Allen and Holberg - CMOS Analog Circuit Design
Short Description
Allen and Holberg - CMOS Analog Circuit Design...
Description
Allen and Holberg - CMOS Analog Circuit Design
I. INTRODUCTION Contents
I.1
Introduction
I.2
Analog Integrated Circuit Design
I.3
Technology Overview
I.4
Notation
I.5
Analog Circuit Analysis Techniques
Page I.0-1
Allen and Holberg - CMOS Analog Circuit Design Organization
Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Opamps
Chapter 9 High Performance Opamps
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
Chapter 3 CMOS Device Modeling
DEVICES
Introduction
Chapter 4 Device Characterization
Page I.0-2
Allen and Holberg - CMOS Analog Circuit Design
Page I.2-1
I.1 - INTRODUCTION
GLOBAL OBJECTIVES • Teach the analysis, modeling, simulation, and design of analog circuits implemented in CMOS technology. • Emphasis will be on the design methodology and a hierarchical approach to the subject.
SPECIFIC OBJECTIVES 1. Present an overall, uniform viewpoint of CMOS analog circuit design. 2. Achieve an understanding of analog circuit design. • Hand calculations using simple models • Emphasis on insight • Simulation to provide second-order design resolution 3. Present a hierarchical approach. • Sub-blocks → Blocks → Circuits → Systems 4. Examples to illustrate the concepts.
Allen and Holberg - CMOS Analog Circuit Design
Page I.2-1
I.2 ANALOG INTEGRATED CIRCUIT DESIGN
ANALOG DESIGN TECHNIQUES VERSUS TIME
FILTERS
AMPLIFICATION
Passive RLC circuits
Open-loop amplifiers
1935-1950 Active-RC Filters Requires precise definition of time constants (RC products)
Feedback Amplifiers Requires precise definition of passive components
1978 Switched Capacitor Filters Requires precise C ratios and clock
Switched Capacitor Amplifiers Requires precise C ratios
1983 Continuous Time Filters Time constants are adjustable
Continuous Time Amplifiers Component ratios are adjustable
1992
?
Digitally assisted analog circuits
?
Allen and Holberg - CMOS Analog Circuit Design
Page I.2-2
DISCRETE VS. INTEGRATED ANALOG CIRCUIT DESIGN
Activity/Item
Discrete
Integrated
Component Accuracy
Well known
Poor absolute accuracies
Breadboarding?
Yes
No (kit parts)
Fabrication
Independent
Very Dependent
Physical
PC layout
Layout, verification, and
Implementation Parasitics
extraction Not Important
Must be included in the design
Simulation
Testing
CAD
Components
Model parameters well
Model parameters vary
known
widely
Generally complete
Must be considered
testing is possible
before the design
Schematic capture,
Schematic capture,
simulation, PC board
simulation, extraction,
layout
LVS, layout and routing
All possible
Active devices, capacitors, and resistors
Allen and Holberg - CMOS Analog Circuit Design
Page I.2-3
THE ANALOG IC DESIGN PROCESS
Conception of the idea
Definition of the design Comparison with design specifications
Implementation
Simulation
Physical Definition
Physical Verification
Parasitic Extraction
Fabrication
Testing and Verification
Product
Comparison with design specifications
Allen and Holberg - CMOS Analog Circuit Design
Page I.2-4
COMPARISON OF ANALOG AND DIGITAL CIRCUITS
Analog Circuits
Digital Circuits
are discontinuous in Signals are continuous in amplitude Signal and can be continuous or discrete in amplitude and time - binary signals have two amplitude states time Designed at the circuit level
Designed at the systems level
Components must have a continuum Component have fixed values of values Customized
Standard
CAD tools are difficult to apply
CAD tools have been extremely successful
Requires precision modeling
Timing models only
Performance optimized
Programmable by software
Irregular block
Regular blocks
Difficult to route automatically
Easy to route automatically
Dynamic range limited by power Dynamic range unlimited supplies and noise (and linearity)
Allen and Holberg - CMOS Analog Circuit Design
Page I.3-1
I.3 TECHNOLOGY OVERVIEW BANDWIDTHS OF SIGNALS USED IN SIGNAL PROCESSING APPLICATIONS
Video Acoustic imaging
Seismic
Radar
Sonar
Audio
AM-FM radio, TV Telecommunications
1
10
100
1k
10k
100k 1M 10M 100M Signal Frequency (Hz)
Microwave
1G
Signal frequency used in signal processing applications.
10G
100G
Allen and Holberg - CMOS Analog Circuit Design
Page I.3-2
BANDWIDTHS THAT CAN BE PROCESSED BY PRESENTDAY TECHNOLOGIES
BiCMOS Bipolar analog Bipolar digital logic MOS digital logic MOS analog Optical GaAs
1
10
100
1k
10k
100k 1M 10M 100M Signal Frequency (Hz)
1G
10G
Frequencies that can be processed by present-day technologies.
100G
Allen and Holberg - CMOS Analog Circuit Design
Page I.3-3
CLASSIFICATION OF SILICON TECHNOLOGY
Silicon IC Technologies
Bipolar
Junction Isolated
Dielectric Isolated
Bipolar/MOS
CMOS
Aluminum gate
MOS
PMOS (Aluminum Gate)
Silicon gate
NMOS
Aluminum gate
Silicon gate
Allen and Holberg - CMOS Analog Circuit Design
Page I.3-4
BIPOLAR VS. MOS TRANSISTORS
CATEGORY
BIPOLAR
CMOS
Turn-on Voltage
0.5-0.6 V
0.8-1 V
Saturation Voltage
0.2-0.3 V
0.2-0.8 V
gm at 100µA
4 mS
0.4 mS (W=10L)
Analog Switch Implementation
Offsets, asymmetric
Good
Power Dissipation
Moderate to high
Low but can be large
Speed
Faster
Fast
Compatible Capacitors
Voltage dependent
Good
AC Performance Dependence
DC variables only
DC variables and geometry
Number of Terminals
3
4
Noise (1/f)
Good
Poor
Noise Thermal
OK
OK
Offset Voltage
< 1 mV
5-10 mV
Allen and Holberg - CMOS Analog Circuit Design
Page I.3-5
WHY CMOS???
CMOS is nearly ideal for mixed-signal designs: • Dense digital logic • High-performance analog
DIGITAL
ANALOG
MIXED-SIGNAL IC
Allen and Holberg - CMOS Analog Circuit Design
I.4
NOTATION
SYMBOLS FOR TRANSISTORS Drain Gate
Drain Bulk Gate
Source Source/bulk n-channel, enhance- n-channel, enhancement, bulk at most ment, VBS ≠ 0 negative supply
Drain Gate
Drain Bulk Gate
Source Source/bulk p-channel, enhance- p-channel, enhancement, bulk at most ment, VBS ≠ 0 positive supply
Page I.4-1
Allen and Holberg - CMOS Analog Circuit Design SYMBOLS FOR CIRCUIT ELEMENTS
Operational Amplifier/Amplifier/OTA
+
-
I
V
+
+
G mV1
AvV1 V1
V1
-
-
VCVS
VCCS I1
I1 Rm I 1
CCVS
Ai I 1
CCCS
Page I.4-2
Allen and Holberg - CMOS Analog Circuit Design
Page I.4-3
Notation for signals
Id id
ID iD
time
Allen and Holberg - CMOS Analog Circuit Design
II. CMOS TECHNOLOGY Contents
II.1
Basic Fabrication Processes
II.2
CMOS Technology
II.3
PN Junction
II.4
MOS Transistor
II.5
Passive Components
II.6
Latchup Protection
II.7
ESD Protection
II.8
Geometrical Considerations
Page II.0-1
Allen and Holberg - CMOS Analog Circuit Design Perspective Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Opamps
Chapter 9 High Performance Opamps
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Page II.0-2
Allen and Holberg - CMOS Analog Circuit Design
Page II.0-3
OBJECTIVE • Provide an understanding of CMOS technology sufficient to enhance circuit design. • Characterize passive components compatible with basic technologies. • Provide a background for modeling at the circuit level. • Understand the limits and constraints introduced by technology.
Allen and Holberg - CMOS Analog Circuit Design
Page II.1-1
II.1 - BASIC FABRICATION PROCESSES
BASIC FABRTICATION PROCESSES
Basic Steps • Oxide growth • Thermal diffusion • Ion implantation • Deposition • Etching Photolithography Means by which the above steps are applied to selected areas of the silicon wafer. Silicon wafer 0.5-0.8 mm 125-200 mm
n-type: 3-5 Ω -cm p-type: 14-16 Ω -cm
Allen and Holberg - CMOS Analog Circuit Design
Page II.1-2
Oxidation The process of growing a layer of silicon dioxide (SiO2)on the surface of a silicon wafer. Original Si surface
tox
SiO 2
0.44 tox
Si substrate
Uses: • Provide isolation between two layers • Protect underlying material from contamination • Very thin oxides (100 to 1000 Å) are grown using dry-oxidation techniques. Thicker oxides (>1000 Å) are grown using wet oxidation techniques.
Allen and Holberg - CMOS Analog Circuit Design
Page II.1-3
Diffusion Movement of impurity atoms at the surface of the silicon into the bulk of the silicon - from higher concentration to lower concentration.
High Concentration
Low Concentration
Diffusion typically done at high temperatures: 800 to 1400 °C. Infinite-source diffusion: N0
ERFC t1 (FC)·φ B (1+FC)1+MJ φB
CBD =
CBS0ABS vBS 1 − (1+MJ)FC + FC , vBS > (FC)·φ B (1+FC)1+MJ φB
and
Allen and Holberg - CMOS Analog Circuit Design
Page III.2-3
Bottom & Sidewall Approximations Polysilicon gate
H
G C
D Source
Drain F E A
B
SiO2 Bulk
Drain bottom = ABCD Drain sidewall = ABFE + BCGF + DCGH + ADHE
CBX =
(CJ)(AX) (CJSW)(PX) + , vBX ≤ (FC)(PB) vBX MJ vBX MJSW 1 − 1 − PB PB
and CBX =
vBX (CJ)(AX) 1 − (1 + MJ)FC + MJ PB (1 − FC)1+MJ +
vBX (CJSW)(PX) , 1 − (1 + MJSW)FC + (MJSW) PB (1 − FC)1+MJSW
vBX ≥ (FC)(PB) where AX = area of the source (X = S) or drain (X = D) PX = perimeter of the source (X = S) or drain (X = D) CJSW = zero-bias, bulk-source/drain sidewall capacitance MJSW = bulk-source/drain sidewall grading coefficient
Allen and Holberg - CMOS Analog Circuit Design
Page III.2-4
Overlap Capacitance
Mask L
Actual L (Leff)
Oxide encroachment
Mask W
LD
Actual W (Weff)
Gate
Source-gate overlap capacitance CGS (C1)
Drain-gate overlap capacitance CGD (C3) Gate
FOX
Source
Drain
FOX
Bulk
Figure 3.2-5 Overlap capacitances of an MOS transistor. (a) Top view showing the overlap between the source or drain and the gate. (b) Side view.
C1 = C3 ≅ (LD)(Weff )Cox = (CGXO)Weff
Allen and Holberg - CMOS Analog Circuit Design
Page III.2-5
Gate to Bulk Overlap Capacitance
Overlap
FOX
C5
Overlap
Gate
C5
Source/Drain
FOX
Bulk
Figure 3.2-6 Gate-bulk overlap capacitances.
On a per-transistor basis, this is generally quite small Channel Capacitance C2 = Weff (L − 2LD)Cox = Weff (Leff )Cox Drain and source portions depend upon operating condition of transistor.
Allen and Holberg - CMOS Analog Circuit Design
Page III.2-6
MOSFET Gate Capacitance Summary:
Capacitance C2 + 2C5 CGS
C1 + _23 C2
CGS, CGD
C1 + _12 C2
CGS, CGD
C1, C3
vDS = constant vBS = 0
CGD CGB
2C5 0 Off
Saturation VT
vDS +VT
NonSaturation
vGS
Figure 3.2-7 Voltage dependence of CGS, CGD, and CGB as a function of VGS with VDS constang and VBS = 0.
iD
v DS = v GS - VT
Non-Sat Region
Saturation Region Cutoff Region 0 0
0.5
1.0
vDS = constant
1.5
2.0
2.5
Allen and Holberg - CMOS Analog Circuit Design CGS, CGD, and CGB
Off CGB = C2 + 2C5 = Cox(Weff )(Leff ) + CGBO(Leff ) CGS = C1 ≅ Cox(LD)(Weff ) = CGSO(Weff ) CGD = C3 ≅ Cox(LD)(Weff ) = CGDO(Weff ) Saturation CGB = 2C5 = CGBO (Leff ) CGS = C1 + (2/3)C2 = Cox(LD + 0.67Leff )(Weff ) = CGSO(Weff ) + 0.67Cox(Weff )(Leff ) CGD = C3 ≅ Cox(LD)(Weff ) = CGDO(Weff ) Nonsaturated CGB = 2C5 = CGBO (Leff ) CGS = C1 + 0.5C2 = Cox(LD + 0.5Leff )(Weff ) = (CGSO + 0.5CoxLeff )Weff CGD = C3 + 0.5C2 = Cox(LD + 0.5Leff )(Weff ) = (CGDO + 0.5CoxLeff )Weff
Page III.2-7
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-1
Small-Signal Model for the MOS Transistor D
rD Cbd
inrD Cgd
G
gbd
gds
gmvgs
B
inD Cgs
gmbsvbs
gbs
inrS
Cgb
Cbs
rS
S
Figure 3.3-1 Small-signal model of the MOS transistor.
gbd =
∂IBD ∂VBD
(at the quiescent point) ≅ 0
and gbs =
∂IBS ∂VBS
(at the quiescent point) ≅ 0
The channel conductances, gm, gmbs, and gds are defined as gm =
∂ID ∂VGS
gmbs =
(at the quiescent point)
∂ID ∂VBS
(at the quiescent point)
and gds =
∂ID ∂VDS
(at the quiescent point)
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-2
Saturation Region gm =
gmbs =
Noting that
(2K'W/L)| ID|(1 + λ VDS) ≅
(2K'W/L)|ID|
−∂ID
∂ID ∂VT = − ∂VSB ∂VT ∂VSB
∂ID −∂ID = , we get ∂VT ∂VGS
gmbs = gm
gds = go =
γ = η gm 2(2|φF| + VSB)1/2
ID λ 1 + λ VDS
≅ ID λ
Relationships of the Small Signal Model Parameters upon the DC Values of Voltage and Current in the Saturation Region. Small Signal DC Current DC Current and DC Voltage Model Parameters Voltage 2K' W gm ≅ (2K' IDW/L)1/2 _ ≅ (V -V ) gmbs gds
L
GS
T
γ (2IDβ)1/2 2(2|φF | +VSB) 1/2
≅ λ ID
γ ( β (VGS −VT) ) 2(2|φF | + VSB)1/2
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-3
Nonsaturation region gm =
∂Id = β VDS ∂VGS
gmbs =
∂ID βγ VDS = ∂VBS 2(2|φF | + VSB)1/2
and gds = β (VGS − VT − VDS)
Relationships of the Small-Signal Model Parameters upon the DC Values of Voltage and Current in the Nonsaturation Region. Small Signal DC Voltage and/or Current Model Parameters Dependence = β VDS gm β γ VDS gmbs 2(2|φF | +VSB)1/2
= β (VGS − VT
gds
Noise 2 4kT i nrD = ∆f rD
(A2)
2 4kT i nrS = ∆f rS
(A2)
and 8kT gm(1+η) (KF )ID 2 2 i nD = + 2∆f (A ) 3 f C L ox
− VDS)
Allen and Holberg - CMOS Analog Circuit Design
Page III.4-1
SPICE Level 3 Model The large-signal model of the MOS device previously discussed neglects many important second-order effects. Most of these second-order effects are due to narrow or short channel dimensions (less than about 3µm). We shall also consider the effects of temperature upon the parameters of the MOS large signal model. We first consider second-order effects due to small geometries. When vGS is greater than VT, the drain current for a small device can be given as Drain Current 1 + fb iDS = BETA vGS − VT − 2 vDE ⋅ vDE
(1)
Weff Weff = µeffCOX BETA = KP L Leff eff
(2)
Leff = L − 2(LD)
(3)
Weff = W − 2(WD)
(4)
vDE = min(vDS , vDS (sat))
(5)
fb = fn +
GAMMA ⋅ fs 4(PHI + vSB)1/2
(6)
Note that PHI is the SPICE model term for the quantity 2φf . Also be aware that PHI is always positive in SPICE regardless of the transistor type (p- or n-channel). fn =
DELTA πεsi Weff 2 ⋅ COX
(7)
1/2 xj LD + wc wp 2 LD fs = 1 − 1− − x xj Leff xj + wp j
(8)
wp = xd (PHI + vSB )1/2
(9)
2⋅εsi 1/2 xd = q ⋅ NSUB
(10)
Allen and Holberg - CMOS Analog Circuit Design wp wp2 wc = xj k1 + k2 − k3 xj xj
Page III.4-2
(11)
k1 = 0.0631353 , k2 = 0.08013292 , k3 = 0.01110777 Threshold Voltage ETA⋅8.15-22 v + GAMMA ⋅ f ( PHI + v )1/2 + f ( PHI + v ) VT = Vbi − s SB n SB C L 3 DS ox eff
(12)
vbi = vfb + PHI
(13)
vbi = VTO − GAMMA ⋅ PHI
(14)
or
Saturation Voltage vgs − VT
vsat =
(15)
1 + fb 1/2
2 2 vDS(sat) = vsat + vC − vsat + vC
vC
=
VMAX ⋅ Leff
µs
(16)
(17)
If VMAX is not given, then vDS(sat) = vsat Effective Mobility
µs =
U0 when VMAX = 0 1 + THETA (vGS − VT)
µeff =
µs when VMAX > 0; otherwise µeff = µs vDE 1+ vC
Channel-Length Modulation When VMAX = 0
(18)
(19)
Allen and Holberg - CMOS Analog Circuit Design
Page III.4-3
1/2
∆L = xd KAPPA (vDS − vDS(sat))
(20)
when VMAX > 0 ∆L = −
ep ⋅ xd 2 2
1/2
ep ⋅ xd 2 2 + 2 + KAPPA ⋅ xd 2 ⋅ (vDS − vDS(sat))
(21)
where ep =
iDS =
vC (vC + vDS(sat)) Leff vDS (sat) iDS 1 − ∆L
(22)
(21)
Weak Inversion Model (Level 3) In the SPICE Level 3 model, the transition point from the region of strong inversion to the weak inversion characteristic of the MOS device is designated as von and is greater than VT. von is given by von = VT + fast
(1)
q ⋅ NFS GAMMA ⋅ fs (PHI + vSB)1/2 + fn (PHI + vSB) kT 1 + + fast = COX 2(PHI + vSB) q
(2)
where
N F S is a parameter used in the evaluation of v on and can be extracted from measurements. The drain current in the weak inversion region, vGS less than von , is given as vGS - von iDS = iDS (von , vDE , vSB) e fast
(3)
where iDS is given as (from Eq. (1), Sec. 3.4 with vGS replaced with von) 1 + fb v ⋅v iDS = BETAvon − VT − 2 DE DE
(4)
Allen and Holberg - CMOS Analog Circuit Design
Page III.4-4
Typical Model Parameters Suitable for SPICE Simulations Using Level-3 Model (Extended Model). These Values Are Based upon a 0.8µm Si-Gate Bulk CMOS nWell Process Parameter Parameter Typical Parameter Value Symbol Description N-Channel P-Channel Units VTO Threshold V 0.7 ± 0.15 −0.7 ± 0.15 UO mobility 660 210 cm2/V-s DELTA Narrow-width threshold 2.4 1.25 adjust factor ETA Static-feedback threshold 0.1 0.1 adjust factor KAPPA Saturation field factor in 0.15 2.5 1/V channel-length modulation THETA Mobility degradation factor 0.1 0.1 1/V NSUB Substrate doping cm-3 3×1016 6×1016 TOX Oxide thickness 140 140 Å XJ Mettallurgical junction 0.2 0.2 µm depth WD Delta width µm LD Lateral diffusion 0.016 0.015 µm NFS Parameter for weak 11 11 cm-2 7×10 6×10 inversion modeling CGSO F/m 220 × 10 −12 220 × 10 −12 CGDO F/m 220 × 10 −12 220 × 10 −12 CGBO F/m 700 × 10 −12 700 × 10 −12 CJ F/m2 770 × 10 −6 560 × 10 −6 CJSW F/m 380 × 10 −12 350 × 10 −12 MJ 0.5 0.5 MJSW 0.38 0.35 NFS Parameter for weak cm-2 7×1011 6×1011 inversion modeling
Allen and Holberg - CMOS Analog Circuit Design
Page III.4-5
Temperature Dependence The temperature-dependent variables in the models developed so far include the: Fermi potential, PHI, EG, bulk junction potential of the source-bulk and drain-bulk junctions, PB, the reverse currents of the pn junctions, IS, and the dependence of mobility upon temperature. The temperature dependence of most of these variables is found in the equations given previously or from well-known expressions. The dependence of mobility upon temperature is given as T BEX UO(T) = UO(T0) T0 where BEX is the temperature exponent for mobility and is typically -1.5. vtherm(T) =
KT q
T2 EG(T) = 1.16 − 7.02 ⋅ 10−4 ⋅ T + 1108.0 EG(T0) T T EG(T) − PHI(T) = PHI(T0) ⋅ − vtherm(T) 3 ⋅ ln + T0 T0 vtherm(T0) vtherm(T) vbi (T) = vbi (T0) +
PHI(T) − PHI(T0) EG(T0) − EG(T) + 2 2
VT0(T) = vbi (T) + GAMMA PHI(T) NSUB PHI(T)= 2 ⋅ vtherm ln ni (T) ni(T) = 1.45 ⋅
3/2
1016
T ⋅ T0
T 1 ⋅ exp EG ⋅ − 1 ⋅ T 0 2 ⋅ vtherm(T0)
For drain and source junction diodes, the following relationships apply. EG(T0) T T EG(T) PB(T) = PB ⋅ − vtherm(T) 3 ⋅ ln + − T0 T v (T v 0 therm 0) therm(T) IS(T) =
IS(T0) EG(T0) EG(T) T ⋅ exp − + 3 ⋅ ln v (T v (T) T N therm 0 therm 0)
where N is diode emission coefficient. The nominal temperature, T0, is 300 K.
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-1
SPICE Simulation of MOS Circuits Minimum required terms for a transistor instance follows: M1 3 6 7 0 NCH W=100U L=1U “M,” tells SPICE that the instance is an MOS transistor (just like “R” tells SPICE that an instance is a resistor). The “1” makes this instance unique (different from M2, M99, etc.) The four numbers following”M1” specify the nets (or nodes) to which the drain, gate, source, and substrate (bulk) are connected. These nets have a specific order as indicated below: M ... Following the net numbers, is the model name governing the character of the particular instance. In the example given above, the model name is “NCH.” There must be a model description of “NCH.” The transistor width and length are specified for the instance by the “W=100U” and “L=1U” expressions. The default units for width and length are meters so the “U” following the number 100 is a multiplier of 10-6. [Recall that the following multipliers can be used in SPICE: M, U, N, P, F, for 10-3, 10-6, 10-9, 10-12 , 10 -15 , respectively.] Additional information can be specified for each instance. Some of these are Drain area and periphery (AD and PD) ← calc depl cap and leakage Source area and periphery (AS and PS) ← calc depl cap and leakage Drain and source resistance in squares (NRD and NRS) Multiplier designating how many devices are in parallel (M) Initial conditions (for initial transient analysis) The number of squares of resistance in the drain and source (NRD and NRS) are used to calculate the drain and source resistance for the transistor.
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-2
Geometric Multiplier: M To apply the “unit-matching” principle, use the geometric multiplier feature rather than scale W/L. This: M1 3 2 1 0 NCH W=20U L=1U is not the same as this: M1 3 2 1 0 NCH W=10U L=1U M=2 The following dual instantiation is equivalent to using a multiplier M1A 3 2 1 0 NCH W=10U L=1U M1B 3 2 1 0 NCH W=10U L=1U
(a)
(b)
(a)M1 3 2 1 0 NCH W=20U L=1U. (b) M1 3 2 1 0 NCH W=10U L=1U M=1. .
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-3
MODEL Description A SPICE simulation file for an MOS circuit is incomplete without a description of the model to be used to characterize the MOS transistors used in the circuit. A model is described by placing a line in the simulation file using the following format. .MODEL MODEL NAME e.g., “NCH” MODEL TYPE either “PMOS” or “NMOS.” MODEL PARAMETERS : LEVEL=1 VTO=1 KP=50U GAMMA=0.5 LAMBDA=0.01 SPICE can calculate what you do not specify You must specify the following • surface state density, NSS, in cm-2 • oxide thickness, TOX, in meters • surface mobility, UO, in cm2/V-s, • substrate doping, NSUB, in cm-3 The equations used to calculate the electrical parameters are VTO = φMS −
KP = UO
(2q ⋅ εsi ⋅ NSUB ⋅ PHI)1/2 q(NSS) + + PHI (εox/TOX) (εox/TOX)
εox TOX
GAMMA =
(2q ⋅ εsi ⋅ NSUB)1/2 (εox/TOX)
and 2kT NSUB PHI = 2φF = ln q ni LAMBDA is not calculated from the process parameters for the LEVEL 1 model.
Allen and Holberg - CMOS Analog Circuit Design
Page III.3-4
Other parameters: IS: Reverse current of the drain-bulk or source-bulk junctions in Amps JS: Reverse-current density in A/m2 JS requires the specification of AS and AD on the model line. If IS is specified, it overrides JS. The default value of IS is usually 10-14 A. RD: Drain ohmic resistance in ohms RS: Source ohmic resistance in ohms RSH: Sheet resistance in ohms/square. RSH is overridden if RD or RS are entered. To use RSH, the values of NRD and NRS must be entered on the model line. The drain-bulk and source-bulk depletion capacitors CJ: Bulk bottom plate junction capacitance MJ: Bottom plate junction grading coefficient CJSW: Bulk sidewall junction capacitance MJSW: Sidewall junction grading coefficient If CJ is entered as a model parameter it overrides the calculation of CJ using NSUB, otherwise, CJ is calculated using NSUB. If CBD and CBS are entered, these values override CJ and NSUB calculations. In order for CJ to result in an actual circuit capacitance, the transistor instance must include AD and AS. In order for CJSW to result in an actual circuit capacitance, the transistor instance must include PD and PS. CGSO: CGDO:
Gate-Source overlap capacitance (at zero bias) Gate-Drain overlap capacitance (at zero bias)
AF: KF:
Flicker noise exponent Flicker noise coefficient
TPG: Indicates type of gate material relative to the substrate TPG=1 > gate material is opposite of the substrate TPG=-1 > gate material is the same as the substrate TPG=0 > gate material is aluminum XQC: Channel charge flag and fraction of channel charge attributed to the drain
Allen and Holberg - CMOS Analog Circuit Design
Page IV.0-1
IV. CMOS PROCESS CHARACTERIZATION Contents IV.1 IV.2 IV.3 IV.4 IV.5
Measurement of basic MOS level 1 parameters Characterization of the extended MOS model Characterization other active components Characterization of resistance Characterization of capacitance
Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Opamps
Chapter 9 High Performance Opamps
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
DEVICES
1
Allen and Holberg - CMOS Analog Circuit Design I.
Page IV.1-1
Characterization of the Simple Transistor Model
Determine V T0(V SB = 0), K', γ, and λ. Terminology: K'S for the saturation region K'L for the nonsaturation region W eff (v GS - V T ) 2 (1 + λ v DS ) iD = K' S 2L eff
2 v DS W eff (v G S - V T ) v D S iD = K' L 2 L eff V T = V T0 + γ
2| φ F | + v S B -
2| φ F |
(1)
(2) (3)
Assume that vDS is chosen such that the λ vDS V T = V T0 . Therefore, Eq. (1) simplifies to W eff (v GS - V T0 ) 2 iD = K’ S 2L eff
(4)
This equation can be manipulated algebraically to obtain the following 1/2 K' S W eff 1/2 1/2 K' S W eff iD = 2L vGS - 2L VT0 eff eff
which has the form y = mx + b 1/2
(5)
(6)
y = iD
(7)
x = v GS
(8)
1
Allen and Holberg - CMOS Analog Circuit Design K' S W eff 1/2 m= 2L eff
Page IV.1-2
(9)
and K' S W eff 1/2 b = − V T0 2L eff 1/2
Plot i D
(10)
versus vGS and measure slope. to get K'S
1/2
When iD = 0 the x intercept (b') is V T0 .
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.1-3
Mobility degradation region
v DS > VDSAT ( iD )
1/2
Weak inversion region
b ′ = VT0
K S′ Weff m= 2 L eff
1/2
v GS (a)
v DS = 0 . 1 V iD K L′ Weff m= L eff
vDS
vGS (b) Figure B.1-1 (a) iD1/2 versus vGS plot used to determine VT0 and K'S. (b) iD versus vGS plot to determine K'L.
Extract the parameter K'L for the nonsaturation region: v D S W eff W eff v DS v GS - K' L v DS V T + iD = K' L 2 L eff L eff
(11)
Plot iD versus vGS as shown in Fig. B.1-1(b), the slope is seen to be 3
Allen and Holberg - CMOS Analog Circuit Design
Page IV.1-4
∆iD W eff vDS m = ∆v = K' L Leff GS Knowing the slope, the term K'L is easily determined to be L eff W eff
K' L = m
1 vDS
(12)
(13)
W eff, Leff, and vDS must be known. The approximate value µo can be extracted from the value of K'L At this point, γ is unknown. Write Eq. (3) in the linear form where y = VT
(14)
2| φ F | + v SB −
(15)
x=
2| φ F |
m=γ
(16)
b = V T0
(17)
2|φF| normally in the range of 0.6 to 0.7 volts. Determine VT at various values of vSB Plot VT versus x and measure the slope to extract γ Slope m, measured from the best fit line, is the parameter γ.
4
Allen and Holberg - CMOS Analog Circuit Design
(i D )
Page IV.1-5
1/2
VT0
VT1
VT2
VT3
v GS Figure B.1-2 iD1/2 versus vGS plot at different vSB values to determine γ.
VSB = 3 V VSB = 2 V VT
VSB = 1 V
m=γ
VSB = 0 V
( vSB + 2 φ F )
0.5
− ( 2 φF )
0.5
Figure B.1-3 Plot of VT versus f(vSB) to determine γ.
We still need to find λ, ∆L, and ∆W. λ should be determined for all device lengths that might be used. Rewrite Eq. (1) is as iD = i' D λ vDS + i' D
(18)
5
Allen and Holberg - CMOS Analog Circuit Design which is in the familiar linear form where y = iD (Eq. (1))
Page IV.1-6
(19)
x = v DS
(20)
m = λ i'D
(21)
b = i'D (Eq. (4) with λ = 0)
(22)
Plot iD versus v DS , and measure the slope of the data in the saturation region, and divide that value by the y-intercept to getλ.
Saturation region Nonsaturation region iD i'D
m = λ i'D
v DS Figure B.1-4 Plot of iD versus vDS to determine λ.
