DESIGN OF TWO-STAGE OP AMPS.pdf

Lecture 23 – Design of Two-Stage Op Amps (3/11/16) Page 23-1
CMOS Analog Circuit Design © P.E. Allen - 2016
LECTURE 23 – DESIGN OF TWO-STAGE OP AMPS
LECTURE OUTLINE
Outline
• Steps in Designing an Op Amp
• Design Procedure for a Two-Stage Op Amp
• Design Example of a Two-Stage Op Amp
• Right Half Plane Zero
• PSRR of the Two-Stage Op Amp
• Summary
CMOS Analog Circuit Design, 3rd
Edition Reference
Pages 286-309

STEPS IN DESIGNING A CMOS OP AMP
Design Inputs
Boundary conditions:
1. Process specification (VT, K', Cox, etc.)
2. Supply voltage and range
3. Supply current and range
4. Operating temperature and range
Requirements:
1. Gain 8. Output-voltage swing
2. Gain bandwidth 9. Output resistance
3. Settling time 10. Offset
4. Slew rate 11. Noise
5. Common-mode input range, ICMR 12. Layout area
6. Common-mode rejection ratio, CMRR
7. Power-supply rejection ratio, PSRR

Outputs of Op Amp Electrical Design
The basic outputs are:
1.) The topology
2.) The dc currents
3.) The W and L values of transistors
4.) The values of components
060625-06
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias
CL
+
-
Cc
W/L ratios
Topology
DC Currents
L
W
Op amp circuit
or systems
specifications
Design of
CMOS
Op Amps
Component
values
C R
50µA

Some Practical Thoughts on Op Amp Design
1.) Decide upon a suitable topology.
• Experience is a great help
• The topology should be the one capable of meeting most of the specifications
• Try to avoid “inventing” a new topology but start with an existing topology
2.) Determine the type of compensation needed to meet the specifications.
• Consider the load and stability requirements
• Use some form of Miller compensation or a self-compensated approach
3.) Design dc currents and device sizes for proper dc, ac, and transient performance.
• This begins with hand calculations based upon approximate design equations.
• Compensation components are also sized in this step of the procedure.
• After each device is sized by hand, a circuit simulator is used to fine tune the
design
Two basic steps of design:
1.) “First-cut” - this step is to use hand calculations to propose a design that has
potential of satisfying the specifications. Design robustness is developed in this
step.
2.) Optimization - this step uses the computer to refine and optimize the design.

A DESIGN PROCEDURE FOR THE TWO-STAGE CMOS OP AMP
Unbuffered, Two-Stage CMOS Op Amp
Notation:
Si =
Wi
Li
= W/L of the ith transistor
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias
+
-
Cc
CL
Fig. 6.3-1

DC Balance Conditions for the Two-Stage Op Amp
For best performance, keep all transistors in
saturation.
M4 is the only transistor that cannot be forced into
saturation by internal connections or external voltages.
Therefore, we develop conditions to force M4 to be in
saturation.
1.) First assume that VSG4 = VSG6. This will cause
“proper mirroring” in the M3-M4 mirror. Also, the
gate and drain of M4 are at the same potential so that
M4 is “guaranteed” to be in saturation.
2.) If VSG4 = VSG6, then I6 =






S6
S4
I4
3.) However, I7 =






S7
S5
I5 =






S7
S5
(2I4)
4.) For balance, I6 must equal I7 
S6
S4
=
2S7
S5
called the “balance conditions”
5.) So if the balance conditions are satisfied, then VDG4 = 0 and M4 is saturated.
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias
+
-
Cc
CL
-
+
VSG6
-
+
VSG4
I4
I5
I7
I6
Fig. 6.3-1A

