Analysis and Design of Analog Integrated Circuits
Lecture 17
Basic Two Stage CMOS Opamp
Michael H. Perrott
April 4, 2012
Copyright © 2012 by Michael H. Perrott
All rights reserved.
M.H. Perrott
Opamps Are Basic Analog Building Blocks
Analog Filters
Current References
Switched Capacitor Circuits
C2
C1
Iref
R1
Vin
Vout
Vin
Vref
Vref
Vout
C1
Vref
Rref
Enable active filters
- Can achieve arbitrary pole/zero placement using only
capacitor/resistor networks around the opamp
Allow accurate voltage to current translation
Provide accurate charge transfer between capacitors
- Extremely useful for switched capacitor circuits used in
analog-to-digital converters and discrete-time analog
filters
M.H. Perrott
2
Key Specifications of Opamps (Open Loop)
Rhuge
For Open Loop Characterization
Vdd
Chuge
Vout
Vin
Vss
CL
Zout
Set Rhuge >> |Zout|
and 1/(RhugeChuge) << wdom
20log(K)
20log Vout/Vin
Vdd
Vout
Vin
Vss
0dB
CL
wdom
w0 wp
w (rad/s)
DC small signal gain: K
Unity gain frequency: w0
Dominant pole frequency: wdom
Parasitic pole frequencies: wp (and higher order poles)
Output swing (max output range for DC gain > Kmin)
M.H. Perrott
3
Key Specifications of Opamps (Closed Loop)
Vdd
Voffset
Vin
Vout
Vss
CL
Offset voltage
Settling time (closed loop bandwidth)
Input common mode range
Equivalent Input-Referred Noise
Common-Mode Rejection Ratio (CMRR)
Ã
!−1
δVoffset
CMRR =
δVin
Power Supply Rejection Ratio (PSRR)
Ã
!−1
Ã
!−1
δVoffset
δVoffset
−
+
PSRR =
PSRR =
δVdd
δVss
M.H. Perrott
4
Slew Rate Issues for Opamps
Vin
Vdd
Vout
Vin
Vss
CL
ideal
Vout
slew-rate limited
Output currents of practical opamps have max limits
- Impacts maximum rate of charging or discharging load
capacitance, C
- For large step response, this leads to the output lagging
L
behind the ideal response based on linear modeling
We refer to this condition as being slew-rate limited
Where slew-rate is of concern, the output stage of the
opamp can be designed to help mitigate this issue
- Will lead to extra complexity and perhaps other issues
M.H. Perrott
5
Basic Two Stage CMOS Op Amp
M8
M7
M5
Iref
Vin-
M1
Vin+
M2
Vout
CL
Rc
M3
M4
Cc
M6
This is a common “workhorse” opamp for medium
performance applications
Provides a nice starting point to discuss various
CMOS opamp design issues
Starting assumptions: W1/L1 = W2/L2, W3/L3 = W4/L4
M.H. Perrott
6
First Stage Analysis
M5
-vid/2
M1
Ibias1
First Stage Two-Port Model
vid/2
M2
vid
Rout1
V1 Zin
Gm1V1
Rout1
Zin2 vout1
vout1
M3
M4
Derive two port model assuming differential input:
1
1
=
Zin1 =
s(Cgs1/2)
s(Cgs2/2)
Gm1 = gm1 = gm2
M.H. Perrott
Rout1 = ro2 ||ro4
7
Derivation of Rout1 (Incorrect Approach)
ro5
-vid/2
M1
vid/2
M2
ro2 (1+gm2 1 )
gm1
ro4
1
gm3
2ro2
M4
Application of Thevenin analysis seems to imply that
Rout1 = 2ro2||ro4
- Why is this incorrect?
