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ECE1371 Advanced Analog CircuitsLecture 1
INTRODUCTION TODELTA-SIGMA ADCS
Richard [email protected]
Trevor [email protected]
1-2
alog circuita modern
peeks at
gems thatw.
ECE1371
Course Goals
• Deepen understanding of CMOS andesign through a top-down study of analog system— a delta-sigma ADC
• Develop circuit insight through brief some nifty little circuits
The circuit world is filled with many littleevery competent designer ought to kno
1-3
eb 16, Apr 6)tion
∆Σ …”nalog IC …”s”
ECE1371
Logistics• Format:
Meet Mondays 1:00-3:00 PM (not F12 2-hr lectures plus proj. presenta
• Grading:30% homework40% project30% exam
• References:Schreier & Temes, “UnderstandingChan-Carusone, Johns & Martin, “ARazavi, “Design of Analog CMOS IC
• Lecture Plan:
1-4
Homework
1: Matlab MOD1&2
2: ∆Σ Toolbox
4 3: Sw.-level MOD2
4 4: SC Integrator
5: SC Int w/ Amp
Project
Due Friday April 10
ECE1371
Date Lecture (M 13:00-15:00) Ref
2015-01-05 RS 1 MOD1 & MOD2 ST 2, 3, A
2015-01-12 RS 2 MODN + ∆Σ Toolbox ST 4, B
2015-01-19 RS 3 Example Design: Part 1 ST 9.1, CCJM 1
2015-01-26 RS 4 Example Design: Part 2 CCJM 18
2015-02-02 TC 5 SC Circuits R 12, CCJM 1
2015-02-09 TC 6 Amplifier Design
2015-02-16 Reading Week– No Lecture
2015-02-23 TC 7 Amplifier Design
2015-03-02 RS 8 Comparator & Flash ADC CCJM 10
2015-03-09 TC 9 Noise in SC Circuits ST C
2015-03-16 RS 10 Advanced ∆Σ ST 6.6, 9.4
2015-03-23 TC 11 Matching & MM-Shaping ST 6.3-6.5, +
2015-03-30 TC 12 Pipeline and SAR ADCs CCJM 15, 17
2015-04-06 Exam Proj. Report
2015-04-13 Project Presentation
1-5
lator
logic
logic
ECE1371
NLCOTD: Level TransVDD1 > VDD2, e.g.
• VDD1 < VDD2, e.g.
• Constraints: CMOS1-V and 3-V devicesno static current
3-V logic 1-V ?
1-V logic 3-V ?
1-6
day)
rs
time
ECE1371
Highlights(i.e. What you will learn to
1 MOD1: 1st-order ∆Σ modulatorStructure and theory of operation
2 Inherent linearity of binary modulato
3 Inherent anti-aliasing of continuous-modulators
4 MOD2: 2nd-order ∆Σ modulator
5 Good FFT practice
1-7
)ith a full-scaleB
stem is thequared response
the integral
e j 2πf ) 2 Sxx f( )⋅
ECE1371
0. Background(Stuff you already know
• The SQNR* of an ideal n-bit ADC wsine-wave input is (6.02 n + 1.76) d
“6 dB = 1 bit.”
• The PSD at the output of a linear syproduct of the input’s PSD and the smagnitude of the system’s frequency
i.e.
• The power in any frequency band isof the PSD over that band
*. SQNR = Signal-to-Quantization-Noise Ratio
H(z)X Y Syy f( ) H(=
1-8
ltering,
bit!), but thes 22 bits.