Summary of the Design Relationships for the Two-Stage Op Amp
Slew rate SR =
I5
Cc
(Assuming I7 >>I5 and CL > Cc)
First-stage gain Av1 =
gm1
gds2 + gds4
=
2gm1
I5(l2 + l4)
Second-stage gain Av2 =
gm6
gds6 + gds7
=
gm6
I6(l6 + l7)
Gain-bandwidth GB =
gm1
Cc
Output pole p2 =
-gm6
CL
RHP zero z1 =
gm6
Cc
60° phase margin requires that gm6 = 2.2gm2(CL/Cc) if all other roots are  10GB.
Positive ICMR Vin(max) = VDD -
I5
b3
- VT03(max) + VT1(min))
Negative ICMR Vin(min) = VSS +
I5
b1
+ VT1(max) + VDS5(sat)

Op Amp Specifications
The following design procedure assumes that specifications for the following parameters
are given.
1. Gain at dc, Av(0)
2. Gain-bandwidth, GB
3. Phase margin (or settling time)
4. Input common-mode range, ICMR
5. Load Capacitance, CL
6. Slew-rate, SR
7. Output voltage swing
8. Power dissipation, Pdiss

Unbuffered Op Amp Design Procedure
This design procedure assumes that the gain at dc (Av), unity gain bandwidth (GB), input
common mode range (Vin(min) and Vin(max)), load capacitance (CL), slew rate (SR),
settling time (Ts), output voltage swing (Vout(max) and Vout(min)), and power dissipation
(Pdiss) are given. Choose the smallest device length which will keep the channel
modulation parameter constant and give good matching for current mirrors.
1. From the desired phase margin, choose the minimum value for Cc, i.e. for a 60° phase
margin we use the following relationship. This assumes that z  10GB.
Cc  0.22CL
2. Determine the minimum value for the “tail current” (I5) from
I5 = SR .Cc
3. Design for S3 from the maximum input voltage specification.
S3 =
I5
K'3[VDD − Vin(max) − VT03(max) + VT1(min)]2
4. Verify that the pole of M3 due to Cgs3 and Cgs4 (= 0.67W3L3Cox) will not be dominant
by assuming it to be greater than 10 GB
gm3
2Cgs3
> 10GB.

Unbuffered Op Amp Design Procedure - Continued
5. Design for S1 (S2) to achieve the desired GB.
gm1 = GB . Cc → S2 =
gm1
2
K'1I5
6. Design for S5 from the minimum input voltage. First calculate VDS5(sat) then find S5.
VDS5(sat) = Vin(min) - VSS-
I5
1
-VT1(max) ≥ 100 mV → S5 =
2I5
K'5[VDS5(sat)]2
7. Find S6 by letting the second pole (p2) be equal to 2.2 times GB and assuming that
VSG4 = VSG6.
gm6 = 2.2gm2(CL/Cc) and
gm6
gm4
=
2KP'S6I6
2KP'S4I4
=
S6
S4
I6
I4
=
S6
S4
→ S6 =
gm6
gm4
S4
8. Calculate I6 from
I6 =
gm62
2K'6S6
Check to make sure that S6 satisfies the Vout(max) requirement and adjust as necessary.
9. Design S7 to achieve the desired current ratios between I5 and I6.
S7 = (I6/I5)S5 (Check the minimum output voltage requirements)

Unbuffered Op Amp Design Procedure - Continued
10. Check gain and power dissipation specifications.
Av =
2gm2gm6
I5(2 + 4)I6(6 + 7)
Pdiss = (I5 + I6)(VDD + VSS)
11. If the gain specification is not met, then the currents, I5 and I6, can be decreased or
the W/L ratios of M2 and/or M6 increased. The previous calculations must be rechecked
to insure that they are satisfied. If the power dissipation is too high, then one can only
reduce the currents I5 and I6. Reduction of currents will probably necessitate increase of
some of the W/L ratios in order to satisfy input and output swings.
12. Simulate the circuit to check to see that all specifications are met.