Calculating ∆L and ∆W. Consider two transistors, with the same widths but different lengths, operating in the nonsaturation region with the same vDS. The widths of the transistors are assumed to be very large so that W ≅ Weff. The large-signal model is given as 2
v K' L W eff DS iD = L (v V )v T 0 D S 2 eff G S
(23)
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.1-7
and K' L W eff ∂ID VD S = gm = ∂V GS L eff
The aspect ratios (W/L) for the two transistors are W1 L1 + ∆L
(24)
(25)
and W2 L2 + ∆L
(26)
Implicit in Eqs. (25) and (26) is that ∆L is assumed to be the same for both transistors. Combining Eq. (24) with Eqs. (25) and (26) gives K' L W (27) gm1 = L + ∆ L v DS 1 and K' L W gm2 = L + ∆ L v DS 2
(28)
where W 1 = W 2 = W (and are assumed to equal the effective width). With further algebraic manipulation of Eqs. (27) and (28), one can show that, L2 + ∆L gm1 = (29) gm1 - gm2 L2 - L1 which further yields (L 2 - L 1 ) g m1 L 2 + ∆L = L eff = g m1 - gm2
(30)
L2 and L1 known gm1 and gm2 can be measured Similarly for W eff : (W 1 - W 2 )g m2 W 2 + ∆W = W eff = g m1 - gm2
(31)
Equation (31) is valid when two transistors have the same length but different widths.
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.1-8
One must be careful in determining ∆L (or ∆W) to make the lengths (or widths) sufficiently different in order to avoid the numerical error due to subtracting large numbers, and small enough that the transistor model chosen is still valid for both transistors.
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.3-1
II. Transistor Characterization for the Extended Model Equations (1) and (2) represent a simplified version of the extended model for a relatively wide MOS transistor operating in the nonsaturation, stronginversion region with VSB = 0.
µsCoxW iD = L
v2 (v DS + γ v V )v T DS 2 DS GS
2| φ F |
2γ − 3 [(v DS + 2| φ F |) 1.5 - (2| φ F |)] 1.5
(1)
where W and L are effective electrical equivalents (dropping the subscript, “eff”, for convenience). (UCRIT)εsi UEXP Cox[v GS - V T - (UTRA)v DS ]
µs = µo
(2)
Eq. (2) holds when the denominator term in the brackets is less than unity. Otherwise, µo = µs. To develop a procedure for extracting µo, consider the case where mobility degradation effects are not being experienced, i.e., µs = µo, Eq. (1) can be rewritten in general as (3) iD = µ o f(C ox , W, L, v GS , V T , v DS , γ , 2|φ F |) This equation is a linear function of vGS and is in the familiar form of y = mx + b (4) where b = 0. Plot iD versus the function, f(Cox, W , L, v GS , V T , v DS , γ , 2|φ F |) and measure the slope = µo. • • •
The data are limited to the nonsaturation region (small vDS ). The transistor must be in the strong-inversion region (vGS > VT). The transistor must operate below the critical-mobility point.
Keep vGS as low as possible without encroaching on the weak-inversion region of operation.
1
Allen and Holberg - CMOS Analog Circuit Design
Page IV.3-2
Region of variable mobility iD
Region of constant mobility
Weak-inversion region
v GS Figure B.2-1 Plot of iD versus vGS in the nonsaturation region.
Once µo is determined, there is ample information to determine UCRIT and UEXP. Consider Eqs. (1) and (2) rewritten and combined as follows. iD = µo[(UCRIT)f2]UEXPf1
(5)
where 2
v C ox W DS f1 = L (v G S - V T )v D S - 2 + γ vDS 2| φ F | 2γ − 3 [(v DS + 2| φ F |) 1.5 - (2| φ F |)] 1.5
(1)
and
2
Allen and Holberg - CMOS Analog Circuit Design f2 =
ε si [v GS - V T - (UTRA)v DS ]C o x
Page IV.3-3
(7)
The units of f 1 and f 2 are FV2/cm2 and cm/V respectively. Notice that f 2 includes the parameter UTRA, which is an unknown. UTRA is disabled in most SPICE models. Equation (5) can be manipulated algebraically to yield iD = log( µo) + UEXP[log(UCRIT)] + UEXP[log(f2)] (8) f1
log
This is in the familiar form of Eq. (4) with x = log(f2)
(9)
iD y = log f 1
(10)
m = UEXP
(11)
b = log( µo) + UEXP[log(UCRIT)]
(12)
By plotting Eq. (8) and measuring the slope, UEXP can be determined. The y-intercept can be extracted from the plot and UCRIT can be determined by back calculation given UEXP, µo, and the intercept, b.
3
Allen and Holberg - CMOS Analog Circuit Design
Page IV.4-1
III. Characterization of Substrate Bipolar Parameters of interest are: β dc , and JS. For v BE >> kT/q, kT iC v BE = q ln JSA E
(1)
and iE βdc = i − 1 (2) B AE is the cross-sectional area of the emitter-base junction of the BJT. iE = iB (β dc + 1)
(3)
Plot iB as a function of iE and measure the slope to determine β dc. Once β dc is known, then Eq. (1) can be rearranged and modified as follows. k T iE β d c kT kT kT − v BE = q ln i ) − ln(J A ) = ln( α dc E S E q q q 1 + β d c ln(JSAE) Plotting ln[iEβ dc/(1 + β dc)] versus vBE results in a graph where kT m = slope = q
(5)
and k T ln(JSA E ) q
b = y-intercept = −
(6)
Since the emitter area is known, JS can be determined directly.
1
Allen and Holberg - CMOS Analog Circuit Design IV.
Page IV.5-1
Characterization of Resistive Components
• Resistors • Contact resistance Characterize the resistor geometry exactly as it will be implemented in a design. Because • sheet resistance is not constant across the width of a resistor • the effects of bends result in inaccuracies • termination effects are not accurately predictable Figure B.5-1 illustrates a structure that can be used to determine sheet resistance, and geometry width variation (bias). Force a current into node A with node F grounded while measuring the voltage drops across BC (Vn) and DE (Vw), the resistors Rn and Rw can be determined as follows Vn Rn = (1) I Rw =
Vw I
(2)
The sheet resistance can be determined from these to be W n - Bias Ln
(3)
W w - Bias Lw
(3)
RS = Rn
RS = Rw where
Rn = resistance of narrow resistor (Ω) Rw = resistance of wide resistor (Ω) R S = sheet resistance of material (polysilicon, diffusion etc. Ω/square) Ln = drawn length of narrow resistor 1
Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-2
Lw = drawn length of wide resistor W n = drawn width of narrow resistor W w = drawn width of wide resistor Bias = difference between drawn width and actual device width Rw Rn A
Wn
Ln B
F
Ww
Lw C
D
E
Figure B.5-1 Sheet resistance and bias monitor.
Solving equations (3) and (4) yields W n - k Ww Bias = 1 - k
(5)
RwLn k=R L n w
(6)
where
and W n - Bias W w - Bias = Rw Ln Lw
RS = Rn
(7)
2
Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-3
Determining sheet resistance and contact resistance
10 squares
RA=220 Ω
20 squares
RB=420 Ω
Figure B.5-2 Two resistors used to determine RS and RC.
R A = R 1 + 2R c;
R1 = N 1RS
(8)
R B = R 2 + 2R c; R2 = N 2RS N1 is the number of squares for R1 RS is the sheet resistivity in Ω/square Rc is the contact resistance. RB - RA RS = N - N 2 1
(9)
and
(10)
and 2R c = R A − N 1R S = R B − N 2R S
(11)
3
Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-4
Voltage coefficient of lightly-doped resistors
R=
V1 − V2 IR
VBIAS =
V1 + V2 2
IR V1
V2
VSS Figure B.5-3 N-well resistor illustrating back-bias dependence.
27.0
Resistance (kΩ)
26.5 26.0 25.5 25.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
Back bias (volts) Figure B.5-4 N-well resistance as a function of back-bias voltage
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-5
Contact Resistance Pad 1
Metal pads
Diffusion or polysilicon
Pad 3
Pad 4
Metal pads
Pad 2
Pad 1
RC
R RC Pad 4 R RC Pad 3 RM RM
Pad 2
5
Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-1
V. Characterization of Capacitance MOS capacitors CGS, CGD, and CGB Depletion capacitors CDB and CSB Interconnect capacitances Cpoly-field, Cmetal-field, and Cmetal-poly (and perhaps multi-metal capacitors SPICE capacitor models C GS0, C GD0, and C GB0 (at V GS = V GB = 0). Normally SPICE calculates CDB and CSB using the areas of the drain and source and the junction (depletion) capacitance, CJ (zero-bias value), that it calculates internally from other model parameters. Two of these model parameters, MJ and MJSW, are used to calculate the depletion capacitance as a function of voltage across the capacitor.
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-2
C GS0 , C GD0 , and C GB0 CGS0 and CGD0, are modeled in SPICE as a function of the device width, while the capacitor CGB0 is per length of the device Measure the CGS of a very wide transistor and divide the result by the width in order to get CGS0 (per unit width).
Source
Drain
Source
Gate Figure B.6-1 Structure for determining CGS and CGD.
Cmeas = W(n)(CGS0 + CGD0)
(1)
where Cmeas = total measured capacitance W = total width of one of the transistors n = total number of transistors
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-3
For very narrow transistors, the capacitance determined using the previous technique will not be very accurate because of fringe field and other edge effects at the edge of the transistor. In order to characterize CGS0 and CGD0 for these narrow devices, a structure similar to that given in Fig. B.6-1 can be used, substituting different device sizes. Such a structure is given in Fig. B.6-3. The equations used to calculate the parasitic capacitances are the same as those given in Eq. (1).
Metal drain interconnect
Drain
Source
Metal source interconnect
Drain
Polysilicon gate
Source
Figure B.6-3 Structure for measuring CGS and CGD, including fringing effects, for transistors having small L.
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-4
CGB0
Drain
Gate overhang Gate Source FOX
Diffusion source
CGB
FOX
Cpoly-field
Figure B.6-4 Illustration of gate-to-bulk and poly-field capacitance.
This capacitance is approximated from the interconnect capacitance Cpoly-field (overhang capacitor is not a true parallel-plate capacitor) Cmeas (F/m2 ) (2) Cpoly-field = L W R R where Cmeas = Cmeas = measured value of the polysilicon strip LR = length of the centerline of the polysilicon strip WR = width of the polysilicon strip (usually chosen as device length) Having determined Cpoly-field, CGB0 can be approximated as CGB0 ≅ 2 (Cpoly-field)(doverhang) = 2C 5 (F/m)
(3)
where doverhang = overhang dimension (see Rule 3D, Table 2.6-1)
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Allen and Holberg - CMOS Analog Circuit Design
Page IV.5-5
C BD and C BS VJ CJ(VJ) = ACJ(0)1 + PB
-MJ
VJ + PCJSW(0)1 + PB
-MJSW (4)
where VJ = the reverse bias voltage across the junction CJ(VJ) = bottom junction capacitance at VJ CJSW(VJ) = junction capacitance of sidewall at VJ A = area of the (bottom) of the capacitor P = perimeter of the capacitor PB = bulk junction potential The constants CJ and MJ can be determined by measuring a large rectangular capacitor structure where the contribution from the sidewall capacitance is minimal. For such a structure, CJ(VJ) can be approximated as VJ -MJ (5) CJ(VJ) = ACJ(0)1 + PB This equation can be rewritten in a way that is convenient for linear regression. VJ log[CJ(VJ)] = (−MJ)log 1 + PB + log[ACJ(0)]
(6)
Plotting log[CJ(VJ)] versus log[1 + VJ/PB] and determine the slope, −MJ, and the Y intercept (where Y is the term on the left), Log[ACJ(0)]. Knowing the area of the capacitor, the calculation of the bottom junction capacitance is straightforward.
5
Allen and Holberg - CMOS Analog Circuit Design
Page V.0-1
V. CMOS SUBCIRCUITS Contents V.1 V.2 V.3 V.4 V.5
V.6
MOS Switch MOS Diode MOS Current Source/Sinks Current Mirrors/Amplifiers Reference Circuits V.5-1 Power Supply Dependence V.5-2 Temperature Dependence Summary
Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Opamps
Chapter 9 High Performance Opamps
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Allen and Holberg - CMOS Analog Circuit Design
Page V.0-2
WHAT IS A SUBCIRCUIT? A subcircuit is a circuit which consists of one or more transistors and generally perfoms only one function. A subcircuit is generally not used by itself but in conjunction with other subcircuits. Example Design hierarchy of analog circuits illustrated by an op amp.
Operational Amplifier Complex Circuits Simple Circuits Biasing Circuits
Current Source
Current Mirror
Input Differential Amplifier
Current Sink
Diff. Amp.
Mirror Load
Second Gain Stage
Inverter
Output Stage
Current Source Current Sink Sink Load Follower Load
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-1
V.1 - MOS SWITCH SWITCH PROPERTIES Ideal Switch RAB(on) = 0Ω
A
B
RAB(off) = ∞
Nonideal Switch CAB
IOFF
ROFF VOFF
RON
+
-
A
B CAC
C +
RA VControl
-
CBC RB
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-2
MOS TRANSISTOR AS A SWITCH Symbol Bulk A
B
A
B
(S/D)
(D/S)
C (G)
On Characteristics of A MOS Switch Assume operation in non-saturation region (vDS < vGS - V T). v D S K’W iD = L (v G S - V T ) - 2 vDS ∂iD K’W ∂vDS = L v G S − V T − v D S Thus,
∂vDS 1 RON = ∂i = K’W D (v G S − V T − v D S ) L
OFF Characteristics of A MOS Switch If vGS < VT, then iD = I OFF = 0 when vDS ≈ 0V. If vDS > 0, then 1 1 ≈∞ ROFF ≈ i λ = I DS OFFλ
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-3
MOS SWITCH VOLTAGE RANGES Assume the MOS switch connects to circuits and the analog signal can vary from 0 to 5V. What are the voltages required at the terminals of the MOS switch to make it work properly?
Bulk (0 to 5V) Circuit 1
(0 to 5V)
(S/D)
(D/S)
Circuit 2
G
• The bulk voltage must be less than or equal to zero to insure that the bulk-source and bulk-drain are reverse biased. • The gate voltage must be greater than 5 + VT in order to turn the switch on. Therefore, VBulk ≤ 0V V G ≥ 5 + VT (Remember that the larger the value of VSB , the larger VT)
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-4
I-V CHARACTERISTICS OF THE MOS SWITCH SPICE ON Characteristics of the MOS Switch
100µA V1 =10V V1
V1=9V
Id
60µA
V1 =8V V1 =7V
V2 -5
20µA Id V1=2V -20µA V1=3V V1=4V -60µA V1 =5V V1 =6V -100µA -1V
-0.6V
-0.2V
0.2V V2
SPICE Input File: MOS Switch On Characteristics M1 1 2 0 3 MNMOS W=3U L=3U .MODEL MNMOS NMOS VTO=0.75, KP=25U, +LAMBDA=0.01, GAMMA=0.8 PHI=0.6 V2 1 0 DC 0.0 V1 2 0 DC 0.0 V3 3 0 DC -5.0 .DC V2 -1 1 0.1 V1 2 10 1 .PRINT DC ID(M1) .PROBE .END
0.6V
1V
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-5
MOS SWITCH ON RESISTANCE AS A FUNCTION OF VGS SPICE ON Resistance of the MOS Switch 100kΩ
MOS Switch On Resistance
W/L = 3µm/3µm 10kΩ W/L = 15µm/3µm W/L = 30µm/3µm 1kΩ
W/L = 150µm/3µm
100Ω 1.0V
1.5V
2.0V
2.5V 3.0V 3.5V Gate-Source Voltage
4.0V
4.5V
5.0V
SPICE Input File: MOS Switch On Resistance as a f(W/L) M1 1 2 0 0 MNMOS W=3U L=3U M2 1 2 0 0 MNMOS W=15U L=3U M3 1 2 0 0 MNMOS W=30U L=3U M4 1 2 0 0 MNMOS W=150U L=3U .MODEL MNMOS NMOS VTO=0.75, KP=25U, LAMBDA=0.01, GAMMA=0.8 PHI=0.6 VDS 1 0 DC 0.001V VGS 2 0 DC 0.0 .DC VGS 1 5 0.1 .PRINT DC ID(M1) ID(M2) ID(M3) ID(M4) .PROBE .END
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-6
INFLUENCE OF SWITCH IMPERFECTIONS ON PERFORMANCE Finite ON Resistance Non-zero charging and discharging rate. φ1
RON + VIN -
VSS
C1
+
+ vC1
C1
VIN -
-
+
vC1
-
Finite OFF Current φ1
φ1
C2 + vIN -
VSS
CHold
+ vOUT -
+ VSS
vOUT -
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-7
EXAMPLES 1.
What is the on resistance of an enhancement MOS switch if VS = 0V, VG = 10V, W/L = 1, VTO = 1V, and K' = 25µA/V2? Assume that vDS ≈ 0V. Therefore, vDS L/W RON ≈ = iD K'(VG-V S -V T) RON =
106 25(10-1) = 4444Ω VG
2.
If V G=10V at t=0, what is the W/L value necessary to discharge C1 to with 5% of its intial charge at t=0.1µS? Assume K'=25µA/V2 and V TO = 1V.
C2=10pF + 5V
+ - C1=20pF
v(t) = 5exp(-t/RC) → 10-7 10-7 = 20 → RC = exp RC ln(20) 10 Therefore, R = 6 x 103Ω Thus,
10x103 L/W L/W = = 6 K'(VG-V S-V T) (2.5x10-5)(9)
W Gives L = 2.67
-
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-8
INFLUENCE OF PARASITIC CAPACITANCES MOSFET Model for Charge Feedthrough Analysis Distributed Model G CGDO
CGSO
D
S CGC=Cox
RCH
Simplified Distributed Model G CGDO D
Cox 2
Cox 2
CGSO S
RCH
CGSO = Voltage independent (1st-order), gate-source, overlap cap. CGDO = Voltage independent (1st order), drain-source overlap cap. CGC = Gate-to-channel capacitance (C ox) RCH = Distributed drain-to-source channel resistance
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-9
Charge Injection Sensitivity to Gate Signal Rate Model: vG
dvG dt
+
vIN
CHold -
Case 1 - Slow Fall Time: • Gate is inverted as vG goes negative . • Channel time constant small enough so that the charge on CHold is absorbed by vIN. • When gate voltage reaches vIN+VT, the device turns off and feedthru occurs via the overlap capacitance. Case 2 - Fast Fall Time: • Gate is inverted as vG goes negative. • Fall rate is faster than the channel time constant so that feedthru occurs via the channel capacitance onto CHold which is not absorbed by v IN. • Feedthru continues when vG reaches vIN+VT. • Total feedthru consists of that due to both the channel capacitance and the overlap capacitances. Other Considerations: • Source resistance effects the amount of charge shared between the drain and the source. • The maximum gate voltage before negative transition effects the amount of charge injected.
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-10
Intuition about Fast and Slow Regimes To develop some intuition about the fast and slow cases, it is useful to model the gave voltage as a piecewise constant waveform (a quantized waveform) and consider the charge flow at each transition as illustrated below. In this figure, the range of voltage at CL illustrated represent the period while the transistor is on. In both cases, the quantized voltage step is the same, but the time between steps is different. The voltage accross CL is observed to be an exponential whose time constant is due to the channel resistance and channel capacitance and does not change from fast case to slow case. vCL
∆V vGATE
Voltage
Time (d)
∆V
Voltage
Time (e) Figure 4.1-10 (a) Illustration of slow ramp and (b) fast ramp using a quantized voltage ramp to illustrate the effects due to the time constant of the channel resistance and capacitance.
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-11
Illustration of Parasitic Capacitances φ1
CGS +
CGD VSS
VIN CBS -
CBD
C1
+
vC1
-
CGS and CGD result in clock feedthrough CBS and CBD cause loading on the desired capacitances Clock Feedthrough Assume slow fall and rise times φ1 ∆φ1
Switch ON Switch OFF φ1
CGS + VIN -
Clock signal couples through CGD on the rising part of signal when switch is off, but VIN charges C1 to the right value regardless.
VSS CBS
Clock signal couples through CGD on the falling part of the signal when the switch is off.
CGD
C1
+
vC1
-
CGD CGD CGD ∆vC1 = -C +C ∆φ 1 ≈ - C ∆φ 1 = - C (v in + VT) 1 GD 1 1
Allen and Holberg - CMOS Analog Circuit Design
EXAMPLE regime)
Switched
Capacitor
Page V.1-12
Integrator
(slow
clock
edge
φ1 Switch ON Switch OFF
vIN+VT φ2 Switch ON Switch OFF
VT T
φ1
t
t
1
2
t
3
t
4
φ2
M1
M2
+
+ VIN -
assuming:
C2
VSS
C1
VSS
vOUT -
CGS1=CGS2=CGD1=CGD2 = CG
Net feedthrough on C1 at t2: CG (VIN + VT) CG+C1 CG CG VC1 = VIN 1 − −V T C1+CG CG+C1
∆V C1 = −
At t3, additional charge has been added due to CGS overlap of M2 as φ2 goes positive. Note that M2 has not turned on yet. CG ∆V C1 (t2-t3) = C +C VT G 1
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-13
Giving at the end of t3 (before M2 turns on): CG VC1 = VIN 1−C +C 1 G
+ Once M2 turns on (at t3 ), all of the charge on C1 is transferred to C2. CG C1 C1 ∆VO = −VC1 = −V IN 1 − C1+CG C2 C2 + Between times at t3 and t4 additional charge is transferred to C1 from the channel capacitance of M2. ∆V O
Cch (Vclk −V T) C2
(t3-t4)= −
The final change in Vout is: CG C1 Cch ∆V O = −V IN 1− − (Vclk − V T) C2 C1+CG C2 C1 so the error due to charge C2
Ideally the output voltage change is −VIN feedthrough is:
C1 CG Cch ∆V O (error) = VIN − (Vclk − V T) C2 C1+CG C2
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-14
Rigorous Quantitative Analysis of Fast and Slow Regimes
Consider the gate voltage traversing from VH to VL (e.g., 5.0 volts to 0.0 volts, respectively) described in the time domain as v G = V H − Ut
(3)
When operating in the slow regime defined by the relationship 2
β V HT 2CL >> U
(4)
where VHT is defined as V HT = V H − V S − V T
(5)
the error (the difference between the desired voltage V S and the actual voltage, VCL) due to charge injection can be described as
Allen and Holberg - CMOS Analog Circuit Design
W · C G D 0 V error = CL
+
C c h 2
Page V.1-15
π U CL W · C G D 0 + (V S + V T − V L ) 2β CL
(6)
In the fast swithing regime defined by the relationship 2
β V HT 2CL > U for slow or 2CL 5 × 108 1.58×10-15 176×10-18 + 2 Verror = 200×10-15 Verror = 10.95 mV
314×10-6 176×10-18 + (5 + 0.887 - 0 ) 220×10-6 200×10-15
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-17
POSSIBLE SOLUTIONS TO CLOCK FEEDTHROUGH 1.) Dummy transistor (MD) φ
φ
W1 L1
WD = W1 LD 2L1
M1
MD
VSS
VSS
Complete cancellation is difficult. Requires a complementary clock. 2.) Limit the clock swing when one terminal of the switch is at a defined potential. vG 0V C +
+
vin > 0 -
vout
VSS
-
vG 3VT 2VT VT t ON
OFF
ON
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-18
CMOS SWITCHES "Transmission Gate" φ
A
VSS VDD
φ
Advantages 1.) Larger dynamic range. 2.) Lower ON resistance. Disadvantages 1.) Requires complementary clock. 2.) Requires more area.
B
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-19
DYNAMIC RANGE LIMITATIONS OF SWITCHES Must have sufficient vGS to give a sufficiently low on resistance Example: VDD 50µ 2µ
A
B VDD
+ VAB
1 µA
50µ 2µ
-
Switch On Resistance
3kΩ 2.5kΩ
VDD = 4V
2kΩ
VDD = 4.5V VDD = 5V
1.5kΩ 1kΩ 0.5kΩ 0kΩ
0V
1V
2V
VAB
3V
4V
5V
SPICE File: Simulation of the resistance of a CMOS transmission switch M1 1 3 2 0 MNMOS L=2U W=50U M2 1 0 2 3 MPMOS L=2U W=50U .MODEL MNMOS NMOS VTO=0.75, KP=25U,LAMBDA=0.01, GAMMA=0.5, PHI=0.5 .MODEL MPMOS PMOS VTO=-0.75, KP=10U,LAMBDA=0.01, GAMMA=0.5, PHI=0.5 VDD 3 0 VAB 1 0 IA 2 0 DC 1U .DC VAB 0 5 0.02 VDD 4 5 0.5 .PRINT DC V(1,2) .END
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-20
“Brooklyn Bridge” Effect If N-channel and P-channel devices are “resistively” scaled (i.e., sized to have the same conductance at equivalent terminal conditions) the resistance versus voltage (common mode) will appear as shown below.
Nch on; Pch on
Nch on Pch off
Pch on Nch off
280 270 5v
260
R
Id
250 5v
240 V
230 220
0.1
210 0
1
2
3
V
4
5
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-21
VOLTAGE DOUBLER USE TO PROVIDE GATE OVERDRIVE Example VDD φA
M1
φA
CPump
M6 M4
M2
+
φA
M7
M8
VDBL
φB
CHold M3
φB
φA
φB
M5
VSS φA φB
Operation: 1. φA low, φB high - C Pump is charged to VDD-VSS. 2. φA high, φB low - CPump transfers negative charge to CHold VDBL ≈ -0.5V DD - V S S 3. Eventually, VDBL approaches the voltage of -VDD + VSS. If VDD = - VSS, then VDBL ≈ - 2VDD.
-
Allen and Holberg - CMOS Analog Circuit Design
Page V.1-22
SUMMARY OF MOS SWITCHES • Symmetrical switching characteristics • High OFF resistance • Moderate ON resistance (OK for most applications) • Clock feedthrough is proportional to size of switch (W) and inversely proportional to switching capacitors. • Complementary switches help increase dynamic range. • As power supply reduces, switches become more difficult to fully turn on. • Switches contribute a kT/C noise which folds back into the baseband.
