ECE145A / 218A Notes : Basic Analysis of Analog Circuits...Basic Analysis of Analog Circuits Mark...

notes, M. Rodwell, copyrighted

ECE145A / 218A Notes :Basic Analysis of Analog Circuits

Mark RodwellUniversity of California, Santa Barbara

[email protected] 805-893-3244, 805-893-3262 fax


Comment

This (2009) is a transitional year:

Next year 145abc will be reorganizedNext year 145abc will be reorganized,145a: fundamentals (devices, analog & RF analysis, models)145cb: RF systems at IC and system level

This year:some students have taken 145c:

l d h d i d lalready have device modelsalready know analog circuit analysis well

some students have notmust cover device modelsmust review some circuit analysis methods

These notes: shortened version (2009 only) of device models


Transistor Circuit Design

This note set

-reviews the basics-starts at the level of a first IC design course-starts at the level of a first IC design course -moves very quickly

This will

establish a common terminology-establish a common terminology-accommodate capable students having minimal background in ICs.


DC d lDC modelsDC bi l iDC bias analysis


Large-Signal Model For Bias Analysis

IcIb

Vbe

+ -

,0 that Provided ceV >

resistanceemitterthetointernalspecifiedisthatnote/ where,/ and )/exp(

,

TcbTbesc

ce

RVqkTVIIVVII === β

d ititttiHBTfti ifiidTh

resistanceemitter thetointernalspecified is that note... exbe

RI

RV

bandwidth.istor peak transfor requirednear that densitiescurrent at operatingHBTsfor t significan is drop The exeRI


DC Bias Example: Current Mirror

refI

1Q2Q

2cI

)()(haveWe 22221111 bb RRIVRRIV ++=++

2exR

2eeR

1exR

1eeR1eI2eI

2bI 1bI

)/ln( ,)/ln( and)()( have We

222111

22221111

sctbesctbe

eeexebeeeexebe

IIVVIIVVRRIVRRIV

==++++

area)emittertheis(thatassume&2 ,1 that Assume 12 eeee

AAARR

==>>β

2and2/impliesThis

area).emitter theis( that assume & 21 EEE

IIRR

AAA

==

=

2/ find which wefrom,2and,2/ implies This

12

2121

cc

ssexex

IIIIRR

===


Simpler DC Model for Bias Analysis

IcIb

IcVbe

Ib

Vbe

==takeinsteadandwithofvariation theignore toanalysis biasin sufficientoften isIt

φbbb VVIV

t h ldd ittd d

. takeinsteadandwith ofvariation , φonbebecbe

V

VVIV

bias,bandwidth peak of 10%~ within densitiescurrent at Biasedy. technologanddensity current upon depends ,onbeV

⎪⎨

⎧= HBTsInGaAs/InPV9.0or 7.0

HBTs Si/SiGeModern V 9.0~φonbeV⎪⎩

⎨HBTs GaAs/GaInP V .41

, φonbe


Simple DC Bias Example

I I 1

Rc

Ib2 Ib1

Q2 Q1

Rb

Ree

-Vee

ee

bbbb

VVVVRI

(SiGe). V 9.0 eApproximat Volts. 0 then drops, eneglect th weIf

21

21

−≅−====

φ

eeeeccc

ee

RVVIIRVVIII

2/)90(/)9.0(2

21

121

21

−−==−−==+

eeeecc RVVII 2/)9.0(21


Efficiently Handling Base Currents In Bias Analysis

:equationsussimultaneosolvecanonet,singifican is drop If bbRI

I Ib1

Rc

/where/)(2

:equationsussimultaneo solve can one

1121

βφ eebbeeccc

IIRRIVIII −−−==+

Ib2 Ib1

Q2 Q1

Rb

it tibfi dQ i k

,/ where 11 βcb II =

Ree

1

2/)( solve 1):iterationby find :Quicker

1 φ eeeec RVI −−=-Vee

2/)( solve to of value thisuse 3)/ solve 2)

11

11

φβ

eebbeecb

cb

RRIVIIII

−−−=≅ ee

bias DChascircuit designed-any well because Works. upondependent y only weakl β


ll i lsmall-signal b b d l ibaseband analysis


Hybrid-π Bipolar Transistor Model

Ccbx

CcbiRbbB CRc

bf τ+τ=τ R Vbe gmVbee-jωτcmbe gR /β=

cbf τ+τ=τ Rbe

R

be

CCb diff =g τ f

gm be

ERexCje

Cbe,diff gmτ f

Accurate model, but too detailed for quick hand analysis


Oversimplified Model for Quick Hand Analysis

bemVg

+

beRbeC bemVg

+

beRB BC C

cbC

beV+

−beV+

−

In most high frequency circuits the node impedance is low and

E E

In most high-frequency circuits, the node impedance is low and Rce is therefore negligible.