Unbuffered Op Amp Design Summary
Step Design Equations Comments
1 Let Cc
≥ 0.2CL PM = 60° and RHP Z=10GB
2 Let I5
≥ SR·Cc
CCL
Assumes SR limited by Cc
0° and RB
3 Maximum input common mode range
4 GB defines the W/L of M1 and M2
5 Minimum input common mode range
6 DC balance conditions
7 PM = 60° and p2
= 2.2GB give gm6
≈ 10gm1
8 Determines the current in M7
9
Check gain and power
dissipation and iterate
if necessary
gm1 = GB×Cc ®
W1
L1
=
W2
L2
=
gm1
2
K1
'
I5
W5
L5
=
2I5
K5
'
VDS5(sat)2
W6
L6
=
gm6
gm4
W4
L4
I6 =
gm6
2
2K6
'
(W6 / L6 )
W7
L7
= max
I6
I5
W5
L5
,
2I7
K7
'
VDS7(sat)2
é
ë
ê
ù
û
ú
Av =
2gm1gm6
I5(l2 +l4 )I6 (l6 +l7 )
and Pdiss = (I5 + I6 )(VDD +|VSS |)
W3
L3
=
W4
L4
=
I5
K3
'
VDD -Vin (max)-|VT 3 |+VT1
[ ]
2

DESIGN EXAMPLE OF A TWO-STAGE OP AMP
Example 23-1 - Design of a Two-Stage Op Amp
If KN’=120µA/V2, KP’= 25µA/V2, VTN = |VTP| = 0.5±0.15V, N = 0.06V-1, and P =
0.08V-1, design a two-stage, CMOS op amp that meets the following specifications.
Assume the channel length is to be 0.5µm and the load capacitor is CL = 10pF.
Av > 3000V/V VDD =2.5V GB = 5MHz SR > 10V/µs
60° phase margin 0.5V<Vout range < 2V ICMR = 1.25V to 2V Pdiss  2mW
Solution
1.) The first step is to calculate the minimum value of the compensation capacitor Cc,
Cc  (2.2/10)(10 pF) = 2.2 pF
2.) Choose Cc as 3pF. Using the slew-rate specification and Cc calculate I5.
I5 = (3x10-12)(10x106) = 30 µA
3.) Next calculate (W/L)3 using ICMR requirements (use worst case thresholds ±0.15V).
(W/L)3 =
30x10-6
(25x10-6)[2.5 - 2 - .65 + 0.35]2 = 30 → (W/L)3 = (W/L)4 = 30

Example 23-1 - Continued
4.) Now we can check the value of the mirror pole, p3, to make sure that it is in fact
greater than 10GB. Assume the Cox = 6fF/µm2. The mirror pole can be found as
p3 ≈
-gm3
2Cgs3
=
- 2K’pS3I3
2(0.667)W3L3Cox
= -1.25x109(rads/sec)
or 199 MHz. Thus, p3, is not of concern in this design because p3 >> 10GB.
5.) The next step in the design is to calculate gm1 to get
gm1 = (5x106)(2)(3x10-12) = 94.25µS
Therefore, (W/L)1 is
(W/L)1 = (W/L)2 =
gm1
2
2K’NI1
=
(94.25)2
2·120·15
= 2.47  3.0  (W/L)1 = (W/L)2 = 3
6.) Next calculate VDS5,
VDS5 = 1.25 -
30x10-6
120x10-6·3
- .65 = 0.31V
Using VDS5 calculate (W/L)5 from the saturation relationship.
(W/L)5 =
2(30x10-6)
(120x10-6)(0.31)2
= 5.16  6 → (W/L)5 = 6

7.) For 60° phase margin, we know that
gm6  10gm1  942.5µS
Assuming that gm6 = 942.5µS and knowing that gm4 = 150µS, we calculate (W/L)6 as
(W/L)6 = 30
942.5x10-6
(150x10-6)
= 188.5  190 (W/L)6 = 190
8.) Calculate I6 using the small-signal gm expression:
I6 =
(942.5x10-6)2
(2)(25x10-6)(188.5)
= 94.2µA  95µA
Calculating (W/L)6 based on Vout(max), gives a value of 15. Since 190 exceeds the
specification and gives better phase margin, we choose (W/L)6 = 190 and I6 = 95µA.
With I6 = 95µA the power dissipation is Pdiss = 2.5V·(30µA+95µA) = 0.3125mW
9.) Finally, calculate (W/L)7
(W/L)7 = 6








95x10-6
30x10-6
= 19  20 → (W/L)7 = 20
Let us check the Vout(min) specification although the W/L of M7 is so large that this is
probably not necessary. The value of Vout(min) is
Vout(min) = VDS7(sat) = (2·95)/(120·20) = 0.281V
which is less than required. At this point, the first-cut design is complete.