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-1
V.2 - DIODES AND ACTIVE RESISTORS MOS ACTIVE RESISTORS Realizations +
i
i
+
v
v
-
-
When the drain is connected to the gate, the transistor is always saturated. vDS ≥ v GS - VT v D - vS ≥ v G - vS - VT ∴ vDG ≥ -V T where V T > 0 Large Signal
I-V Characteristics i AC DC
K'W i = iD = ( ) [ vGS - VT ]2 2L β = 2 ( vGS - VT ) 2 , ignore λ or v
v = vDS = vGS = VT +
Small signal i G
D + v
S
gm v
gmbs vbs
rds
-
S
v If VBS = 0 , then ROUT = i If V BS ≠ 0? Note:
1 1 = g +g ≈ g M DS M
Generally, gm ≈ 10 gmbs ≈ 100 gds
2iD β
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-2
VOLTAGE DIVISION USING ACTIVE RESISTORS Objective : Derive a voltage Vout from VSS and VDD VDD M2 Vout M1 VSS
Equating iD1 to iD2 results in :
vDS1 =
β2 β1 v DS2 - V T 2 + VT1
where vGS1 = vDS1
and
v GS2 = vDS2
Example : If VDD = -VSS = 5 volts, Vout = 1 volt, and ID1 = ID2 = 50 µamps, then use the model parameters of Table 3.1-2 to find W/L ratios. β iD1 = ( vGS - VT ) 2 2 β 1 = 4.0 µA/V2 β 2 = 11.1 µA/V2 K'n = 17 µA/V2 K'p = 8 µA/V2 1 then ( W/L )1 = 4.25
and ( W/L )2 = 1.34
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-3
EXTENDED DYNAMIC RANGE OF ACTIVE RESISTORS Concept: I VC +
I1 M1
+ vDS
I I2 M2
-
+ VC -
R
Consider : Assume both devices are non-saturated 2 v DS I1 = β 1 (v DS + V C - V T )v DS 2 2 v DS I2 = β 2 (V C - V T )v DS - 2 2 2 v DS v DS I = I1 + I2 = β v D S 2 + (V C - V T )v D S - 2 + (V C - V T )v D S - 2
I = 2β(VC - VT)vDS
1 R = 2β(V - V ) C T
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-4
Implementation : VDD + V - C
M3A
S
M2A
M1
+ VGS
G- i
M3B
+ VGS
D
S -
M2 vDS +
R VSS
i
G
S M2B
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-5
NMOS Parallel Transistor Realization : + i1
VC + G1
i
D1
-
+ D2 i2
i G2 +
M1
v
VSS
S1
M2
r ac
v
VC -
S2 _
Voltage-Current Characteristic : 2mA
Vc=7V 6V 5V
I(VSENSE)
1mA
4V
W=15u L=3u VBS=-5.0V
3V
0 -1mA
-2mA
-2
-1
0
VDS
1
NMOS parallel transistor realization M1 2 1 0 5 MNMOS W=15U L=3U M2 2 4 0 5 MNMOS W=15U L=3U .MODEL MNMOS NMOS VTO=0.75, KP=25U, LAMBDA=0.01, GAMMA=0.8 PHI=0.6 VC 1 2 E1 4 0 1 2 1.0 VSENSE 10 2 DC 0 VDS 10 0 VSS 5 0 DC -5 .DC VDS -2.0 2.0 .2 VC 3 7 1 .PRINT DC I(VSENSE) .PROBE .END
2
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-6
P-Channel Extended Range Active Resistor Circuit vAB
iAB
V DD M2B
M2A + VC -
M1A
+ VC M1B
M3A
+ V C
M3B
-
V SS Voltage Current Characteristics
100uA 4V 5V 60uA 3V 20uA
Vc=2V
i AB - 20uA
- 60uA
- 100uA
-4
-3
-2
-1
0 1 VAB
2
3
4
P-Channel Extended Range Active Resistor M1A 3 4 5 10 MPMOS W=3U L=3U M1B 3 6 5 10 MPMOS W=3U L=3U M2A 10 3 4 4 MNMOS W=3U L=3U M2B 10 5 6 6 MNMOS W=3U L=3U M3A 4 7 0 0 MNMOS W=3U L=3U M3B 6 7 0 0 MNMOS W=3U L=3U VSENSE 1 3 DC 0V VC 7 0 VAB 1 5 VDD 10 0 DC 5V .MODEL MNMOS NMOS VTO=0.75, KP=25U + LAMBDA=0.01, GAMMA=0.8 PHI=0.6 .MODEL MPMOS PMOS VTO=-0.75 KP=8U +LAMBDA=0.02 GAMMA=0.4 PHI=0.6 .DC VAB -4.0 4.0 0.2 VC 2 5 1 .PRINT DC I(VSENSE) .PROBE .END
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-7
THE SINGLE MOSFET DIFFERENTIAL RESISTOR VC i1
i1
R
v1
v2 v2
- v1 i2
+
v1
-
- v1
v2 v2
+ -
i2
R
VC Assume the MOSFET's are in the non-saturation region 1 i1 = β(V C - v 2 -V T )(v 1 - v 2 ) - 2 (v 1 - v 2 ) 2 1 i2 = β(V C - v 2 -V T )(-v 1 - v 2 ) - (-v 1 - v 2 ) 2 2 Rewrite as 1 i1 = β(V C - v 2 -V T )(v 1 - v 2 ) - (v 1 2 - 2v 1 v 2 + v 2 2 ) 2 1 i2 = β(V C - v 2 -V T )(-v 1 - v 2 ) - (v 1 2 + 2v 1 v 2 + v 2 2 ) 2 1 i1 - i2 = β(V C -v 2 -V T )(2v 1 ) - (v12-2v1v2+v22-v12-2v1v2-v22) 2 i1 - i 2 = 2β [ (VC - VT)v1 - 2v1v2 + 2v1v2 ] 2R =
v1-(-v1) 2v1 2v1 1 = = = i1-i2 2β(VC-VT)v1 β(VC-VT) i1-i2
or R= v1 ≤ V C - V T
1 W 2K L (VC-VT)
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-8
Single-MOSFET, Differential Resistor Realization VC r ac/2
i1
i1
v2
v1
v2
v1 VCC
r ac/2 - v1
v2 i2
- v1
v2 i2
R
VC Voltage-Current Characteristics 1.0mA
VC= 7V 6V
0.6mA
5V 4V 0.2mA ID(M1)
3V
- 0.2mA
- 0.6mA
- 1.0mA
-2
-1
0
V1 Single MOSFET Differential Resistor Realization M1 1 2 3 4 MNMOS1 W=15U L=3U M2 5 2 3 4 MNMOS1 W=15U L=3U VC 2 0 VCC 4 0 DC -5V V1 1 0 E1 5 0 1 0 -1 .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 .DC V1 -2.0 2.0 0.2 VC 3 7 1 .PRINT DC ID(M1) .PROBE .END
1
2
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-9
The Double MOSFET Differential Resistor VC1
i1
v1 iD2 iD1 i1
R
v
v1
+
v2
-
v
+
VC2 i2
R
v
v iD3
iD4
v2 VC1
i2
1 iD1 = β(V C1 -v-V T )(v1 -v) - 2(v1-v)2 1 iD2 = β(V C2 -v-V T )(v1 -v) - (v1-v)2 2 1 iD3 = β(V C1 -v-V T )(v2 -v) - (v2-v)2 2 1 iD4 = β(V C2 -v-V T )(v2 -v) - (v2-v)2 2 1 1 i1=iD1+iD3=β(VC1-v-VT)(v1-v)- (v1-v)2+(VC2-v-VT)(v2-v)- (v2-v)2 2 2 1 1 i2=iD2+iD4=β(VC2-v-VT)(v1-v)- (v1-v)2+(VC1-v-VT)(v2-V)- (v2-v)2 2 2 i1 - i 2 = β[(VC1-v-VT)(v 1-v) + (VC2-v-VT)(v 2-v) - (V C2-v-VT)(v1- v) - (VC1-v-VT)(v2-v)] = β[v1(VC1-VC2) + v 2(VC2-VC1)] = β(VC1-VC2)(v1-v2) v1-v2 v1-v2 1 = = i1-i2 β(VC1-VC2)(v1-v2) KW (VC1-VC2) L 1 v1 ,v 2 ≤ min [(V C1-VT),(VC2-VT)] R i n = KW (VC1-VC2) L Rin =
or
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-10
Double-MOSFET, Differential Resistor Realization VC1 iD1
i1
M1
v3
v1 i1
iD2
r ac/2
v1
VSS
v3
M2
r ac/2
v2
VC2
v4 i2
R
iD3
M3
iD4
VSS
v2
i2
M4
v4
VC1 Voltage-Current Characteristics 150uA
VC2 = 6V 5V
100uA
VBC =-5V V3 =0V VC1 =7V
50uA I(VSENSE)
Double MOSFET Differential Resistor Realization M1 1 2 3 4 MNMOS1 W=3U L=3U M2 1 5 8 4 MNMOS1 W=3U L=3U M3 6 5 3 4 MNMOS1 W=3U L=3U M4 6 2 8 4 MNMOS1 W=3U L=3U VSENSE 3 8 DC 0 VC1 2 0 DC 7V VC2 5 0 VSS 4 0 DC -5V V12 1 6 .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 .DC V12 -3 3 0.2 VC2 2 6 1 .PRINT DC I(VSENSE)) .PROBE .END
4V 3V 2V
0
- 50uA - 100uA - 150uA
-3
-2
-1
0
V1-V2
1
2
3
Allen and Holberg - CMOS Analog Circuit Design
Page V.2-11
SUMMARY OF ACTIVE RESISTOR REALIZATIONS
AC Resistance Realization
Linearity
How Controlled
Restrictions
Single MOSFET
Poor
V GS or W/L
vBULK < Min (vS, v D)
Parallel MOSFET
Good
VC or W/L
v ≤ (VC - VT)
Single-MOSFET, differential resistor
Good
VC or W/L
Double-MOSFET, Very Good VC1 - V C2 or W/L differential resistor
|v1| < VC - VT vBULK < -v1 Differential around v1 v1, v2 < min(VC1-VT, VC2-VT) vBULK < min(v1,v2) Transresistance only
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-1
V.3 - CURRENT SINKS & SOURCES CHARACTERIZATION OF SOURCES & SINKS 1). Minimum voltage (vMIN) across sink or source for which the current is no longer constant. 2). Output resistance which is a measure of the "flatness" of the current sink or source. CMOS Current Sinks & Sources
VDD
VG = V GG
iD iD VGG
+ v
-
vMIN
v
0
iD
VDD
VG = V GG
VGG iD + v
1 rOUT = λI D vMIN = vDS(SAT.) = vON
vMIN
0
VDD
0
where vON = vGS - VT
v
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-2
SMALL SIGNAL MODEL FOR THE MOSFET
D G
B S
gm =
gmbs =
2K'WID L gm γ 2 2φ F + |V BS |
1 1 r ds ≈ g = λI ds D
G +
B +
vgs
vbs
S
-
D
gmvgs
gmbsvbs
rds
S
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-3
INCREASING THE ROUT OF A CURRENT SOURCE MOS Circuit
Small-Signal Model i OUT
+ iout +
M2 r ds2 gm2 vgs2 vOUT
+ -
r VGG
-
gmbs2 vbs2 + v S2 -
r
Loop equation: vout = [iout - (g m2vgs2 + gmbs2vbs2)]r ds2 + iout r But, v gs2 = -vs2 and vbs2 = - vs2. vout = [iout + gm2vs2 + gmbs2vs2]rds2 + iout r Replace vs2 by i outrvout = iout [ rds2 + gm2 rds2r + gmbs2rds2r + r ] Therefore, rout = rds2 + r [1 + gm2 rds2 + gmbs2rds2] MOS Small Signal Simplifications Normally, g m ≈ 10g mbs ≈ 100g d s Continuing rout ≅ rgm2 rds2 r out ≈ r x (voltage gain of M2 from source to drain)
vout -
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-4
CASCODE CURRENT SINK MOS
Small-Signal Model
Circuit iOUT
IREF
iout
+ M2
M4
+ vOUT gmbs2vbs2
M3 M1
r ds2
gm2vgs2
vout
-
+ vS2 -
r ds1 gm1vgs1
-
vout = [iout - (g m2vgs2 + g mbs2vbs2)]rds2 + ioutrds1 vout = iout[rds2 + gm2rds2r(1 + η2) + rds1] rout = rds2 + r[1 + g m2rds2(1 + η2)] ≅ r ds1gm2rds2(1 + η2) Note : v MIN = VT + 2VON ≅ 0.75 + 1.5 = 2.25 (assuming V ON ≈ VT)
NMOS Cascode1mA Slope = 1/R o iO
0.75mA
+ + vGS2 -
iO 0.5mA 0.25mA 0mA 0V
VMIN 2V
4V
vO
6V
vO
+ vGS1 - 8V
10V
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-5
Gate-Source Matching Principle iD2 + iD1
iD2 +
M1 -
+
vGS1
vGS2
vGS2
-
M2
iD1
+
S = W/L
M2
vGS1
M1 -
Assume that M1 and M2 are matched but may not have the same W/L ratios. 1). If vGS1 = vGS2, then iD1 = (S1/S2)iD2 a). v GS1 may be physically connected together , or b). v GS1 may be equal to vGS2 by some other means. 2). If i D1 = iD2, then a). vGS1 = VT +
S2/S1(vGS2 - VT) , or
b). If S1 = S2 and VS1 ≈ V S2 then vGS1 = vGS2 Strictly speaking, absolute matching requires that vDS be equal for two matched devices.
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-6
Reduction of VMIN or VOUT (sat) High-Swing Cascode Method 1 for Reducing the Value of vOUT(sat) IREF
IOUT
2VT + 2VON
IOUT VOUT(sat)
M4
M3
+ VT + VON -
+
M2
VOUT
M1
+ VT + VON
+ VT + VON -
-
0
VOUT
VT + 2VON
Standard Cascode Sink :
vGS = VON + VT =
Part of v G S to achieve drain current
Part of v G S to enhance the channel
+
∴ vDS(sat) = v GS - V T = (VON + VT) - VT = VON iD
ID
VT
VT+VON
Above is based on the Gate-Source matching principle.
vGS
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-7
Circuit Which Reduces the Value of Vout(sat) of the Cascode Current Sink
iREF
iREF
M6
M4 1/4
+ -
+ 1/1
+
VT + 2VON
iOUT
M2
+
-
vOUT
VT + VON
-
M3
+ 1/1
1/1
VT + VON
2VT + 3VON
M5
iout
VT + 2VON
M1
+ VON
1/1
-
VT + VON
1/1
-
iD vOUT(sat)
-
W=1 L 1
W =1 L 4
VT + VON
VT + 2VON
ID
0
vOUT
2V ON
K'W iD = 2 (v - VT ) 2 L GS
=
0
K'W 2 2 L (V ON )
vGS VT
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-8
Method 2 for Reducing VMIN for MOS Cascode Sink/Source
iREF
iREF
iO 1
M5
M4
+ 1/4
M1
VT + VON
M2
+ 1
+ VON
VT + VON
1
-
-
Assume (W/L)1 = (W/L)2 = (W/L)4 = 4(W/L)5 values are identical and ignore bulk effects. Let I = IO REF VGS1 =
2I REF + VT = VON + VT W1 K’ L 1
and V GS5 =
2I REF + VT W5 K’ L 5
Since (W/L)1 = 4(W/L)5
V GS5 =
2I REF + VT = 2 W1 K’ 4L 1
2I REF W1 K’ L 1
+ V T = 2 VON + VT
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-9
Since V GS3 = VGS4 = VON +VT VDS1 = V DS2 = V ON which gives a minimum output voltage while keeping all devices in saturation of v M I N = 2 VO N Output Plot: 1000µA
ID(M4)
750µA
500µA
250µA
0µA 0V
1V
2V
3V VOUT
4V
5V
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-10
Matching Improved by Adding M3 iREF
iREF
VT
+ M3 1
+ -
M4 +
V T + VON
1/4
M1
VON
1
M5
+
-
iO
V T + 2VON
VT + V ON
M2
V ON
+ 1
V T + V ON
-
+ -
1
What is the purpose of M3? The presence of M3 forces the VDS1 = VDS2 which is necessary to guarantee that M1 and M2 act alike (e.g., both will have the same VT).
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-11
CMOS REGULATED CASCODE CURRENT SOURCE Circuit Diagram VDD
VDD iOUT IB2
RB2 M3
iD3 +
vGS3
IB1 M4
vOUT
Iout
M2
M1
-
VDS3 (min) VDS3 (sat)
vDS3
Principle of operation: As v OUT decreases, M3 will enter the non-saturation region and iOUT will begin to decrease. However, this causes a decrease in the gatesource voltage of M4 which causes an increase in the gate voltage of M3. The minimum value of vOUT is determined by the gate-source voltage of M4 and Vdsat of M3. Assume that all devices are in saturation.
vOUT(min) =
2IB2 K'(W/L)4 +
2Iout K'(W/L)3 + VT4
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-12
CMOS REGULATED CASCODE CURRENT SOURCE - CONT. Small Signal Model r ds3 iout + vgs3 +
RB2
r ds4
gm4vgs4
vgs4
gm3vgs3 r ds2
-
+
vout -
(Ignore bulk effects)
iout = gm3vgs3 + g ds3(vout - v gs4) vgs4 = ioutrds2 vgs3 = vg3 - vs3 = -gm4(rds4||RB2)v gs4 - vgs4 = -rds2[1 + gm4(r ds4||RB2)]iout ∴
iout = -gm3rds2 [1 + gm4(rds4 ||RB2)]iout + g ds3 vout - gds3 rds2 iout
Solving for vout, vout = rds3[1 + gm3rds2 + gds3rds2 + gm3rds2gm4(r ds4||RB2)]iout vout rout = i = rds3[1 + gm3rds2 + gds3rds2 + gm3rds2gm4(r ds4||RB2)] out g m 2r3 r out = r ds3 g m3 r ds2 g m4 (r ds4 ||R B2 ) = 2 Example K' N = 25µA/V2, λ = 0.01, IB1 = IB2 = 100µA, all transistors with minimum geometry (W = 3µm, L=3µm), and RB2 = rds, we get rds = 1MΩ and gm = 70.7µmho rout ≈ (1MΩ)(70.7µmho)((1MΩ)(70.7µmho)(1MΩ||1MΩ)= 2.5GΩ!!!
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-13
CMOS REGULATED CASCODE CURRENT SOURCE - CONT. SPICE Simulation 160µA
IB1=150µA
140µA 120µA
IB1=125µA
100µA
IB1=100µA
iOUT 80µA
IB1=75µA
60µA 40µA
IB1=50µA
20µA 0µA 0
1
2
3
vOUT SPICE Input File CMOS Regulated Cascode Current Sink VDD 6 0 DC 5.0 IB1 6 4 DC 25U VOUT 1 0 DC 5.0 M1 4 4 0 0 MNMOS1 W=15U L=3U M2 3 4 0 0 MNMOS1 W=15U L=3U M3 1 2 3 0 MNMOS1 W=30U L=3U M4 2 3 0 0 MNMOS1 W=15U L=3U M5 5 4 0 0 MNMOS1 W=15U L=3U M6 5 5 6 6 MPMOS1 W=15U L=3U M7 2 5 6 6 MPMOS1 W=6U L=3U .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 .MODEL MPMOS1 PMOS VTO=-0.75 KP=8U +LAMBDA=0.02 GAMMA=0.4 PHI=0.6 .DC VOUT 5 0 0.1 IB1 50U 150U 25U .OP .PRINT DC ID(M3) .PROBE
.END
4
5
Allen and Holberg - CMOS Analog Circuit Design
Page V.3-14
SUMMARY OF CURRENT SINKS/SOURCES
Current Sink/Source Simple Cascode High-Swing Cascode Regulated Cascode
rOUT
Minimum Voltage
rd s
VON
≈ gm2 rds2rds1 ≈ gm2 rds2rds1
VT + 2 VON
≈ gm 2rds3
VT + 2 VON
2 V ON
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-1
V.4 - CURRENT MIRRORS/AMPLIFIERS What Is A Current Mirror/Amplifier ? Rout
R in CURRENT
iI
iO
MIRROR/
+ vI Ideally, iO = A I iI Rin ≈ 0
+ vO
AMPLIFIER
-
Rout ≈ ∞
Graphical Characterization iI
iO
slope = 1/Rin
slope = 1/Rout
II
IO
vI
vMIN INPUT
vO
vMIN OUTPUT
iO
AI 1
iI TRANSFER
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-2
CURRENT MIRROR AND CURRENT AMPLIFIERS Sources of Errors iI CURRENT
iI
iO
MIRROR/ AMPLIFIER
+
+ vDS1 M1 -
In general, iO W2L1 v GS - V T 2 2 iI = W1L2v GS - V T 1
iO
1+λvDS2 µo2Cox2 1+λvDS1 µo1Cox1
If the devices are matched, iO W2L1 1+λvDS2 = iI W1L2 1+λvDS1
If vDS1 = vDS2, W 2L 1 iO iI = W1L2 Therefore the sources of error are: 1). v DS1 ≠ vDS2 2). M1 and M2 not matched (∆β and ∆VT)
+ vGS -
M2
vDS2 -
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-3
Simple Current Mirror With λ ≠ 0
DS1
M2
+ vGS -
+ v DS2 -
( 1 + λv
M1
v SS
Ratio Error
vDS1 -
10 λ = 0.02 8
1 + λv DS2
iO
iI +
- 1)x100%
Circuit -
λ = 0.015
6 4 λ = 0.01 2 0 0V
1V
2V 3V VDS1-VDS2
4V
Ratio error (%) versus drain voltage difference Used to measure λ iO 1+ λ v D S 2 S1 = iI 1+ λ v D S 1 S2 If S 1 = S 2, v DS2 = 10V, vDS1 = 1V, and i O/iI = 1.501, then iO 1+10λ ∴ = 1.501 = iI 1+ λ
0.5 ---> λ = 8.5 = 0.059
5V
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-4
Matching Accuracy of MOS Current Mirrors Neglect λ effects iD1
iD2 +
M1 Define:
+
vGS1
vGS2
M2 (vDS2 > vGS2 - VT1)
-
∆β = β 2 - β 1
iO = iD2 = β2(vGS2-VT2)2 iI iD1 β1(vGS1-VT1)2
and
∆VT = V T2 - V T1 and
β1 + β2 2 VT1+VT2 VT = 2 β=
∆VT ∆β ∆β , β = β + , V = V 2 T1 T 2 2 2 ∆VT and V T2 = V T + 2
∴ β1 = β -
Thus, ∆V T2 ∆β β+ v GS - vT iO 2 2 ∆V T 2 iI = ∆β β- v GS - v T + 2 2
=
2 T 1-2(v∆V-V GS T) ∆VT ∆β 1 1+ 2β 2(vGS-VT)
∆β 1 + 2β
∆VT ∆VT 2 iO ∆β ∆β iI ≈ 1+ 2β 1+ 2β 1-2(vGS-VT)1-2(vGS-VT) 2∆VT iO ∆β iI ≈ 1 + β - (v GS - V T ) , ∆VT ∆β ± ≈ 5% , (v = ± 10% β GS - V T ) iO ∴ iI ≈ 1 ± 0.05 - (± 0.2) = 1 ± 0.15 = 1 ± 0.25 if β and VT are correlated
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-5
Matching Accuracy - Continued
RATIO ERROR (
iO - 1)100% iI
Illustration
7 6
II = 1uA
5 4 3 II = 5uA
2
II = 10uA 1
II = 50uA 1
2
3
4
5
6
∆VT(mV)
7
8
9
10
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-6
Layout Techniques to Remove Layout Error Layout without correction technique iI
iO
iI M1
iO
M2
M2
V SS
VSS
Layou
t with correction technique iI
iO
iI
M1
M2 a
M2 b
M2 c
M2 d
M2 e
M1
iO M2 a
M2 b
VSS
VSS
M2 c
M2 d
M2 e
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-7
Practical Current Mirrors/Amplifiers • Simple mirror • Cascode current mirror • Wilson current mirror Simple Current Mirror iI = 60uA
60uA
Current mirrors and amplifiers .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 M1 1 1 0 0 MNMOS1 W=3U L=3U M2 3 1 0 0 MNMOS1 W=3U L=3U IIN 0 1 VOUT 3 0 .DC VOUT 0 5 0.1 +IIN 0 60U 10U .PRINT DC ID(M2) .PROBE .END
iI = 50uA iI = 40uA
40uA
iI = 30uA
iOUT
iI = 20uA
20uA
iI = 10uA
0 0
1
2
3
4
5
vOUT
iI
iO 3u 3u
3u 3u
M1
M2
+
vO -
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-8
Cadcode Current Mirror CIRCUIT
SPICE
iOUT
iI M3
60uA
M4
40uA
iOUT
mproved current mirror .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 M1 1 1 0 0 MNMOS1 W=3U L=3U M2 2 1 0 0 MNMOS1 W=3U L=3U M3 3 3 1 0 MNMOS1 W=3U L=3U M4 4 3 2 0 MNMOS1 W=3U L=3U IIN 0 3 VOUT 4 0 .DC VOUT 0 5 0.1 + IIN 10U 60U 10U .PRINT DC ID(M4) .PROBE .END
iI = 60uA iI = 50uA iI = 40uA iI = 30uA iI = 20uA
20uA
M1
M2
iI = 10uA
0
VSS
0
1
2
3
4
5
vOUT
Example of Small Signal Output Resistance Calculation ii io + v3 -
+ rds3
gm3 v3
v4 -
+ rds4
v1 =v3 =0 + v1 -
1). 2). 3).
gm4 (v3 +v1 -v2 ) io
+ rds1
gm1 v1
v2 -
rds2
gm2 v1
gmbs v2 vo
-
vo = v4 + v2 = rds4 [i o - gm4(v 3 + v1 - v2) + gmbs4v2] + rds2(io - gm2v1) v 2 = iords2 v o = io [rds4 + (gm4 rds2)rds4 + (rds2gmbs4)rds4 + rds2] vo 4). rout = io = rds4 + rds2 + rds2rds4(gm4 + gmbs4 )
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-9
Wilson Current Mirror Circuit and PerformanceIin = 80uA
iI
M3
70uA
65.5uA
iO
60uA 50uA
45.0uA M1
M2
VSS
40uA
iO
30uA 22.5uA
20uA 10uA
0 0 Wilson Current Source M1 1 2 0 0 MNMOS W=15U L=3U M2 2 2 0 0 MNMOS W=15U L=3U M3 3 1 2 0 MNMOS W=15U L=3U R 1 0 100MEG .MODEL MNMOS NMOS VTO=0.75, KP=25U, +LAMBDA=0.01, GAMMA=0.8 PHI=0.6 IIN 0 1 VOUT 3 0 .DC VOUT 0 5 0.1 IIN 10U 80U 10U .PRINT DC V(2) V(1) ID(M3) .PROBE .END
1
2
3
4
5
vOUT
Principle of Operation: Series negative feedback increase output resistance 1. Assume input current is constant and that there is high resistance to ground from the gate of M3 or drain of M1. 2. A positive increase in output current causes an increase in vGS2 . 3. The increase in vGS2 causes an increase in vGS1 . 4. The increase in vGS1 causes an increase in iD1. 5. If the input current is constant, then the current through the resistance to ground from the gate of M3 or the drain of M1 decreases resulting in a decrease in vGS3. 6. A decrease in v GS3 causes a decrease in the output current opposing the assumed increase in step 2.
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-10
Output Impedance of the Wilson Current Source iout +
iin=0
v3
+ rds3
gm3(v1-v2)
-
gm1v2
rds1
+ v1 -
rds2
gm2v2
+ v2 -
gmbs3v2 vout -
vout = rds3[iout - gm3v1 + gm3v2 + gmbs3v2] + v2 vout = rds3iout - gm3rds3(-gm1rds1v2) + gm3rds3v2 + gmbs3rds3v2 + v2 rds2 v2 = iout 1 + g m2 r d s 2 vout = ioutrds3 + [gm3rds3 + gmbs3rds3 + gm1rds1gm3rds3]v2 + v2 1 + g m3 r ds3 + g mbs3 r ds3 + g m1 r ds1 g m3 r ds3 1+gm2rds2
rout = rds3 + rds2
r ds2 g m1 r ds1 g m3 r ds3 r out ≈ ≈ r ds1 × (g m 3 r ds3 ) if g m 1 = g m 2 gm2rds2
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-11
Improved Wilson Current Mirror
Iout
Iin
M1
M4
M3
Additional diode-connected transistor equalizes the drain-source voltage drops of transistors M2 and M3
M2
SPICE simulation
improved Wilson current source .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 M1 1 2 3 0 MNMOS1 W=12U L=3U M2 3 3 0 0 MNMOS1 W=12U L=3U M3 5 3 0 0 MNMOS1 W=12U L=3U M4 2 2 5 0 MNMOS1 W=12U L=3U .DC VOUT 0 5 0.2 IIN 10U 80U 10U R 2 0 100MEG IIN 0 2 VOUT 1 0 .PROBE .PRINT DC ID(M1) .END
77.5uA
45.0uA
Iout
Iin = 80uA 70uA 60uA 50uA 40uA 30uA 20uA
22.5uA
10uA 0
1
2
3
Vout
4
5
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-12
Regulated Cascode Current Mirror VDD
iOUT IB
M4
+
IIN M3
vOUT
+ vGS4 -
M1
M2 -
VSS
Small Signal Equivalent Model (gmbs effects ignored) iout + rds4
gm4vgs4
v4 + g m3 v3
rds3
rds2
vout
v3 -
-
vout = (iout - gm4vgs4)rds4 + ioutrds2 vgs4 = v 4 - v 3 v3 = ioutrds2 v4 = -gm3v3rds3 vout = ioutrds4 - gm4(-gm3ioutrds2rds3 - ioutrds2)r ds4 + ioutrds2 rout = rds4 + gm4 g m3 r ds2rds3rds4 + rds2 + gm4 rds2rds4 vOUT(min) =
2IB K'(W/L)4 +
2Iout K'(W/L)3 + VT4
Allen and Holberg - CMOS Analog Circuit Design
Page V.4-13
SUMMARY OF CURRENT MIRRORS
Accuracy
Output Resistance
Minimum Voltage
Simple
Poor (Lambda)
r ds
VON
Cascode
Excellent
g m rds2
VT + 2V ON
Wilson
Excellent
g m rds2
2VON
Good
gm 2rds3
VT + 2VON
Current Mirror
Regulated Cascode
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-1
V.5 - REFERENCE CIRCUITS Introduction What is a Reference Circuit? A reference circuit is an independent voltage or current source which has a high degree of precision and stability.
Requirements for a Reference Circuit
1.) Output voltage/current should be independent of power supply. 2.) Output voltage/current should be independent of temperature. 3.) Output voltage/current should be independent of processing variations.
V-I Characteristics of an Ideal Reference i Voltage Reference Iref
Current Reference
Vref
v
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-2
Concept of Sensitivity Definition Sensitivity is a measure of dependence of Vref (Iref) upon a parameter or variable x which influences Vref (Iref).
Vref
S x
∂Vref Vref x ∂Vref = = ∂x Vref ∂x x
where x = VDD or temperature
Application of Sensitivity V ∂Vref ref ∂x Vref = x x
S
For example, if the sensitivity is 1, then a 10% change in x will cause a 10% change in V ref.
Vref Ideally,
S x
→0
Allen and Holberg - CMOS Analog Circuit Design
V.5-1 - SIMPLE REFERENCES Objective is to minimize,
Vref
S V DD
=
∂Vref Vref ∂VD D VDD
Types of references include, 1. Voltage dividers - passive and active. 2. MOS diode reference. 3. PN junction diode reference. 4. Gate-source threshold referenced circuit. 5. Base-emitter referenced circuit.
Page V.5-3
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-4
Passive Divider Accuracy is approximately equivalent to 6 bits (1/64). VDD
RA V1 I
RB V2 RC
VSS
Active Dividers I V3 M3
VDD
VDD
V2 M2 V1 M1
M2
M2
+ M1 Vref -
+ M1 Vref -
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-5
PN Junction Voltage References VCC kT I I VREF = VBE = q ln = Vt ln Is I s
I
R
V CC - V B E VCC ≈ R If I = R
+
VREF
VCC V REF ≈ V t ln RI
-
s
Sensitivity: VREF
S
=
V CC
1 VCC ln RI s VREF
If VCC = 10V, R = 10 kΩ, and Is = 10-15A, then
S
= 0.0362.
V CC
Modifying the Value of VREF
VCC
R IR1
R1
If β >> 1, then VREF ≈ IR1 (R 1+R2)
I +
VREF R2
-
VBE replacing IR1by R1 gives, R 1 +R 2 VREF ≈ V BE R1 or R1+R2 VCC V REF ≈ Vt ln R1 RIs
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-6
Gate-Source Referenced Circuits (MOS equivalent of the pn junction referenced circuit) VDD
2(VDD-V REF) βR
VREF = VGS = VT + I
R
1 V REF = VT − βR
+
+
2(VDD-VT) 1 + 2 2 βR β R
VREF Sensitivity:
-
V REF
S VDD
VDD 1 = V 1 + βR(V REF REF - V T )
If V DD = 10V, W/L = 10, R = 100kΩ and using the results of Table 3.1-2 gives VREF
VREF = 1.97V.and
S
= 0.29.
