ECE1371 Advanced Analog CircuitsLecture 2

MODN and the ∆Σ Toolbox

Richard Schreierrichard.schreier@analog.com

Trevor Caldwelltrevor.caldwell@utoronto.ca

2-2

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

2-3

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

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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

2-4

-Flop

y?

Q

Q

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NLCOTD: Dynamic Flip• Standard CMOS version

• Can the circuit be simplified?Is a complementarty clock necessar

D

CK

2-5

day)

box

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Highlights(i.e. What you will learn to

1 Nth-order modulator (MOD N )

2 High-level design with the ∆Σ Tool

2-6

ystem

DAC

V

E

) + NTF (z )E (z )

sfer functionsfer functionerror

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0. Review: A ∆Σ ADC S

∆ΣModulator

DigitalDecimator

U

V

W

LoopFilter

U

V (z ) = STF (z )U( z

STF (z ): signal tranNTF (z ): noise tranE (z ): quantization

desired

shaped

Nyquist-ratePCM Data

noise

f s 2⁄f B

f s 2⁄f B

f B

signal

Y

2-7

z( ) 1 z 1––( )=z( ) 1=

0.4 0.5

2

ency ( f )

π2σe2

3------------- OSR( ) 3–

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Review: MOD1

QUE

NTFSTFz-1

z-1

V

0 0.1 0.2 0.30

1

2

3

4NTF ej 2πf( )

Normalized Frequ

ω2≅

NTF poles & zeros:

IQNP =

2-8

z( ) 1 z 1––( )2=z( ) z 1–=

0.3 0.4 0.5

j 2πf ) 2

uency ( f )

π4σe2

5------------- OSR( ) 5–=

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Review: MOD2

Q1z−1

zz−1

U V

E

NTFSTF

0 0.1 0.20

8

16

NTF e(

Normalized Freq

ω4≅

NTF poles & zeros:

IQNP

2-9

e

.eedback.

too

f.

ff.

f B )

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Review Summary• ∆Σ works by spectrally separating th

quantization noise from the signalRequires oversampling.Achieved by the use of filtering and f

• A binary DAC is inherently linear,and thus a binary ∆Σ modulator is

• MOD1-CT has inherent anti-aliasing

• MOD1 has NTF (z) = 1 – z–1

⇒ Arbitrary accuracy for DC inputs;9 dB/octave SQNR-OSR trade-of

• MOD2 has NTF (z) = (1 – z–1)2⇒ 15 dB/octave SQNR-OSR trade-o

OSR f s 2(⁄≡

2-10

es]

OD1’s NTF

Q1 V

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1. MODN[Ch. 4 of Schreier & Tem

• MODN’s NTF is the Nth power of M

1z−

zz−1U z

z−1

NTF z( ) 1 z 1––( )N=STF z( ) z 1–=

N integrators(N–1) non-delaying, 1 delaying

2-11

)10

-1

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NTF ComparisonN

TF

ej2

πf(

)(d

B)

Normalized Frequency ( f10

-310

-2–100

–80

–60

–40

–20

0

20

40

MOD1

MOD2

MOD3

MOD4M

OD5

2-12

ce

N+1)th powerOSR trade-off

ee f( ) fd

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Predicted Performan• In-band quantization noise power

• Quantization noise drops as the (2of OSR— (6N+3) dB/octave SQNR-

IQNP NTF e j 2πf( ) 2 S⋅

0

0.5 OSR⁄

∫=

2πf( )2N 2σe2⋅ fd

0

0.5 OSR⁄

∫≈

π2N

2N 1+( ) OSR( )2N 1+-------------------------------------------------------σe

2=

2-13

nce–on the

of

i which minimize

22)2 xd , n = 4

12)2 xd , n = 3

12)2 xd , n = 2

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Improving NTF PerformaNTF Zero Optimizati

• Minimize the integral of overpassband

Normalize passband edge to 1 for easecalculation:

NTF 2

…

–1 1a–a

H f( )2

Need to find the a

x 2 a12–( )2 x 2 a–(

1–1

∫

x 2 x 2 a–(1–

1∫

x 2 a–(1–

1∫

fs/fB

the integral

2-14

= 8SQNR

Improvement

0 dB

3.5 dB

8 dB

13 dB

18 dB

23 dB

28 dB

9 34 dB

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Solutions Up to OrderOrder

Optimal Zero PlacementRelative to fB

1 0

2

3 0,

4

5 0, [Y. Yang]

6 ±0.23862, ±0.66121, ±0.93247

7 0, ±0.40585, ±0.74153, ±0.94911

8 ±0.18343, ±0.52553, ±0.79667, ±0.9602

1 3⁄±

3 5⁄±

3 7⁄ 3 7⁄( )2 3 35⁄–±±

5 9⁄ 5 9⁄( )2 5 21⁄–±±

2-15

n

1z−11

-g

g + Delaying (LDI Loop)

e roots ofz1)2

----------- 0=

the unit circle,θcos 1 g 2⁄–=,

f the value of g.