Neglecting Rbb in high-frequency analysis is a poor approximationNeglecting Rbb in high frequency analysis is a poor approximation but is nevertheless common in introductory treatments.

The "textbook" analyses which follow use this oversimplified model. These introductory treatments will later be refined. y


Common Emitter Stage: Basics

LeqRCLeqC IRV δδ ⋅−= )2

/)5 βδδ b II =

q

)(/ )7 , eebbTin RrIVR +== βδδ/)5 βδδ cb II

ce II δδ ≅ )1+mebe gIV / )4 δδ =

δδ)3

−

eEe IRV δδ ⋅=)3

)()( )6 eebeeeb RrIRrIV +⋅=+⋅= βδδδ EReebeeeb

)/1/()/( / / )8 meLeqeeLeqBoutinout gRRRrRVVVV +−=+−== δδδδ

)( is Transistor ; )/1/(- isGain EeinmeLeq RrRgRR ++ β


Emitter Follower Stage: Basics

/)4 βδδ cb II =

)(/ )6 , LeqebbTin RrIVR +== βδδ/) βδδ cb

ce II δδ ≅ )1+mebe gIV / )3 δδ =

IRV δδ)2

−

eLeqe IRV δδ ⋅=)2

)()( )5 LeqebLeqeeb RrIRrIV +⋅=+⋅= βδδδ LeqRqq

)/1/()/( / / )7 mLeqLeqLeqeLeqEoutinout gRRRrRVVVV +=+== δδδδ

)( is Transistor ; )/1/( isGain EeinmLeqLeq RrRgRR ++ β


Common-Base Stage: Basics

CLeqout IRV δδ ⋅=)2)/( )6 βδδ beein RrIV +⋅=

βδδ //)7 RIVR βδδ // )7 , BeeeTin RrIVR +==

mebe gIV / )5 δδ =ce II δδ ≅ )1

+−

LeqRBR

/ )3 βδδ cb II = / )4 βδδ bcb RIV ⋅=

cb

)//( / )7 βδδ beLeqinout RrRVV +=

ββ / is Transistor ; )//( isGain beinbeLeq RrRRrR ++


Emitter Follower Output ImpedanceECE145C /218C notes, M. Rodwell, copyrighted

Common-Base Stage: Basics

CLeqout IRV δδ ⋅= )2)/( )6 βδδ beein RrIV +⋅=

βδδ // )7 , BeeeTin RrIVR +==

BRemitteroutR ,

ampoutR ,

mebe gIV / )5 δδ =ce II δδ ≅ )1

LeqR

+−

BR

B

/ )3 βδδ cb II =

)//( / )7 βδδ beLeqinout RrRVV +=

/)4 βδδ bcb RIV ⋅=

EERββ /isTransistor;)//(isGain beinbeLeq RrRRrR ++EE

impedanceinput C.B. as problem same is impedanceoutput E.F.

ββ //1/, BmBeemitterout RgRrR +=+=

RRR EEemitteroutampout RRR ,, =


Including Bias Circuit Resistances

CRLCLeq RRR =1BR

TinR ,

1BRinTBBAmpin RRRR 21, =

LRgenR inV

1BRV

t i tith thll lidd d)(t i i llTh

1BRgenV

.impedancescircuit net thedetermine toimpedances terminalrtransistowith theparallelin added )(trivially are These

etc. , )/(/ which,From ,, genampinampingenin RRRVV +=


Baseband Analysis Of Multistage Circuits

Q Q

For baseband analysis of multi-stage circuits, simply break into individual stages.

Q2Q1 Q4

Q3 Load impedance of the Nth stage includes the input impedance of the (N+1)th stage

Analysis is then trivialAnalysis is then trivial...