M.H. Perrott
8
Derivation of Rout1 (Correct Approach)
ro5
M1
i1
i1
M2
vtest
i 1=
2ro2
i2
1
gm3
ro4
itest = i1 + i2
vtest
M4
i2
2ro2
vtest
i1 + r
o4
Correct approach includes the impact of the current
mirror feedback
vtest
vtest
vtest
itest = i1 + i2 = i1 + i1 +
=2
+
ro4
2ro2
ro4
⇒ Rout1 = ro2||ro4
M.H. Perrott
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Derivation of Gm1
incremental ground
-vid/2
M1
i1
vid/2
M2
iout = i1 + i2
i2
i1
1
gm3
M4
For differential input, we can approximate the source
of M1 and M2 as being at incremental ground
gm1
vid
i1 = −gm1(−vid /2) =
2
gm2
gm1
i2 = gm2(vid/2) =
vid =
vid
2
2
⇒ iout = gm1vid ⇒ Gm1 = gm1 = gm2
M.H. Perrott
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Derivation of Zin
incremental ground
Cgs2
Cgs1
-vid/2
M1
M2
vid/2
-vid/2
vid/2
Cgs1
-vid/2
1
gm3
M4
M.H. Perrott
1
sCgs1/2
=
vid/2
Cgs1
For differential input, we can
simplify the input capacitance
calculation through the steps
shown at the right
⇒ Zin1 =
Cgs2
1
Cgs2
-vid/2
vid/2
Cgs1/2
vid
Cgs1/2
sCgs2/2
11
Second Stage Analysis
M7
Second Stage Two-Port Model
Ibias2
Vout
vin2
V2 Zin2 Gm2V2
Rout2
CL vout
CL
Vin2
M6
Two port model derivation is straightforward
- This is a common source amplifier
1
Zin2 =
sCgs6
Gm2 = gm6
M.H. Perrott
Rout2 = ro6 ||ro7
12
Overall Opamp Model
First Stage Two-Port Model
vid
V1
Cgs1
gm1V1
2
ro2||ro4
Second Stage Two-Port Model
V2
Cgs6 gm6V2
ro6||ro7
vout
CL
Overall transfer function
vout (s)
K
H(s) =
=
vid(s)
(1 + s/wp1)(1 + s/wp2)
- DC gain
- Poles
K = gm1(ro2||ro4 )gm6(ro6||ro7 )
wp1 =
1
(ro2||ro4 )Cgs6
1
wp2 =
(ro6||ro7 )CL
In general, wp2 << wp1 since CL >> Cgs6
M.H. Perrott
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Consider The Dominant Pole To Be wp2
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
0dB
1
wp2 =
(ro6||ro7)CL
w0
w (rad/s)
K
gm1(ro2||ro4 )gm6(ro6||ro7 )
H(s) =
=
1 + s/wp2
1 + s(ro6||ro7 )CL
At frequencies >> wp2
gm1(ro2||ro4 )gm6
H(s) ≈
sCL
gm1(ro2||ro4 )gm6
⇒ wo ≈
CL
We want wp1 > w0 for good phase margin with unity gain feedback
M.H. Perrott
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Key Issue for Achieving Adequate Phase Margin
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
0dB
wp2 =
1
(ro6||ro7)CL
w (rad/s)
w0
1
wp1 =
(ro2||ro4)Cgs6
gm1(ro2||ro4 )gm6
wo ≈
CL
To achieve wp1 > w0
1
wp1 =
> wo ⇒ CL > gm1gm6(ro2 ||ro4 )2Cgs6
(ro2||ro4 )Cgs6
- We need a very large value of C
relative to Cgs6
This will generally be impractical!
M.H. Perrott
L
15
Pole Splitting Using a Compensation Capacitor
Cc
CM
vid
V1
Cgs1
gm1V1
2
ro2||ro4
V2
Cgs6 gm6V2
ro6||ro7
vout
CL
Consider placing capacitor Cc across the second stage
- Load capacitance seen by stage 1 becomes roughly
CM = (1 + gm6(ro6||ro7 ))Cc ≈ gm6(ro6||ro7 )Cc
This large Miller capacitance now causes wp1 to become
dramatically lower such that it forms the dominant pole
1
1
wp1 ≈
≈
(ro2 ||ro4 )CM
(ro2||ro4 )gm6(ro6||ro7 )Cc
We will see that wp2 actually increases in frequency!