DigitalOut
(to digitalfilter)
ECE1371
1. What is ∆Σ?• ∆Σ is NOT a fraternity
• Simplified ∆Σ ADC structure:
• Key features: coarse quantization, fifeedback and oversampling
Quantization is often quite coarse (1 effective resolution can still be as high a
LoopFilter
CoarseADC
DAC
LoopFilter
CoarseADC
DAC
AnalogIn
1-9
g?n required
in theample rate
le
pression.ed digitally.
f s 2f B=
f s 2f B( )⁄
ECE1371
What is Oversamplin• Oversampling is sampling faster tha
by the Nyquist criterionFor a lowpass signal containing energyfrequency range , the minimum srequired for perfect reconstruction is
• The oversampling ratio is
• For a regular ADC,To make the anti-alias filter (AAF) feasib
• For a ∆Σ ADC,To get adequate quantization noise supSignals between and ~ are remov
0 f B,( )
OSR ≡
OSR 2 3–∼
OSR 30∼
f B f s
1-10
AAF
f is very close
f
d
Alias far away
ECE1371Oversampling Simplifies
f s 2⁄
DesiredSignal
UndesiredSignalsOSR ~ 1:
First alias band
f s 2⁄
OSR = 3: Wide transition ban
1-11
Work?tization error.it resolution?
uantizatione-shaping
n digitaloutputn@2fB
Nyquist-ratePCM Data
w
f B
ECE1371
How Does A ∆Σ ADC • Coarse quantization ⇒ lots of quan
So how can a ∆Σ ADC achieve 22-b
• A ∆Σ ADC spectrally separates the qerror from the signal through nois
∆ΣADC
u v DecimatioFilter
analog1 bit @fs
desiredshaped
1
–1t
noise
input
t
f s 2⁄f B f s 2⁄f B
undesiredsignals
signal
1-12
stemand drives aand noise is
er.
ion analogoutput
analogoutput
w
f B f s
ECE1371
A ∆Σ DAC System
• Mathematically similar to an ADC syExcept that now the modulator is digitallow-resolution DAC, and that the out-of-bhandled by an analog reconstruction filt
∆ΣModulator
u v ReconstructFilter
digital
1 bit @fs
signal shaped
1
–1t
noise
input(interpolated)
f B f s 2⁄ f B f s 2⁄
1-13
ay?
ry high
he anti-alias
rate
R.
ECE1371
Why Do It The ∆Σ W• ADC: Simplified Anti-Alias Filter
Since the input is oversampled, only vefrequencies alias to the passband.A simple RC section often suffices.If a continuous-time loop filter is used, tfilter can often be eliminated altogether.
• DAC: Simplified Reconstruction FilteThe nearby images present in Nyquist-rreconstruction can be removed digitally.
+ Inherent LinearitySimple structures can yield very high SN
+ Robust Implementation∆Σ tolerates sizable component errors.
1-14
odulatores]
ntizerit)
back
v
y
v’
v
1
–1
e, linear.
ECE1371
2. MOD1: 1st-Order ∆Σ M[Ch. 2 of Schreier & Tem
z-1
z-1
QU VY
Qua
DAC
(1-b
FeedDACV’
“ ∆” “ Σ”
Since two points define a lina binary DAC is inherentl y
1-15
ut ther as an
Y V
E
1–z-1)E(z)
V
ECE1371
MOD1 Analysis• Exact analysis is intractable for all b
simplest inputs, so treat the quantizeadditive noise source:
z-1
z-1
Q
⇒(1–z-1) V(z) = U(z) – z-1V(z) + (
U Y
V(z) = Y(z) + E(z)Y(z) = ( U(z) – z-1V(z) ) / (1–z-1)
V(z) = U(z) + (1–z–1)E(z)
1-16
n (NTF)F(z)•E(z)
pe!
t the noise
0.4 0.5
) 2
ency ( f /fs)
ECE1371
The Noise Transfer Functio• In general, V(z) = STF(z)•U(z) + NT
• For MOD1, NTF(z) = 1–z–1
The quantization noise has spectral sha
• The total noise power increases, bupower at low frequencies is reduced
0 0.1 0.2 0.30
1
2
3
4NTF ej 2πf(
Normalized Frequ
ω2 for ω 1«≅
Poles & zeros:
1-17
ower
er is
SR increases
σe2
e2--- ω2dω
0
ωB
∫SR ) 3–
ECE1371
In-band Quant. Noise P• Assume that e is white with power
i.e.• The in-band quantization noise pow
• Since ,
• For MOD1, an octave increase in OSQNR by 9 dB
“1.5-bit/octave SQNR-OSR trade-off.”