VDD
Modifying the Value of VREF V DD
R IR1
R1
R 1 +R 2 V GS R2
VREF ≈
I +
V REF R2 -
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-7
Bootstrapped Current Source VDD
M4 M5
M3
RB
ID1 = I1
iOUT
ID2 = I2
M6 M2 M8
M1 R
I2
Principle: If M3 = M4, then I 1 = I2
(1)
also, VGS1 = VT1 +
2I1 KNS1 = I2R
therefore, VT1 1 I2 = R + R
2I1 KNS1
Desired operating point
Eq. (2)
Eq. (1)
(2) Undesired operating point
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-8
Bootstrapped Current Sink/Source - Continued An examination of the second-order effects of this circuitThe relationship between M1 and R can be expressed as, I2R = VT1 +
2I1 β1
Instead of assuming that I1 = I2 because of the current mirror, M3-M4, let us consider the effects of the channel modulation which gives 1 + λPV G S 4 I 2 = I1 1 + λ P (V D D - V DS1 ) Solving for I1 from the above two expressions gives I1R(1 + λPVGS4) = [1 + λP(V DD-V DS1)]
2I1 β1
Differentiating with respect to VDD and assuming the VDS1 and VGS4 are constant gives (IOUT = I1), 20VDD λPV T1 +
IOUT
S VDD
=
IOUTR(1 + λ P V GS4 )
2I1 β1
1 + λ P (V D D - V DS1 ) 2β1I1
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-9
Bootstrapped Current Sink/Source - Continued Assume that VDD =5V, K N ' = 23.6 µA/V2, VTN=0.79V, γN=0.53V0.5, φ P = 0.590V, λN=0.02V-1, K P ' = 5.8µA/V2, VTP=-0.52V, γP=0.67V0.5, φ P = 0.6V, λN=0.012V-1. Therefore, VGS4 = 1.50V, VT2 = 1.085V, VGS2 = 1.545V, and VDS1 = 2.795V which gives IOUT
S VDD
∆IOUT/IOUT = 0.08 = ∆V /V DD DD ∆V DD
= 3.2µA If ∆VDD = 6V - 4V = 2V, then ∆IOUT = 0.08 I OUT V DD SPICE Results: 120µA 100µA
∆IOUT ≈ 2.8µA for ∆VDD 4V →6V
80µA
IOUT
60µA 40µA 20µA 0µA 0V
2V
4V
VDD
6V
8V
10V
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-10
Base-Emitter Voltage Referenced Circuit VDD
M4 M5
M3
RB
ID1 = I1
ID2 = I2
M6 M1 M8
I2 ≈
VBE1 R = I5
V = I2R ≈ V BE1
Q1
M2
R
iOUT
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-11
V.5-2 - TEMPERATURE DEPENDENCE Objective Minimize the fractional temperature coefficient which is defined as 1 ∂Vref T CF = V ∂T parts per million per °C or ppm/°C ref Temperature Variation of References PN Junction: v i ≈ I s exp Vt -V GO I s = KT 3 exp V t
∂I I1s ∂Ts
=
∂(ln I s ) 3 V GO V GO = T + TV ≈ TV ∂T t t
dvBE V BE - V G O = -2mV/°C at room temperature dT ≈ T (VGO = 1.205 V and is called the bandgap voltage)
Gate-Source Voltage with constant current (Strong Inversion): dVGS dVT dT = dT +
2L d WCox dT
ID µ o
µo = KT-1.5 ; VT = VT0 - αT or VT(T) = VT(To) - α(T-To) dV G S 3 = α + dT 4
V GS - V T T
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-12
PN Junction Voltage Reference VDD
R
I
+ VREF = VBE = kT ln I = Vt ln I q Is Is -
V DD - v B E V DD I= ≈ R R
TCF =
------->
VDD V REF = V t ln RI s
dV REF V REF - V G O VREF dT = TVREF
1
Vt dR - V REF RdT
Assume VREF = 0.6 volts and that R is a polysilicon resistor dR = +1500 ppm/° C gives a RdT 0.6 - 1.205 0.026 TC F = (300K)(0.6) - 0.6 (0.0015) = -0.003361 - 0.000065 = -3426 ppm/°C
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-13
Gate - Source Referenced Circuits MOS Equivalent of the PN Junction Referenced Circuit VDD
R
I +
VREF -
1 V R E F = V T − βR +
1 dV REF = TCF = V REF dT
2(VDD−VT) 1 + 2 2 βR β R
1 VREF
V DD − V REF 1.5 1 d R − 2βR R dT T 1 1 + 2βR (V DD − V REF )
−α +
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-14
Example W = 2L, VDD = 5V, R = 100 KΩ , K’=110 µ, VT = 0.7, T = 300 K, α = 2.3 mV/°C Solving for VREF gives VREF = 1.281 V dR = +1500 ppm/°C RdT
−2.3× 10-3 + 1 dV REF 1 TCF = 1.281 dT = 1.281
TCF = - 928 ppm/°C
1+
5 − 1.281 1.5 − 1500 × 10-6 300 2 × 2110×10-6 × 100K 1 2 × 2110×10-6 × 100K (5 - 1.281)
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-15
Bootstrapped Current Source/Sink VDD
M4 M5
M3
RB
I D1 = I1
I D2 = I2
iOUT
M6 M2 M8
M1 R
2ID1 L + VT VGS1 V ON + V T K'W ID2 = R = = = ID1 = IOUT R R Assuming that VON is constant as a function of temperature because of the bootstrapped current reference, then ∴ TCF =
1 dVT 1 dR -α 1 dR = VT dT R dT VT - R dT
If R is a polysilicon resistor, then -2.3 x 10 -3 - 1.5x10 -3 = -3800 ppm/°C 1 If R is an implanted resistor, then TC F =
TC F =
-2.3 x 10 -3 - 0.4x10-3 = -2700 ppm/°C 1
Allen and Holberg - CMOS Analog Circuit Design
Page V.5-16
Base-Emitter Voltage - Referenced Circuit VDD
M4 M5
M3
RB
ID1 = I1
ID2 = I2
iOUT
M6 M1 M8
Q1
vBE1 I2 ≈ R
M2
R
1 dv BE 1 dR -----> TCF = v - R dT BE dT
Assuming VBE = 0.6 volts and a polysilicon resistor gives 1 TCF = 0.6 (-2x10-3) - (1.5x10-3) = -4833 ppm/°C
Allen and Holberg - CMOS Analog Circuit Design
Page V.6-1
V.6 - SUMMARY • The circuits in this chapter represent the first level of building blocks in analog circuit design. • The MOS transistor makes a good switch and a variable resistor with reasonable ranges of linearity in certain applications. • Primary switch imperfection is clock feedthrough. In order for switches to be used with lower power supplies, VT must be decreased. • The primary characteristics defining a current sink or source are VMIN and Rout. V MIN → 0 and Rout → ∞. Typically the product of VMIN times Rout is a constant in most designs. • Current mirrors are characterized by: Gain accuracy Gain linearity VMIN on output Rout Rin • Reasonably good power supply independent and temperature independent voltage and current references are possible. These references do not satisfy very stable reference requirements.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.0-1
VI. CMOS AMPLIFIERS Contents VI.1 Simple Inverting Amplifier VI.2 Differential Amplifiers VI.3 Cascode Amplifier VI.4 Output Amplifiers VI.5 Summary Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS OTA's
Chapter 9 High Performance OTA's
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-1
VI.1 SIMPLE INVERTING AMPLIFIERS CHARACTERIZATION OF AMPLIFIERS We shall characterize the amplifiers of this Chapter by the following aspects: • Large Signal Voltage Transfer Characteristics • Maximum Signal Swing Limits • Small Signal Midband Performance Gain Input resistance Output resistance • Small Signal Frequency Response • Other Considerations Noise Power Etc.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-2
VI.1.1 - CMOS INVERTERS Types
VDD VGG2 M2
M2
+ M1
+ vIN -
M2
+ M1
vOUT + vIN Active Current Load Source Inverter Inverter
+
vOUT
vIN
-
-
+ M1
vOUT Push-pull inverter
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-3
ACTIVE LOAD INVERTER - VOLTAGE TRANSFER CURVE CMOS Active Load Inverter VDD 3 0 DC 5.0
5V 1.2 v IN=5V
M2
iD (mA)
0.8
vIN =4V
0.6
vIN =3.5V
0.4
vIN=3V
W1 15µm = L1 3µm M1
v IN -
vIN=1V v IN=2V vIN =1.5V
0
1
2
v OUT
3
4
+ vOUT
+
vIN=2.5V
0.2 0
W2 15µm = L2 3µm
vIN=4.5V
1.0
-
5 5 4 3 vOUT 2 1 0 0
1
2
v IN
3
4
SPICE Input File: VIN 1 0 DC 0.0 M1 2 1 0 0 MNMOS1 W=15U L=3U M2 2 2 3 3 MPMOS1 W=15U L=3U .MODEL MNMOS1 NMOS VTO=0.75 KP=25U LAMBDA=0.01 GAMMA=0.8 PHI=0.6 .MODEL MPMOS1 PMOS VTO=-0.75 KP=8U LAMBDA=0.02 GAMMA=0.4 PHI=0.6 .DC VIN 0 5 0.1 .OP .PRINT DC V(2) .PROBE .END
5
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-4
Active Load CMOS Inverter Output Swing Limits Maximum: vIN =0 ⇒ iD=0 ⇒ vSD2=|VT2 |
VDD M2 iD
∴ v OUT (max) ≈ V DD − |V TP | Minimum: Assume v IN = VDD, M1 active, M2 saturated, and VT1 = VT2 = VT. vDS1 2 M1: iD = β1(vGS1 −VT)vDS1 − 2
vOUT vIN
M1 VSS
(v OUT−VSS)2 = β1 (VDD−VSS−VT)(vOUT−VSS)− 2
β2 β2 β2 2 2 M2: iD = 2 (vGS −VT) = 2 (VDD−vOUT−VT) = 2 (vOUT+VT−VDD)2 β2 iD = 2 (vOUT −VSS+VSS+VT−VDD)2 β2 = 2 (vOUT−VSS)−(VDD −VSS−VT)2 Define vOUT' = vOUT − V SS and VX = VDD − V SS − V T (vOUT')2 ∴ iD = β1V X vOUT '− (M1) 2 β2 iD = 2 v OUT ' − V X 2 (M2) Equate currents v OUT '2 β2 2 2 2 v OUT ' − 2V X v OUT ' + V X = β1V X v OUT ' − 2 β2 2 2 2 or β1v OUT ' − 2V X v OUT ' + V X = 2VX vOUT' − v OUT ' β2 β2 β2 1 + vOUT '2 − 2VX 1 + V X2 = 0 vOUT ' + β β β 1 1 1 β2/β1 2 vOUT' − 2VX vOUT' + 1+β /β VX2 = 0 2 1
Allen and Holberg - CMOS Analog Circuit Design
β2/β1 1 − 1+β2/β1 = VX 1 − V D D − V SS − V T v O U T (min.) = V D D − V T − 1 + β 2 /β 1
∴
vOUT' = VX1 ±
Page VI.1-5
1 1 + β 2 /β 1
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-6
Interpretation of vOUT(min.) vOUT(min.) = (VDD − V SS − V T)1 −
1
β2 1 + β 1
+ VSS
VDD = −VSS = 5V VT = 1V
vOUT 5
3
v MAX
1 0
0.1
1
10
-1 v MIN -3
-5
Gain ~
β1 β2
β1 β2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-7
Active Load Inverters Small Signal Characteristics Model: 5V S2=B2 gm2 vgs2
M2
+
M1
v OUT + vIN -
rds2
D1=D2=G2 G1 + v in gm1 vin
r ds1
v out
v in g m1 v in
-
-
+
+
+
r ds1
g m2 v out
rds2
v out -
-
S1=B1
-
Small Signal Voltage Gain vout = −g m1 v in + g m2 v outr ds1 || r d s 2 −gm1 −g m1 vout = ≈ vin g ds1 + g ds2 + g m 2 gm2 = −
W1 2KN L I1 1 W2 2KP L I2
= −
2
vout vin = −
If
K N ' W1L2 =− K P ' W2L1
µNO W1L2 µPO W2L1
vout W 1/L1 = 20, then W2/L2 vin = −6.67 using the parameters of Table 3.1-2
Small Signal Output Resistance 1 1 rout = g ≈ gm2 ds1 + g ds2 + g m 2
β1 β2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-8
High Gain CMOS Inverters VDD
VGG
VDD
M2
M2
I
I
vIN
vOUT vIN
vOUT
M1
M1
V SS
V SS
Inverter with current source load
Push-pull, inverter I
M2 vIN = V SS
M1 vIN = VDD
.8V SS .8V DD .6V SS
.6V DD
.4VSS .2VSS 0 .2VDD .4VDD VSS
.4V DD I J K
.2VDD H
G
F=F'
0 D'
E' E D G' K' J' I' H'
.5VSS
0
.5VDD
C=C' .2V SS B=B'.4V SS .6V SS A=A' V DD
vOUT
Large signal transfer characteristics of inverter with a current source and push pull inverter
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-9
High Gain, CMOS Inverter Large Signal Transfer Characteristics vOUT C' A=A'
B ≅ B'
M1 sat. M2 non-sat.
C E'
D VSD2 > VSG2 -VT2 VDD - v OUT > VDD - VGG - VT2 v OUT < VGG + VT2
.8VDD
vOUT =vin - VT1
M1 & M2 saturated
.6VDD
E
M1 non-sat. M2 sat.
.4VDD
M2 non-sat.
vout = VGG + VT2
.2V DD
M2 sat.
VSS
vOUT =vin + VT2
VDD
D'
VGG = 0
F=F'
.8VSS .6VSS
.4VSS
.2VSS
0
.2VDD
.4VDD .6VDD .8VDD VDD
v IN
.2VSS .4VSS
G H
I
.6VSS .8VSS VDS1 > VGS1 - VT1 v OUT - VSS > v IN - VSS - VT1 v OUT > v IN - VT1
VSS
J
K
G' H' I'
J'
K'
Advantages: 1. High gain. 2. Large output signal swing. 3. Large current sink and source capability in push pull inverter.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-10
CMOS Inverter Characteristics Circuit: VDD
M2 I
vIN
vOUT M1
VSS PSPICE Characteristics: 6
M2 linear
4
2
v OUT
M2 saturated
Current in M1 or M2
100 µA
M1 linear voltage transfer curve
0
-2
vIN + VT2 M1 off
M2 linear
CMOS inverter DC current and sweep VIN 1 0 VDD 3 0 DC 5.0 VSS 4 0 DC -5.0 M1 2 1 4 4 MNMOS1 W=3U L=3U M2 2 1 3 3 MPMOS1 W=9U L=3U .MODEL MNMOS1 NMOS VTO=0.75 KP=25U +LAMBDA=0.01 GAMMA=0.8 PHI=0.6 .MODEL MPMOS1 PMOS VTO=-0.75 KP=8U +LAMBDA=0.02 GAMMA=0.4 PHI=0.6 .DC VIN -5.0 5.0 0.1 .PRINT DC V(2) ID(M1) .PROBE .END
M1 saturated M1 linear vIN - VT2
-4
0 µA M2 off
-6 -5
-3
-1
0
vIN
1
3
5
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-11
Current Source Inverter - Output Swing Limits VDD
VGG
M2 I
vOUT vIN
M1
VSS
v OUT(max.) ≈ V D D
β2 VDD−VGG−VT2 1 − β V −V −V SS T1 1 DD
vOUT(min.) = VDD −V T1−(V DD −V SS−V T1)
CMOS Push - Pull Inverter - Output Swing Limits VDD
M2 I
vIN
vOUT M1
v OUT(max.) ≈ V D D v OUT(min.) ≈ V SS
VSS
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-12
High Gain, CMOS Inverters Small Signal Characteristics Model +
+ gm1vin
vin
r ds1
gm2vin
r ds2
vout
-
-
Small Signal Voltage Gain: vOUT = −g m1 v in + g m2 v inr ds1 || r ds2 OR vout − g m1 + g m 2 vin = g ds1 + g ds2 =
−
2 I D
W1 KN' L + 1 λ1 + λ2
W2 KP' L 2
=
K !!! ID
Set g m2 = 0 for the current source inverter W1 W2 Assume that iD = 1 µA and L1 = L2 , using the values of Table 3.1-2 gives vOUT vin = −328 = −194
for the push-pull inverter (L=10 µm) for the current source inverter (L=10 µm)
Small Signal Output Resistance: rout =
1 g ds1 + g ds2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-13
High Gain, CMOS Inverters Dependence of Gain upon Bias Current log AV
1000
100 weak inversion strong inversion
10
1
0.1µA
1µA
10µA
100µA
1000µA
ID
Limit is the subthreshold current where square law characteristic turns into an exponential characteristic. Assume that the level where subthreshold effects begin is approximately 0.1µA, the maximum gains of the CMOS inverters become: The CMOS inverters become: Push-Pull: -1036 Current source load: -615 Current sink load: -422
W L = 1, L=10 µm
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-14
Frequency Response of CMOS Inverters General Configuration X = vOUT ; Active Load CMOS Inverter (gm = gm1 ) X = VGG ; CMOS Inverter with a Current Source Load (g m = gm1) X = v IN ;
CMOS Push Pull Inverter (gm = gm1 + gm2) VDD
C GS2
C GD1
X C GD2
M2 C BD2
+
vin CGD1
v in
C BD1
CL
+ g m vin
R
CT
v out
-
-
M1
VSS (b)
(a)
(a) General configuration of an inverter illustrating parasitic capacitances. (b) Small signal model of (a) CGD1 and CGD2 are overlap capacitances CBD1 and C BD2 are the bulk-drain capacitances CL is the load capacitance seen by the inverter Frequency Response vOUT −g m Rω 1(1− s/z) , vIN = s + ω 1
1 ω 1 = RC
gm and z = C
GD1
1 (gm2 = 0 for push pull and current source inverters) R = g +g +g ds1 ds2 m2 C ≈ CGD1+CGS2+CBD1+CBD2+CL (Active load inverter) C ≈ CGD1+CGD2+CBD1+CBD2+CL (Current source & push-pull inverter) if g mR >> 1
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-15
Frequency Response of CMOS Inverters Dependence of Frequency Response on Bias Current When g m2 ≠ 0
(active load inverter):
W 2K' 1 L ID R≈g or ω -3dB = ~ ID C m2 When gm2 = 0 (push pull and current source inverter): (λ 1 + λ 2 ) I D 1 R = (λ or ω = ~ ID -3dB C 1 + λ 2 ) ID Example: Find the −3dB frequency for the CMOS inverter using a current source load and the CMOS push pull inverter assuming that iD = 1µA, CGD1=CGD2=0.2pF and CBD1=CBD2=0.5pF W1 W2 Using the parameters of Table 3.1-2 and assuming that L1 = L2 =1 Gives, For the active load CMOS inverter, gm2 ω-3dB = C = 3.124x10 -6 rads/sec or 512KHz For the push pull or current source CMOS inverter, g gd1 + g ds2 ω -3dB = = 14.3x10 3 rads/sec or 2.27 KHz C gm1 z=C = 29.155 Mrads/sec or 4.64 MHz GD1 The reason for the difference is the higher output resistance of the push pull or current source CMOS inverters
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-16
NOISE IN MOS INVERTERS Noise Calculation
RL RS
i 2d
v IN +
RL
i 2L
RS
-
2
We wish to determine the equivalent input noise voltage, vn as shown below:
RL v2n
RS
8 2 id = KTg m (A2/Hz) 3 4KT 2 2 iL = RL (A /Hz) Comments: 1.) 1/f noise has been ignored. 2.) Resistors are noise-free, they are used to show topological aspects. Can repeat the noise analysis for the resistors if desired.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-17
Noise in an Active Load Inverter
VDD e2n2 M2
2
eout
vOUT vIN
eout
e2n1
2 eeq =
vIN=0
2 gm1 2 2 + en2 gm2
= en1
2
M1
gm2 2 2 2 2 eeq = g = en1 +g en2 m1 2 m1
VSS
2 en1 1
gm2
2 gm2 2 en2 + g m1 e2 n1
Sec 3.2, Eq (15) B 2 en = fWL ; B=constant for a process
1/f noise:
Sec. 3.3, Eq (6) 2K' W gm = ID L
So
2
eeq =
2 en1 1
' W2 2KP ID L2 BP fW1L1 + fW L B W 2 2 N 1 ' 2KN I D L1
2 2 eeq = en1 1 +
KP' BP L12 K ' BNL2 N
To minimize 1/f noise 1). L2 >> L 1 2
2). en1
small
----->
Gain = −
' KNW1 ' K PW2
L2 L1
Allen and Holberg - CMOS Analog Circuit Design
Noise in An Active Load Inverter - (Cont'd) Suppose the noise is thermal - Sec. 3.2, Eq,(13) 2 en
8kT(1+η) 3gm 8kT(1+η1) 2 eeq = 1 + 3gm1 =
gm2 2 (1+η2)gm1 2 gm1 (1+η1)gm2
W 2 1/2 K ' P L 8kT(1+η (1+η ) ) 2 1 2 2 eeq = 1+ 3gm1 (1+η1)KN'W1 L1 or 2
eeq =
8kT(1+η 1) 1+η2 gm2 1 + 3gm1 1+η1 gm1
To minimize thermal noise gm1 1. Maximize gain gm2 2. Increase gm1 =
2KNW1 L1 ID
Page VI.1-18
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-19
Noise in Other Types of Inverters Current Source Load Inverter -> same as active load inverter Push-Pull InverterVDD e2n2 M2
vOUT
vIN M1 e2n1
VSS
1 rout = g ds1 + g ds2 2 2 2 eout = (gm1rout)2 en1 + (gm2rout)2 en2 vout = −(gm1 + gm2 )rout vin gm1 gm2 2 2 2 2 2 eeq = g e + en2 n1 m1 + g m 2 g m1 + g m 2
2 eeq =
2 gm2 2 en2 1 + g 2 m1 en1 2 en1 2 g 1 + m 2 gm1
2
2
=
To minimize noise - Reduce en1 and en2 .
KP' BP L1 2 1 + ' L2 K NBN 2 en1 KP' W 2L 1 2 1 + K ' W L N 1 2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-20
SUMMARY OF MOS INVERTERS Inverter Type AC Voltage Gain
AC Output Resistance
Bandwidth (CGB=0)
gm2
p-channel active load sinking inverter
-gm1 gm2
1 gm2
n-channel active load sinking inverter
-gm1 gm2+gmb2
1 gm2 +g mb2
Current source load sinking inverter
-gm1 gds1+gds2
1 gds1+gds2
-(gm1 +gm2 ) gds1+gds2
1 gds1+gds2
Push-Pull inverter
CBD1 +C GS1 +C GS2 +C BD2
gm2+gmb2 CBD1 +CGD1 +CGS2 +CBD2
gds1+gds2 CBD1 +CGD1 +CGS2 +CBD2
gds1+gds2
Equivalent, input-referred,meansquare noise voltage g v2n1 gm1
2
+v2n2
g v2n1 gm1
2
+v2n2
g v2n1 gm1
2
+v2n2
m2
m2
v2n1gm1 CBD1 +CGD1 +CGS2 +CBD2 g +g m1 m2
m2
2
v2 g + g n2+gm2 m1 m2
2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.1-21
KEY MOSFET RELATIONSHIP USEFUL FOR DESIGN Assume MOSFET is in saturation.
1.)
KW iD = 2L (vGS − VT ) 2
2.)
vDS (sat) =
3.)
gm =
2iD KW/L
2IDKW L
or
vGS =
2iD KW/L − VT
or
KW iD(sat) = 2L vDS (sat)2
or
KW gm = L (VGS − V T)
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-1
VI.2 - DIFFERENTIAL AMPLIFIERS Definition of a Differential Amplifier
v1
+ Differential Amplifier
v2
vOUT
-
v 1 + v 2 2
vOUT = AVD(v1 − v2) ± A VC
(100) Differential voltage gain = AVD Common mode voltage gain = AVC (1) AVD Common mode rejection ratio = (1000) AVC VOS(out) Input offset voltage = VOS(in) = (2-10mV) AVD Common mode input range = VICMR (VSS+2V 1, gm3 = gm4 , and g m1 = gm2 = gmd, then iout' ≈ g m1vgs1 − gm2vgs2 = gmd(v gs1 − vgs2) = gmdvid or i out ' ≈ g m d v id =
K N'WISS vi d L
Unloaded Differential Voltage Gain (RL = ∞) gmd 2 v = v out ≈ g i d (λ N + λ P ) ds2 + g ds4
K N'W v ISSL i d
Example If all W/L ratios are 3µm/3µm and ISS = 10µA, then gmd(N-channel) = (17x10-6)(10x10-6) = 13 µA/V gmd (P-channel) = (8x10-6)(10x10-6) = 8.9 µA/V and vout 2(13x10-6) (N-channel) = = 86.67 vid (0.01+0.02)10x10-6 vout 2(8.9x10-6) (P-channel) = = 59.33 vid (0.01+0.02)10x10-6
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-13
INTUITIVE SMALL SIGNAL ANALYSIS OF MOSFET CIRCUITS Principle: Consider only small changes superimposed on the dc conditions. Technique: Identify the transistor(s) that convert input voltage to current (these transistors are called the active devices). Trace the currents to where they flow into the resistance seen from a given node and multiply this resistance times the currents to find the voltage at this node. Example - Differential Amplifier VDD gm1vin 2 M3
gm1vin 2 M4 Rout
gm1vin 2
gm2vin 2 M1
+
M2
vin 2 -
VGG
v + in -
vOUT
v + 2in
M5
VSS Current flowing into the output node (drains of M2 and M4) is gm1vin gm2vin iout = 2 + 2 Output resistance, R out, seen at this node is 1 R out = rds2||rds4 = g +g ds2 ds4 Therefore, the open circuit voltage gain is v out gm1+gm2 gm1 gm2 vin = 2(gds2+gds4) = gds2+gds4 = gds2+gds4
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-14
CMOS DIFFERENTIAL AMPLIFIER Common Mode Gain The differential amplifier that uses a current mirror load should theoretically have zero common mode gain. For example: VDD M1-M3-M4
M3
M4
vOUT vG1
M1
vG2
M2 M1-M2
VSS
Total Common mode output due to v I C
Common mode = output due to M1-M3-M4 path
Common mode − output due to M1−M2 path
Therefore, the common mode gain will approach zero and is nonzero because of mismatches in the gain between the two paths.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-15
CMOS DIFFERENTIAL AMPLIFIER Consider the following differential amplifier VDD M3
M4
vO1
vO2
vIC
M1
VGG5
vIC
M2
M5
VSS Use of symmetry to simplify gain calculations VDD M3
M4
vO1
vO2
vIC
M1
VGG
1 xM5 2
M2
1 xM5 2
VSS
vIC
VGG
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-16
CMOS DIFFERENTIAL AMPLIFIER Small signal model gmbs1 vbs1
+
rds1
+ vIC=vg1 -
vs1 -
2rds5
+
gm1 vgs1
vo1
rds3 gm3 vo1
-
gm1 +gmbs1 vs1
gm1 vg1
rds1
r ds1
+ vIC=vg1 2rds5 -
+ vs1 -
+
gm1 +gmbs1 vs1 r ds3
1 gm1 +gmbs1
1 gm3
gm1 vg1
vo1 -
Writing nodal equations 0.5g ds + g ds1 + g mbs1 vs1 − gds1 vo1 = gm1 vIC −g ds1 + g m1 + g m b s 1 v o1 + g ds1 + g ds3 + g m 3 vo1 = −gm1 vIC
vo1 Solving for v gives, IC −0.5gm1gds5 vo1 = vIC gds3+gm30.5g ds + g m1 + g mbs1 + g ds1 + 0.5g ds1 g ds5 or
−0.5gm1gds5 −gds5 vo1 vIC ≈ gm3g m1 + g mbs1 ≈ 2gm3
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-17
COMMON MODE REJECTION RATIO (CMRR) Differential mode gain Avd CMRR = Common mode gain = Avc For the previous example,
|CMRR| =
gm1 gm3
2(g m1 + g mbs1) 2gm1 ≈g gds5 ds5
gm1gds5 2g m3 g m1 + g mbs1
=
Therefore, current sinks/sources with a larger output resistance(rds5) will increase the CMRR. Example Let all W/L ratios be unity, ISS = 100µA, and use the values of Table 3.1-2 to find the CMRR of a CMOS differential amplifier. gm1 = 2x17(µA/V2)x100µA = 58.3µS gds5 = 0.01V-1 x 100µA =1µS Therefore, |CMRR| = 116
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-18
CMOS DIFFERENTIAL AMPLIFIERS Parasitic Capacitances VDD M3
M4 Cgd4
Cout
vOUT
CM
vG1
M1
vG2
M2
CT ISS
VSS CT = tail capacitor (common mode only) CM = mirror capacitor = Cdg1 + Cdb1 + Cgs3 + Cgs4 + Cdb3 COUT = output capacitor ≈ Cbd4 + Cbd2 + Cgd2 + CL Small Signal Model +
+
g m1vgs1
v gs1 = vgs2 = vid /2 -v id /2 -
-
+ v gs3
rds3
+
C gd4 rds2
rds1
1
g m3 -
CM
gm4v gs3 g m2v gs2
rds4
C OUT
v out -
We will examine the frequency response of the differential amplifier in more detail later.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-19
SLEW RATE Slew rate is defined as an output voltage rate limit usually caused by the current necessary to charge a capacitance. dV dT
i.e. i = C
For the CMOS differential amplifier shown, VDD M3
M4 CL
+
M1
+ v - O
M2
vIN ISS
VSS ISS Slew rate = C L where CL is the total capacitance seen from the output node to ground. If C L = 5pF and ISS = 10µA, then the SR = 2V/µS
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-20
CMOS DIFFERENTIAL AMPLIFIERS NOISE Assumption: Neglect thermal noise(low frequency) and ignore the thermal noise sources of rd and rs . Therefore: 2 KF ind = iD AF (AF = 0.8 and KF = 10-28 ) fCoxL2
or 2
2
vnd =
ind
KF = i D(AF-1) 2 2 gm 2fµoCox WL
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-21
VDD
IDD
v2eq1
v2eq2 M1
M2
i2o + M3
M4
v2eq3
v2eq4
vOUT
-
VSS
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-22
CMOS DIFFERENTIAL AMPLIFIERS NOISE Total output noise current is found as, 2
iod
2
2
2
2
= gm1 2 v1 + g m2 2 v2 + gm32 v3 + g m4 2 v4
2
Define vneq as the equivalent input noise voltage of the differential amplifier. Therefore, 2
2
iod = gm12 vneq or gm3 2 2 2 2 2 2 vneq = veq1 + veq2 + veq3 + veq4 gm1 Where gm1 = gm2 and gm3 = gm4 VDD I DD
v2neq M1
M2
i2o
M3
+ M4 -
vOUT VSS
It is desirable to increase the transconductance of M1 and M2 and decrease the transconductance of M3 and M4. (Empirical studies suggest p-channel devices have less noise)
Allen and Holberg - CMOS Analog Circuit Design
Page VI.2-23
CMOS DIFFERENTIAL AMPLIFIERS Minimization of Noise gm3 2 2 2 2 2 2 vneq = veq1 + veq2 + g veq3 + veq4 m1
In terms of voltage spectral-noise densities we get, gm3 2 2 2 2 2 2 eeq = en1 + en2 + g en3 + en4 m1
1/f noise Let
KF B 2 en = = 2fCoxWLK' fWL 2
2
assume en1 = en2 ∴
and
2
2
en3 = en4
K N 'B N L12 2BP 2 eeq (1/f) = fW L 1 + K 'B L 1 1 P P 3 1) Since BN ≈ 5BP use PMOS for M1 and M2 with large area. K P'BP L1 1 2 ≈ 12.5 so that eeq (1/f) ≈ 2) Make L < K 'B 3 N N
Thermal Noise 16KT(1+η1)
2
eeq (th) = 3
W1 2K P'I 1 L 1
1) Large value of gm1. L1 2) L3 < 1.
K N ' WL 3 1 + 3 K P'W1 L1
2BP fW1L1
Page VI.3-1
Allen and Holberg - CMOS Analog Circuit Design
VI.3 - CASCODE AMPLIFIERS VI.3.1-CMOS CASCODE AMPLIFIERS Objective Prevent Cgd of the inverter from loading the previous stage. Gives very high gain. Cascode Amplifier Circuit
VDD
VGG2
Miller effect:
M3 vout
VGG1
M2
Cgd1 + vin
M1 v 1 VSS
Large Signal Characteristics When V GG1 designed properly, vout(min) = V on1 + V on2
Inverter Cin ≈ Gain x Cgd1 Cascode Cin ≈ 3Cgd1 v1 ≈ 2 vin
Page VI.3-2
Allen and Holberg - CMOS Analog Circuit Design
CASCODE AMPLIFIER-CONTINUED Small Signal Model + rds2
gm2 v1
gmbs2 v1
+
+ vin
-
rds1
v1
gm1 vin
vout
rds3
-
(gm2 +gmbs2 )v1
C1
rds2
+
+ C2
vin gm1vin
-
rds1
v1
-
1 gm2 (1+η2 )
+ C3
rds3
gm2( 1+η2 )v1
vout
-
Nodal Equations: (gm1 − sC1)vin + (g m2 + g mbs2 + g ds1 + g ds2 + sC1 + sC2)v1 − (gds2)vout = 0 −(gds2 + gm2 + gmbs2)v1 + (gds2 + gds3 + sC3)vout = 0 Solving for v out/vin gives (sC1-gm1)gm2(1+η) ≅ 2 s (C3C1+C3C2)+s(C 1+C2)(g ds2+g ds3) +C3gm2 (1+η)+gds3gm2(1+η)
Page VI.3-3
Allen and Holberg - CMOS Analog Circuit Design
Small Signal Characteristics Low-frequency Gains: vout −g m1 (g ds2 + g m2 + g mbs2 ) = vin g ds1 g ds2 + g ds3 (g m2 + g mbs2 + g ds1 + g ds2 ) ≈
−g m 1 gds3 =
2K'(W1/L 1)ID1 λ3ID3
Also (see next page), v1 −2gm1 = vin g m2 (1 + η 2 ) Gain Enhancement: VDD VGG2
M4
M3
vout VGG1
M2 I4
I2 I1 v in
M1
vout −gm1 vin ≈ gds3 2K'W1 I1 vout L1 vin ≈ λI2 But I1 = I2 + I4
VSS
I4 = 24I2 ⇒ x5 Gain enhancement
Page VI.3-4
Allen and Holberg - CMOS Analog Circuit Design
Voltage Gain of M1: v1 −gm1 = vin gm2 ? What is the small signal resistance looking into the source of M2? Consider the model below: i1 + v1 = vs2
gm2 vgs2 r ds2
r ds3
-
vs2 = (i1 + gm2vgs2)r ds2 + i1rds3 = rds2i1 + gm2(−vs2)r ds2 + i1rds3 or vs2(1 + g m2 rds2) = i1(rds2 + rds3) Therefore, r ds2 + r ds3 r ds2 + r d s 3 r d s 3 vs2 1 ≈ g r = g 1 + r R= i =1 + g r 1 m2 d s 2 m2 ds2 m2 ds2 Some limiting cases: 1 rds3 = 0 ⇒ R = g m2 2 rds3 = rds2 ⇒ R = g m2 and rds3 rds3 >> rds2 ⇒ R = gm2rds2 Therefore, the gain vin to v1 is −g m1 (g ds2 + g ds3 ) −2gm1 −2gm1 v1 vin ≈ (g m2 + g mbs2 )g ds3 ≈ g m2 + g mbs2 ≈ gm2
Page VI.3-5
Allen and Holberg - CMOS Analog Circuit Design
CASCODE AMPLIFIER-CONTINUED High Resistance Driver for the Inverter M1-M2
VDD M2 M4
VGG2 Cgd1
ro = g
1
ds3 +gds4
vout
iin + M1 v1
M3
CΜ
VSS C2
iin
R1
C1
R1 = (gds3 + g ds4 )-1
+ v1 -
+ gmv1
C3
R3
vout -
R3 = (gds1 + g ds2 )-1
C1= Cgs1 + Cbd3 + Cbd4 + Cgd3 + Cgd4 C2 = Cgd1 vout(s) = iin(s)
C3 = Cbd1 + Cbd2 + Cgd2 + CL
−gm1 gm1 1−s G1G3 C2 1+R (C +C )+R (C +C )+g R R C s+(C C +C C +C C )R R s2 1 2 1 3 2 3 1 3 1 1 3 3 2 3 m1 1 3 2
Page VI.3-6
Allen and Holberg - CMOS Analog Circuit Design
Note: s s 1 s2 1 d(s) = 1 + as + bs2 = 1 − 1 − = 1 − s + + p 1 p2 p2 p1p2 p1 If |p2| >> |p 1| , then s s2 d(s) ≈ 1 − p + p p 1 1 2
or
1 p1 = − a
and
a p2 = − b
Using this technique we get, p1 ≈
−1 −1 ≈ R1(C1+C3)+R3(C2+C3)+gm1R1R3C2 gm1R1R3C2
(Miller effect on C2 causes p1 to be dominant; CM ≈ gm1R2Cgd1) −gm1C2 p2 ≈ C C +C C +C C 1 2 1 3 2 3
Page VI.3-7
Allen and Holberg - CMOS Analog Circuit Design
CASCODE AMPLIFIER - CONTINUED How does the Cascode Amplifier solve this problem? VDD M5 VGG5 M2
vout
VGG2 M4 1 ro =g +g ds3 ds4
Cgd1 iin M1
M3 VSS Cgd1
iin
r1
+ v1 -
rds2
C2 gmv1
+ v2 r2 - gm2(1+η)v2
+ C3
r3
r1 = ro = (gds3 + gds4)-1 C2 = Cgs2 + Csb2 + Cdb1 + Cgd1 r 2 = g ds1 + g m 2 (1 + η )
-1
1 ≈ g m2
C3 = Cgd2 + Cdb2 + Cgd5 + Cdb5 + CL gm2 -1 1 r3 ≈ g + g ≈ d s 5 gds5 ds1 g ds2
vout -
Page VI.3-8
Allen and Holberg - CMOS Analog Circuit Design
Cascode amplifier with higher gain and output resistance
VDD
io
VGG4 M4 VGG3 M3 VGG2
gm2 v1 Vout
r ds2
gmbs2 v1
G1 M2
Vin
-
g m3 v4
gmbs3 v4
r ds3
vout
D1=S2
+
+ vin
M1 VSS
+
D2
gm1 v in
v1
-
r ds1
r ds4
+ v4 -
' =rds3
-
Allen and Holberg - CMOS Analog Circuit Design
Page VI.4-1
VI.4 - OUTPUT AMPLIFIERS Requirements 1. Provide sufficient output power in the form of voltage or current. 2. Avoid signal distortion for large signal swings. 3. Be efficient. 4. Provide protection from abnormal conditions. Types of Output Stages 1. Class A amplifier. 2. Source follower. 3. Push-Pull amplifier ( inverting and follower). 4. Substrate BJT. 5. Negative feedback (OP amp and resistive).