M.H. Perrott
16
Pole Splitting Using a Compensation Capacitor (Part 2)
CM
vid
V1
Cgs1
gm1V1
2
Cc
ro2||ro4
V2
Cgs6 gm6V2
Rth_C
ro6||ro7
vout
L
CL
Assuming wp1 forms the dominant pole, we can
approximate Cc as a short when calculating wp2
1
Rth CL ≈
gm6
gm6
1
=
⇒ wp2 ≈
(1/gm6)(Cgs6 + CL)
Cgs6 + CL
- Note: we must have C
c
>> Cgs6 for this to be accurate
The inclusion of capacitor Cc has led to wp2
increasing in frequency
M.H. Perrott
17
Impact of Pole Splitting using Compensation Cap
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
w (rad/s)
wp1 =
1
(ro2||ro4)gm6(ro6||ro7)Cc
1
wp2 =
(ro6||ro7)CL
gm6
wp2 =
Cgs6+CL
1
wp1 =
(ro2||ro4)Cgs6
Pole splitting allows the dominant pole frequency to
be dramatically decreased and the main parasitic pole
to be dramatically increased
- We can achieve higher unity gain frequency with
improved phase margin and with reasonable area
M.H. Perrott
18
Unity Gain Frequency with Compensation Cap
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
0dB
w (rad/s)
wp1 =
1
(ro2||ro4)gm6(ro6||ro7)Cc
w0
gm6
wp2 =
Cgs6+CL
K
gm1(ro2 ||ro4 )gm6(ro6 ||ro7 )
H(s) =
=
1 + s/wp1
1 + s(ro2||ro4)gm6(ro6||ro7 )Cc
At frequencies >> wp1
gm1(ro2||ro4 )gm6(ro6||ro7 )
gm1
H(s) ≈
⇒ wo ≈
s(ro2 ||ro4)gm6(ro6||ro7 )Cc
Cc
We want wp2 > w0 for good phase margin with unity gain feedback
M.H. Perrott
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Key Constraints for Achieving Adequate Phase Margin
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
0dB
w (rad/s)
wp1 =
w0
1
(ro2||ro4)gm6(ro6||ro7)Cc
To achieve wp2 > w0
gm6
wp2 =
Cgs6+CL
gm1
wo ≈
Cc
gm6
gm1
wp2 =
> wo ⇒ Cc >
(Cgs6 + CL)
Cgs6 + CL
gm6
- Note: we must have C
M.H. Perrott
c
>> Cgs6 for this to be accurate
20
More Accurate Calculations Related to Phase Margin
20log(gm1(ro2||ro4)gm6(ro6||ro7))
20log Vout/Vid
0dB
wp1 =
w0
1
(ro2||ro4)gm6(ro6||ro7)Cc
To achieve wp2 > w0
w (rad/s)
gm6Cc
wp2 =
Cgs6CL+Cc(Cgs6+CL)
gm1
wo ≈
Cc
Ã
gm1 Cgs6CL
wp2 > wo ⇒ Cc >
+ Cgs6 + CL
gm6
Cc
M.H. Perrott
!