See ω( ) σe2 π⁄=
IQNP H ejω( ) 2See ω( )dω
0
ωB
∫=σπ
---≅
OSR πωB-------≡ IQNP
π2σe2
3------------- O(=
1-18
Time
400 500
ECE1371
A Simulation of MOD1—
0 100 200 300–1
0
1
Sample Number
1-19
Freq.
10–1
cade
R = 128
NBW = 5.7x10–6
ECE1371
A Simulation of MOD1—
10–3 10–2–100
–80
–60
–40
–20
0
Normalized Frequency
20 dB/de
SQNR = 55 dB @ OS
Full-scale test tone
Shaped “Noise”
dBF
S/N
BW
1-20
OD1ycted by the
ched
CK
D Q
DFFQB
v
arator
ECE1371
CT Implementation of M• Ri/Rf sets the full-scale; C is arbitrar
Also observe that an input at fs is rejeintegrator— inherent anti-aliasing
LatIntegrator
clock
yu
C
Ri
Rf
Comp
1-21
s
nd –1
ainsrift
10 15 20
= 0.06
Time
ECE1371
MOD1-CT Waveform
• With u=0, v alternates between +1 a
• With u>0, y drifts upwards; v contconsecutive +1s to counteract this d
0 50 5 10 15 20
0
u = 0
v
y
u
Time
–1
1
0
v
y
–1
1
1-22
z 1–
s-----------
cellation
k πi
ECE1371
MOD1-CT STF =Recall
1 –------z es=
Pole-zero can@ s = 0
Zeros @ s = 2
s-plane ω
σ
1-23
ponses
3
z–1
ECE1371
MOD1-CT Frequency Res
0 1 2–50
–40
–30
–20
–10
0
10
Frequency (Hz)
dB
InherentAnti-Aliasing
Quant. NoiseNotch
NTF = 1–z–1
STF = 1–s
1-24
e
.
e
too
off.
f B )
ECE1371
Summary• ∆Σ works by spectrally separating th
quantization noise from the signalRequires oversampling.
• Noise-shaping is achieved by the usof filtering and feedback
• A binary DAC is inherently linear,and thus a binary ∆Σ modulator is
• MOD1 has NTF (z) = 1 – z–1
⇒ Arbitrary accuracy for DC inputs.1.5 bit/octave SQNR-OSR trade-
• MOD1-CT has inherent anti-aliasing
OSR f s 2(⁄≡
1-25
3V
3V
V
ECE1371
NLCOTD
3V → 1V:3V
1V
1V → 3V:
1
3V
3V
3V
3V
1-26
odulatores] another:
1–z–1)E(z)
V
ECE1371
3. MOD2: 2nd-Order ∆Σ M[Ch. 3 of Schreier & Tem
• Replace the quantizer in MOD1 withcopy of MOD1 in a recursive fashion
V(z) = U(z) + (1–z–1)E1(z), E1(z) = (
⇒V(z) = U(z) + (1–z–1)2E(z)
z-1
Q
z-1
z-1
z-1
UE1
E
1-27
ms
z( ) 1 z 1––( )2=z( ) z 1–=
z( ) 1 z 1––( )2=z( ) z 2–=
ECE1371
Simplified Block Diagra
Q1z−1
zz−1
U V
E
NTFSTF
Q1z−1
1z−1
U V
E
-2-1 NTFSTF
1-28
10–1
y
as muchOD1
ECE1371
NTF Comparison
10–3 10–2–100
–80
–60
–40
–20
0
NT
Fe
j2πf
()
(dB
)
Normalized Frequenc
MOD1
MOD2MOD2 has twice attenuation as Mat all frequencies
1-29
ower
and
ses MOD2’s
ee ω( )dω
ECE1371
In-band Quant. Noise P• For MOD2,
• As before,
• So now
With binary quantization to ±1, and thus .
