD-VSS)
VSS= -VDD, Voutmax=VDD-Vdssat

5. SOURCE FOLLOWER
or VSS+VT
VB=VSS

6. Push-pull

7. Push-pull inverting amp
Gain = ?

8. Implementation
vin range:
Gain = ?

10. PUSH-PULL SOURCE FOLLOWER

11. If Vo = 0
Bias current well controlled
Vo swing is very limited

12. Very small

13. If A(Vosp-Vosn)<<Vov
If Vosp=Vosn=0

14. How do we implement this error amp?
Simplest solution:
use G and S as the + and – input
and D as the output

15. Super source follower DVo => DI1 =(gm1+gmb1)DVo
DVGS2= ro1(gm1+gmb1)DVo
DI2 = gm2ro1(gm1+gmb1)DVo
VDD I1
go=gm2ro1(gm1+gmb1)
+(gm1+gmb1)+go2
≈gm2ro1(gm1+gmb1) M1 Vo M2
Vin
Gm ≈gm1+gm1ro1gm2 I2
gm1
AV=Gm/go≈
gm1 + gmb1
HW: rework these when I1 and I2 have finite ros.

16. VDD
If we re-arrange with
a flipped version, we
get this push-pull
super source follower I1 M4 M2 M1 Vo M3
Vin
HW: provide a transistor
level implementation.
Comment on power
efficiency. Calculate your Ro and Vo range. I2

18. Output stage for differential amps:
Vxx M4 M7 M7
Vyy M3 M3
Vbb M2 M2 M6 M6 M1
Vzz M1

19. Second stage push pull: Monticelli style
Vxx M4 M7 M7
Vyy M3 M3
Vbp
Vbn
Vbb M2 M2 M6 M6 M1
Vzz M1
Requires: VDD-VSS > Vgs6+Vgs7+Vdssat_floating_CS
D. M. Monticelli, “A quad CMOS single-supply Op Amp with rail-to-rail output swing,” IEEE J.Solid-State Circuits, no. 6, pp. 1026–34, Dec

20. Don’t do M4 Vxx M4 M7 M7 Vyy M3 M3 Vbb M2 M2 M6 M6 M1 Vzz M1
Unpredictable current in second stage

21. These circuits can come from the same biasing circuit for the main amplifier.
So, no extra current, power consumption, noise, and offset introduced.
Vxx
Vbn
Vbp
Vzz
So, Vg6Q = Vzz
This sets Id6Q.
So, Vg7Q = Vxx
This sets Id7Q.

22. Floating CS do not change ro1 or DC gain
The impedance looking down from Vo1+ is Rn
To find impedance looking up from Vo1+, inject a test current i up.
V’o1+ = i*Rp
i_gdsn/p = (i - gmnvo1++gmpv’o1+)
Vo1+=V’o1+ + i_gdsn/p /(gdsn + gdsp)
Vo1+=i*Rp+(i - gmnvo1++gmpi*Rp) /(gdsn+gdsp)
Vo1+(1+gmn /(gdsn+gdsp))
=i*{Rp[1+gmp/(gdsn+gdsp)] +1/(gdsn+gdsp)}
Vo1+/i =Rp gmp/gmn
gds4 Rp
gds3
gm3vcp
V’o1+
gmnvo1+
gdsn
gdsp
-gmpv’o1+
vo1+
gm2avcn
+gmb2avcn
gds2a Rn
gds1
So, size them so that gmp ≈ gmn
Note: gmn and gmp include possible body effects.

23. To the two gate terminals of M6 and M7, the two floating CS appears as a voltage source providing a voltage offset between the gates.
The impedance seen by the two gate terminals can be calculated by:
Rs = (Vgp – Vgn)/(current through the floating CS)
= (Vgp – Vgn)/(gmp*Vgp – gmn*Vgn + (Vgp – Vgn)*(gdsn+gdsp))
≈ 1/(gmp + gdsn+gdsp)
≈ 1/gmp
The above assumed that the NMOS and PMOS are sized to have the same gm.
Also, the calculation is only valid when both NMOS and PMOS are fully on and both in saturation.
When Vo is experiencing large swings, these conditions are not met. And the voltage difference between the two gate terminals no longer remain constant.

24. HW: summarize the above discussion into a design procedure for given desired bias currents in output stage Io and the cascade stage Ic. Suppose a Iref is given.
HW: find Vo swing range, if all transistors remain in saturation. Find io range if all transistors remain in saturation, where io = i7 – i6. When the floating current source transistors are allowed to go into triode but all other transistors need to remain in saturation, what is the io range?

25. Low Voltage Push-pull output
Main goal: make Vo swing from Vss to Vdd.
Equivalent to double the output stage gm, ↑gain, GB
Make slew rate higher than Io/CL
1:M
Io/M Io
1:M
Make output Vebo small. When Vo1 swings, say by 2.1Vebo, ICL can be 10xIo!
But quiescent Io depends on Vo1Q, and uncontrolled.