Page VI.4-2
Allen and Holberg - CMOS Analog Circuit Design
CLASS A AMPLIFIER VDD
VGG2
M2 IQ
Vin
Iout
M1
Vout
CL
RL
VSS
KnW1 2 Iout 2L1 (VDD − VSS − VT1) − IQ KpW2 Iout = 2L (VDD − V GG2 − |VT2|) 2 < Iout+ 2 +=
|Iout| determined by: dvout dt = CL (slew rate) vout(peak) 2. |I out| = RL
1. |Iout| = C L
Vout(peak) 2 Efficiency = Psupply = (V DD + V SS )
PRL
1 1 = rout = g 2λID ds1 + g ds2
≤ 25%
(typically large)
Page VI.4-3
Allen and Holberg - CMOS Analog Circuit Design
SOURCE FOLLOWER N-Channel
Push Pull VDD
vIN
VDD
M1
M1 vOUT
VGG
vIN
M2 VSS
Large Signal Characteristics vOUT = vIN − vGS1 Maximum Output Swing Limits vOUT (MAX) = VDD − VT1 (VT1 greater than V T0 because of v BS) Single Channel Follower: vOUT(MIN) = V SS Push Pull Follower: vOUT(MIN) = VSS + |VT2| (VT2 greater than VT0 because of v BS)
vOUT
M2 VSS
Page VI.4-4
Allen and Holberg - CMOS Analog Circuit Design
SOURCE FOLLOWERS Small Signal Characteristics Single Channel Follower (Current source and active load): C1
+
+
gm1 vin
vin
gm1 vout
-
rds1 gmbs1 vout
rds2
C2
gm2 vgs2
vout
-
Small Signal Voltage Transfer Function: gm1 vout = vin gds1+gds2+gm1+gmbs1+gm2 where g m2 = 0 if v GS2 = V G G Example:
W 10 µm If VDD= −VSS =5V, vOUT = 0V, iD = 100µA, and L =10 µm ,
then; vout 41.23 = vin 1+1+41.23(1+0.2723)+41.23 = 0.4309 when vGS2 = vOUT vout 41.23 = vin 1+1+41.23(1+0.2723) = 0.751 when vGS2 = VGG vout ≈ 0.786 (gds1= gds2 ≈ 0) Approximation gives v in Output Resistance: 1 r out = gds1+gds2+gm1+gmbs1+gm2 where g m2 = 0 if v GS2 = V G G rout = 10.5 KΩ (vGS2 = vOUT) and rout = 18.4 KΩ (vGS2 = VGG)
Page VI.4-5
Allen and Holberg - CMOS Analog Circuit Design
SOURCE FOLLOWERS Push Pull Source Follower Model: C1
M1
+
gm1vin
vin
gm1vout
M2
+
gm2 vin gm2vout
rds1
1
gmbs1
rds2
1 gmbs2
C2 v out
-
-
Small Signal Voltage Transfer Function: g m1 + g m 2 vout vin = g ds1 + g ds2 + g m1 + g mbs1 + g m2 + g m b s 2 Example: W 10µm If VDD = −VSS = 5V, vOUT = 0V, iD = 100µA, and L = 10µm then, vout 41.23 + 28.28 = vin 1 + 0.5 + 41.23(1 + 0.2723) + 28.28(1 + 0.1268) = 0.81 Output Resistance: 1 r out = g ds1 + g ds2 + g m 1 + g mbs1 + g m 2 + g m b s 2 = 11.7KΩ
Page VI.4-6
Allen and Holberg - CMOS Analog Circuit Design
PUSH-PULL INVERTERING CMOS AMPLIFIER ConceptVDD
VTR2 vIN VTR1
M2
+
Iout
+ -
Vout
M1
CL
RL
VSS
ImplementationVDD
M5
M6
M1
M3
VGG3
M2
M4
VGG4
Iout
Vout
vIN
M7 VSS
CL
M8
RL
Page VI.4-7
Allen and Holberg - CMOS Analog Circuit Design
PUSH-PULL SOURCE FOLLOWER VDD
M2
Iout
VTR Vout
vIN
CL
M1
RL
VSS VDD VGG6
M6 M2 M5
Iout Vout
M4 M1
vIN
M3
VSS
CL
RL
Page VI.4-8
Allen and Holberg - CMOS Analog Circuit Design
USE OF NEGATIVE FEEDBACK TO REDUCE ROUT
VDD
error amplifier
M2
-
+
Iout
Vout
vIN -
+ CL
M1
RL
VSS
Use of negative feedback to reduce the output resistance of Fig. 6.3-4. VDD
M2
R1 vIN
R2
Iout
M1
Vout
CL
RL
VSS
Use of resistive feedback to decrease the output resistance of Fig.6.3-4.
Allen and Holberg - CMOS Analog Circuit Design
VI.5 - SUMMARY • Analog Amplifier Building Blocks Inverters - Class A Push-Pull - Class AB or B Cascode - Increased bandwidth Differential - Common mode rejection, good input stage Output - Low output resistance with minimum distortion
Page VI.5-1
Allen and Holberg - CMOS Analog Circuit Design
SECTION 7 - COMPARATORS
Page VII.0-1
Allen and Holberg - CMOS Analog Circuit Design
Page VII.0-1
VII. COMPARATORS Contents VI.1 VI.2 VI.3 VI.4 VI.5
Comparators Models and Performance Development of a CMOS Comparator Design of a Two-Stage CMOS Comparator Other Types of Comparators Improvement in Comparator Performance A. Hysteresis B. Autozeroing VI.6 High Speed Comparators Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS OTA's
Chapter 9 High Performance OTA's
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-1
VII.1 - CHARACTERIZATION OF COMPARATORS What is a Comparator? A comparator is a circuit which compares two analog signals and outputs a binary signal based on the comparsion. (It can be an op amp without frequency compensation.) Characterization of Comparators We shall characterize the comparator by the following aspects: • Resolving capability • Speed or propagation time delay • Maximum signal swing limits • Input offset voltage • Other Considerations Noise Power Etc.
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-2
VOLTAGE COMPARATORS
Definition of a Comparator VA VB
+
VOUT -
Noninverting VOUT VOH
VOUT =
V O H V OL
when VA ≥ VB
VA - VB when V A < V B
VOL
Inverting VOUT
VOUT =
V O L V OH
when VA ≥ VB
VOH
when V A ≤ V B
VA - VB VOL
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-3
COMPARATOR PERFORMANCE
1. Speed or propagation time delay. The amount of time between the time when VA - V B = 0 and the output is 50% between initial and final value. 2. Resolving capability. The input change necessary to cause the output to make a transition between its two stable states. 3. Input common mode range. The input voltage range over which the comparator can detect V A = VB . 4. Output voltage swing (typically binary). 5. Input offset voltage. The value of V OUT reflected back to the input when VA is physically connected to V B.
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-4
APPROACHES TO THE DESIGN OF VOLTAGE COMPARATORS Open Loop Use of a high-gain differential amplifier. V OH - V O L Gain = resolution of the comparator Regenerative Use of positive feedback to detect small differences between two voltages, VA and VB. I.e., sense amplifiers in digital memories. Open Loop - Regenerative Use of low gain, high speed comparator cascaded with a latch. Results in comparators with very low propagation time delay. Charge Balancing Differential charging of a capacitor. Compatible with switched capacitor circuit techniques.
Type
Offset Voltage (Power supply)
Resolution
Speed (8 bit)
Open-loop
1-10 mV
300µV (±5V)
10 MHz
Regenerative Charge Balancing
0.1 mV
50µV (±5V)
50 MHz
0.1 mV
5mV (5V)
30 MHz
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-5
COMPARATOR MODELS - OPEN LOOP Zero Order Model
VOUT VOH
+
-
VP - VN
VOL
Model VP +
+
+ fo VP - VN
-
-
VN
fo( V P
V O H - VN ) = V OL
VO
for ( V P - V N ) ≥ 0
for ( V P - V N ) ≤ 0
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-6
COMPARATOR MODELS - CONT'D
First Order Model
Transfer Curve VOUT VOH VIL
VP - VN
VIH VOL
Model VP +
+
+ f1 VP - VN
VN
VO
-
V O H for ( V P - V N ) ≥ VIH f1( V P - V N ) = AV( V P - V N ) f o r V I L ≤ ( V P - V N ) ≤ V I H V OL for ( V P - V N ) ≤ VIL
Allen and Holberg - CMOS Analog Circuit Design
Page VII.1-7
COMPARATOR MODELS - CONT'D First Order Model with Offset Transfer Curve VOUT VOH
VOS VIL
VP - VN
VIH VOL First Order Model with Offset +-VOS VP
+
-
V'P +
+
+ f1 V'P - V'N -
-
-
V'N
VN
Time Response of Noninverting, first order model VOH v = VOH + VOL 2
VOUT VOL VIH VP - VN
VO
v = VIH + VIL 2
tP VIL Time
Allen and Holberg - CMOS Analog Circuit Design
Page VII.2-1
VII.2 - DEVELOPMENT OF A CMOS COMPARATOR SIMPLE INVERTING COMPARATOR VDD vN
M2 I2 IB M1
VBIAS
vO
VSS Fig. 7.2-1 Simple inverting comparator ∆VIN
VDD
vO
vN
VTRP
Fig. 7.2-2 DC transfer curve of a simple comparator
Low gain ⇒ Poor resolution VTRP = f V D D + process parameters
Allen and Holberg - CMOS Analog Circuit Design
Page VII.2-2
CALCULATION OF THE TRIP POINT, VTRP vO
t.
VDD
VDD
vO = V IN + VT2
c 2a
M
t.
a 2s
M
VIN
M2 vO
VBIAS
M1
M1 sat. M1 act.
VBIAS - VT1
VSS Operating RegionsvDS1 ≥ v GS1 - VT
VSS VSS
VIN vN
VTRP
VDD
‘ vO - VSS ≥ V BIAS - VSS - VT1 v O ≥ V BIAS - V T 1
vSD2 ≥ vSG2 - VT2
‘
V DD - vO ≥ VDD - vIN - VT2 v O ≤ v I N + VT2
Trip PointAssume both M1 and M2 are saturated, solve and equate drain currents for VTRP. Assume λ ≈ 0. K N W1 2 iD1 = 2 L V BIAS - V SS - V T 1 1 K P W2 2 iD2 = 2 L V D D - v I N - VT2 2
iD1=iD2 ‘
vIN = VTRP = VDD- VT2 -
KN( W1/L1) KP( W2/L2) ( V BIAS - V SS - V T 1)
W1 W2 I.e. V DD = -VSS = 5V, VBIAS = -2V and KN L = KP L 1 2 VTRP = 5-1-(-2+5-1) = 4-2 = 2V
Allen and Holberg - CMOS Analog Circuit Design
Page VII.2-3
COMPARATOR USING A DIFFERENTIAL AMPLIFIER VDD
M3
M4 vO M1
vP
M2
VBIAS
vN
M5
vO VOH = VDD VOH' M1 & M2 in saturation VOL' VOL VSS -1
+1
Gain is still low for a comparator
vP - vN Av
Allen and Holberg - CMOS Analog Circuit Design
Page VII.2-4
DERIVATION OF OUTPUT SWING LIMITS VDD
vP > vN
M3 I1 vP
M4 I2
M1
M2
vO vN
1. Current in M1 increases and current in M2 decreases. 2. Mirroring of M3-M4 will cause vO to approach VDD . 3. VOH' = VDD - VDS4(sat)
ISS
VBIAS
VOH ' = VDD -
I4 β4
VOH ' = VDD -
I5 Kp'( W4/L4)
M5 VSS
vP < vN
V O H ' = VD D -
I5 Kp'( W3/L3)
Assume vN is a fixed DC voltage 1. vO starts to decrease, M3-M4 mirror is valid so that I1 = I2 = ISS/2 . 2. VOL ' = vN - VGS2 + VDS2 when M2 becomes non-sat. we have VDS2(sat) = VGS2 - VT so that
4. Finally, vO ‘ V DD causing the mirror M3-M4 to no longer be valid and V OH ≈ V DD. (I2 = I4 = 0 , I3 = I1 = I5)
V OL ' = v N - V T 2 3. For further decrease in vO, M2 is nonsat and therefore the VGS2 can increase allowing the sources of M1 and M2 to fall(as v P falls). 4. Eventually M5 becomes non-sat and I5 starts to decrease to zero. M2 becomes a switch and v O tracks V S2(VDS5) all the way to VSS. ∴ V OL = V SS .
I 1 still equals I2 due to mirror
Allen and Holberg - CMOS Analog Circuit Design
Page VII.2-5
TWO-STAGE COMPARATOR Combine the differential amplifier stage with the inverter stage. • Sufficient gain. • Good signal swing.
VDD M3
M4 M6
vN
M1
M2
I8
M8
M5
VSS
vP
vO
M7
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-1
VII.3 - DESIGN OF A TWO-STAGE CMOS COMPARATOR DC BALANCE CONDITIONS FOR TWO-STAGE COMPARATOR •
Try to keep all devices in saturation - more gain and wider signal swings.
•
Based on gate-source and DC current relationship. I.e. if M1 and M2 are two matched devices and if VGS1 = VGS2, then ID1 = I D2 or vice versa. W1 Let S1 = L , 1 M1 and M2 matched gives S 1 = S2. M3 and M4 matched gives S 3 = S4. also, I 1 = I2 = 0.5I5. From gate-source matching, we have S7 S6 VGS5 = VGS7 ‘ I7 = I5 and I 6 = I4 ← Assume S5 S4 VGS4 =VGS6 For balance conditions, I6 must be equal to I7, thus I 5 S7 S6 . I4 S5 = S4 Since
I5 I4 = 2, then DC balance is achieved under the following: S6 S . 7 ‘ VDG4 = 0 ‘ M4 is saturated. = 2 S4 S5
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-2
SYSTEMATIC OFFSET ERROR VDD =10V +
KN = 24.75 µA/V2 KP = 10.125 µA/V2 VTN = -VTP = 1V λN = 0.015V -1 λP = 0.020V -1
+ 2V 20 2V 10 M3 - M6 40 M4
20 10
I8
20 10
vN M1
Find VOS to make i6 = i 7
vP
M5
VSS =0V
(2) Find how much vGS6 must be reduced to make i6 = i 7 ∆vGS6 = vGS6(2.115i4) - vGS6(2.057i4) 2L6 KPW6 i 4 2.115 - 2.057 = 14.11 mV
(3) Reflecting ∆vGS6 into the input KN( W2/L2) 2 = 89.9 I5 λ 2 + λ 4 ∆vGS6 14.1 mV = 0.157 mV ∴ VOS = A (diff) = 89.9 v A v(diff) =
vO =5V
+
10 10 3V
M7
(1) Find the mismatch between i6 and i 7 i7 1 + λ N v D S 7 W7/L7 1 + (0.015)(5) i5 = 1 + λ N v D S 5 W5/L5 = 1 + (0.015)(3) (1) = 1.029 i6 1 + λ P v D S 6 W6/L6 1 + (0.02)(5) i4 = 1 + λ P v D S 4 W4/L4 = 1 + (0.02)(2) (2) = 2.115 i5 = 2i4 ∴ i7 = (1.029)(2)i4 = 2.057i4 and i6 = 2.115i4
∆vGS6 =
i6 i7
M2 20µA
M8
10
20 10
10 10
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-3
DESIGNING FOR COMMON MODE INPUT RANGE VDD + VSG3 M3 I5/2
+
VG1
vG1 (min) = VSS + VDS5 + VGS1
VDG1 -
+ M1 VGS1 -
v G1 (min) = V SS + V DS5 + V T1 (max) +
+ VDS1 -
vG1 (max) = VDD - VSG3 - VDG1(sat) +
I5
VBIAS
v G1 (max) = V D D -
VDS5 VSS
I5 2β1
M5
I5 2β3 - VT3(max) + V T1 (min)
where V DG1(sat) = -VT1
-
Example Design M1 through M4 for a CM input range 1.5 to 9 Volts when VDD = 10 V, ISS = 40µA, and VSS = 0V. Table 3.1-2 parameters with |VTN,P| = 0.4 to 1.0 Volts, I5 vG1(min) = VSS + VDS5 + β1 + VT1(max) 40µA + 1 (assumed VDS5 ≈ 0.1V- it probably more 1.5 = 0 + 0.1 + β1 reasonable to assume β1 is already defined and find β5) β1 =
KNW 1 2 L1 = 250 µA/V ‘
vG1(max) = VDD β3 =
W1 W 2 L1 = L2 = 14.70
I5 β3 - |VT3(max)| + VT1(min)
K PW 3 2 L3 = 250 µA/V ‘
W3 W 4 L3 = L4 = 31.25
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-4
GAIN OF THE TWO-STAGE COMPARATOR +
+ gm1vid
r ds2
r ds4
v1 gm6v1 -
r ds6
r ds7
-
vid = vP - vN gm1 ds2 + g ds4
Av = g
2 Av =
gm6 g ds6 + g ds7
W1 W6 KNKP L L 1 6
( λ2 + λ4) ( λ6 + λ7)
vout
I1I6
W6 W1 Using L = 5, L = 5, λN = 0.015V-1 , λP = 0.02V-1 1 6 and Table 3.1-2 values; 2 (17)(8)(5)(5) . -6 95199.10-6 Av = 10 = (0.015+0.02)2 I1I6 I1I6 Assume I1 = 10 µA and I6 = 100 µA Av = 3010 V OH - VOL = Resolution = 5 mV (assume) Av 5 . then VOH - V OL = 1000 3000 = 15 Volts
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-5
PROPAGATION DELAY OF THE TWO-STAGE COMPARATOR VDD signal swing less than the M4 output
M3
vN
VGS6 + -
M1
M2
vP CL1
VBIAS
M6 i6 key node vO i7 CL2
M5
M7
i5 VSS V GS6 = VDD - v P + V D G 2 dv iC = C dt , ∆t =
∆v CI
∆t2+ = C L2
K P W6 V 2 L6 ( D D
∆t+2
VTRP3 V VDD SS VTRP3
V TRP3 - V S S - v P - V D G 2 - |V T6 | ) 2 -
V DD - V TRP3 ∆t2- = CL2 W L 7 5 i L7 W5 5
∆t-2
Slew rate =
isource/sink CLi
I7
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-6
CALCULATION OF COMPARATOR PROPAGATION DELAY Find the total propagation delay of the comparator shown when the input vP goes from -1 to +1 in 2ns. Assume the trip point of the output(next stage) is zero. Total delay = 1st stage + 2nd stage delay delay
+5V 10 10
M3
40 10 C L1=0.3pF
M4 vDO
vN
M1
M2 20 10
vP
VTRP2 = VDD - VGS6, VGS6 = |VT6 | +
CL2= 10pF
I7 =40µA
I5=20µA
∆t = ∆t1 + ∆t2 ( v DO (t 0 ) - V TRP2) ∆t1 = CL1 , I5 vDO(t0) = 5 because vP = -1V
M6 I6
-5V
2I7 KP'( W6/L6)
2.40 = 2.58 V ‘ VTRP2 = 5 - 2.58 = 2.42 V 8.4 0.3pF = 38.7ns ∴ ∆t1 = (5 - 2.42) 20µA CL2 CL2 ∆t2 = v O (t 0 ) - 0 =5 I 6 - I 7 I 6 - I 7 KP6' W6 I6 = 2 L V D D - V D O (min) - VT6 2 6 [VDO(min) is an optimistic assumption based on vDS2 ≈ 0] VGS6 = 1 +
VDO(min) ≈ vDS2(≈0) - vGS1 + vN = -VT1 -
I5 = -1.77 KN.2
8.10-6 2 2 (4)(5 - (-1.77) -1) = 533 µA 10 pF ∴ ∆t2 = 5 (533 - 40) µA = 101 ns
I6 =
∆t = ∆t1 + ∆t2 ≈ 139 ns Second order consideration: Charging of Csb of M1 and M2
vO
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-7
SIMULATION OF THE PROPAGATION DELAY 5v +5V 10 10
M3
M4 (6)
vN
M1
3v 2.42v
M8
20 10
M2
M6
CL1
40 10 (9)
vP
vO
CL2
10 10
M7 -5V 1v
vP
0v
tprop=167 ns
20 10
V(9)
Actual -1 v V(6)
-1.54v
COMPARATOR PROPAGATION DELAY VDD 10 0 DC 5V VSS 11 0 DC -5V VN 1 0 DC 0V VP 2 0 PULSE(-1 1 0N 1N 1N 500N 1U) M1 3 1 5 5 MNMOS W=20U L=10U M2 6 2 5 5 MNMOS W=20U L=10U M3 3 3 10 10 MPMOS W=10U L=10U M4 6 3 10 10 MPMOS W=10U L=10U M5 5 8 11 11 MNMOS W=10U L=10U M6 9 6 10 10 MPMOS W=40U L=10U M7 9 8 11 11 MNMOS W=20U L=10U M8 8 8 11 11 MNMOS W=10U L=10U CL1 6 0 0.3PF CL2 9 0 10PF IS 0 8 DC 20UA .MODEL MNMOS NMOS VTO=1 KP=17U +LAMBDA=0.015 GAMMA=0.8 PHI=0.6 .MODEL MPMOS PMOS VTO=-1 KP=8U +LAMBDA=0.02 GAMMA=0.4 PHI=0.6 .TRAN 2N 300N .PRINT TRAN V(6) V(9) V(2) .PROBE .END
Approx.
-3 v
-5 v 0ns
50ns
100ns
150ns
Time
200ns
250ns
300ns
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-8
SMALL SIGNAL PERFORMANCE
+
vin -
+ gm1 -
gm2 R1
C1
+
R2
C2
vout -
vout(s) A oω p1ω p2 = vin(s) ( s + ω p 1) ( s + ω p 2) 1 ω p1 = R C 1 1 1 ω p2 = R2C2 Ao = gm1gm2R1R2
Example - (Fig 7.3-4) 1 1 = 10µA = 3.33MΩ ds2 + g ds4 1 ω p1 = (0.3pF)(3.33MΩ) = 1Mrps
I5 = 20µA ‘ R1 = g
1 1 = 40µA(.03) = 833KΩ ds6 + g ds7 1 ω p2 = (10pF)(833KΩ) = 120Krps
I7 = 40µA ‘ R2 = g
g m1 = 26µs, gm2 = 50.6µs ‘ A o = 1099
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-9
TWO-STAGE, CMOS COMPARATOR General Schematic VDD M3
M4 M6 M1
vN
M2
I8
M5
M8
vP
vO
M7
VSS Key Relationships for Design: β i D = (v G S - V T ) 2 2 or v DS (sat) = Also, gm = where KW β= L
2βI D
2iD(sat) β
β ⇒ iD (sat) = 2 [vDS(sat)]2
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-10
COMPARATOR DESIGN PROCEDURE 1. Set the output current to meet the slew rate requirements. dV i = C dt 2. Determine the minimum sizes for M6 and M7 for the proper ouput voltage swing. vDS (sat) =
2ID β
3. Knowing the second stage current and minimum device size for M6, calculate the second stage gain. A2 =
-g m6 g ds6 + g ds7
4. Calculate the required first stage gain from A2 and gain specifications. 5. Determine the current in the first stage based upon proper mirroring and minimum values for M6 and M7. Verify that Pdiss is met. 6. Calculate the device size of M1 from A1 and I DS1. A1 = g
-g m1 ds1 + g ds3
and
gm1 =
2K'W/L IDS1
7. Design minimum device size for M5 based on negative CMR requirement using the following (IDS1 = 0.5IDS5): vG1(min) = VSS + VDS5 + where VDS5 =
IDS5 β1 + VT1(max)
2IDS5 β5 = VDS5(sat)
8. Increase either M5 or M7 for proper mirroring. 9. Design M4 for proper positive CMR using: vG1(max) = VDD -
IDS5 β3 - VTO3 (max) + VT1
10. Increase M3 or M6 for proper mirroring. 11. Simulate circuit.
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-11
DESIGN OF A TWO-STAGE COMPARATOR Specifications: Lambda = 0.05V-1 (L = 5 µm)
Avo > 66 dB Pdiss < 10 mW
VDD = 10 V
CL = 2 pF
VSS = 0 V
tprop < 1 µs
K'W Recall that β = L
CMR = 4-6 V Output swing is VDD - 2V and VSS + 2V 1). For t prop vDS7(sat) =
2I7 β7 =
W7 2(200µA) → L7 > 5.88 17.0µA/V 2( W7/L7)
M6: 2V > vDS6(sat) =
2( IOUT+I7) = β6
-g m6 -1 = 3). A 2 = g ds6 + g ds7 λ N + λ P
W6 2(400µA) → L6 > 12.5 8.0µA/V 2( W6/L6)
2KP'W6 I6L6 ≈ -10
4). A vo = A 1A2 = 66 dB ≈ 2000 → A1 = 200
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-12
COMPARATOR DESIGN - CONT'D S4 5). Assuming vGS4 = v GS6, then I4 = S6 I6 1 choose S 4 = 1 which gives I4 = 12.5 (200µA) = 16.0 µA S5 200µA Assume S5 = 1 which gives I 5 = I = 7 S7 5.88 = 34 µA 1 and I4 = I5 = 17 µA 2 W to keep L ratios greater than 1. Choose I 4 = 17 µA W4 W 6 17 ∴ I5 = 34 µA L4 = L6 200 = 1.06 ≈ 1.0 Pdiss = 10( I 7 + I 5 ) = 2.34 mW < 10 mW 1 6). A1 = λ + λ 1 4 W1 ∴ L = 200 1
2KN'W1 W1 I 2 4 = 200 → = (λ + λ )A [ 1 4 1 ] I4L 1 L1 2KN' (Good for noise)
7). V DS5 = vG1(min) - VSS V DS5 = 4 - 0 VDS5 =
8). S5 =
2I5 β5 =
I5 β1 - VT1(max)
(34) -1 = 2.90 V 2(17.0)(200) W5 2(34µ) → L > 0.48 (17µ)S5 5
I5 W5 34 S = (5.88) = 1.0 → 7 I7 200 L5 = 1 . 0
Allen and Holberg - CMOS Analog Circuit Design
Page VI.3-13
COMPARATOR DESIGN - CONT'D
9). VG1(max) = VDD β3 =
I5 β3 - VTO3 (max) + VT1(min) I5
V D D - V G1 (max) - VTO3 (max) + V T1 (min) 2
34 µA = 2.76.10 -6 ( 1 0 - 6 - 1 + 0 . 5) 2 W3 (2.76)(2) W3 W 4 ∴L = = 0.69 8 L3 = L4 > 0.69 3 W4 (Previously showed L > 1.06 so no modification is necessary) 4 =
10). Summary W Wdrawn = (L - 1.6) L Design Ratios
W1 L1 W3 L3 W5 L5 W6 L6 W7 L7
W2 = L = 200 2 W4 = L = 1.0 4 = 1.0 = 12.5 = 5.88
W1 L1 W3 L3 W5 L5 W6 L6 W7 L7
Actual Values with 5µm
Proper Mirroring
minimum geometry
and LD = 0.8µm
W 2 1000 = L = 5 2 W4 5 = L =5 4 = 1.0 62.5 = 5 30 = 5
(Need to adjust for proper mirroring)
⇒
680 5 3.4 5 ‘5 5 3.4 5 ‘5 5 60 5 30 5 ↑ S6 S7 = 2 S4 S5
Allen and Holberg - CMOS Analog Circuit Design
Page VII.4-1
VII.4 - OTHER TYPES OF COMPARATORS FOLDED CASCODE CMOS COMPARATOR Circuit Diagram VDD
MP3
MP4
MP12
MP13
MP8
MP6
MN25
MN1
MN2
vOUT
MN10
MN11
MN9
MN5
v1
v2
MN24
MN7
V SS
Small Signal Model 1 gm12 gm1 v2
i1
1 gm13
+ i2 i2
gm2 v1
i1
rout
vout -
where R out ≈ (rds5gm11rds11)||((rds4||rds2)gm13rds13) = =g
1 ds5gds11 (gds2+gds4)gds13 gm11 + gm13
The small signal voltage gain is vout = r out (i2-i1) = (gm2 +gm1 )Rout vin = g
where vin = v1 - v2.
gm1 +gm2 vin ds5gds11 (gds2+gds4)gds13 gm11 + gm13
Allen and Holberg - CMOS Analog Circuit Design
Page VII.4-2
FOLDED CASCODE CMOS COMPARATOR - CONTINUED Frequency Response Small signal modelC1
i1
1
gm1 v2
gm12 gm2 v1
+
i2
C2 1
gm13
C3
i2
i1
rout
vout -
where C1 = C GS12 + C BS12 + C DG3 + C BD3 C2 = C GS13 + C BS13 + C DG4 + C BD4 and
C3 = CDG11 + CBD11 + CDG13 + CBD13 + CLoad AVD0ω3 AVD(s) ≈ s + ω 3
where 1 ω3 = routC3 Typical performanceW 1 W 2 W 11 W 13 ID1 = ID2 = 50µA and ID3= I D4 = 100µA, L1 = L2 = L11 = L13 =1, assume C 3 ≈ 0.5pF, and using the values of Table 3.1-2 gives: gm1 = gm2 = gm11 =41.2µS gds5 = gds11 = 0.5µS
gm13 = 28.3µS
gds4 = gds13 = 0.25µS
Therefore, rout = 121MΩ, ω3 = 16.553krps, and AVD0 = 4,978 resulting in a gain-bandwidth of 13.11MHz. C3∆V 0.5pFx10V Delay = ∆T = I = 100µA = 50nS max
Allen and Holberg - CMOS Analog Circuit Design
Page VII.4-3
OPEN LOOP COMPARATOR - MC 14575
BIAS
M1
M6 M8
M10 vO
-
M2
+
M3
M9
M11
M7 M4
M5
Performance (ISET = 50 µA) Rise time = 100 ns into 50 pF Fall time Propagation delay = 1 µs Slew rate = 2.7 Volts/µs Loop Gain = 32,000
Comments The inverter pair of M8-M9 and M10-M11 are for the purpose of providing an output drive capability and minimizing the propagation delay.