ECE1371

Topological Implicatio• Feedback around pairs integrators:

zz−

1z−1

1z−1

-g

2 Delaying Integrators Non-delayinIntegrators

Poles are the roots of

1 g 1z 1–------------

2+ 0=

i.e. z 1 j g±=

Not quite on the unit circle,but fairly close if g<<1.

Poles are th1 g

z –(---------+

i.e. z e jθ±=

Precisely onregardless o

2-16

dulatortizerantizer

er input getsven if the

tizer levels are a small input.

small.

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Problem: A High-Order MoWants a Multi-bit Quan

E.g. MOD3 with an Infinite Quand Zero Input

0 10 20 30 40-7

-5

-3

-1

1

3

5

7

Sample Number

v

Quantizlarge, e

6 quanused by

input is

2-17

1btizer)

o input!

40

40

HUGE!

Longstringsof +1/–1

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Simulation of MOD3-(MOD3 with a Binary Quan

• MOD3-1b is unstable, even with zer

0 10 20 30–1

0

1

0 10 20 30–200

–100

0

100

200

Sample Number

v

y

2-18

roblem

we lose the

tiation) error is

duces thettenuated.

ible

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Solutions to the Stability PHistorical Order

1 Multi-bit quantizationInitially considered undesirable becauseinherent linearity of a 1-bit DAC.

2 More general NTF (not pure differenLower the NTF gain so that quantizationamplified less.Unfortunately, reducing the NTF gain reamount by which quantization noise is a

3 Multi-stage (MASH) architecture

• Combinations of the above are poss

2-19

n1 is

all times,

,

n N-bitthan ∆/2 = 1.

that MOD N isale

1 h i( )i 0=∞∑=

0…} ai 0>

1 h 1– 1=

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Multi-bit QuantizatioA modulator with NTF = H and STF =guaranteed to be stable if atwhere and

• In MODN , so

and thus

• impliesMODN is guaranteed to be stable with aquantizer if the input magnitude is less This result is quite conservative.

• Similarly, guaranteesstable for inputs up to 50% of full-sc

u u max<u max nlev 1 h 1–+= h

H z( ) 1 z 1––( )N=h n( ) 1 a1– a2 a3– … 1–( )N aN, , , , ,{=

h 1 H 1–( ) 2N= =

nlev 2N= u max nlev +=

nlev 2N 1+=

2-20

ntizer)mid-tread

y

v

M +1

v: even integersfrom – M to +M

e ∆ 2⁄≤ 1=

ECE1371

M-Step Symmetric Qua∆ = 2, (nlev = M + 1

• No-overload range: ⇒

M even: M odd: mid-rise

e = v – ye = v – y

v

y1

M

–M

2

M

–M

–M –1 M +1 –M –1

v: odd integersfrom – M to +M

y nlev≤

2-21

iterion

Hypothesis]

0

u max )

h 1 1–

ECE1371

Inductive Proof of Cr• Assume STF = 1 and

• Assume for .[Induction

Then⇒⇒

• So by induction for all i >

h 1

n∀( ) u n( ) ≤(e i( ) 1≤ i n<

y n( ) u n( ) h i( )e n i–( )i 1=∞∑+=

u max h i( ) e n i–( )i 1=∞∑+≤

u max h i( )i 1=∞∑+≤ u max +=

u max nlev 1 h 1–+=y n( ) nlev≤e n( ) 1≤

e i( ) 1≤

2-22

h ,

d must be

B z( ) z n=

poles away toward z = 0 gain of theach unity.

ECE1371

More General NTF• Instead of wit

use a more generalRoots of B are the poles of the NTF aninside the unit circle.

NTF z( ) A z( ) B z( )⁄=B z( )

Moving thefrom z = 1makes theNTF appro

2-23

bility

the maximum the infinity-

tability?than FS)

ure stability?,

H ∞ 2≤

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The Lee Criterion for Stain a 1-bit Modulator:

[Wai Lee, 1987]

• The measure of the “gain” of H is magnitude of H over frequency, akanorm of H:

Q: Is the Lee criterion necessar y for sNo. MOD2 is stable (for DC inputs less but .