10.) Now check to see that the gain specification has been met
Av =
(94.25x10-6)(942.5x10-6)
15x10-6(.06 + .08)95x10-6(.06 + .08)
= 3,180V/V
which barely exceeds the specifications. Since we are at 2xLmin, it won’t do any good to
increase the channel lengths. Decreasing the currents or increasing W6/L6 will help.
The figure below shows the results of the first-cut design. The W/L ratios shown do
not account for the lateral diffusion discussed above. The next phase requires simulation.

RIGHT-HALF PLANE ZERO
Controlling the Right-Half Plane Zero
Why is the RHP zero a problem?
Because it boosts the magnitude but lags the phase - the worst possible combination for
stability.
Solution of the problem:
The compensation comes from the feedback path through Cc, but the RHP zero
comes from the feedforward path through Cc so eliminate the feedforward path!
150129-013
jw
s
jw1
jw2
jw3
q1
q2
q3
180 > q1 > q2 > q3
z1
Loop
Gain
0dB log10w
log10w
0°
180°
RHP Zero Boost
RHP Zero Lag
Loop
Phase
Shift

Elimination of the Feedforward Path through the Miller Capacitor
1.) Removing the feedforward path.
Roots: Dominant pole and output pole.
2.) Controlling the RHP zero location using a
nulling resistor†
.
Roots:
a.) Dominant pole (Miller pole)
b.) Output pole
c.) Pole due to Rz and first stage output capacitance, p4 ≈
-1
RzCI
d.) Controllable zero, z1 =
1
Cc(1/gmII - Rz)
Note that z1 can be placed anywhere on the real axis.
†
W.J. Parrish, “An Ion Implanted CMOS Amplifier for High Performance Active Filters”, Ph.D. Dissertation, 1976, Univ. of CA, Santa Barbara.
120523-01
Cc
vOUT
+1
Inverting
High-Gain
Stage
Feedback Only
120523-02
Cc
vOUT
Inverting
High-Gain
Stage
Rz

A Design Procedure that Allows the RHP Zero to Cancel the Output Pole, p2
We desire that z1 = p2 in terms of the previous notation.
Therefore,
1
Cc(1/gmII - Rz)
=
-gmII
CII
The value of Rz can be found as
Rz =






Cc + CII
Cc
(1/gmII)
With p2 canceled, the remaining roots are p1 and p4(the pole due to Rz) . For unity-gain
stability, all that is required is that
p4  Av(0)p1 =
Av(0)
gmIIRIIRICc
=
gmI
Cc
and (1/RzCI)  (gmI/Cc) = GB
Substituting Rz into the above inequality and assuming CII >> Cc results in
Cc 
gmI
gmII
CICII
This procedure gives excellent stability for a fixed value of CII ( CL).
Unfortunately, as CL changes, p2 changes and the zero must be readjusted to cancel p2.

Using the Nulling Resistor in the Miller Compensated Two-Stage Op Amp
Circuit:
We saw earlier that the roots were:
p1 = -
gm2
AvCc
= -
gm1
AvCc
p2 = -
gm6
CL
p4 = -
1
RzCI
z1 =
-1
RzCc - Cc/gm6
where Av = gm1gm6RIRII.
(Note that p4 is the pole resulting from the nulling resistor compensation technique.)
Design of the Nulling Resistor (M8)
VDD
VSS
IBias
CL
Cc
CM vout
VB
VA
M1 M2
M3 M4
M5
M6
M7
M9
M10
M11
M12
vin+
vin-
M8
Fig. 160-03
VC

For the zero to be on top of the second pole (p2), the following relationship must hold
Rz =
1
gm6 





CL + Cc
Cc
=






Cc+CL
Cc
1
2K’PS6I6
The resistor, Rz, is realized by the transistor M8 which is operating in the active region
because the dc current through it is zero. Therefore, Rz, can be written as
Rz =
vDS8
iD8