Allen and Holberg - CMOS Analog Circuit Design
Page VII.4-4
CLAMPED CMOS VOLTAGE COMPARATOR
VDD
VDD
M6
M8
BIAS
VPB
M1
vO -
+ M3
M2 M9
M4
VNB
M5
M7
VSS
Drain of M2 and M3 clamped to the gate voltages of M4 and M5.
M6 and M7 provide a current, push-pull output drive capability similiar to the current , push-pull CMOS OP amp.
Comparator is really a voltage comparator with a current output.
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-1
VII.5 - COMPARATORS WITH HYSTERESIS HYSTERESIS Why Hysteresis? Eliminates "chattering" when the input is noisy. Comparator with no Hysteresis vin
Comparator threshold
Time
Comparator output
Comparator with Hysteresis vin vout VTRP+ VTRPTime vin VTRP-
VTRP+ comparator output
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-2
VOLTAGE COMPARATORS USING EXTERNAL FEEDBACK Inverting vOUT VOH
-
vB vA
vOUT
+
VREFR2 R1 +R2 VOHR1 R1 +R2
R2
R1 + V - REF
VOL
vB
VOLR1 R1 +R2
Noninverting vOUT R2 vIN
VOH
R1
vA + vB
-
VREF R1 +R2 R2
vOUT
vIN
+ - VREF
R1 V R2 OL
VOL R1 V R2 OH
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-3
COMPARATORS WITH INTERNAL FEEDBACK Cross-Coupled Bistable VDD M3
M10 M11
M4
M8
M6
M1
M2 vO
BIAS
M5
M9
M7 VSS
(1). Positive feedback gives hysteresis. (2). Also speeds up the propagation delay time.
1.0V -600m
2.0V
3.0V
4.0V
5.0V
6.0V
-400m
-200m
0m
200m
EXAMPLE 7.4-1 COMPARATOR WITH HYSTERESIS
400m
600m
Allen and Holberg - CMOS Analog Circuit Design Page VII.6-4
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-5
AUTO ZEROING OF VOLTAGE COMPARATORS Model of the Comparator Including Offset
+ -
IDEAL
+
-
VOS
Auto Zero Scheme-First Half of Cycle
+ -
+
IDEAL
-
CAZ
VOS
Auto Zero Scheme-Second Half of Cycle
VIN
+ VOS
+
+ VOS
IDEAL
-
+ 0V -
+ V - OS
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-6
GENERALIZED AUTO ZERO CONFIGURATION φ1
φ2
vIN+
+ φ1
vIN-
φ2
VOS
+ CAZ
IDEAL +
-
-
VOS
φ1
Good for inverting or noninverting when the other terminal is not on ground.
vOUT
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-7
Noninverting Auto-Zeroed Comparator φ1
φ2
φ1
vOUT
vIN
+ CAZ φ2
φ1
Inverting Auto-Zeroed Comparator φ1
φ2 CAZ -
vIN
vOUT φ1
Use nonoverlapping, two-phase clock.
+
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-1
VII.6 - HIGH SPEED COMPARATORS Concept Question: For a given input change, what combination of first-order openloop comparators and a latch gives minimum propagation delay?
+ vIN -
C1
C2
C3
Cn
Latch
Q Q
n first-order, open-loop comparators with identical gains, A Concept: voltage High Output Level
∆ = input voltage change Latch
n tn-1 )e-t/τ ] v out = A [1 - (1 + (n-1)! ∆
vout = e t/τ ∆
A5∆ A4∆ A3 ∆
5 4 3 v out = A2[1 - (1 + t)e -t/τ )] ∆ 2 v out = A(1-e-t/τ )∆ n=1
A2∆ A∆ t3
tL
Time
Propagation delay time = t3 + tL for n=3 Answer: tp(min) occurs when n=6 and A=2.72=e Implementation: n=3 and A≈6 gave nearly the same result with less area. [Ref: Doernberg et al., “A 10-bit 5 MSPS CMOS Two-Step FLASH ADC”JSSC April 1989 pp 241249]
Allen and Holberg - CMOS Analog Circuit Design
Page VII.6-2
HIGH SPEED COMPARATORS-CONT'D Conceptual Implementation-
+ vIN -
+
-
+
-
+
-
-
+
-
+
-
+
Latch
Q Q
VDD
Q FB
Reset Q FB
VB1
Offset and level shifting-
vIN
VB2 VSS
vIN-VOS + -
_
+ VOS
LATCH
-
+
Allen and Holberg - CMOS Analog Circuit Design
Page VII.7-1
VII.7 - COMPARATOR SUMMARY • Key performance parameters: Propagation time delay Resolving capability Input common mode swing Input offset voltage • Types of comparators: Open loop Regenerative Open loop and regenerative Charge balancing • Open loop comparator needs differential input and second stage • Systemative offset error is offset (using perfectly matched transistors) that is due to current mirror errors. • For fast comparators, keep all node swings at a minimum except for the output (current comparators?). • Key design equations: iD =
KW 2 2L (vGS-VT) ,
vDS(sat) =
2iD K(W/L) , and gm =
2KWID L
• Positive feedback is used for regenerative comparators. • Use autozeroing to remove offset voltages (charge injection is limit). • Fastest comparators using low-gain, fast open loop amplifiers cascaded with a latch.
Allen and Holberg - CMOS Analog Circuit Design
Page VII.0-1
VIII. SIMPLE CMOS OPERATIONAL AMPLIFIERS (OP AMPS) AND OPERATIONAL TRANSCONDUCTANCE AMPLIFIERS (OTA'S) Contents VIII.1 VIII.2 VIII.3
Design Principles OTA Compensation Two-Stage CMOS OTA Design
Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Op Amps
Chapter 9 High Performance OTA's
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Allen and Holberg - CMOS Analog Circuit Design
Op Amp Characteristics Non-ideal model for an op amp V
1
R
I b2
icm
CMRR
2
e V
R
2
2
V os V
n
R
I n
id
C
Ideal
id +
1
R
icm
I b1
Boundary Conditions Process Specification Supply Voltage Supply Current Temperature Range Typical Specifications Gain Gainbandwidth Settling Time Slew Rate Input CMR CMRR PSRR Output Swing Output Resistance Offset Noise Layout Area
Requirement See Tables 3.1-1 and 3.1-2 +5 V ±10% 100 µA 0 to 70°C ≥ 80 dB ≥ 10 MHz ≤ 0.1 µsec ≥ 2 V/µsec ≥ ±2 V ≥ 60 dB ≥ 60 dB ≥ 2 VP-P Capacitive load only ≤ ±5 mV ≤ 50nV/ Hz at 1KHz ≤ 10,000 square µm
out
Allen and Holberg - CMOS Analog Circuit Design
Frequency Response Av0 A v (s) = s s s (p − 1 ) (p − 1 ) (p − 1 ) . . . 1 2 3
-6dB/oct
Gain, dB
GB
0 dB
ω1
Frequency
180
Phase (degrees) 90 Phase margin 0 Frequency
-90
ω2
ω3
Allen and Holberg - CMOS Analog Circuit Design
Power supply rejection ratio (PSRR): vout vin (v ps =0) A vd(s) ∆VDD PSRR = · Avd(s) = Aps(s) = vout ∆vOUT vps (v in=0) Common-mode input range (ICMR). Maximum common mode signal range over which the differential voltage gain of the op amp remains constant. Maximum and minimum output voltage swing. Slew rate: ∆vOUT Slew rate = max ∆t
10V
5V Output Voltage 0V
-5V
-10V 0µs
Input Voltage 2µs
4µs Time
6µs
8µs
10µs
Allen and Holberg - CMOS Analog Circuit Design
Settling Time 1.4 1.2
Upper tolerance 1
Vout(t)
Lower tolerance
0.8 0.6
Settling time
0.4 0.2 0 0
2
4
6
Time (sec)
8
10
12
14
Allen and Holberg - CMOS Analog Circuit Design
Design Approach
Design
Specifications
Iterate
Analysis Simulation
Modify
Specifications: • Gain
• Bandwidth
• Output voltage swing
• PSRR
• Settling time
• CMRR
• Power dissipation
• Noise
• Supply voltage
• Common-mode input range
• Silicon area
Allen and Holberg - CMOS Analog Circuit Design
Design Strategy
The design process involves two distinct activities: Architecture Design • Find an architecture already available and adapt it to present requirements • Create a new architecture that can meet requirements Component Design • Design transistor sizes • Design compensation network If available architectures do not meet requirements, then an existing architecture must be modified, or a new one designed. Once a satisfactory architecture has been obtained, then devices and the compensation network must be designed.
Allen and Holberg - CMOS Analog Circuit Design
Op Amp Architecture VDD
M3
M4 M6
M1
M2
-
+
Compensation
IBias M5
M7
vOUT
Allen and Holberg - CMOS Analog Circuit Design
Compensation In virtually all op amp applications, feedback will be applied around the amplifier. Therefore, stable performance requires that the amplifier be compensated. Essentially we desire that the loop gain be less than unity when the phase shift around the loop is greater than 135˚
β
+ IN
Σ
A
OUT A = IN 1 + Aβ Goal: 1 + Aβ > 0 Rule of thumb: arg[Aβ] < 135˚ at mag[Aβ] = 1
OUT
Allen and Holberg - CMOS Analog Circuit Design
Graphical Illustration of Stability Requirements β=1
|Aβ| (dB)
-6dB/oct
GB 0 dB
ω1
Frequency
180o Arg[Aβ] 90o
0o Frequency
ω2
-12dB/oct
Allen and Holberg - CMOS Analog Circuit Design
Step Response of Two-Pole System Impact of placing ω2 at different locations:
stability Date/Time run: 04/08/97 19:45:36
Temperature: 27.0
1.5V
3 2 1 1.0V
ω1 = 1000 rps Case 1: ω2 = 1 x 106 rps
0.5V
Case 2: ω2 = 0.5 x 106 rps Case 3: ω2 = 0.25 x 106 rps 0V 0s
5us
10us
v(5) Time
15us
20us
Allen and Holberg - CMOS Analog Circuit Design
Types of Compensation 1. Miller - Use of a capacitor feeding back around a high-gain, inverting stage. • Miller capacitor only • Miller capacitor with an unity-gain buffer to block the forward path through the compensation capacitor. Can eliminate the RHP zero. • Miller with a nulling resistor. Similar to Miller but with an added series resistance to gain control over the RHP zero. 2. Self compensating - Load capacitor compensates the op amp (later). 3. Feedforward - Bypassing a positive gain amplifier resulting in phase lead. Gain can be less than unity.
Allen and Holberg - CMOS Analog Circuit Design
Miller Compensation VDD
M3
M4
CM
M1
M2
-
M6
Cc
+
v OUT
IBias
C1 CL
M7
M5
Small-signal model 1 gds2+g ds4 -v gm1 in 2
CM r ds1
+ v1
-
gm4v 1 1 gm3
1 gds6 +gds7
Cc
+ v g m2 in 2
+
v2
C1
-
CL vout
gm6 v2
-
Simplified small-signal model 1 gds2+gds4
+ vin
-
1 gds6+gds7
Cc
+ gm1vin
v2
-
+ C1
gm6v2
CL vout
-
Allen and Holberg - CMOS Analog Circuit Design
Analysis (gmI)(gmII)(RI)(RII)(1 - sCc/gmII) Vo(s) = Vin(s) 1 + s[RI(C1 + Cc) + RII(CL + Cc) + gmIIRIRIICc] + s2RIRII[C1CL + Cc(C1+ CL)] p1 ≅ g
-1 mII RI RII Cc
-gmIICc p2 ≅ C C + C C + C C 1 L L c 1 c p2 ≅
-gmII CL
z1 =
gmII Cc
where gmI = gm1 = gm2
RI = g
1 ds2+gds4
gmII = gm6
RII = g
1 ds6+gds7
Allen and Holberg - CMOS Analog Circuit Design
Miller Compensation
β=1 |Aβ| (dB)
-6dB/oct Before compensation
GB
After compensation ω1
Frequency
ω2
-12dB/oct
180o Before compensation Arg[Aβ] 90o
0o
After compensation Phase margin Frequency
Allen and Holberg - CMOS Analog Circuit Design
Conditions for Stability • Unity-gainbandwith is given as: 1 gmI = C c gmIIRIRIICc
GB = Av(0)·|p1| = ( gmIgmIIRIRII) ·
• The requirement for 45° phase margin is: ω ω ω - tan-1 - tan-1 = 45° z |p1| |p2|
Arg[Aß] = ±180° - tan-1
Let ω = GB and assume that z ≥ 10GB, therefore we get, GB GB GB = 45° - tan-1 - tan-1 z |p1| |p2|
±180° - tan-1 or
GB GB 135° ≈ tan-1(Av(0)) + tan-1 + tan-1(0.1) = 90° + tan-1 + 5.7° |p2| |p 2 | GB GB 39.3° ≈ tan -1 | ⇒ |p | = 0.818 ⇒ |p 2 | ≥ 1.22GB |p2 2
• The requirement for 60° phase margin: | p 2 | ≥ 2.2GB if z ≥ 10GB
Allen and Holberg - CMOS Analog Circuit Design
• If 60° phase margin is required, then the following relationships apply: gmII 10gmI Cc > Cc
⇒
g m II > 10g m I
Furthermore, gmII 2.2gmI C2 > Cc which after substitution gives: C c > 0.22C 2 Note: gmI = gm1 = gm2
and
gmII = gm6
Allen and Holberg - CMOS Analog Circuit Design
Phase margin = 45 degrees Phase margin = 60 degrees
Parasitic pole, ω2, held constant while dominant pole, ω1, is moved.
Allen and Holberg - CMOS Analog Circuit Design
Eliminating RHP Zero M3
M4
M1
M6
Cc
M2 RZ
V OUT
CII Vbias
M7 M5
RZ + vin
-
Cc
+ gmI vin
v2
+ RI
CI
-
gmII v2
VI sCc gmIVin + R + sCIVI + (VI − Vo) = 0 I 1 + sCcRz Vo sCc gmIIVI + R + sCIIVo + (Vo − VI) = 0 II 1 + sCcRz These equations can be solved to give Vo(s) a{1 − s[(Cc/gmII) − RzCc]} Vin(s) = 1 + bs + cs2 + ds3 where a = gmIgmIIRIRII b = (CII + Cc)RII + (CI + Cc)RI + gmIIRIRIICc + RzCc
RII
CII vout
-
Allen and Holberg - CMOS Analog Circuit Design c = [RIRII(CICII + CcCI + CcCII) + RzCc(RICI + RIICII)] d = RIRIIRzCICIICc If Rz is assumed to be less than RI or RII and the poles widely spaced, then the roots are p1 ≅
p2 ≅
−1 −1 ≅g R RC (1 + gmIIRII)RICc mII II I c −gmIICc
−gmII ≅ C CICII + CcCI + CcCII II
−1 p3 = R C z I and z1 =
1 Cc(1/gmII − Rz)
By setting Rz = 1/gmII The RHP zero moves to infinity
Allen and Holberg - CMOS Analog Circuit Design
Implementing Compensation Resistor M3
M4
M1
M2
M6
MZ
Cc V OUT
CII Vbias
M7 M5
Allen and Holberg - CMOS Analog Circuit Design
Two-Stage Operational Amplifier Design
VDD M6 M3
M4
Cc vout
vin +
M1
CL
M2
+ VBias -
M7
M5
VSS Figure 6.3-1 Schematic of an unbuffered, two-stage CMOS op amp Important with relationships: an n-channel input pair. gm1 = gm2 = gmI, gm6 = gmII, gds2 + gds4 = GI, and gds6 + gds7 = GII. Slew rate SR =
I5 Cc
First-stage gain Av1 =
(1) gm1 gds2 + gds4
Second-stage gain Av2 = Gain-bandwidth GB =
Output pole p2 = RHP zero z1 =
=
gm6 gds6 + gds7
2gm1 I5(λ 2 + λ 4) =
gm6 I6(λ 6 + λ 7)
gm1 Cc
(3)
(4)
−gm6 CL
(5)
gm6 Cc
Positive CMR Vin(max) = VDD −
(2)
(6) I5
β3
− |VT03|(max) + VT1(min))
(7)
Allen and Holberg - CMOS Analog Circuit Design I5
Negative CMR Vin(min) = VSS +
Saturation voltageVDS(sat) =
β1
+ VT1(max) + VDS5(sat)
2IDS
(8)
(9)
β
All transistors are in saturation for the above relationships. The following design procedure assumes that specifications for the following parameters are given. 1. 2. 3. 4. 5. 6. 7.
Gain at dc, Av(0) Gain-bandwidth, GB Input common-mode range, ICMR Load Capacitance, CL Slew-rate, SR Output voltage swing Power dissipation, Pdiss
Choose a device length to establish of the channel-length modulation parameter λ. Design the compensation capacitor Cc. It was shown that placing the loading pole p2 2.2 times higher than the GB permitted a 60° phase margin (assuming that the RHP zero z1 is placed at or beyond ten times GB). This results in the following requirement for the minimum value for Cc. Cc > (2.2/10)CL Next, determine the minimum value for the tail current I5, based upon slew-rate requirements. Using Eq. (1), the value for I5 is determined to be I5 = SR (Cc) If the slew-rate specification is not given, then one can choose a value based upon settlingtime requirements. Determine a value that is roughly ten times faster than the settling-time specification, assuming that the output slews approximately one-half of the supply rail. The value of I5 resulting from this calculation can be changed later if need be. The aspect ratio of M3 can now be determined by using the requirement for positive input common-mode range. The following design equation for (W/L)3 was derived from Eq. (7). S3 = (W/L)3 =
I5 (K'3) [VDD − Vin(max) − |VT03|(max) + VT1(min)]2
If the value determined for (W/L)3 is less than one, then it should be increased to a value that minimizes the product of W and L. This minimizes the area of the gate region, which
Allen and Holberg - CMOS Analog Circuit Design in turn reduces the gate capacitance. This gate capacitance will affect a pole-zero pair which causes a small degradation in phase margin. Requirements for the transconductance of the input transistors can be determined from knowledge of Cc and GB. The transconductance gm2 can be calculated using the following equation gm1 = GB(Cc) The aspect ratio (W/L)1 is directly obtainable from gm1 as shown below g2m1 S1 = (W/L)1 = (K' )(I ) 2 5 Enough information is now available to calculate the saturation voltage of transistor M5. Using the negative ICMR equation, calculate VDS5 using the following relationship derived from Eq. (8). I5 VDS5 = Vin(min) − VSS − β 1
1/2
− VT1(max)
If the value for VDS5 is less than about 100 mV then the possibility of a rather large (W/L)5 may result. This may not be acceptable. If the value for VDS5 is less than zero, then the ICMR specification may be too stringent. To solve this problem, I5 can be reduced or (W/L)1 increased. The effects of these changes must be accounted for in previous design steps. One must iterate until the desired result is achieved. With VDS5 determined, (W/L)5 can be extracted using Eq. (9) in the following way S5 = (W/L)5 =
2(I5) K'5(VDS5)2
For a phase margin of 60°, the location of the loading pole was assumed to be placed at 2.2 times GB. Based upon this assumption and the relationship for |p2| in Eq. (5), the transconductance gm6 can be determined using the following relationship gm6 = 2.2(gm2)(CL/Cc) Since S3 is known as well as gm6 and gm3, assuming balanced conditions, gm6 S6 = S3 g m3 I6 can be calculated from the consideration of the “proper mirroring” of first-stage the current mirror load of Fig. 6.3-1. For accurate current mirroring, we want VSD3 to be equal to VSD4. This will occur if VSG4 is equal to VSG6. VSG4 will be equal to VSG6 if
Allen and Holberg - CMOS Analog Circuit Design (W/L)6 S 6 I6 = (W/L) I1 = S I1 4 4 Choose the larger of these two values for I6 (Eq. 19 or Eq. 20). If the larger value is found in Eq (19), then (W/L)6 must be increased to satisfy Eq. (20). If the larger value is found in Eq. (20), then no other adjustments must be made. One also should check the power dissipation requirements since I6 will most likely determine the majority of the power dissipation. The device size of M7 can be determined from the balance equation given below I6 I6 S7 = (W/L)7 = (W/L)5 = S5 I5 I5 The first-cut design of all W/L ratios are now complete. Fig. 6.3-2 illustrates the above design procedure showing the various design relationships and where they apply in the two-stage CMOS op amp.
Max. ICMR and/or p3 VSG4 -
M3
vin +
VDD
+
+
VSG6 -
M4
Cc
Cc ≈ 0.2CL (PM = 60°)
M2
Min. ICMR
I5
+ VBias -
M5
M6 I6
g GB = m1 Cc
M1
Vout(max)
I5 = SR·Cc
gm6 or Proper Mirroring VSG4=VSG6 vout CL
Vout(min)
M7 VSS
Figure 6.3-2 Illustration of the design relationships and the circuit for a two-stage CMOS op amp. At this point in the design procedure, the total amplifier gain must be checked against the specifications. Av =
(2)(gm2)(gm6) I5(λ2 + λ3)I6(λ6 + λ7)
If the gain is too low, a number of things can be adjusted. The best way to do this is to use the table below, which shows the effects of various device sizes and currents on the
Allen and Holberg - CMOS Analog Circuit Design different parameters generally specified. Each adjustment may require another pass through this design procedure in order to insure that all specifications have been met. Table 6.3-2 summarizes the above design procedure.
Dependencies of device performance on various parameters
Increase DC Gain Increase GB Increase RHP Zero Increase Slew Rate Increase CL
Drain Current I5 I7 (↓)1/2 (↓)1/2
M1 and M2 W/L L (↑)1/2 ↑
(↑)1/2
(↑)1/2 (↑)1/2
↑
M3 and M4 W L ↑
Inverter Inverter Comp. Load Cap.. W6/L6 W 7 L7 Cc ↑ (↑)1/2 (↑)1/2
↓ ↓ ↓ ↓
Allen and Holberg - CMOS Analog Circuit Design
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. 1. 2.
Choose the smallest device length which will keep the channel modulation parameter constant and give good matching for current mirrors. 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
3.
Determine the minimum value for the “tail current” (I5) from the largest of the two values. I5 = SR .Cc VDD + |VSS| I5 ≅ 10 . 2 Ts
4.
Design for S3 from the maximum input voltage specification. S3 =
5.
I5 K'3[VDD − Vin(max) − |VT03|(max) + VT1(min)]2
≥1
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.
6.
Design for S1 (S2) to achieve the desired GB. 2
gm1 = GB . Cc ⇒ S2 = 7.
gm2 K'2I5
Design for S5 from the minimum input voltage. First calculate VDS5(sat) then find S5. VDS5(sat) = Vin(min) − VSS − S5 =
2I5 K'5[VDS5(sat)]2
I5
β1
− VT1(max) ≥ 100 mV
Allen and Holberg - CMOS Analog Circuit Design 8.
Find gm6 and S6 by the relationship relating to phase margin, load, and compensation capacitors, and the balance condition. gm6 = 2.2gm2(CL/Cc) gm6 S6 = S3 g m3
9.
Calculate I6 : I6 = (S6/S4)I4 = (S6/S4)(I5/2)
10. Design S7 to achieve the desired current ratios between I5 and I6. S7 = (I6/I5)S5 11. Check gain and power dissipation specifications. Av =
2gm2gm6 I5(λ2 + λ3)I6(λ6 + λ7)
Pdiss = (I5 + I6)(VDD + |VSS|) 12. 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 have been 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. 13. Simulate the circuit to check to see that all specifications are met.
Allen and Holberg - CMOS Analog Circuit Design
Example: Design of a Two-Stage Op Amp Using the material and device parameters given in Tables 3.1-1 and 3.1-2, design an amplifier similar to that shown in Fig. 6.3-1 that meets the following specifications. Assume the channel length is to be 1µm. Av > 3000V/V
VDD = 2.5V
VSS = -2.5V
GB = 5MHz
CL = 10pF
SR > 10V/µs
Vout range = ±2V
ICMR = -1 to 2V
Pdiss ≤ 2mW
Solution Calculate the minimum value of the compensation capacitor Cc, Cc > (2.2/10)(10 pF) = 2.2 pF Choose Cc as 3pF. Using the slew-rate specification and Cc calculate I5. I5 = (3 × 10-12)(10 × 106) = 30 µA Next calculate (W/L)3 using ICMR requirements. (W/L)3 =
30 × 10-6 = 15 (50 × 10-6)[2.5 − 2 − .85 + 0.55]2
gm3 =
2 × 50x10-6 x15x10-6 × 15 = 150µS
Therefore (W/L)3 = (W/L)4 = 15 Check the value of the mirror pole, p3, to make sure that it is in fact greater than 10GB. Assume the Cox = 0.4fF/µm2. The mirror pole can be found as -gm3 - 2K’pS3I3 p3 ≈ 2C = 2(0.667)W L C = 15.75x109(rads/sec) gs3 3 3 ox or 2.98 GHz. Thus, p3, is not of concern in this design because p3 >> 10GB. The next step in the design is to calculate gm1 gm1 = (5 × 106)(2π)(3 × 10-12) = 94.25µS Therefore, (W/L)1 is gm12 (94.25)2 (W/L)1 = (W/L)2 = 2K’ I = 2·110·15 = 2.79 ≈ 3.0 N 1 Next calculate VDS5
Allen and Holberg - CMOS Analog Circuit Design
VDS5 = (−1) − (−2.5) −
30 × 10-6 - .85 = 0.35V 110 × 10-6·3
Using VDS5 calculate (W/L)5 from Eq. (16) (W/L)5 =
2(30 × 10-6) = 4.49 ≈ 4.5 (50 × 10-6)(0.35)2
From Eq. (20) of Sec. 6.2, we know that gm6 ≥ 10gm1 ≥ 942.5µS Assuming that gm6 = 942.5µS (W/L)6 = 15
942.5 × 10-6 = 94.25 150 × 10-6
Using the equations for proper mirroring, I6 is determined to be I6 = (15 × 10-6)(94.25/15) = 94.25 µA Finally, calculate (W/L)7 94.25 × 10-6 (W/L)7 = 4.5 ≈ 14.14 30 × 10-6 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 Vmin(out) = VDS7(sat) =
2 × 94.25 = 0.348V 110 × 14.14
which is much less than required. At this point, the first-cut design is complete. Examining the results shows that the large value of M7 is due to the large value of M5 which in turn is due to a tight specification on the negative input common mode range. To reduce these values the specification should be loosened or a different architecture (i.e. p-channel input pair) examined. Now check to see that the gain specification has been met Av =
(2)(94.25 × 10-6)(942.5 × 10-6) = 19,240 30 × 10-6(.04 + .05)38 × 10-6(.04 + .05)
which meets specifications.
Allen and Holberg - CMOS Analog Circuit Design
IX. HIGH PERFORMANCE CMOS AMPLIFIERS Contents IX.1 IX.2 IX.3 IX.4 IX.5 IX.6 IX.7
Improving The Two-Stage Architecture Two-stage Cascode Architecture Folded Cascode Architecture Differential Output Architecture (Class AB) Low power amplifiers Dynamically biased amplifiers
Organization Chapter 10 D/A and A/D Converters
Chapter 11 Analog Systems
SYSTEMS
Chapter 7 CMOS Comparators
Chapter 8 Simple CMOS Op Amps
Chapter 9 High Performance OTA's
COMPLEX
CIRCUITS Chapter 5 CMOS Subcircuits
Chapter 6 CMOS Amplifiers
SIMPLE
Chapter 2 CMOS Technology
DEVICES
Chapter 3 CMOS Device Modeling
Chapter 4 Device Characterization
Allen and Holberg - CMOS Analog Circuit Design
IX.1 IMPROVING THE TWO-STAGE ARCHITECTURE Amplifiers Using an MOS Output Stage PUSH-PULL CMOS OTA This amplifier is a simple extension of the seven-transistor OTA studied in Section 8. VDD
M9
VDD
M7
VDD vOUT
M1 -
VDD
M2 +
Cc CL
M8 M6 M4
M3
small-signal equivalent circuit: Cc
vout
+ gm1 vi
C1
r1
v1
gm6 v1
CL
r2
-
1 1 where gm1 = gm2, r1 = g , r = 2 g ds6 + g ds7 ds2 + g ds4
gm1 Ai vi 2
Allen and Holberg - CMOS Analog Circuit Design
Amplifiers Using an MOS Output Stage - Continued
Network equations: [g1 + s(C1 + CL)]v1 - sCcv2 = gm1vi gm1AIvi 2 i7 AI is the current gain from M1 to M7: AI = i 1 -g m6 z= AI Cc 1 2 -g1g2 p1 ≈ gm6Cc -g m6 p2 ≈ C L gm1gm6 AV ≈ g g 1 2 [g5 + sCc]v1 + [g2 + s(Cc + CL)]v2 =
To guarantee that the zero stays in the left-half plane, AI > 2
Allen and Holberg - CMOS Analog Circuit Design
Amplifiers Using an MOS Output Stage - Continued
Example: VDD
VDD
VDD
M9
-
M7
M1
VDD
M2
VDD
M9
VDD
+
M7
M1
-
Cc
+
M2
Cc CL
CL
M8 M6 M3
M6
M4
M4
M3
Push-Pull (AI ≈ 3)
Standard (AI = 0)
140 120 100 Gain(db)
80 Push-Pull phase
60
Standard gain 40 Push-Pull gain
20 Standard phase
0 -20 1
10
2
10
3
10
VDD
4
10
5
10
6
10
Frequency
7
10
8
10
9
10
Allen and Holberg - CMOS Analog Circuit Design
IX.2 Two-Stage Cascode Architecture Why Cascode Op Amps? • Control the frequency behavior • Increase PSRR • Simplifies design
Where is the Cascode Technique Applied? • First stage Good noise performance Requires level translation to second stage Requires Miller compensation • Second stage Self compensating Reduces the efficiency of the Miller compensation Increases PSRR
Allen and Holberg - CMOS Analog Circuit Design
Power Supply Rejection Ratio (PSRR)
Definition: + vdd + vin
+
-
-
+
+
VDD vout
-
vss -
VSS
-
vout vin (vdd=0) Av(vdd=0) + PSRR = A (v =0) = v dd in out vdd (vin=0)
Calculation of PSRR: + vdd
+
Addvdd
VDD
v1
-
-
vout +
VSS
=>
vout
v2
Av (v1 -v2 )
K
vout = Addvdd + A v(v1-v2) = A ddvdd - Avvout vout(1 + Av) = A ddvdd
Extends bandwidth beyond GB ↓
vout Add A dd 1 = ≈ = vdd 1 + Av Av PSRR+
vout KAdd A dd = ≈ vdd 1+KAv Av
Allen and Holberg - CMOS Analog Circuit Design
Intuitive Interpretation of Positive PSRR for the Two-Stage OTA
+
+ vdd
VDD
-
-
1.) The M7 current sink causes VGS6 to act like a battery.