Q: Is the Lee criterion sufficient to ensNo. There are lots of counter-examplesbut often works.

H ∞ maxω 0 2π,[ ]∈

H ejω( )( )≡

H ∞ 4=

H ∞ 1.5≤

2-24

SR = 32H ∞

5 2

5 2 H ∞

H ∞

um

cliff!

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Simulated SQNR vs.5th-order NTFs; 1-b Quant.; O

1 1.25 1.5 1.750

70

90

1 1.25 1.5 1.7–20

–10

umax

(dB

FS

)S

QN

R(d

B)

0

Stable input limit dropsas increases.H ∞

SQNR has a broad maxim

2-25

ulation

N =

3

N = 2

N = 1

512 1024

ECE1371

SQNR Limits— 1-bit Mod

OSR

Pea

k S

QN

R (

dB)

N =

4

N =

6N

= 5

N =

7N

= 8

4 8 16 32 64 128 2560

20

40

60

80

100

120

140

2-26

ulators

N = 2

N = 1

6 512 1024

ECE1371

SQNR Limits for 2-bit Mod

N = 4 N = 3N =

6N

= 5

N =

7N

= 8

OSR

Pea

k S

QN

R (

dB)

4 8 16 32 64 128 250

20

40

60

80

100

120

140

2-27

ulators

N = 2

N = 1

6 512 1024

ECE1371

SQNR Limits for 3-bit Mod

N = 4 N = 3N =

6N

= 5

N =

7N

= 8

OSR

Pea

k S

QN

R (

dB)

4 8 16 32 64 128 250

20

40

60

80

100

120

140

2-28

Σ ADCtizer:

STF / NTF

F L0 NTF⋅=

F E⋅ , where

ECE1371

Generic Single-Loop ƥ Linear Loop Filter + Nonlinear Quan

E

L1

L0 Y VU

Inverse Relations:L1 = 1 – 1/NTF, L0 =

Y L0U L1V+=V Y E+=

NTF 11 L1–---------------= & ST

V STF U⋅ NT+=

2-29

nge

x”

ABCD: state-space descriptionof the modulator.

scaleABCD

D

chreier & Temes)

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∆Σ Toolboxhttp://www.mathworks.com/matlabcentral/fileexcha

Search for “Delta Sigma Toolbo

NTF (and STF)available.

Specify OSR,lowpass/bandpass,

no. of Q. levels.

synthesizeNTF

realize-

predictSNR,

Parameters fora specifictopology.

stuffABCD

mapABC

Time-domainsimulation andSNR measure-

ments.

simulateDSM,

calculateTF

simulateSNR

mapCtoDsimulateESLdesignHBF

Also

Manual is delsig.pdf (App. B of S

NTF

2-30

odel

M = 3

Mid-rise quantizer;: odd integers [ – M,+M ]

NTF 11 L1–---------------=

STFL0

1 L1–---------------=

4–4y

v3

1

-1

-3

e

n

ECE1371

∆Σ Toolbox Modulator M

L1

L0 Y VU

Quantizer:M = 1 M = 2

Mid-tread quantizer;v: even integers [ – M,+M ] v

Modulator:

[ -M, +M ][ -M, +M ]

∆ = 2

yv

y

v1

–1

2

–2

2

ee

–2 3–3

Loop filter can be specified by NTF orby ABCD, a state-space representatio

2-31

quency

thesband

in some are not

is maximallyes.)

0 h 1( )z 1– …+ +

ECE1371

NTF SynthesissynthesizeNTF

• Not all NTFs are realizableCausality requires , or, in the fredomain, . Recall

• Not all NTFs yield stable modulatorsRule of thumb for single-bit modulators:

[Lee].

