ee 501 fall 2009 design project 1
DESCRIPTION
EE 501 Fall 2009 Design Project 1. Fully differential multi-stage CMOS Op Amp w ith Common Mode Feedback a nd Compensation for high GB. Major components. Input differential pair Cascoding for boosting DM gain, CMRR Second gain stage Compensation Common mode feedback - PowerPoint PPT PresentationTRANSCRIPT
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EE 501 Fall 2009
Design Project 1
Fully differential multi-stage CMOS Op Amp
with Common Mode Feedbackand Compensation for high GB
Major components• Input differential pair• Cascoding for boosting DM gain, CMRR• Second gain stage• Compensation• Common mode feedback• Biasing circuits• VDD independent current reference (T
independence not required for this project but is for next project)
• Start up circuit2
Input differential pair• NMOS input vs PMOS input
– Transconductance gain gm requirement– Noise consideration
• Area, mismatch, and offset voltage issues• Common mode rejection
– Tail current source: cascode or not– Effects on CMRR, ICMR
• Input common mode range– Choice of telecopic vs folded cascode– Choice of VEB’s and Vbias for cascode 3
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VSS
M1 M2VIN
M5
VIN
Vzz
VSS
M1 M2VIN
M5c
VIN
VzzM5
What are the advantages/disadvantages of cascoding M5? Which one offers better differential signal virtual ground at S1/S2?Which one offers better Vicm-min?Which one offers better Vicm rejection?Which one offers better power supply rejection?Which one is more area efficient?
5
VDD
M1 M2VIN
M5
VIN
Vzz
What are the advantages/disadvantages of NMOS input vs PMOS input? Which one offers better differential signal transconductance?Which one offers better 1st stage – 2nd stage current split considerations?Which one offers better 1/f noise contribution?Which one offers better input referred thermal noise?Which one offers better off set voltage?
VDD
M1 M2VIN
M5c
VIN
Vzz M5
Cascoding
• Telescopic cascode• Folded cascode
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7
TelescopicN-input
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TelescopicP-input
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Folded cascode
P-input
Second gain stage
• NMOS input• PMOS input• Current source load
– Cascode load
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Compensation
• Miller compensation• Rz-Cc compensation
– Implementation– Process tracking bias
• Cc feeding back to cascode• Cc feeding back to triode node
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Common mode feedback
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Biasing circuit
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VDD independent reference
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Startup circuit
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VDD
VSS
M1 M2
M3 M4
VIN
M7
VIN
Vbb
Vxx
Vyy
Vzz
M3c M4c
M2cM1c
Vo1-Vo1+
First stage bias constraints:
Vbb = Vdd – |Vtp| – Veb3Vxx < Vdd – Vsdsat3 – |Vtp| – Veb3c
Vyy < Vdd – Vsdsat3 – Vsdsat3c + Vtn1c
Vicm > Vicmmin = Vss + Vdssat7 + Vtn1 + Veb1Vicm < Vicmmax = Vyy – Vdssat1c
Vicmr = Vicmmax – Vicmmin <Vdd – Vss – Vsdsat3 – Vsdsat3c – Vdssat7 – Vdssat1c + Vtn1c – Vtn1 – Veb1
This must > specification
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VDD
VSS
M6
CL
VO+
M7
CC
Mz
VO1-
Vzz
Second stage bias constraints:
Vo+ < Vdd – Vsdsat6Vo+ > Vss + Vdssat7
Vzz does not have to be the sameas that in first stage
Vo1- is not fixed, Vsdsat5 is varying with signal.
Vo1 needs to be able to swing high enough to turn off M6, so Vo1max >= Vdd – |Vtp|
Vo1 needs to be able to swing low enough for M6 to provide 2*I7=2*I6Q for charging CL, so Vo1min <= Vdd – |Vtp| – sqrt(2)Veb6Q
Vo1DSW = Vo1max – Vo1min >= sqrt(2)Veb6Q
This Vo1 swing range should be subtracted from the max V_ICMR
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Inter-stage bias constraints: Vo1 range must accommodate Vg6 needs, and Vg6 ranges from Vdd – |Vtp| to Vdd – Vtp| – sqrt(2)*Veb6Q
So worst case Vsdsat6: sqrt(2)*Vsdsat6QVo+ should be clear of this.If Vo range is symmetric, this gives enough room to put a cascode on M7 to reduce ro and Co.
Therefore we need: Vsdsat3+Vsdsat3c < |Vtp| so that when Vg6 = Vdd–|Vtp| M3 and M3c are still in saturation.