M3
2.) Therefore, vdd
M4
couples from the source
M6 M1
to gate of M6.
Cc
M2
-
vout
+
3.) The path to the output is through any capacitance
M7
M5
from gate to drain of M6.
+ VBias
VSS
+ -
4.) Resultant circuit
model-
+ vdd
vout Cc
Rout
-
Must reduce C c !
Vout Vdd 0dB
-60 to -80dB
1 Rout Cc
ω
Other sources of PSRR beside C c
Allen and Holberg - CMOS Analog Circuit Design
Intuitive Interpretation of the Negative PSRR for the Two-Stage OTA +
VDD
-
M3
M4 M6
M1
Cc
M2
-
+
vout M7
M5
+ VBias
+
?
vss VSS
+-
Two mechanisms of vss injection: iss
M5 or M7
+
+ VBias
-
M5
+ vss VSS
VBias
-
+-
vss
-
VSS
Transconductance injection
+ -+ -
Capacitance injection
Allen and Holberg - CMOS Analog Circuit Design
Intuitive Interpretation of the Negative PSRR for the Two-Stage OTA Continued Transconductance injection: Vout Vss
Path through the input stage: Not important as long as CMRR is high.
20 to 40 dB
Path through the output stage: vout ≈ issRout = gm7 vssRout 0dB
1 Rout Cout
vout vss = g m7Rout Frequency dependence 1 Rout → Rout|| sCout Capacitance injection: + vss
vout Cgd7
Rout
Vout Vss 0dB
-60 to -80dB
Reduce Cgd7!
1 Rout C gd7
ω
First stage transconductance injection
ω
Allen and Holberg - CMOS Analog Circuit Design
Problems with the two-stage OTA: • Insufficient gain • Poor stability for large load capacitance • Poor PSRR These problems can be addressed using various cascode structures.
We will consider several approaches: • Cascoding the first stage • Cascoding the second stage • Folded cascode
Allen and Holberg - CMOS Analog Circuit Design
First Stage Cascode VDD
M3
VDD
VBP
M4
VDD
MC2
VDD
VBN MC3
VO1
MC1
+
M1
M2
VBIAS VSS
ro1 ≈ (gmc2rdsc2)rds4 || (gmc1rdsc1)rds2 Gain ≈ gm2ro1
• Overall gain increased by ≈
gmcrdsc 2
• Requires voltage translation to drive next stage • Requires additional biasing for cascode devices • Common-mode problem at drains of M1 and M2
Allen and Holberg - CMOS Analog Circuit Design
First Stage Cascode - Continued Common-mode improvement: VDD
M3
VDD
M4
VDD
ICM VBP
MC2
MC3
VDD
MC1
M9
+
VO1
-
M1
M2
VBIAS VSS
Common-mode circuitry (M9) maintains Vds of M1 and M2 A V = gm1ro1 ro1 ≈ (gmc2rdsc2)rds4 || (gmc1rdsc1)rds2 -1 p1 ≈ C r L o1 GB ≈ A V|p1| ≈
gm1 CL
Output range of this amplifier is poor when used by itself. It needs an output stage to be practical.
Allen and Holberg - CMOS Analog Circuit Design
First Stage Cascode - Continued Implementation of ICM VDD
M4
M3
ICM
ICM VBP
VDD MC2 VSS VSS
VSS MC1
MC3 vin 2
-vin vin 2 2 VSS
VSS
-vin 2
VSS M2
M1
I5 +ICM
VSS
Allen and Holberg - CMOS Analog Circuit Design
Level Translator for First Stage Cascode
VDD M4 MT2 M6
MC2 MT1 MC1
Vo VSS
M2 VBIAS M5 VSS
M7
Allen and Holberg - CMOS Analog Circuit Design
Improved PSRR For Two-Stage OTA
Use cascode to reject Cc feedforward VDD
M3
M6
M8
VB1
-
M4
Cc
M9
+
M2
M1
vOUT
M5 VB2
M7 VSS
+PSRR is reduced by M9
Disadvantage Miller pole is larger because R1 ≈
1 gm9
positive input common mode range is restricted
Allen and Holberg - CMOS Analog Circuit Design
Complete Two Stage Cascode
VDD M4
M3
MT2 VBP
M6
MC2 MT1
ICM MC3
MC1
Vo VSS
M9 M1
M2 M7
VBIAS M5 VSS
Allen and Holberg - CMOS Analog Circuit Design
Second Stage Cascode
VDD M3
M4
M6
VBP
Comp
M2
M1
VBN
MC6
Vo
MC5
M7
VBIAS M5 VSS
Allen and Holberg - CMOS Analog Circuit Design
LOAD COMPENSATED CASCODE AMPLIFIER
VDD VDD
M9
M3
M4
VDD M6
VDD
MC2 VBP
MC3 VBP
VDD
-
+ M2
M1 M5
I5
VDD
VDD
MC1 VBN
VSS
VBIAS
CL
M8
M7
VSS
gm 2 A V1 = g m4 1 A V2 = 2 (g m6 + g m9 )R o where
gm2 AV = 2(g ) (g m6 + gm9) Ro m4
Ro ≈ (g mc2rdsc2)rds6 || (gmc1rdsc1)rds7 and M7 = M8 Or, gm1 +g m2 KR o 2
AV = where
K=
W6/L6 W 9/L 9 W4/L4 = W3/L3
Allen and Holberg - CMOS Analog Circuit Design
Design Example Pertinent design equations: iOUT SR = C L AV =
gm2 2(gm4) (gm6 + gm7 ) ro
GB =
g m2 (g m6 + g m7 ) 2(gm4)CL
Vin(max) = VDD -
I5 ß3 - |VT3|(max) + V T1(min)
Vin(min) = VSS + VDS5 + Specifications: VDD = -VSS = 5V SR = 5V/µs into CL = 50pf GB = 5 MHz AV > 5000 CMR = ±3V Output swing = ±3V
I5 ß1 + VT1(max)
Allen and Holberg - CMOS Analog Circuit Design
Design Procedure 1.) Design for maximum source/sink current Isource/sink = C L(SR) = 50pf(5V/µs) = 250 µA 2.) Note that S6 Max. IOUT (source) = S I5 4 Max. IOUT (Sink) = Max. IOUT (source) if S3 = S 4, S9 = S6 and
S7 = S 8
3.) Choose I 5 = 100 µA ∴ S 9 = S 6 = 2.5 S 4 = 2.5 S 3 4.) Design for ± 3V output capability a.) Negative peak Let VDSC1(sat.) = V DS7(sat.) = 1V under negative peak conditions, IC1 = I 7 = 250 µA Divide 2V equally, ∴ 2V =
2I7 + KN'S7
∴ S 7 = S C1 = 29.4
2IC1 =2 KN'SC1
2I7 KN'S7 = 2
-> S 8 = S 7 = 29.4
500 µA 17 µA/V 2 S 7
Allen and Holberg - CMOS Analog Circuit Design
b.) Positive peak, divide voltage equally, VSD6 = VSDC2 = 1V , --> 2V = ∴ S 6 = S C2 = 62.5
2I6 + KP'S6
2IC2 =2 KP'SC2
2I6 KP'S6
--> S 3 = S 4 = 25
5.) Design of VBP and V BN a.) VBN (Assume max. I OUT (sink) conditions) IOUT(sink) = 250 µA MC1 VBN
-3 -5V
VDSC1 = VGSC1 - V TC1 (ignoring bulk effects)
-4
1 = VGSC1 -1 --> VGSC1 = 2V ∴ V BN = -2V
M7 -5
b.) VBP (Assume max. IOUT (source) conditions) +5V
+4
VDSC2 = VGSC2 - |V TC2| (ignoring bulk effects) VSGC2 = 2V ∴ V BP = +2V
VBP MC2
IOUT(source)= 250 µA +3V
Allen and Holberg - CMOS Analog Circuit Design
6.) Check max. Vin influence on S3 (S4) I5 ß3 - |VT03|max + VT1(min)
Vin (max) = VDD +3 = +5 S3 =
100 µA - 1.2 + 0.8 KP'S3
100 µA = 4.88 (Use S 3 = S4 = 25) µA 8 2 (1.6V) 2 V
With S 3 = 25, V in (max) = 3.89V which exceeds the specification. 7.) Find gm1 (gm2) a.) AV specification gm1 g m6 + g m 7 R II AV = gm4 2 gm4 =
2I4Kp'S4 = 141.1 µs
gm6 =
2I6KP'S6 = 353.5 µs
gm7 =
2I7KN'S7 = 353.5 µs
gmc1 = gm7 gmc2 = gm6 1 rds6 = rdsc2 = I λ = 0.4 MΩ 6 P 1 rds7 = rdsc1 = I7λΝ = 0.8 MΩ RII ≈ (gmc1rdsc1rds7) || (gmc2rdsc2rds6) = 45.25 MΩ gm1 707 µs ( 226.24 MΩ || 56.56 MΩ) > 5000 V/V ∴ 141.1 2
∴
gm1 > 44 µs
Allen and Holberg - CMOS Analog Circuit Design
b.) GB specification GB =
g m1 (g m6 + g m7 ) (50pF) = 10π.10 6 rps 2gm4
gm1 =
(10π.106)(141.1.10-6)(50.10-12) = 627 µS 707.10-6/2
gm 2 = 231 ∴ S 1 = S2 = I5K N ' gm1gm6+gm7 627 707µS R II = gm4 2 141.1 2 (45.25MΩ) = 71,080 8.) Find S5 from Vin (min) AV =
Vin (min) = VSS + VDS5 + -3 = -5 + VDS5 +
V DS5 = 0.8 -
100 µA + 1.2 µA 17 2 S 1 V
100 = 0.8 - 0.1596 = 0.641 (17)(231)
VDS5 (sat) = 0.641 =
S5 =
I5 ß1 + VT1(max)
2(100 µA) µA (17 2 )S5 V
2(100µA) = 28.6 17µA/V 2(0.641)2
9.) VBIAS KN'.28.6 I5 = ( V BIAS + 5 -1) 2 = 100 µA 2 VBIAS = 0.411 - 4 = -3.359V
Allen and Holberg - CMOS Analog Circuit Design
10.) Summary of design S1 = S2 = 231 S3 = S4 = 25 S5 = 28.6 S6 = S9 = SC2 = SC3 = 62.5
S7 = S8 = SC1 = 29.4 VBP = 2V VBN = -2V VBIAS = -3.359V
11.) Check on power dissipation Pdiss = 10(I8 + I5 + I7) = 10(125µA + 100µA + 125µA) = 3.5mW 12.) Design W's for lateral diffusion and simulate
Allen and Holberg - CMOS Analog Circuit Design
X.3 FOLDED CASCODE ARCHITECTURE Principle VDD
1.5 I
1.5 I
VBP
M3
M4
I
v+IN
M1 M2
v-IN
vout =gm1 Rout vin I
M5
M6
M7
M8
I
VSS
Currents in upper current sinks must be greater than I to avoid zero current in the cascode mirror (M5-M8). Advantages Good input CMR. Good frequency response. Self compensating.
Allen and Holberg - CMOS Analog Circuit Design
Folded Cascode OP Amp VDD VB3 M4
M3
+
-
VB3
M14
M2
M1
V OUT M9
M8
Cc M16
VB1
M5 M15 VB2
M11
M10
M13
M12 VSS
High gain, High speed, cascode amp GB ≈ 10 MHz, AVDC ≈ 100 dB
Allen and Holberg - CMOS Analog Circuit Design
XI. 4 DIFFERENTIAL OUTPUT OTA'S Implementation Using Two Differential-In, Singled-Ended Op Amps
+ -
-
+
+
R
R -
+
Conceptual Implementation of Differential In-Out OP Amp
VDD
VDD
+
VDD
VBIAS
Vin -
-
Vout
+
CL
CL Common mode feedback
VSS
Allen and Holberg - CMOS Analog Circuit Design
Schematic of a Fully Differential In-Out, FoldedCascode Op Amp
VDD = +5V MP1A VBIAS1
MP1
MP2A VDD
VDD
VBIAS2
MP2 vOUT +
CL +
CL
vIN VSS
-
MN2A
VSS
VBIAS3
MN2 MN1 VSS
VSS
MN1A VSS
VBIAS4
MN3 MN3A VSS = -5V
Allen and Holberg - CMOS Analog Circuit Design
Evolution of Class AB Amplifier VDD
VDD
VB2
v+IN
v-IN
vOUT
vOUT
v+IN
v-IN
VB1
VSS
M12
M11
VB2
v+IN
M1
M2
v-IN
v+IN
M3
M4
v-IN
VB1
M10
M17
M18
VSS
combine
M9
M14
M13
Problem: DC levels of input voltages incompatible
Allen and Holberg - CMOS Analog Circuit Design
M17
M12
M14
M11
M5 v+OUT
M6
v+IN M7
M1
M2
M3
M4
v-IN
v-OUT
M8
I
I M18
M10
M13
M9
DC problem solved, but amplifier has low gain and requires CM feedback
M17
M12
M14
M11 VB3
VB1 M5 v+OUT
M6
v+IN
M1
VB2 M7
M3
M4
VB4
M8
v-OUT M15
I
I M18
v-IN
M2
M16
M10
Gain improved using cascode
M9
M13
Allen and Holberg - CMOS Analog Circuit Design
IX.5 LOW POWER AMPLIFIERS General Objective is to minimize the dc power dissipation. Typical applications are: 1. Battery powered circuits. 2. Biomedical instrumentation. 3. Low power analog "VLSI." Weak Inversion or Subthreshold Operation Drain current qvGS W iD = I D exp nkT ( 1 + lv D S) L
Small signal parameters qiD gm = nkT ,
rds ≈ (λiD )-1
Device characteristics iD
iD
square law
100nA 100nA weak inversion
vGS < VT
1
2
vDS
exponential
VT
vGS
Allen and Holberg - CMOS Analog Circuit Design
Op Amp Operating in Weak Inversion
Consider the two-stage op amp with reduced currents and power supplieds, AV =
gm2 gm6 1 = n2n6( kT/q) 2( l2+l4) ( l6+l7) ( gds2+gds4) ( gds6+gds7)
where, gm1 ID1 GB = C = (n kT/q)C 1
and
2ID1 n1kT SR = C = 2GB q
VDD
M3
M4 M6 C
M1
M2
M7
M5
VSS
Allen and Holberg - CMOS Analog Circuit Design
Design Example
Calculate the gain, unity-gain bandwidth, and slew rate of the previous two-stage op amp used in weak inversion if: ID5 = 200nA
nP = 1.5
λP = 0.02V-1
L = 10 µm
nN = 2.5
λN = 0.01V-1
C = 5pF
T = 27˚C
1 AV = (1.5)(2.5)(0.026)(2)(0.1+0.02)(0.01+0.02) = 5698 GB =
100.10-9 = 307.69Krps or 48.97KHz (2.5)(0.026)(5.10-12)
SR = 2(153.85.103)(2.5)(0.026) = 0.04V/µs If V DD = -V SS = 2.5, the power dissipation is 0.2µW assuming ID7 = I D5.
Allen and Holberg - CMOS Analog Circuit Design
Push-Pull Micropower Op Amp
First stage clamped (low gain, low bias current)VDD M3
M4
M8
M6 M2
M1
CC
VB
M5 M7
M9
VSS Gain enhancement for Push-Pull Micropower Op Amp M11 M3
M10
M12
M4
M8
M6 M2
M1
CC
VB
M5 M13 M7
M9
VSS
Allen and Holberg - CMOS Analog Circuit Design
Push-Pull Cascode Micropower Op Amp
VDD
M3
M4
M8
M6
M1
M2
VB1
VDD M10
VB2 VB
M5
VSS M11 M7
M9
VSS
1 1 + n nN P AV = 2 ≈ 10,000 2 Vt ( λ P n P + λ N 2 n N 2 ) self-compensating Low power 8 bits Requires a high input impedance buffer at output Integral linearity depends on the resistor ratios
111
Input
Allen and Holberg - CMOS Analog Circuit Design
Page X.2-3
3-BIT VOLTAGE SCALING D/A CONVERTER WHICH MINIMIZES THE SWITCHES Require time for the logic to perform
b0
b1
b2
+VREF R/2 8
3-to-8 Decoder
R 7 R 6 R 5 R 4 R 3 R 2 R 1 R/2
vOUT
Allen and Holberg - CMOS Analog Circuit Design
Page X.2-4
Accuracy Requirements of a Voltage Scaling D/A Find the accuracy requirements for the voltage scaling D/A converter as a function of the number of bits N if the resistor string is a 5 micron wide polysilicon strip. If the relative accuracy is 2%, what is the largest number of bits that can be resolved to within ±0.5 LSB? Assume that the ideal voltage to ground across k resistors is kR V k = N VREF 2 R The worst case variation in Vk is found by assuming all resistors above this point in the string are maximum and below this are minimum. Therefore, V k' =
kRminVREF ( 2N-k) R max + kR min
The difference between the ideal and worst case voltages is, Vk Vk' kRmin kR = VREF VREF 2NR ( 2N-k) R max + kR min Assuming that this difference should be less than 0.5 LSB gives, kRmin kR 0.5 < 2NR ( 2N-k) R max + kR min 2N Expressing Rmax as R+0.5∆R and Rmin as R-0.5∆R and assuming the worst case occurs midway in the resistor string where k=0.5( 2N) and assuming that 5 micron polysilicon has a 2% relative accuracy gives, 0.5(R- 0.5∆R) 1 ∆R 1 0 . 5 - 0.5(R + 0.5∆R) + 0.5(R- 0.5∆R) = 4 R < 2 2-N ⇒
∆R 1 < N-1 R 2
or
0.25(0.02) < 0.5( 2-N) ⇒ N = 6
Allen and Holberg - CMOS Analog Circuit Design
Page X.2-5
R-2R LADDER DAC's Configuration: R
A 2R
2R
R
B 2R
R
2R
b1
b0
R
C
2R
b2
2R
b N-1
bN + -
Equivalent circuit at A: R
A
b0 VREF + 2 -
Equivalent circuit at B: R
+ -
b0 VREF 2
R
R
B 2R + b1 VREF -
+
( b4 + b2 )V 0
= -
B 1
REF
Finally, the equivalent circuit at Q:
+ -
(
b0 2N
Q R b1 b2 b + N-1 + N-2 +...+ N-1 VREF 2 2 2
vOUT R -
)
sign bit
+
R
bN
+
VREF
VREF
Allen and Holberg - CMOS Analog Circuit Design
Page X.3-1
X.3 CHARGE SCALING D/A CONVERTER Binary weighted capacitor array: C 2
C
φ
C 2N-2
C 4
C 2N-1
v OUT
C 2
N-1
+
1
SN-1
φ
2
SN-2
φ
2
SN-3
S1
φ
2
φ
S0
2
φ
2
Terminating capacitor
VREF
Operation: 1.) During φ1, all capacitors are discharged. 2.) During φ 2 , capacitors with bi = 1 are connected to VREF and capacitors with bi = 0 are grounded. 3.) The resulting output voltage is, bN-1C b N-2 C/2 b N-3 C/4 b 0 C / ( 2N-1) vOUT = VREF 2C + 2C + 2C + ... + 2C
If Ceq. is defined as the sum of all capacitances connected to VREF, then Ceq. vOUT = 2C VREF
+
Ceq. vOUT
VREF -
2C-Ceq.
Allen and Holberg - CMOS Analog Circuit Design
Page X.3-2
Other Versions of the Charge Scaling D/A Converter Bipolar Operation: Charge all capacitors to V REF. If bi = 1, connect the capacitor to ground, if bi = 0, connect the capacitor to VREF. Will require an extra bit to decide whether to connect the capacitors initially to ground or to VREF.
Four-Quadrant Operation: If VREF can have ±values, then a full, four quadrant DAC can be obtained.
Multiplying DAC: If VREF is an analog signal (sampled and held), then the output is the product of a digital word and an analog signal and is called a multiplying DAC (MDAC).
Allen and Holberg - CMOS Analog Circuit Design
Page X.3-3
Influence of Capacitor Ratio Accurcy on No. of Bits Use the data of Fig.2.4-2 to estimate the number of bits possible for a charge scaling D/A converter assuming a worst case approach and the worst conditions occur at the midscale (1 = MSB). The ideal output of the charge scaling DA converter is, vOUT Ceq. VREF = 2C The worst case output of the charge scaling DA converter is, Ceq.(min) v'OUT VREF = 2C - C eq. (max) + C eq. (min) The difference between the ideal output and the worst case output is, vOUT v 'O U T Ceq.(min) 1 = VREF VREF 2 ( 2C - C eq.) (max) ± C eq.(min) Assuming the worst case condition occurs at midscale, then Ceq. = C ∴
vOUT v 'O U T VREF - VREF
C(min) 1 = 2 - C (max) - C (min)
If C(max) = C + 0.5∆C and C(min) = C - 0.5∆C, then setting the difference between the ideal and worst case to 0.5LSB gives, 0.5( C (max) ± C (min)) - C (min) ≤ 0.5( 1/2N) C (max) + C (min) or 1 C(max) - C (min) ≤ N ( C (max) + C (min)) 2 or ∆C ≤
1 2C ⇒ 2N
∆C 2C
≤ 2-N ⇒
∆C 1 ≤ C 2N-1
A 50µm x 50µm unit capacitor gives a relative accuracy of 0.1% and N = 11 bits. It is more appropriate that the relative accuracy is a function of N. For example, if ∆C/C ≈ 0.001 + 0.0001N, then N=9 bits.
Allen and Holberg - CMOS Analog Circuit Design
Page X.3-4
Increasing the Number of Bits for a Charge Scaling D/A Converter Use a capacitive divider. For example, a 13-bit DACLSB Array
MSB Array -
1.016pF (Attenuating capacitor)
+ 1pF
2pF
8pF
4pF
16pF
32pF 1pF
2pF
4pF
8pF
16pF
32pF
v OUT
64pF
1pF b0
b1 b0
b2 b1
b3 b2
b4 b3
b5 b4
b6 b6
b5
b8
b7
b9
b7
b8
b 11
b 10 b9
b 10
b12 b 11
b 12
+VREF -VREF
5
VL =∑ i=0
±b iCi VREF 64
12
±bi Ci VREF
i=6
127
VR = ∑
An equivalent circuit1 64 1 + C = 1 ⇒ C = 63 ≈ 1.016 64
+ 1.016pF 64pF
vOUT
127pF
+
+
VR
VL -
-
1 127 = 128 V R + 128 VL
-
∑ 12
VR =
±biVREFCi 127
∑ 5
and VL=
i=6
or
63 1 1 + 64 64 127 = 1 V + R 63 1 63 1 VL 1 + 64 + 127 + 64 + 127 64 64
12 ±VREF ∑ vOUT = 128 i=6 b iC i
±biVREFCi 64
i=0
biCi + ∑ 64 5
i=0
Allen and Holberg - CMOS Analog Circuit Design
Page X.3-5
Removal of the Amplifier Input Capacitance Effects Use the binary weighted capacitors as the input to a charge amplifier. Example of A Two-Stage Configuration:
VREF φ b b 1 0 0
φ b b φ b b φ b b φ b b φ b b 1 1 1 1 2 2 1 3 3 1 1 5 5 4 4
φ b b 1 6 6
φ b b 1 7 7
C 8
C 4
C 2
C 4
C 2
C 8
C
C 8
φ
1
2C
+
v OUT
C
Allen and Holberg - CMOS Analog Circuit Design
Page X.4-1
X.4 - VOLTAGE SCALING-CHARGE SCALING DAC'S VREF
R(2M-1) SF v OUT
R(2M-2)
R(2M-3)
Ck-1
C k-2
2k-1 C
k-2
2
C1 2C
C0 C
C
C
M=4
k=8 SA R1 SB
A S(k-1)A
S(k-2) A
S1A
S0A
S(k-1)B
S(k-2) B
S1B
S0B
R0
B
Advantages: • Resistor string is inherently monotonic so the first M bits are monotonic. • Can remove voltage threshold offsets. • Switching both busses A and B removes switch imperfections. • Can make tradeoffs in performance between the resistors and capacitors. • Example with 4 MSB's voltage scaling and 8 LSB's charge scaling:
Allen and Holberg - CMOS Analog Circuit Design
Page X.4-2
Voltage Scaling, Charge Scaling DAC - Cont'd Operation: 1.) SF, SB, and S1B through Sk,B are closed discharging all capacitors. If the output of the DAC is applied to any circuit having a nonzero threshold, switch SB could be connected to this circuit to cancel this threshold effect. 2.) Switch SF is opened and buses A and B are connected across the resistor whose lower and upper voltage is V'REF and V'REF + 2-MVREF respectively, where b1 b2 bM - 1 b0 V'REF = V ref M + M-1 + M-2 + ···· + 2 2 2 21 + 2k-1 C
REF
-
B
2C
C
C
vOUT
A
+ 2-MV
2k-2 C
S k,A
S k-1,A
S 2A
SA
S k,B
S k-1,B
S 2B
SB
-
+ V 'REF
-
3.) Final step is to determine whether to connect the bottom plates of the capacitors to bus A (bi=1) or bus B (bi=0). Ceq. 2-MVREF
-
2KC - Ceq.
vOUT -
+ V 'REF -
M+K-1
+
+
vOUT =
∑bi2-(M+K-i) VREF i=0
Allen and Holberg - CMOS Analog Circuit Design
Page X.4-3
Charge Scaling, Voltage Scaling DAC Use capacitors for MSB's and resistors for LSB's vOUT 2N-1C
2 N-2C
2 N-3C
VREF 4C
2C
C
C
R
VREF
Switch network
MSBs
R
R LSBs
R
R
•
Resistors must be trimmed for absolute accuracy.
•
LSB's are monotonic.
Allen and Holberg - CMOS Analog Circuit Design
Page X.5-1
X.5- OTHER TYPES OF D/A CONVERTERS CHARGE REDISTRIBUTION SERIAL DAC Precharge to VREF if bi = "1" Redistribution switch S1 S4
S2 VREF S3
+ VC1 -
C1
Precharge C1 to ground if bi = "0"
C2
+ VC2 -
Initially discharge S4
C1 = C2 b0 = LSB b1 = NLSB . . . bN = MSB
Conversion sequence: 4 Bit D/A Converter INPUT WORD: 1101 1
13/16
3/4
VC1 /VREF 1/2 1/4 0
2
4
6
8
1
13/16
3/4
VC2 /VREF 1/2 1/4 0
2
4
6
8
Close S4: VC2 = 0 Start with LSB firstClose S2 (b0=1): VC1 = VREF VREF Close S1: VC1 = 2 = VC2 Close S3 (b1=0): VC1 = 0 VREF Close S1: VC1 = VC2 = 4 Close S2 (b2=1): VC1 = VREF 5 Close S1: VC1 = VC2 = 8 VREF Close S2 (b3=1): VC1 = VREF 13 Close S1: VC1 = VC2 = 16 VREF
Comments: • LSB must go first. • n cycles to make an n-bit D-A conversion. • Top plate parasitics add error. • Switch parasitics add error.
Allen and Holberg - CMOS Analog Circuit Design
Page X.5-2
ALGORITHMIC SERIAL DAC Pipeline Approach to Implementing a DAC: 1/2
1/2 ∑
0
1/2
1/2 ∑
-1
z
b0
∑
-1
z
b1
LSB
bN-1
MSB
VREF
bN-2 b0 bN-1 vOUT (z) = 2 z-1 + 4 z -2 +.... + N z-N VREF 2 where bi = 1 or 0
Approaches: 1.) Pipeline with N cascaded stages. 2.) Algorithmic. vOUT(z) =
bi z-1VREF 1 - 0.5z - 1
z
-1
vOUT
Allen and Holberg - CMOS Analog Circuit Design
Page X.5-3
Example of an Algorithmic DAC Operation Realization using iterative techniques: A +VREF
(Bit "1") B
0
+
∑ +
Sample and hold
VOUT
(Bit "0")
1 2
Assume that the digital word is 11001 in the order of MSB to LSB. The steps in the conversion are: 1.) VOUT(0) is zeroed. 2.) LSB = 1, switch A closed, VOUT (1) = VREF . 3.) Next LSB = 0, switch B closed, VOUT(2) = 0 + 0.5VREF VOUT (2) = 0.5V REF . 4.) Next LSB = 0, switch B closed, VOUT (3) = 0 + 0.25VREF VOUT (3) = 0.25VREF. 5.) Next LSB = 1, switch A closed, VOUT(4) = VREF + (1/8)VREF VOUT(4) = (9/8)VREF . 6.) Finally, the MSB is 1, switch A is closed, and VOUT(5) = VREF + (9/16)VREF VOUT(5) = (25/16)VREF 7.) Finally, the MSB+1 is 0 (always last cycle), switch A is closed, and VOUT(6) = (25/32)VREF
Allen and Holberg - CMOS Analog Circuit Design
Page X.6-1
X.6 - CHARACTERIZATION OF ANALOG TO DIGITAL CONVERTERS General A/D Converter Block Diagram
Digital Processor
x(t) Filtering
Sampling
A/D Converter Types 1.) Serial. 2.) Medium speed. 3.) High speed and high performance. 4.) New converters and techniques.