• “An octave increase in OSR increaSQNR by 15 dB (2.5 bits)”
H e jω( ) 2 ω4≈
IQNP H e jω( ) 2S0
ωB∫=
See ω( ) σe2 π⁄=
IQNPπ4σe
2
5------------- OSR( ) 5–=
∆ 2= σe2 ∆2 12⁄ 1 3⁄= =
1-30
ele
50 200
ECE1371
Simulation ExamplInput at 75% of FullSca
0 50 100 1–1
0
1
Sample number
1-31
Dle
10–1
cade
Theoretical PSD(k = 1)
NBW = 5.7×10−6
ECE1371
Simulated MOD2 PSInput at 50% of FullSca
10–3 10–2–140
–120
–100
–80
–60
–40
–20
0
SQNR = 86 dB@ OSR = 128
40 dB/de
Simulated spectrum
Normalized Frequency
dBF
S/N
BW
(smoothed)
1-32
tude 256
–20 0
MOD1
NR
ECE1371
SQNR vs. Input AmpliMOD1 & MOD2 @ OSR =
–100 –80 –60 –400
20
40
60
80
100
120
Input Amplitude (dBFS)
SQ
NR
(dB
)
MOD2Predicted SQNR
Simulated SQ
1-33
256 512 1024
ve assumesinput)
1
optimistic.tic.
ECE1371
SQNR vs. OSR
4 8 16 32 64 1280
20
40
60
80
100
120
SQ
NR
(dB
)
(Theoretical curve assumes-3 dBFS input)
(Theoretical cur0 dBFS
MOD
MOD2
Predictions for MOD2 areBehavior of MOD1 is erra
1-34
MOD2
ECE1371
Audio Demo: MOD1 vs. [dsdemo4]
MOD1
MOD2
SineWave
SlowRamp
Speech
1-35
aryck toantization
ng.al decimationd by a digital
a DAC.
nearity
R trade-off.
r.
ECE1371
MOD1 + MOD2 Summ• ∆Σ ADCs rely on filtering and feedba
achieve high SNR despite coarse quThey also rely on digital signal processi∆Σ ADCs need to be followed by a digitfilter and ∆Σ DACs need to be precedeinterpolation filter.
• Oversampling eases analog filteringrequirements
Anti-alias filter in an ADC; image filter in
• Binary quantization yields inherent li
• MOD2 is better than MOD115 dB/octave vs. 9 dB/octave SQNR-OSQuantization noise more white.Higher-order modulators are even bette
1-36
cemes]
the record.
BFS peak.
N⁄ )) 2⁄
0 N0
1
ECE1371
4. Good FFT Practi[Appendix A of Schreier & Te
• Use coherent samplingI.e. have an integer number of cycles in
• Use windowingA Hann windowworks well.
• Use enough pointsRecommend .
• Scale (and smooth) the spectrumA full-scale sine wave should yield a 0-d
• State the noise bandwidthFor a Hann window, .
w n( ) 1 2πn(cos–(=
N 64 OSR⋅=
NBW 1.5 N⁄=
1-37
ampling
ro FFT bin
ge”
0.4 0.5
ECE1371
Coherent vs. Incoherent S
• Coherent sampling: only one non-ze
• Incoherent sampling: “spectral leaka
0 0.1 0.2 0.3–300
–200
–100
0
100
dB
Normalized Frequency
Incoherent
Coherent
1-38
t mean that
ite record:
in
75 0.5
indow:
pear in-band
ECE1371
Windowing• ∆Σ data is usually not periodic
Just because the input repeats does nothe output does too!
• A finite-length data record = an infinmultiplied by a rectangular window
,Windowing is unavoidable.
• “Multiplication in time is convolution frequency”
w n( ) 1= 0 n≤ N<
0 0.125 0.25 0.3–100
–90–80–70–60–50–40–30–20–10
0Frequency response of a 32-point rectangular w
Slow roll-off ⇒ out-of-band Q. noise may apdB
1-39
ster= 256
0.5f
l ∆Σ spectrum
wed spectrum
se
w 2⁄
ECE1371
Example Spectral DisaRectangular window, N
0 0.25–60
–40
–20
0
20
40
Normalized Frequency,
dB
Actua
Windo
Out-of-band quantization noiobscures the notch!