We also need: Vyy < Vdd – |Vtp| – sqrt(2)*Veb6Q + Vtn1c so that when Vg6 is the lowest M1c is still in saturation.
Vicmmax = Vyy – Vdssat1c < Vdd – |Vtp| – sqrt(2)*Veb6Q – Vdssat1c + Vtn1c
Vicmr <Vdd – Vss – |Vtp| – sqrt(2)*Veb6Q – Vdssat7 – Vdssat1c + Vtn1c – Vtn1 – Veb1
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Slew rate constraints
To charge and discharge CL and Cc, current goes from Vo node into or out of these caps.
Cc dVo/dt + CL dVo/dt = I6–I7 max|dVo/dt| <= I7/(CL+Cc)
The other end of Cc goes to Vo1 node. Cc dVo/dt + I3 = I1 max |dVo/dt| <= I1Q / Cc
If SR is limiting factor, we don’t want to waste any current, and we should have SR = I6Q/(CL+Cc) = I1Q / Cc I6Q : I1Q = (CL+Cc) : Cc
This significantly limits freedom in Cc and current split choices.
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If SR is not the performance limiter, can have the Cc charge discharge slower than CL.
In this case: SR = I1/Cc
If we have put most of the current into 2nd stage, and chosen Cc to be large compared to CL (CL/Cc ratio small), SR can be small.
This leads to constant slope slewing in large size step response and in large magnitude sinusoidal output at near GB frequencies.
This in turn leads to increased nonlinearity at high frequencies and high signal magnitudes.
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Design steps for max GB1. From total power allowance, compute Itot
Itot = Ptot/(VDD-VSS)
2. Assume that about 5 to 10 % of the current budget will go to reference and biasing circuits, 90 to 95% is available for 1st and 2nd stages. As a starting point, take I6 = ¾ Itot
3. Sizing of M6– For high speed, take small L6, e.g. L6 = L_min or 1.5L_min– For maximum GB, we maximize |p2|
W6 = CL/(2C_j L_drain + C_ox L6)– Compute: Cdb6 = Cj L_drain W6 + 2C_jsw(L_drain+W6)
Cgs6 = C_ox W6 L6
gm6 = sqrt(2 I6 uC_ox W6/L6)
|p2| = gm6 / (CL + Cgs6 + 2Cdb6) ≈ gm6 / 2CL
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Design steps for max GB4. Select Cc: Cc <= min{0.5 CL, 0.4 I_tot / SR}
e.g. take Cc = 0.2 CL
5. Compute gm1 and I5I5/2 >= SR * Cc, and watch I5 + I6 to be about 0.9 I_tot
if I5 too large, reduce Cc
Let GB = |p2|, gm1/Cc = gm6/2CL
gm1 = gm6 * Cc/2CL, 0.1gm6
6. Size M1/M2:Take L1 to be 1.5 or 2 L_min
W1 = gm1^2 L1 /{uC_ox I7}
If 1/f is concern, increase L1 and W1 together.
7. In the cascoded case, Vds3 and Vds4 are matched. Size (W/L)3 so that Veb3 to be about 1/3 |VTP|
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Design steps for max GB8. M7, M5 and M0 all share the same Vgs that
comes from the biasing circuitSelect a suitable VEB of about 0.3 V for them
Select the same L for all three
W5 = 2 I5 / VEB^2 / uC_ox
W7 = 2 I7 / VEB^2 / uC_ox
Take W0 to be about 5% W6 so that its current is about 5%
9. Sizing M8, M9, RzLet M8 have the same L as M5
W8 to W5 ratio to be the same as W0 to W6 ratio
Let M9 have the same W/L as M8, Rz have the same L
Perform parametric scan of W for Rz, to achieve best PM
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M2c
M2Vin-
CL
M4
M4c
M1c
M1Vin+
M3
M3c
Vyy
M5
VDD VDD
M7
CC vo
VDD
M6
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Design steps for max GB10. You can size M3c and M4c to be about the same as M3 and M4
Vdssat3c,4c are also about Veb3
Vdd-Vd3c is about 3 or 4 time Veb3
This makes it easy to ensure both M3 and M3c to be in saturation
11. Size the resistor and watch how Vd3 changesChoose the resistor value so that Vd3 is right in the middle of Vdd and
Vd3c
This gives Vds3 = Vds3c = ½ Vgs3 = ½ (1 + 1/3)VTP = 2/3 VTP
Both M3 and M3c are in saturation
Since Vds4 = Vds3, M4 is in saturation
Vds4c = Vgs5 – Vds4 = Vtp+Veb5 -2/3Vtp =1/3Vtp +Veb5
As long as Veb5 > 0, M4c is in saturation
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M2c
Vin-
CL
M4
M4c
M1c
Vin+
M3
M3c
M5
VDD VDD
M7
CC vo
VDD
M6
1B 2B
3B
4B
5B