Quantization
Digital Coding
y(kTN)
Allen and Holberg - CMOS Analog Circuit Design
Page X.6-2
Characterization of A/D Converters Ideal Input-Output Characteristics for a 3-bit ADC
111
Ideal A/D conversion Output digital number
110
Normal quantized value (± 1/2 LSB)
101
100
Ideal transition
011
010
Ideally quantized analog input
1 LSB 001 0
000
1 8
1 FS 8
2 8
3 8
4 8
5 8
6 8
7 8
2 FS 3 FS 4 FS 5 FS 6 FS 7 FS 8 8 8 8 8 8 Normal quantized value (± LSB)
FS
Allen and Holberg - CMOS Analog Circuit Design
Page X.6-3
Nonideal Characteristics of A/D Converters
Offset error 111
111 110
Ideal
Digital output code
Digital output code
110 101 100 011
101 100 011
010
010
001
001
000
1 FS 4
1 FS 2
3 FS 4
Gain error
000 FS
Ideal
1 FS 4
1 FS 2
3 FS 4
Analog input value
Analog input value
Offset Error
Scale factor (gain) error
111
111
110
110
FS
Digital output code
Digital output code
Ideal 101 100
Ideal
011
Nonlinearity
010
100
Missed codes due to excessive differential nonlinearity
011 010
001 000
101
001 1 FS 4
1 FS 2
3 FS 4
Analog input value
Integral Nonlinearity
000 FS
1 FS 4
1 FS 2
3 FS 4
Analog input value
Differential Nonlinearity
FS
Allen and Holberg - CMOS Analog Circuit Design
Page X.6-4
Sampled Data Aspect of ADC's
S-H command Sample Hold
Hold
Amplitude
ta
ts
Output valid for A/D conversion
S*
S(t) S* S(t) t
Tsample = ts + ta ta = acquisition time ts = settling time tADC = time for ADC to convert analog input to digital word. Conversion time = ts + ta + tADC. kT Noise = C V 2 (rms)
Allen and Holberg - CMOS Analog Circuit Design
Page X.6-5
Sample and Hold Circuits Simple
-
φ
vO
A1 +
vI CH
Improved
φ
φ
-
φ
-
vO
A2 +
A1 +
vI
CH
Waveforms
v0(t) Volts
v1(t),v0(t) v1(t) Switch closed (sample)
Switch open (hold)
Switch closed (sample)
t
Allen and Holberg - CMOS Analog Circuit Design
Page X.7-1
X.7 - SERIAL A/D CONVERTERS Single-Slope, A/D Converter
vIN* NT
NT
+ Ramp generator
VREF
vr
Output counter
-
vr vIN*
Reset
Output 0
NT
t
Interval counter f= 1 T clock
•
Simplicity of operation
•
Subject to error in the ramp generator
•
Long conversion times
Allen and Holberg - CMOS Analog Circuit Design
Page X.7-2
Dual Slope, A/D Converter Block Diagram: 1 Vin* 2
Positive integrator
-VREF
vint Vth
+ -
Digital control
Counter
Binary output
Operation: 1.) Initially vint = 0 and vin is sampled and held (Vin* > 0). 2.) Reset by integrating until vint(0) = Vth. 3.) Integrate Vin* for Nref clock cycles to get, NrefT * ⌠ * vint(t1) = vint (NrefT) = k ⌡Vin dt + vint(0) = kNrefTVin + Vth 0 * 4.) The Carry Output on the counter is used to switch the integrator from Vin to -V REF. Integrate until vint is equal to Vth resulting in NoutT + t1 ⌠ vint(t1 + t2) = vint(t1) + k ⌡-VREFdt = V th t1 Nout * * = V in ∴ kNref TVin + Vth - kVREFNoutT= Vth ⇒ VREF N ref
Allen and Holberg - CMOS Analog Circuit Design
Page X.7-3
Waveform of the Dual -Slope A/D Converter
vin V'''in V''in V'in
Vth 0
t
0
t1 = NrefT Reset
t0 (start)
t'2 t''2 t'''2 t2 = Nout T
•
Very accurate method of A/D conversion.
•
Requires a long time -2( 2N) T
Allen and Holberg - CMOS Analog Circuit Design
Page X.7-4
Switched Capacitor Integrators Noninverting: C1 + v1 (t)
C2
φ1
φ2 φ2
φ1
+ +
v2 (t)
fsignal 0?
Bit i
1
0.4
Yes
1
2
2(0.4000)-1 = -0.200
No
0
3
2(-0.200)+1 = +0.600
Yes
1
4
2(+0.600)-1 = +0.200
Yes
1
Results for various values ov Vin. Vin
b(i)
v(i+1)
b(i+1)
v(i+2)
b(i+2)
v(i+3)
b(i+3)
v(i+4)
b(i+4)
v(i+5)
-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1 1 1 1 1 1 1 1
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
-1 -1 -1 -1 -1 1 1 1 1 1 -1 -1 -1 -1 -1 1 1 1 1 1 1
-1 -0.6 -0.2 0.2 0.6 -1 -0.6 -0.2 0.2 0.6 -1 -0.6 -0.2 0.2 0.6 -1 -0.6 -0.2 0.2 0.6 1
-1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1 1
-1 -0.2 0.6 -0.6 0.2 -1 -0.2 0.6 -0.6 0.2 -1 -0.2 0.6 -0.6 0.2 -1 -0.2 0.6 -0.6 0.2 1
-1 -1 1 -1 1 -1 -1 1 -1 1 -1 -1 1 -1 1 -1 -1 1 -1 1 1
-1 0.6 0.2 -0.2 -0.6 -1 0.6 0.2 -0.2 -0.6 -1 0.6 0.2 -0.2 -0.6 -1 0.6 0.2 -0.2 -0.6 1
-1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 -1 1 1 -1 -1 1
-1 0.2 -0.6 0.6 -0.2 -1 0.2 -0.6 0.6 -0.2 -1 0.2 -0.6 0.6 -0.2 -1 0.2 -0.6 0.6 -0.2 1
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-10
Output Voltage for a 4-stage Converter
1 0.8
Inputs (Volts)
0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1
-0.8 -0.6 -0.4 -0.2
0 0.2 VIN (Volts)
0.4
0.6
0.8
1
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-11
RESOLUTION LIMITS OF THE 1-BIT/STAGE PIPELINE ADC 1st-Order Errors of The 1-Bit/Stage Pipeline ADC • Gain magnitude and gain matching (k1) • Offset of the X2 amplifier and the sample/hold (k2) • Comparator offset (k3) • Summer magnitude and gain matching (k4) • Summer offset (k 5) Illustration: + -
±k3 Vi-1
2
±k2 + + ∑ +
bi Vref
+
∑
±k1
+ ∑ + ±k4
Vi = AiVi-1 + VOSi - biAsiVref where
+1 if Vi-1 > ±k3 = ±VOCi bi = -1 if V < ±k = ±V i-1 3 OCi
Ai = all gain related errors of the ith stage VOSi = system offset errors of the ith stage VOCi = the comparator offset of the ith stage Asi = the gain of the summing junction of the ith stage
+
±k5 + ∑
Vi
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-12
Generalization of the First-Order Errors Extending the ith stage first-order errors to N stages gives: N-1 N A V + ∏ i in ∑ ∏AjVOSi + VOSN i=1 i=1 j=i+1 N
VN =
N-1 N - Vref ∑ ∏Aj Asibi + A sNbN i=1 j=i+1 Assuming identical errors in each stage gives:
VN = A NVin +
N
∑( AN-i) VOS i=1
N N-i - V ref∑( A ) Asbi i=1
Assuming only the first stage has errors: N
VN = A12N-1Vin + 2N-1VOS1 - Vref2N-1As1b1 - Vref
∑( 2N-i) bi i=2
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-13
Identification of Errors 1. Gain Errors 2N( ∆A/A) < 1
⇒ N=10 ⇒
∆A 1 < A 1000
Illustration of gain errors Vi Vref
2∆A A 2∆A 1A 1+
1
bi+1 =+1
-1
0
1
bi+1 =-1
-1 bi =-1
bi =+1
Vi-1 Vref
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-14
Identification of Errors - Cont'd 2. System Offset Errors VOS <
Vref 2N
For N=10 and V ref = 1V, VOS < 1mV
Illustration of system offset error Vi Vref 1+VOS
1
1+VOS bi+1 =+1
-1
0
1
bi+1 =-1
-1 bi =-1
bi =+1
Vi-1 Vref
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-15
Identification of Errors - Cont'd 3. Summing Gain Error ∆A s 1 < A s 2N ∆A 1 For N=10, A < 1024 Vi Vref 1+∆As /As 1-∆As /As
-1
-0.5
0.5
1
Vi-1 Vref
-1+∆As /As -1-∆As /As
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-16
Identification of Errors - Cont'd 4. Comparator Offset Error The comparator offset error is any nonzero value of the input to a stage where the stage bit is caused to change. It can be expressed as: Vi = 2Vi-1 - biVref where +1 if Vi-1 > VOCi bi = -1 if V < V i-1 OCi
Illustration of comparator offset error: Vi Vref
bi+1 =+1 -1-1
-0.5
0.5
bi+1 =-1
bi =-1
bi =+1 VOC = ?
1
Vi-1 Vref
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-17
SUMMARY
1.) The 1-bit/pipe, pipeline converter which uses standard components including a sample and hold, an amplifier, and a comparator would be capable of realizing at most an 8 or 9 bit converter. 2.) The accuracy of the gains and offset of the first stage of an N-Bit converter must be within 0.5LSB. 3.) The accuracy of the gains and offset of a stage diminishes with the remaining number of stages to the output of the converter. 4.) Error correction and self-calibrating techniques are necessary in order to realize the potential resolution capability of the 1-bit/'stage pipeline ADC.
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-18
Cyclic Algorithmic A/D Converter The output of the ith stage of a pipeline A/D converter is Voi = (2Vo,i-1 - biVREF )z-1 If Voi is stored and feedback to the input, the same stage can be used for the conversion. The configuration is as follows: Comparator Voi
X2
+1
Sample and hold
∑
+Vref
+1
-Vref
Practical implementation: Comparator Va
X2
+Vref
Sample and hold
Vb S1 Vin
+1
∑
-1
+ -
Vo Vo = "1" +Vref Vo = "0"
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-19
Algorithmic ADC - Example Assume that Vin* = 0.8VREF. The conversion proceeds as; 1.) 0.8VREF is sampled and applied to the X2 amplifier by S1. 2.) Va(0) is 1.6VREF (b1=1) which causes -VREF to be subtracted from Va(0) giving Vb(0) = 0.6V REF 3.) In the next cycle, Va(1) is 1.2VREF (b2=1) and Vb(1) is 0.2VREF. 4.) The next cycle gives Va(2) = 0.4VREF (b3=0) and Vb(2) is 0.4VREF. 5.) The next cycle gives Va(3) = 0.8VREF (b4=0) and Vb(3) is 0.8VREF. 6.) Finally, V a(4) = 1.6V REF (b5=1) and V b(4) = 0.6V REF. ∴ The digital word is 11001. ⇒
Vanalog = 0.78125VREF.
Va/VREF
Vb /VREF
2.0
2.0
1.6
1.6
1.2
1.2
0.8
0.8 0.6
0.4
0.4 0.2
0
0
1
2
3
4
5
t T
1
2
3
4
5
t T
Allen and Holberg - CMOS Analog Circuit Design
Algorithmic A/D Converters-Practical Results • Only one accurate gain-of-two amplifier required. • Small area requirements • Slow conversion time - nT. • Errors: Finite op amp gain, input offset voltage, charge injection, capacitance voltage dependence.
Practical Converter 12 Bits Differential linearity of 0.019% (0.8LSB) Integral linearity of 0.034% (1.5LSB) Sample rate of 4KHz.
Page X.8-20
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-21
SELF-CALIBRATING ADC's S1 MAINDAC COMP CN
CN-1
C1B
C1A
CCAL
TO SUCCESIVE APPROXIMATION REGISTER
Vref GND REGISTER
CALIBRATION DAC SUBDAC
SUCCESSIVE APPROXIMATION REGISTER
ADDER DATA REGISTER
CONTROL LOGIC
VεN
VεN-1
DATA OUTPUT
Main ADC is an N-bit charge scaling array. Sub DAC is an M-bit voltage scaling array. Calibration DAC is an M+2 bit voltage scaling array. This is an voltage-scaling, charge-scaling A/D converter with (N+M)- bits resolution.
Allen and Holberg - CMOS Analog Circuit Design
Page X.8-22
Self-Calibration Procedure During calibration cycles, the nonlinearity factors caused by capacitor mismatching are calibrated and stored in the data register for use in the following normal conversion cycles. The calibration procedure begins from MSB by connecting CN to VREF and the remaining capacitors CNX to GND, then exchange the voltage connection as follows: VX
CN
VX
CNX
CN
VREF
CNX
VREF
where C NX = C1B + C1A + ... + C N-1 The final voltage VX after exchanging the voltage connections is VX = VREF
CNX - CN CNX + C N
If the capacitor ratio is accurate and CNX = CN ⇒ VX = 0, otherwise VX ≠ 0. This residual voltage VX is digitized by the calibration DAC. Other less significant bits are calibrated in the same manner. After all bits are calibrated, the normal successive-approximation conversion cycles occurs. The calibrated data stored in the data register is converted to an analog signal by calibration DAC and is fed to the main DAC by CCAL to compensate the capacitor mismatching error.
Allen and Holberg - CMOS Analog Circuit Design
Self-Calibrating ADC Performance Supply voltage ± 5V Resolution of 16 bits Linearity of 16 bits Offset less than 0.25 LSB Conversion time for 0.5 LSB linearity: 12 µs for 12 Bits 80 µs for 16 Bits. RMS noise of 40 µV. Power dissipation of 20 mW (excludes logic) Area of 7.5 mm (excludes logic).
Page X.8-23
Allen and Holberg - CMOS Analog Circuit Design
X.9 - HIGH SPEED ADC's Conversion Time ≈ T (T = clock period) • Flash or parallel • Time interleaving • Pipeline - Multiple Bits • Pipeline - Single Bit
Page X.9-1
Allen and Holberg - CMOS Analog Circuit Design
Page X.9-2
FLASH A/D CONVERTER
VREF
V*in = 0.7 VREF
R 7 VREF 8
+ R
1
-
6 VREF 8
+ -
R 5 VREF 8
+ R
0
-
4 VREF 8
+
+ R
0
-
2 VREF 8
+
+ R
0
-
R 1 VREF 8
0
-
R 3 VREF 8
1
-
• Fast conversion time, one clock cycle • Requires 2N-1 comparators • Maximum practical bits is 6 or less • 6 bits at 10 MHz is practical
0
Digital decoding network
Output digital word 101
Allen and Holberg - CMOS Analog Circuit Design
Page X.9-3
Time-Interleaved A/D Converter Array Use medium speed, high bit converters in parallel. T1 S/H
N-bit A/D
T2 S/H
Vin
N-bit A/D . ..
Digital word out
TM S/H
N-bit A/D
A/D Converter No.1 A/D Converter No.2 A/D Converter No.M T1
T2
TM
T1 + TC T2 + TC
TM + TC
t
Allen and Holberg - CMOS Analog Circuit Design
Page X.9-4
Relative Die Size vs. Number of Bits
320
160
Relative die size
FLASH 80
5 Succ. Approx. Array (m- WAY)
4
40 3
m
20
10
4
5
6
7
# of bits
8
9
Allen and Holberg - CMOS Analog Circuit Design
Page X.9-5
2M-BIT, PARALLEL-CASCADE ADC • Compromise between speed and area • 8-bit, 1M Hz. Gain = 2M +
V*in
∑
V*in
Vref
Vref +
+
-
-
+
+
-
-
+ -
Digital decoding network
+ -
+
+
-
-
Digital decoding network
D/A Converter
2M - 1 equal resistors and comparators
M MSB's
2M - 1 equal resistors and comparators
M LSB's
Allen and Holberg - CMOS Analog Circuit Design
Page X.9-6
Conversion of Digital back to Analog for Pipeline Architectures Use XOR gates to connect to the appropriate point in the resistor divider resulting in the analog output corresponding to the digital output. Vref Analog Out
V*in 1
+ 0
1
+ 1 +
0
0 +
0
0
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-1
X.10 - OVERSAMPED (∆-∑) A/D CONVERTER NYQUIST VERSUS OVERSAMPLED A/D CONVERTERS Oversampling A/D converters use a sampling clock frequency(fS) much higher than the Nyquist rate(fN). Conventional Nyquist ADC Block Diagram: Digital Processor
x(t) Filtering Sampling Oversampling ADC Block Diagram
x(t) Filtering
Sampling
Quantization
Digital Coding
Modulator
Decimation Filter
Quantization
Digital Coding
The anti-aliasing filter at the input stage limits the bandwidth of the input signal and prevents the possible aliasing of the following sampling step. The modulator pushes the quantization noise to the higher frequency and leaves only a small fraction of noise energy in the signal band. A digital low pass filter cuts off the high frequency quantization noise. Therefore, the signal to noise ratio is increased.
y(kTN)
y(kTN)
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-2
ANTI-ALIASING FILTER The anti-aliasing filter of an oversampling ADC requires less effort than that of a conventional ADC. The frequency response of the anti-aliasing filter for the conventional ADC is sharper than the oversampling ADC. Conventional ADC's Anti-Aliasing Filter
fB
f fN/2
fS=fN
Oversampling ADC's Anti-Aliasing Filter
fB
f fN/2
fN
fS/2
fS
fB : Signal Bandwidth fN : Nyquist Frequency, fN = 2fB fS : Sampling Frequency and usually fS >> fN fS M : Oversampling ratio, M = f N So the analog anti-aliasing filter of an oversampling ADC is less expensive than the conventional ADC. If M is sufficiently large, the analog anti-aliasing filter is simply an RC filter.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-3
QUANTIZATION Conventional ADC's Quantization The resolution of conventional ADCs is determined by the relative accuracy of their analog components. For a higher resolution, self-calibration technique can be adopted to enhance the matching accuracy. Multilevel Quantizer: output (y) 5 3
ideal curve -6
-4
-2
1 -1
2
-3
4
6
input (x)
y=Gx+e
-5
e 1
x -1
The quantized signal y can be represented by y = Gx + e where, G = gain of ADC, normally = 1 e = quantization error
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-4
Conventional ADC's Quantization - Cont'd The mean square value of quantization error, e, is ∆/2 2 1 ⌠ e(x)2 dx = ∆ e2rms = ∆ ⌡ 12 -∆/2 where VREF ) 2N When a quantized signal is sampled at fS (= 1/τ), all of its noise power folds into the frequency band from 0 to fS/2. If the noise power is white, then the spectral density of the ∆ = the quantization level of an ADC (typically
sampled noise is 2 E(f) = erms f S
1/2 = erms 2τ
where τ = 1/fS and fS = sampling frequency The inband noise energy no is fB 2 e 2f 2 B 2 ⌡ 2 rms 2 no = ⌠ E (f)df = e rms (2fBτ) = erms f = M S 0 erms no = M fS The oversampling ratio M = 2f B Therefore, each doubling of the sampling frequency decreases the in-band noise energy by 3 dB, and increases the resolution by 0.5 bit. This is not a very efficient method of reducing the inband noise.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-5
OVERSAMPLING ADC fS
analog fB input
∑∆ MODULATOR
fD
LOW-PASS
2fB
FILTER
digital PCM
DECIMATOR
Oversampling ADCs consist of a ∑∆ modulator, a decimator (down-sampler), and a digital low pass filter. ∑∆ modulator Also called the noise shaper because it can shape the quantization noise and push majority of the noise to high frequency band. It modulates the analog input signal to a simple digital code, normally is one bit, using a sampling rate much higher the Nyquist rate. Decimator Also called the down-sampler because it down samples the high frequency modulator output into a low frequency output. Low-pass filter Use digital low pass filter to cut off the high frequency quantization noise and preserve the input signal.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-6
Sigma-Delta (∑∆) Modulator First Order ∑∆ Modulator The open loop quantizer in a conventional ADC can be modified by adding a closed loop to become a ∑∆ modulator. fS
x(t)
+
+
Integrator
yi
A/D
-
D/A Modulator Output
input signal Amplitude (V)
modulator output
0
0.5 Time (ms)
1
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-7
First-Order ∑∆ Modulator gn
+
xn
wn+1
Delay
wn
yn
en Quantizer
Accumulator yn = wn + en (1)
wn+1 = gn + wn = xn - yn + wn = xn -(wn + en) + wn = xn - en (2) Therefore, wn = xn-1 - en-1, which when substituted into (1) gives yn = xn-1 + ( en - en-1) The output of ∑∆ modulator yn is the input signal delayed by one clock cycle xn-1, plus the quantization noise difference en - en-1. The modulation noise spectrum density of en - en-1 is
signal baseband
ωτ ωτ N(f) = E(f) 1 - z-1 = E(f) 1 - e-jωτ = 2E(f) sin 2 = 2e rms 2τ sin 2 Plot of Noise Spectrum 2
N(f)/E(f)
1.5
1
N(f)
E(f)
0.5
0 0
fB
Frequency
fS 2
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-8
First Order ∑∆ Modulator-Cont’d The noise power in the signal band is fB
fB
⌠ ωτ 2 2 2 df 2e no = ⌠ N(f) 2τ sin df = ⌡ rms 2 ⌡ 0 0 fB 2 ⌠ 2 no = ⌡ 2erms 2τ (πfτ) df 0 ωτ 2πf πf where sin 2 = sin = sin ≈ πfτ 2fS f S if fS >> f Therefore, fB
2 ermsπ2(2τfB)3 2 2 ⌠f 2 df = no ≈ (2τ)3π2 erms ⌡ 3 0 where , fS >> fB Thus, no = erms
π π 2fBτ 3/2 = erms M-3/2 3 3
Each doubling of the oversampling ratio reduces the modulation noise by 9 dB and increase the resolution by 1.5 bits.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-9
Oversampling Ratio Required for a First-Order ∆ Σ Modulator A block diagram for a first-order, sigma-delta modulator is shown in the z-domain. Find the magnitude of the output spectral noise with VIN(z) = 0 and determine the bandwidth of a 10-bit analog-to-digital ∆ = rms value of converter if the sampling frequency, fS, 12 quantization is 10 MHz. noise + Vin (z) + Vout (z) Solution + 1 Σ Σ 1 z-1 Vout(z) = e rms + z-1 [Vin(z) Vout(z)] or z-1 Vout(z) = z erms if Vin(z) = 0 → Vout(z) = (1-z-1)erms ωτ ωτ 2 |N(f)| = E(f) 1 - e-jωτ = 2E(f) sin 2 = 2 erms sin 2 fS The noise power is found as fB fB ⌠ 2πfτ 2 2 2 ⌡|N(f)|2 df = 2 no(f) = ⌠ erms sin2 2 df fS ⌡ 0 0 2πfτ Let sin 2 ≈ πfτ if fS >> fB. Therefore, fB 2 2 2 8π2 2 fB 3 ⌡f2 df = no(f) = 4 f erms (πτ)2 ⌠ 3 erms fS S 0 or 3/2 VREF fB 8 ≤ 10 no(f) = π·erms f 3 2 S Solving for fB/fS gives (using ∆ in erms term is equal to VREF) fB 12 3 1 2/3 = [0.659x10-3]2/3 = 0.007576 = fS 8 π 210 fB = 0.007576·10MHz = 75.76kHz.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-10
Decimator (down-sampling) The one-bit output from the ∑∆ modulator is at very high frequency, so we need a decimator (or down sampler) to reduce the frequency before going to the digital filter. fS
analog fB input
∑∆
fD
MODULATOR Removes the modulation noise
from modulator xn
LOW-PASS
2fB
FILTER
digital PCM
DECIMATOR
Removes the out-of-band components of the signal to low-pass filter
+
yn REGISTER
fS fS 1 N yn = N ∑ xn , where N = = down-sampling ratio fD N=0 The transfer function of decimator is N-1 1 1 - z-N Y(z) 1 H(z) = X(z) = N ∑ z-i = N 1 - z-1 i=0 sinc(πfNτ) H(ejωτ) = sinc(πfτ)
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-11
Frequency Spectrum of the Decimator sinc(πfNτ) H(ejωτ) = sinc(πfτ) 10 0
Quantization Noise Signal Bandwidth
Magnitude (dB)
-10
-20 -30 -40 -50
fD
Frequency
fS 2
fD = intermediate decimation frequency When the modulation noise is sampled at fD, its components in the vicinity of fD and the harmonics of fD fold into the signal band. Therefore, the zeros of the decimation filter must be placed at these frequencies.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-12
Digital Lowpass Filter fS
analog fB input
∑∆
fD
MODULATOR
FIR or IIR digital low pass filter 10
Magnitude (dB)
-20
-50
-80
-110
LOW-PASS
2fB
FILTER
digital PCM
DECIMATOR
4000
Frequency (Hz)
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-13
After Digital Low Pass Filtering 0
Magnitude (dB)
signal -50
-150 -100
quantization noise
0
1000
Frequency (Hz)
Bit resolution From the frequency response of above diagram, the signal-to-noise ratio (SNR) signal SNR = 10log 10 f (dB) B
∑ noise(f)
f=0 and Bit resolution (B) ≈
SNR(db) 6dB
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-14
digital PCM FILTER
Frequency
2fB
Frequency Frequency
Time Time
MODULATOR
analog fB input
∑∆
fS
DECIMATOR
fD
LOW-PASS
System block in time domain and frequency domain
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-15
Second-Order ∑∆ Modulator Second order ∑∆ modulator can be implemented by cascading two first order ∑∆ modulators. INTEGRATOR 2 INTEGRATOR 1 xn DELAY A/D + ∑ + ∑ + ∑ + ∑ + + DELAY D/A QUANTIZER yn = xn-1 + (en - 2en-1 + en-2) The output of a second order ∑∆ modualtor yn is the input signal delayed by one clock cycle xn-1, plus the quantization noise difference en - 2en-1 + en-2. The modulation noise spectrum density of en - 2en-1 + en-2 is ωτ 2 2 N(f) = E(f) 1 - z-1 = E(f) 1 - e-jωτ = 4E(f) sin2 2 Noise Spectrum
signal baseband
4
N(f)/E(f)
3
2
2nd order N(f)
first order N(f)
E(f) 1
0
0
fB
Frequency
fS 2
yn
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-16
Second-Order ∑∆ Modulator- Cond’d The noise power in the signal band is ∆ 2 π2 -5/2 ∆π π2 -5/2 no = e rms M = M = ( M) -5/2 , fs >> fo 12 5 5 2 15 Each doubling of the oversampling ratio reduces the modulation noise by 15 dB and increase the resolution by 2.5 bits. Higher-Order Σ−∆ Modulators Let L = the number of loops. The spectral density of the modulation can be written as ωτ L | NL(f)| = e rms 2τ 2sin 2 The rms noise in the signal band is given approximately by πL no ≈ erms (2fBτ) L+0.5 2L+1 This noise falls 3(2L+1) dB for every doubling of the sampling rate providing L+0.5 extra bits. Decimation Filter sinc(πfNτ) L+1 is close to being optimum for decimating the A filter function of sinc(πfτ) signal from an Lth-order ∆−Σ modulator. Stability For orders greater than 2, the loop can become unstable. Loop configuration must be used that provide stability for order greater than two.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-17
The modulation noise spectral density of a second-order, 1-bit ∆Σ modulator is given as |N(f)| =
4∆ 12
ωt 2 sin2 4 fs
where ∆ is the signal level out of the 1-bit quantizer and fs = (1/τ) = the sampling frequency and is 10MHz. Find the signal bandwidth, fB, in Hz if the modulator is to be used in an 18 bit oversampled ADC. Be sure to state any assumption you use in working this problem.
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-18
Circuit Implementation of A Second Order ∑∆ Modulator V+REF V-REF
S1
S2 C1
C2
C2
S3 S4
IN
S1 -
S2 C1
+
S3 S4
+ S1
C1 S2
S4 S3
+
+ S1
C2
-
C1 S2
OUT S4 S3 C2 V-REF V+REF
Fully differential, switched-capacitor integrators can reduce charge injection effect. Circuit Tolerance of a Second Order ∑∆ Modulator 1. 20% variation of C1/C2 has only a minor impact on performance. 2. The Op Amp gain should be comparable to the oversampling ratio. 3. The unity-gain bandwidth of Op Amp should be at least an order of magnitude greater than the sampling rate.
Allen and Holberg - CMOS Analog Circuit Design SOURCES OF ERRORS IN Σ ∆ A/D CONVERTERS 1.
Quantization in time and amplitude Jitter and hysteresis
2.
Linear Errors Gain and delay
3.
Nonlinear Errors Harmonic distortion Thermal noise
Page X.10-19
Allen and Holberg - CMOS Analog Circuit Design
Page X.10-20
Comparison of the Various Examples Discussed Type of Converter
ResoluNo. of No. of Dependent tion Cycles/ Compar on Passive Components Conver- ators sion
Speed
INL/DNL (LSB's)
Area
Power (mW)
Flash
1
2N-1
Yes
Low
High
N/A
Largest
Largest
Two-Step Flash
2
31
Yes
10 bits
5Ms/s
±3/±0.6
350
1 after initial delay
3
Yes
13 bits
250ks/s ±1.5/±0.5
54k mils2 3600 mils2
64
3
Yes
16 bits
24kHz
75.3k mils2
110
Pipeline Oversampling
91dB
15
Allen and Holberg - CMOS Analog Circuit Design
Page X.11-1
X.11 -FUNDAMENTAL LIMITS OF SAMPLING A/D CONVERTERS kT/C Noise Assume that the ON resistance of a switch is R and the sampling capacitor is C and that the time to charge the capacitor fully is 1 T = f ≈ 10RC c
(1)
Set the value of the LSB = Vref = 2N
V ref equal to kT/C noise of the switch, 2N
kT C
(2)
Solve for C of (1) and substitute into (2) to get Vref N 2
2
= 10kTRfc ⇒
2N f c =
Vref 10kRT
(3)
Taking the log of both sides of (3) gives N = -1.67 log(fc) + 3.3 log(Vref) - 1.67 log(10kRT) or N = 32.2 + 3.33 log(V ref ) - 1.67 log(Rf c ) (At room temperature) kT/C Noise Comparison of high-performance, monolithic A/D converters in terms of resolution versus sampling frequency with fundamental limits due to kT/C noise superimposed.
Allen and Holberg - CMOS Analog Circuit Design
Page X.11-2
Fundamental Limits of Sampling A/D Converters - Continued Maximum Sample Rate Assume that the maximum sample rate is determined by the time required for the amplifiers and/or sample-hold circuits to settle with the desired accuracy for high resolution. Further assume that the dynamics of these circuits can be modeled by a second-order system with a transfer function of ω n2 A(s) = A(0) s 2 + 2ζω n s + ω n 2 If ω n ≈ GB of the circuit and if the system is underdamped, then the step response is given as e-ζGBt vo(t) = 1 sin 1-ζ 2 GB·t + φ A(0) 2 1-ζ
This response looks like the following, 2 1.5 +ε 1 -ε 0.5 0
Settling Time 0
4
8
ωn t
12
16
20
If we define the error (±ε) in vo settling to A(0) as the multiplier of the sinusoid, then an expression for the settling time can be derived as e-ζGBt 1 1 2πζGB ⇒ fsample = = ts = 2πζGB ln ts 1 1 - ζ2 ln ε 1-ζ2
For reasonable values of ζ, fsample can be approximated as πGB G B f sample ≈ 10 = 3
Allen and Holberg - CMOS Analog Circuit Design
Page X.11-3
Aperature Uncertainty (Jitter) A problem in all clocked or sampled A/D converters. vin Clock
Analog In
Analog-todigital converter
Digital Out
∆V
∆T
dvin ∆V = slope x ∆T = dt ∆T Vref/2N ∆V ∆T = Aperature uncertainty = dV = dv /dt in in dt Assume that vin(t) = Vp sin ωt dvin dt = ωV p max
∆T = Therefore, ∆T =
Vref Vref 1 1 x ≈ = ωV p 2NωVref 2 N ω 2N
1 1 = 2πf2N πf2N+1
Suppose f = 100kHz and N = 8, ∆T =
1 = 6.22ns 200πKx29
6.22ns 622ppm Clock accuracy = 10,000ns = 0.06% = ?
Slope = dvin dt t
Allen and Holberg - CMOS Analog Circuit Design
Page X.12-1
X.12 - SUMMARY OF A/D CONVERTERS Typical Performance Characteristics
A/D Architecture
Typical Performance Characteristics
Serial 1 = 2N T fc
1-100 conversions/sec., 12-14 bit accuracy, requires no element-matching, a stable voltage reference is necessary
Successive Approximation 1 fc ≈ NT
10,000-100,000 conversions/sec., 8-10 bits of untrimmed or uncalibrated accuracy, 12-14 bits of trimmed or calibrated accuracy
High Speed 1 T < f < NT c
1 to 40 megaconversions/sec., 7-9 bits of accuracy, 10-12 bits of accuracy with error correction and other techniques
Oversampling
8,000-600,000 conversions/sec., 12-16 bits accuracy, requires linear integrators but no precision passive components, minimizes noise and offsets
1
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