Rin4Rin3Rin2

Vout2

Q2Q1 Q4

Q3Vin1

in4

Vout1

out2Vout3

R

Rin3

Vin2=Vout1 Rin4

Vout2=Vin3Vout3=Vin4

Rin2Vin2 Vout1 Rin4


ll i lsmall-signal b b d l ibaseband analysis


High-Frequency Analysis: The General Problem

fF4.6=cbxCfF3.7=cbiCΩ= 49bbR

Ω= 6500cR

B C

ebmVg '

cbiΩ49bbR

Ω= 3.4exR

ebV '

B C

=diffCfF182

=jeCfF38

=πRΩ350

E fF4.6=cbxCfF3.7=cbiCΩ= 49bbR

Ω= 6500cR

VV

B C

=R

Ω= 3.4exR

ebmVg 'ebV '

E

=diffCfF182

=jeCfF38

=πRΩ350

Q2Q1

VgenE

Analyzing frequency response is difficult: cannot separate stage-by-stage

Method #1: nodal analysis: accurate, general, tedious.y , g ,

Method #2: method of time constants: accurate, limited applicabilty, quick & intuitive


N d l A l iNodal Analysis


Tutorial: Transfer Function Analysis: Nodal Analysis I

amplifier.emitter -common :examplefamiliar very & SimpleVin

Vout

Rgen

RLCLRb

Vgen

CB C

Rgen

Vin VoutCcb

ER

Vgen

RLCL

Rb gmVbeRbeCbe


Tutorial: Transfer Function Analysis: Nodal Analysis II

:circuit Reduced Vin Vout

)||||( bbegeni RRRR =

RLCLRiIi=Vgen/Rgen gmVbe

Cbe

KCLfrom EquationsNodal Write :1 Step

⎥⎤

⎢⎡

=⎥⎤

⎢⎡⎥⎤

⎢⎡ −++ iincbcbbei IVsCsCsCG

⎥⎦

⎢⎣

⎥⎦

⎢⎣⎥⎦

⎢⎣ ++− 0outcbLLcbm VsCsCGsCg


Tutorial: Transfer Function Analysis: Nodal Analysis III

inout sDsNIV = )(/)(/: EquationsNodal Solve :2 Step

cbmcbm

cbbei sCgsCg

sCsCGsN −−=

−++

= )(01

)(

cbLLcbm

cbcbbei

sCsCGsCgsCsCsCG

sD++−

−++=)(

cbLLcbmg

( )( ) ( )( ))( cbcbmcbLLcbbei sCsCgsCsCGsCsCGsD −−−++++= ( )( ) ( )( )

ofpowersinOrganize:3Step

)( cbcbmcbLLcbbei

s

g

()(

ofpowers in Organize :3 Step

Li

CgCGCGCGCGsGGsD

s

+++++=

)(

( 2

cbcbcbcbLcbcbbeLbe

cbmcbLbeLcbiLi

CCCCCCCCCCs

CgCGCGCGCGs

−++++

+++++


Tutorial: Transfer Function Analysis: Nodal Analysis IV

.function..transfer thefromgainsandconstants separating form, spolynomial-of-ratio essdimensionl into Separate :4 Step

==)()(

/

gp g

gengen

out

in

out

sDsN

RVV

IV

⎟⎞

⎜⎛ +++++

−−=

()(

)(

bbLbLbiLiLi

cbm

gengenin

CgCGCGCGCGsGGsCg

⎟⎟⎠

⎞⎜⎜⎝

⎛

+++

+++++

)(

(2

LcbcbbeLbe

cbmcbLbeLcbiLiLi

CCCCCCs

CgCGCGCGCGsGG

/)( −− genLicbmout RRRsCgV

)(

(1

2 ⎟⎟⎠

⎞⎜⎜⎝

⎛

+++

+++++=

LiLcbcbbeLbe

cbLimcbibeicbLLL

g

gen

out

RRCCCCCCs

CRRgCRCRCRCRsV

)( ⎠⎝ LiLcbcbbeLbe


Tutorial: Transfer Function Analysis: Nodal Analysis V

)||()||()||||(

that note,

,

+=

+==

genAmpin

Ampin

genbbe

bbe

gen

genbbe

gen

i

RRR

RRRRR

RRRR

RR

so...,

⎟⎞

⎜⎛

gpggg

RV ( )