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A More Accurate Transfer Function Model
Cc
vid
V1
Cgs1
gm1V1
2
ro2||ro4
V2
Cgs6 gm6V2
ro6||ro7
vout
CL
K(1 + s/wz )
vout (s)
=
H(s) =
vid(s)
(1 + s/wp1)(1 + s/wp2)
K = gm1(ro2||ro4 )gm6(ro6||ro7 )
1
wp1 =
(ro2||ro4 )gm6(ro6||ro7 )Cc
gm6Cc
wp2 =
Cgs6CL + Cc(Cgs6 + CL)
µ
¶
gm6
Right half plane (RHP) zero
wz = −
causes potential stability issues
Cc
M.H. Perrott
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Plotting the Magnitude of a RHP Zero
Plot the magnitude response of right half plane wz
20 log |Az (w)| = 20 log |1 − jw/wz |
- For w << |w |:
- For w >> |w |:
z
20 log |Az (w)| ≈ 20 log |1| = 0
z
20 log |Az (w)| ≈ 20 log |w/wz |
20log|Az(ω)|
20 dB/decade
0 dB
ωz
ω
Magnitude response is the same as for left half plane zero
M.H. Perrott
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Plotting the Phase of a RHP Zero
Plot the phase response of right half plane wz
6
Az (w) = 6 (1 − jw/wz ) = arctan (−w/wz )
- For w << |w |:
- For w = |w |:
- For w >> |w |:
6
z
z
6
z
0o
6
Az (w) ≈ arctan (0) = 0◦
Az (w) ≈ arctan (−1) = −45◦
Az (w) ≈ arctan (−∞) = −90◦
ωz/10
ωz
ωz∗10
ω
Az(ω) -45o
-90
o
Phase response is negative rather than positive (similar to pole)
M.H. Perrott
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Phase Margin Degradation Due to RHP Zero
20log Vout/Vid
0dB
w (rad/s)
gm6
gm1
|wz| =
w0 =
Cc
Cc
gm6
wp2 =
Cgs6+CL
Since the RHP zero adds negative phase (similar to
pole), it reduces phase margin
- We want:
|wz | À wo ⇒ gm6 À gm1
This is not a desirable constraint
M.H. Perrott
25
Adding a Compensation Resistor
Cc
Rc
vid
V1
Cgs1
gm1V1
2
ro2||ro4
V2
ro6||ro7
Cgs6 gm6V2
vout
CL
K(1 + s/wz )
vout (s)
=
H(s) =
vid(s)
(1 + s/wp1)(1 + s/wp2)
K = gm1(ro2||ro4 )gm6(ro6||ro7 )
1
wp1 =
(ro2||ro4 )gm6(ro6||ro7 )Cc
gm6Cc
wp2 =
Cgs6CL + Cc(Cgs6 + CL)
µ
¶
gm6
1
wz = −
Cc 1 − gm6Rc
M.H. Perrott
RHP zero effectively
removed if Rc = 1/gm6
Improved phase
margin possible with
Rc > 1/gm6
- See Johns&Martin,
pp. 242-244
26
Implementing Rc with a Triode Device
M8
M7
M5
Iref
Vin-
M1
Vin+
M2
Vout
CL
M9
M3
M4
Cc
M6
More compact implementation than a poly resistor
Triode channel resistance can somewhat track 1/gm6
across process and temperature variations
Key issue: supply sensitivity
- See pp. 246-248 of Johns&Martin for solutions to this issue
M.H. Perrott
27
Calculations for Triode Compensation Resistor
Vdd
M9
M4
Cc
M6
Triode resistance calculated as
1
Rc =
μnCox(W9/L9)(Vgs9 − VT H )
1
=
μn Cox(W9/L9)(Vdd − Vgs6 − VT H )
Assuming square law, 1/gm6 is calculated as
1
1
=
gm6
μnCox(W6/L6)(Vgs6 − VT H )
Depending on Vdd, Rc can track 1/gm6 across process/temp
M.H. Perrott
28
Summary
Basic two-stage CMOS opamp is a workhorse for many
moderate performance analog applications
- Relatively simple structure with reasonable performance
Key issue: two-stages lead to two poles that are
relatively close to each other
- This leads to very poor phase margin unless very large
CL is used
Inclusion of a compensation capacitor across the
second stage leads to pole splitting such that stable
performance can be achieved with reasonable area
- A compensation resistor is also desirable to help
eliminate the impact of a RHP zero that occurs due to
compensation
We will use the basic two stage CMOS opamp structure
to explore various opamp specifications in the next lecture
M.H. Perrott
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