VDS8=0
=
1
K’PS8(VSG8-|VTP|)
The bias circuit is designed so that voltage VA is equal to VB.
 VGS10 − VT = VGS8 − VT VSG11 = VSG6 






W11
L11
=






I10
I6 





W6
L6
In the saturation region
VGS10 − VT =
2(I10)
K'P(W10/L10)
= VGS8 − VT
 Rz =
1
K’PS8
K’PS10
2I10
=
1
S8
S10
2K’PI10
Equating the two expressions for Rz gives






W8
L8
=






Cc
CL + Cc
S10S6I6
I10

Example 23-2 - RHP Zero Compensation
Use results of Ex. 23-1 and design compensation circuitry so that the RHP zero is
moved from the RHP to the LHP and placed on top of the output pole p2. Use device
data given in Ex. 23-1.
Solution
The task at hand is the design of transistors M8, M9, M10, M11, and bias current
I10. The first step in this design is to establish the bias components. In order to set VA
equal to VB, then VSG11 must equal VSG6. Therefore,
S11 = (I11/I6)S6
Choose I11 = I10 = I9 = 15µA which gives S11 = (15µA/95µA)190 = 30.
The aspect ratio of M10 is essentially a free parameter,
and will be set equal to 1. There must be sufficient supply
voltage to support the sum of VSG11, VSG10, and VDS9. The
ratio of I10/I5 determines the (W/L) of M9. This ratio is
(W/L)9 = (I10/I5)(W/L)5 = (15/30)(6) = 3
Now (W/L)8 is determined to be
(W/L)8 =








3pF
3pF+10pF
1·190·95µA
15µA
= 8
VDD
VSS
IBias
VB
VA
M4
M9
M10
M11
M12
M8
100327-03
VC
I9
Cc
M6
M5
I5

It is worthwhile to check that the RHP zero has been moved on top of p2. To do this, first
calculate the value of Rz. VSG8 must first be determined. It is equal to VSG10, which is
VSG10 =
2I10
K’PS10
+ |VTP| =
2·15
25·1
+ 0.5 = 1.595V
Next determine Rz.
Rz =
1
K’PS8(VSG10-|VTP|)
=
106
25·8(1.595-.7)
= 4.564k
The location of z1 is calculated as
z1 =
-1
(4.564 x 103)(3x10-12) -
3x10-12
950x10-6
= -94.91x106 rads/sec
The output pole, p2, is
p2 = -
950x10-6
10x10-12 = -95x106 rads/sec
Thus, we see that for all practical purposes, the output pole is canceled by the zero
that has been moved from the RHP to the LHP.
The results of this design are summarized below where L = 0.5µm.
W8 = 4µm W9 = 1.5µm W10 = 0.5µm and W11 = 15 µm

An Alternate Form of Nulling Resistor
To cancel p2,
z1 = p2 → Rz =
Cc+CL
gm6ACC
=
1
gm6B
Which gives
gm6B = gm6A






Cc
Cc+CL
In the previous example,
gm6A = 950µS, Cc = 3pF
and CL = 10pF.
Choose I6B = 10µA to get
gm6B =
gm6ACc
Cc + CL
→
2KPW6BI6B
L6B
=








Cc
Cc+CL
2KPW6AID6
L6A
or
W6B
L6B
=








3
13
2 I6A
I6B
W6A
L6A
=








3
13
2








95
10
(190) = 96.12 → W6B = 48µm
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias
+
-
Cc
CL
M11 M10
M6B
M8 M9
Fig. 6.3-4A