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Design steps for max GB12. Size M1c and M2c to be about the same as M1,2
13. Size M1B and M2B to have W/L ratio that is about 5 to 10% of M1
Total current in M1B M2B is about 5 to 10% of I7
Vdd-Vd3c is about 3 or 4 time Veb3
This makes it easy to ensure both M3 and M3c to be in saturation
14. Match M3B and M4BTheir sizes are non-critical
But use reasonable Veb
15. Size M5B to create the correct bias for M1c,2cChoose its W/L ratio so that Veb5b is 3 to 4 times Veb1
This guarantees saturation of M1,2 and M1c,2c
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Bias generator
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Design steps for max GB16. For simplicity, you can have all M1 through M4 in the
bias circuit to be the same size • M6 is the tail current source in first stage• M5 is not needed in the self biasing design• Let W/L ratio of M1-4 to be about 1/10 of (W/L)_tail• This creates a current of about 1/10 I7 • M4 and M5 has better match than M6 and M1• You could flip up down for better matching
17. Size resistor• Scan resistor value to achieve I1, I2 that you want
18. Size shaded part so that • Current in M8 is tiny• Vg8 =Vg7 > 3Vtn but < 3Vtn + Veb1 + Veb2
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M2c
M2Vin-
CL
M4
M4c
M1c
M1Vin+
M3
M3c
Vyy
M5
VDD VDD
M7
CC vo
VDD
M6
Can feed to here also
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The Miller capacitor Cc sees an amplifier consisting of a common gate amplifier M4c followed by a common source amplifier M6.This amplifier’s low frequency gain is
4 1 6( )m c o m og r g r
The miller capacitance seen at D4 is very large:
4 1 6( )m c o m o Cg r g r C
The impedance at this node is: 1/gm4c
This node gives the lowest frequency pole which determines the bandwidth of the amplifier:
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4 1 6 1 6
1( )
m c
m c o m o C o m o C
gBW pg r g r C r g r C
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The DC gain of the amplifier is
0 1 1 6( )( )m o m oA g r g r
The gain bandwidth product is then
10 1 1 1 6
1 6
1| | ( )( ) mm o m o
o m o C C
gGB A p g r g rr g r C C
The max rate of Vcc can change is when all of first stage current goes to charge CC. Hence, slew rate is
1
C
ISRC
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M2c
M2Vin-
CL
M4
M4c
VDD
M7
CC vo
VDD
M6
To calculate zero:
Set Vo = 0 and all DC voltage to gnd
Vd4 = VccVg6 = Vcc*gm4c/g01
Icc + gm6 Vg6 =0
sCc + gm6gm4c/g01 =0
Z1= – gm6gm4c/(g01Cc)
At extremely high frequency!Cannot be used to cancel a pole near unity gain frequency!
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M2c
M2Vin-
CL
M4
M4c
VDD
M7
CC vo
VDD
M6
Alternative compensation for removing right-half plane zero:
Noncascode version:
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VDD
VSS
M1 M2
M3 M4 M6
CL
VIN
VOUT
M7M5
VIN
CC
M8
M9
M0
Mz
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VDD
VSS
M1 M2
M3 M4 M6
CL
VIN
VOUT
M7M5
VIN
CC
M8
M9
M0
Mz
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VDD
VSS
M1 M2
M3 M4 M6
CL
VIN
VOUT
M7M5
VIN
CC
M8
M9
M0
Mz
B1 B2
M2cM1c
M3c M4c
B3 B4
B5
M6
Mz
CL
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VDD
VSS
M1 M2
M3 M4 M6
CL
VIN
VOUT
M5
VIN
CC
M8
M9
M0
Mz
B1 B2
M2cM1c
M3c M4c
B3 B4
B5
M7
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VDD
VSS
M1 M2
M3 M4 M6
CL
VIN
VOUT
M7M5
VIN
CC
M8
M9
M0
Mz
M6
CL
VOUT
M7
CC
Mz
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VDD
VSS
VBP
VBN
IREF
IREF
1. Need to add start-up circuit
2. Add MOSCAPs between VBP and VDD, and between VBN and VSS
3. NMOS W ratio and R determines current value
4. Cascode to improve supply sensitivity
5. Or use a regulate amp
6. VBN and VBP may be directly used as biasing voltage for non-critical use
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VSS
IREF IREF
1st stage
tail
Folded stageNMOS
2nd stage load
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