W f( )
1-40
N = 16)
0.375 0.5
f
ECE1371Window Comparison (
0 0.125 0.25–100
–90
–80
–70
–60
–50
–40
–30
–20
–10
0
Normalized Frequency,
(dB
)
Rectangular
Hann2
Hann
Wf()
W0(
)----
--------
----
1-41
f tones locateddic.”
Hann2
5
35N/128
3N/8
35/18N
n-----------
12πnN
-----------cos–
2--------------------------------
2
ECE1371
Window PropertiesWindow Rectangular Hann†
†. MATLAB’s “hann” function causes spectral leakage oin FFT bins unless you add the optional argument “perio
,
( otherwise)1
Number of non-zeroFFT bins
1 3
N 3N/8
N N/2
1/N 1.5/N
w n( )n 0 1 … N 1–, , ,=
w n( ) 0=
12πN
------cos–
2--------------------------
w 22 w n( )2∑=
W 0( ) w n( )∑=
NBWw 2
2
W 0( )2----------------=
1-42
N bins to
all fraction
⇒
.
..
N 30 OSR⋅>
ECE1371
Window Length,• Need to have enough in-band noise
1 Make the number of signal bins a smof the total number of in-band bins
<20% signal bins ⇒ >15 in-band bins
2 Make the SNR repeatable yields std. dev. ~1.4 dB yields std. dev. ~1.0 dB
yields std. dev. ~0.5 dB
• is recommended
N 30 OSR⋅=N 64 OSR⋅=N 256 OSR⋅=
N 64 OSR⋅=
1-43
A.
,ather than 0.
j 2πknN
--------------
k( ) X M k 1+( )≡
f–
ECE1371
FFT Scaling• The FFT implemented in MATLAB is
• If †, then
⇒ Need to divide FFT by to get
†. f is an integer in . I’ve defined since Matlab indexes from 1 r
X M k 1+( ) x M n 1+( )e–
n 0=
N 1–
∑=
x n( ) A 2πfn N⁄( )sin=
0 N 2⁄,( ) Xx n( ) x M n 1+( )≡
X k( )AN2
--------- , k = f or N
0 , otherwise
=
N 2⁄( )
1-44
ingg 1 sample filters:
ral estimate
2πkN
-----------n, 0 n N<≤
0 , otherwise
ECE1371
The Need For Smooth• The FFT can be interpreted as takin
from the outputs of N complex FIR
⇒ an FFT yields a high-variance spect
x h0 n( )
h1 n( )
hk n( )
hN 1– n( )
y 0 N( ) X 0( )=
y 1 N( ) X 1( )=
y k N( ) X k( )=
y N 1– N( ) X N 1–( )=
hk n( ) ej
=
1-45
ng
nction
ctrumsmooth()
lly-increasingnsity ofmake a nice
ECE1371
How To Do Smoothi1 Average multiple FFTs
Implemented by MATLAB’s psd() fu
2 Take one big FFT and “filter” the speImplemented by the ∆Σ Toolbox’s logfunctionlogsmooth() averages an exponentianumber of bins in order to reduce the depoints in the high-frequency regime andlog-frequency plot
1-46
ectra
y
aw FFTgsmooth
ECE1371
Raw and Smoothed SpdB
FS
Normalized Frequenc10
–210
–1–120
–100
–80
–60
–40
–20
0
Rlo
1-47
OD2)
10–1
y
ancy)
ECE1371
Simulation vs. Theory (M
10–4
10–3
10–2
–160
–140
–120
–100
–80
–60
–40
–20
0
20
Normalized Frequenc
Simulated SpectrumNTF = Theoretical Q. Noise?
dBF
S
“Slight”Discrep(~40 dB
1-48
? a full-scale) comes out
rror signal.
rse.
nsityt to
in individual noise have! of the FFT.
ECE1371
What Went Wrong1 We normalized the spectrum so that
sine wave (which has a power of 0.5at 0 dB (whence the “dBFS” units)
⇒ We need to do the same for the ei.e. use .