• Can optimize NTF zeros to minimizemean-square value of H in the pas

• The NTF and STF share poles, and modulator topologies the STF zerosarbitraryRestrict the NTF such that an all-pole STFflat. (Almost the same as Butterworth pol

h 0( ) 1=H ∞( ) 1= H z( ) h 0( )z=

H ∞ 1.5<

2-32

emo1 ]t DC

;esizeNTF(5););

ce(0,0.5);i*pi*f);valTF (H,z);(H_z));Gain (H,0,0.5/OSR)

ECE1371

Lowpass Example [ dsd5th-order NTF, all zeros a

• Pole/Zero diagram:

OSR = 32H = synthplotPZ (H

f = linspaz = exp(2H_z = eplot(f,dbvg = rms

2-33

0.5

1/64

fs)

ECE1371

Lowpass NTF

0–80

–60

–40

–20

0

rms in-band

0–60

–50

–40

Out-of-band gain = 1.5

gain = –47 dB

dB

Normalized Frequency ( f /

2-34

ass NTF=32

TF (5,OSR,1);

pread acrossd-of-interest toe the rms valueTF.

optimizationflag

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Improved 5 th-Order LowpZeros optimized for OSR

OSR = 32;H = synthesizeN...

Zeros sthe banminimizof the N

2-35

0.5

1/64

/fs)

ECE1371

Improved NTF

0–80

–60

–40

–20

0

0–80

–70

–60rms gain = –66 dB

dB

Normalized Frequency ( f

2-36

nter frequency

[] or NaN meansuse default value,i.e. Hinf = 1.5

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Bandpass ExampleOSR = 64;f0 = 1/6;H=synthesizeNTF (6,OSR,1,[], f0 );...

ce

2-37

TF

0.5fs)

same poles as

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Bandpass NTF and S

0–60

–40

–20

0

NTFdB

Normalized Frequency ( f /

all-pole STF with∆Σ toolbox NTF

2-38

tionr may work

ulti-bit

e

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Summary: NTF Selec• If OSR is high, a single-bit modulato

• To improve SQNR,Optimize zeros,Increase , orIncrease order.

• If SQNR is insufficient, must use a mdesign

Can turn all the above knobs to enhancperformance.

• Feedback DAC assumed to be ideal

H ∞

2-39

sdemo2 ]

c

2048FT)

NBW=1.5 bins

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NTF-Based Simulation [ d

• In mex form; 128K points in < 0.1 se

order=5; OSR=32;ntf = synthesizeNTF (order,OSR,1);N=2^17; fbin=959; A=0.5; % 128K pointsinput = A*sin(2*pi*fbin/N*[0:N-1]);output = simulateDSM (input,ntf);spec = fft(output.* ds_hann (N)/(N/4));plot( dbv (spec(1:N/(2*OSR))));

0 1024–200

–150

–100

–50

0

FFT Bin Number (128K-point F

dBF

S/N

BW

2-40

nt’d

0.5

100

W = 1.8x10–4 fs-point FFT)

/fs)

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Simulation Example Co

0–120

–100

–80

–60

–40

–20

0

dBF

S/N

BW

0 50

–1

0

1

Time (sample number)

NB(8k

u

v

Normalized Frequency ( f

2-41

lateSNR

0

Peak SNR is 85 dB.Max. input is –3 dBFS.

ECE1371

SNR vs. Amplitude: simu

Input Amplitude (dBFS)

SQ

NR

(dB

)

–100 –80 –60 –40 –200

20

40

60

80

100

[snr amp] = simulateSNR (ntf,OSR);plot(amp,snr,'b-^');

OSR=32

2-42

-01-19)o2 to:

ptimized fore poles/zeros.

dulator with

forms.oise curve. *

urve and

answer it.

ll-scale is M.

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Homework #2 (Due 2015A. Extract code from dsdemo1 & dsdem

1 Create a 3 rd-order NTF with zeros oOSR = 32 and . Plot thand frequency response of your NTF

2 Simulate an 8-step (9-level) ∆Σ mothis NTF.Plot example input and output wavePlot a spectrum and the predicted nPlot the SQNR vs. input amplitude cnote the maximum stable input.

B. Compose your own short question and

*. Beware that with an M-step modulator the fu

NTF ∞ 2=

2-43

day solidify

box

ECE1371

What You Learned ToAnd what the homework should

1 Nth-order modulator (MOD N )

2 High-level design with the ∆Σ Tool

2-44

hase

ECE1371

NLCOTD: True Single-PDynamic FF

+ Clock not inverted anywhere

+ Small

+ Fast

D

C Q

2-45

Q

Can drop inverter.(Handy if makinga divider.)

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TSPFF Operation

D

C

C

C

X

Y Z

DCXYZQ

d

~d

d~d

d

2-46

ped.mic nodes at

ECE1371

TSPFF Gotchas• Leakage:

Won’t work if clock is too slow.Possible high current if clock is stop

Need to add devices that hold the dynaa safe value.

• No positive feedbackVulnerable to metastability.