26. In order for the output transistors to have well defined quiescent current, we have to bias the circuit so that at Q, Vd12 = VgsnQ to establish correct quiescent current in nMOS. This must be through current mirror ratio.
Problem:
Vd12 is a high impedance node, small current mismatch in M10 and M6 leads to significant voltage change at vd12, which in turn changes the biasing current in the output stage.
Solution:
use feedback to stabilize common mode of Vd12.
Vzz

27. Vzz
Vd12
Vd11
Feedback to
M4 or part of it
Since Vd12 and Vd11 are normally nearly constant. We do not need to worry about the input range accommodation for this circuit.
Size the circuit so that, when vid =0, Io remain near desired level over all process variations in M10 and M6.

28. Output quiescent current control by local CMFB.
Vo1 CMFB
1:M
Itail
Io/M
Io/M Io
1:M
1:1
match

29. Output quiescent current control by local CMFB.
Vo1 CMFB
1:M
Itail
Io/M
Io/M Io
1:M
1:1
match

30. Can passive CM detector be use?

31. M1a+M1b forms diff pair with M2
Vo1+/- swing is about +- 2.# * Vebo, or about to 0.4 V, so M1a and M1b should have Veb about equal to 2*Vebo.
Vb3 should be selected so that M12 is still in saturation when Vo1 drops to a little below Vthn-Vebo.
The Veb of M12, M10, and the size of diode connected CMFB1 transistor should be such that M 10 is guaranteed to be in saturation.
Note that there is only one high impedance node (Vo1 node) in this CMFB loop, hence the loop UGF can be made high and still maintain good stability.

32. Vo+ and Vo- also need common mode stabilization
M17t and M21t are in triode
M21t
M17t
M21
M17
M18
Vocmd
M19
M15 R R C C
M20
M16
Choose R to a couple times bigger than Ro
Choose C to be near or a couple times larger than Cgs of CMFB circuit.

33. M:1
1:M
Vxx
Vxx
M:1
1:M
Vocmd R R C C

34. Why RC in common mode detector
KCL at V:
(V1 – V)/R + (V2 – V)/R = V * sCgs
(V1 + V2)/R = V(sCgs + 2/R)
V = (V1 + V2)/2 * 1/(1 + sRCgs/2)
When |s| =|jw| << 2/RCgs,
V ≈ (V1 + V2)/2
Otherwise V is not close to common mode
To have CM detector work up to GB,
R << 2/(2pGB*Cgs) V1 R V
Cgs R V2

35. Why RC in common mode detector
KCL at V:
(V1 – V)(1/R +sC) + (V2 – V) (1/R +sC)
= V * sCgs
(V1 + V2) (1/R +sC) = V(sCgs + 2/R +2sC)
V = (V1 + V2)/2 *
(1 + sRC)/(1 + sRC + sRCgs/2)
As long as Cgs/2 < C, at all freq:
V ≈ (V1 + V2)/2 V1 R C V
Cgs C R V2
Hence, the RC network acts as a better CM detector

36. Why CMBF to M17t instead of first stage:
For CM behavior, assume DM=0.
Vo1+=Vo1-, and Vo+=Vo-=Vocm.
Without CMFB effect, at Q, Vo+ will be equal to Vg, which may be far below desired Vocm level.
With CMFB connected, the feedback effect will drive Vo1 so as to move Vo+ up to the desired Vocm level.
Since Vo1+ and Vo1- have a competing action on Vo+, it may take quite bit Vo1 movement to achieve the desired Vo+ movement, causing the biasing current in the second stage to be much larger than what is intended.
Vo1+
Vo1-
Vo+ Vg

37. Compensation for diff signal path’s closed-loop stability
Vxx M4 M7 M7
Vyy M3 M3 CC CC
Vbb M2 M2 M6 M6 M1
Vzz M1
Standard lead compensation

38. At relatively low frequency:
Because of gain from Vo1 to Vo, small signal Vo1 is much smaller than small signal Vo.
Small signal current in compensation network is approximately Vo/(1/gmz+1/sCc).
This current is injected to the Vo1 node.
Alternatively, a similar current can be injected:
Impedance looking into a cascode node is about 1/gm
Connecting Cc to a cascode node generates a current of the form Vo/(1/gm +1/sCc)
Because the base transistor is a current source, this small signal current goes to the Vo1 node
Even at high frequency, the current form is still valid.

39. Alternative compensation: Ahuja
Vxx M4 M7 M7
Vyy M3 M3 CC CC
Vbb M2 M2 M6 M6 M1
Vzz M1
Also called indirect compensation.

40. In the Cc+Mz connection,
Bias voltage of Mz can be matched to track bias voltage of M6
 robustness to process and temperature variations
Size of Mz can be parameter scanned so as to place zero to cancel the secondary pole of the amplifier
In the Cc to cascode connection,
Bias voltage can still be derived using current mirrors from a single current source,  still have process and temperature tracking
But size of cascode transistor is determined based on folded cascode stage design
Cannot arbitrarily choose its size without considerations for output impedance at Vo1, gain of op amp, and so on.

41. With Vicm at “sweet spot”, sweep Vin near Vicm with very fine steps (uV)
Vo+
Vo-
Vicm
Vin
Vin
d(Vo+-Vo-)
dVin
Vo+-Vo-
Vin