Increasing the Magnitude of the Output Pole†
The magnitude of the output pole,
p2, can be increased by introducing
gain in the Miller capacitor feedback
path as shown where,
A ≈ gm8(rds8||rds9||rds2||rds4).
The roots become,
1.) The dominant pole increased slightly because RI (output of first stage) is decreased.
2.) The output pole is increased by a
factor of A to get new p2 ≈
-Agm6
CII
3.) The pole at the source of M8 (-gm8/Cc)
becomes a zero on the negative real axis.
Roots:
†
B.K. Ahuja, “An Improved Frequency Compensation Technique for CMOS Operational Amplifiers,” IEEE J. of Solid-State Circuits, Vol. SC-18,
No. 6 (Dec. 1983) pp. 629-633.
VDD
VSS
VBias
Cc
M6
M7
M8
M9
M10
M12
M11
vOUT
120523-03
VDD
VSS
Cc
M6
M7
M9
M10
M12
M11
vOUT
A
VDD
Cc
rds7
vout
M6 CII
GB·Cc
1 » 0
VDD
vout
M6
CII
M8
120523-04
rds7
A
jw
s
gm6
Cgd6
-gm8
Cc
-Agm6
C2
-1
gm6rds
2Cc
120523-05

Issues with the Previous Method†
The previous technique assumed that the gate-source capacitance of M8 could be
neglected. Unfortunately, this assumption ignores a pair of complex poles near the unity
gain frequency. Below is the small signal model with the capacitance that causes this
that includes Cgs8.
The solution proposed in the reference below is to decrease the impedance at the source
of M8 by using a negative feedback loop. Below is a possible solution that will have
better phase margin.
†
Uday Dasgupta, “Issues with ‘Ahuja’ Frequency Compensation Technique,” Proc. of IEEE Inter. Symposium on Radio Frequency Integration
Technology, Jan. 9, 2009, pp. 326-329.
Iin R1 R2 C2
gm8Vs8
Vout
V1
+
-
+
-
+
-
Vs8
1
gm8
Cc
gm6V1
Cgd6
Cgs8
160311-01
VDD
VSS
Cc
M6
M7
M8
M9
M10
M12
M11
vOUT
M14
M13
M15
M16
160311-02

POWER SUPPLY REJECTION RATIO OF THE TWO-STAGE OP AMP
What is PSRR?
PSRR =
Av(Vdd=0)
Add(Vin=0)
How do you calculate PSRR?
You could calculate Av and Add and divide, however
Vout = AddVdd + Av(V1-V2) = AddVdd - AvVout → Vout(1+Av) = AddVdd

Vout
Vdd
=
Add
1+Av

Add
Av
=
1
PSRR+ (Good for frequencies up to GB)

Approximate Model for PSRR+
1.) The M7 current sink causes VSG6 to act like a battery.
2.) Therefore, Vdd couples from the source to gate of M6.
3.) The path to the output is through any capacitance from gate to drain of M6.
Conclusion:
The Miller capacitor Cc couples the positive power supply ripple directly to the output.
Must reduce or eliminate Cc.

Approximate Model for PSRR-
What is Zout?
Zout =
Vt
It
 It = gmIIV1 = gmII






gmIVt
GI+sCI+sCc
Thus, Zout =
GI+s(CI+Cc)
gmIgMII

Vout
Vss
=
1+
rds7
Zout
1
=
s(Cc+CI) + GI+gmIgmIIrds7
s(Cc+CI) + GI
 Pole at
-GI
Cc+CI
The negative PSRR is much better than the positive PSRR.
M1 M2
M3 M4
M5
M6
M7
Vout
VDD
VSS
VBias
Cc
CII
Vss
CI
Fig. 180-11
VBias connected to VSS
rds7
vout
Zout
Vss
rds7
Path through Cgd7
is negligible
150131-01
Vt
rds6||rds7
Vout
CI
Cc
+
-
V1
+
-
gmIVin
gmIIV1
RI
CII+Cgd7
It

SUMMARY
• The output of the design of an op amp is
- Schematic
- DC currents
- W/L ratios
- Component values
• Design procedures provide an organized approach to creating the dc currents, W/L
ratios, and the component values
• The right-half plane zero causes the Miller compensation to deteriorate
• Methods for eliminating the influence of the RHP zero are:
- Nulling resistor
- Increasing the magnitude of the output pole
• The PSRR of the two-stage op amp is poor because of the Miller capacitance, however,
methods exist to eliminate this problem
• The two-stage op amp is a very general and flexible op amp

DESIGN OF TWO-STAGE OP AMPS.pdf

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DESIGN OF TWO-STAGE OP AMPS.pdf