But this makes the discrepancy 3 dB wo
2 We tried to plot a power spectral detogether with something that we waninterpret as a power spectrum
• Sine-wave components are located FFT bins, but broadband signals liketheir power spread over all FFT bins
The “noise floor” depends on the length
See f( ) 4 3⁄=
1-49
+ Noise
0.5
–3 dB/octave
NR = 0 dB
ECE1371
Spectrum of a Sine Wave
Normalized Frequency, f
(“dB
FS
”)S
x′f(
)
0 0.25–40
–30
–20
–10
0
N = 26
N = 28
N = 210
N = 212
0 dBFS 0 dBFSSine Wave Noise
⇒ S
1-50
by the
er all binspower there
in by a factor
s not
ECE1371
Observations• The power of the sine wave is given
height of its spectral peak
• The power of the noise is spread ovThe greater the number of bins, the lessis in any one bin.
• Doubling N reduces the power per bof 2 (i.e. 3 dB)
But the total integrated noise power doechange.
1-51
oise?k
bandwidthpower at
idth (NBW) ofer in each
,H f( )
H f( )f
ECE1371
So How Do We Handle N• Recall that an FFT is like a filter ban
• The longer the FFT, the narrower theof each filter and thus the lower the each output
• We need to know the noise bandwthe filters in order to convert the powbin (filter output) to a power density
• For a filter with frequency response
NBWH f( ) 2 fd∫H f 0( )2
----------------------------=
NBW
f0
1-52
th
seval]
) j– 2πfn( )exp
N
1N----=
ECE1371
FFT Noise BandwidRectangular Window
,
,
[Par
∴
hk n( ) j 2πkN
-----------n exp= H k f( ) hk n(
n 0=
N 1–
∑=
f 0kN----= H k f 0( ) 1
n 0=
N 1–
∑= =
H k f( ) 2∫ hk n( ) 2∑ N= =
NBWH k f( ) 2 fd∫H k f 0( )2
------------------------------- NN 2-------= =
1-53
t
10–1
y
pectrumQ. Noise
= 2×10–5
N = 216
F f( ) 2 NBW⋅
ECE1371
Better Spectral Plo
10–4 10–3 10–2–160
–140
–120
–100
–80
–60
–40
–20
0
20dB
FS
/NB
W
Normalized Fre quenc
Simulated STheoretical
NBW = 1.5 / N
43--- NTpassband for
OSR = 128
1-54
-01-12) MOD1’st samplesing ways.
equals the[–1,1].
scale sine- Include theand list theOSR = 128.
er it.
ECE1371
Homework #1 (Due 2015A. Create a Matlab function that computes
output sequence given a vector of inpuand exercise your function in the follow
1 Verify that the average of the outputinput for a few random DC inputs in
2 Plot the output spectrum with a half-wave input. Use good FFT practice.theoretical quantization noise curve theoretical and simulated SQNR for
B. Repeat with MOD2.
C. Compose your own question and answ
1-55
VE
Qn( ) v n( )
x 1 n 1+( )u n( )
ECE1371
MOD2 Expanded
Q1z−1
zz−1
U
z-1
u n( )z-1
x 1 n( )
x 1 n 1+( ) x 2 n 1+( ) x 2
Difference Equations:v n( ) Q x 2 n( )( )=
x 2 n 1+( ) x 2 n( ) v n( )– +=x 1 n 1+( ) x 1 n( ) v n( )– +=
1-56
de
ECE1371
Example Matlab Cofunction [v] = simulateMOD2(u)
x1 = 0; x2 = 0; for i = 1:length(u) v(i) = quantize( x2 ); x1 = x1 + u(i) - v(i); x2 = x2 + x1 - v(i);
endreturn
function v = quantize( y ) if y>=0 v = 1; else v = -1; endreturn
1-57
day solidify
rs
time
ECE1371
What You Learned ToAnd what the homework should
1 MOD1: 1st-order ∆Σ modulatorStructure and theory of operation
2 Inherent linearity of binary modulato
3 Inherent anti-aliasing of continuous-modulators
4 MOD2: 2nd-order ∆Σ modulator
5 Good FFT practice