)/1(

,

, −⎟⎟⎠

⎞⎜⎜⎝

⎛

+= Lm

genAmpin

Ampin

gen

out

C

RgRR

RVV

1a1b

)(

)(1)/1(

2 ⎟⎟⎞

⎜⎜⎛ +++++

−−×

cbLimcbibeicbLLL

mcb

CRRgCRCRCRCRsgsC 1a

)( 2 ⎟

⎠⎜⎝ +++ LiLcbcbbeLbe RRCCCCCCs

2a

11

)()(

21

sasasb

VV

sVsV outout

+++

=⇒

1)( 21 sasaVsVbandmidgengen ++

−


Tutorial: Transfer Function Analysis: Nodal Analysis VI

11)( 11 sbVsbVsV ++

polynomial theofzeros)&(polesroots the Find:5 Step

)/1)(/1(1

11

)()(

21

12

21

1

ppbandmidgen

out

bandmidgen

out

gen

out

sssssb

VV

sasasb

VV

sVsV

−−+

=++

+=

−−

?polesthefindingofmethodsefficientarewhat ?poles thefindingofmethodsefficient arewhat


Fi di P lFinding Poles f T f F tifrom Transfer Functions


Finding Poles and Zeros

1)(

: FormPolynomial-of-Ratio2sbsbVsV +++

...1...1*

)()(

221

21

band-midat

m

gen

out

gen

out

sasasbsbs

VV

sVsV

++++++

=

:Zeros and Poles

).../1)(/1)(/1().../1)(/1)(/1(*

)()(

111

111

ppp

zzzm

gen

out

gen

out

sssssssssssss

VV

sVsV

−−−−−−

=))()(()( 111band-midat pppgengen


Finding Poles: Complex Poles

++++++

=out

sasasasbsbk

VV

1...1

32

221

⎟⎟⎞

⎜⎜⎛ +++

≈

<<

+++

out

gen

sbsbkV

saaa

sasasaV

...1

and sfrequencie moderateat theignore can we then/ If

1

221

3223

321

⎟⎟⎠

⎜⎜⎝ ++

≈gen

out

sasak

V

thencomplexarethisofrootstheIf

1 221

21

nωdjω

⎟⎟⎠

⎞⎜⎜⎝

⎛+++++

=⎟⎟⎠

⎞⎜⎜⎝

⎛++

+++=

nngen

out

sssbsbk

sasasbsbk

VV

ωωζ /)/2(1...1

1...1

thencomplex, are thisof roots theIf

22

221

221

221

nζω

⎟⎟⎟⎟⎞

⎜⎜⎜⎜⎛

⎟⎞

⎜⎛⎟⎞

⎜⎛

+++=out

sssbsbk

VV ...1 2

21 )1( 222 ζωω −= nd

⎟⎟⎠

⎜⎜⎝

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

+⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−dndn

gen

js

jsV

ωζωωζω11


Finding Poles: Separated Pole Approximation

then,)/( e.g. separated widely are roots theIf

112 aaa <<

1...1

,)(g

2

221

112

sasasbsbk

IVout

⎟⎟⎠

⎞⎜⎜⎝

⎛++

+++=

ωj

...1

12

21

21

sbsbkI

V

sasaI

out

in

⎞⎛ ⎞⎛+++

≅

⎠⎝ ++

σ12 / aas −=

( ) 111

21 s

aasa

Iin

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛++

σ1as −=

pole.dominant theis 1a


I t d tIntroductory Ci it D iCircuit Design: summary


Gain Stages: Elementary Bandwidth Analysis

Using the oversimplified device model below, with Cpi denoting the sum of base-emitter depletion and diffusion capacitances, bandwidth of CE/CB/CC stages can be found….g

VRC C

V

bemVg

+

beRbeCB C

cbC

beV−

EE


CE Stage: Elementary Bandwidth Analysis

RgenRL

CL

Rc

RL

Rin

Ri is the parallel combination of Rgen, Rin, and Rpi

RLeq is the parallel combination of RL, Rc, and Ro

Note in the dominant pole (a1) the miller-multiplication of the collector base capacitancecapacitance


CC Stage: Elementary Bandwidth Analysis

RgenCL

RinREE

RL

Ri is the parallel combination of Rgen, and Rin,

RLeq is the parallel combination of Ree and RL

Note that the frequency response is a mess. Given CL, the transfer function very often has complex poles and may show strong gain peaking hencevery often has complex poles, and may show strong gain peaking, hence ringing in the pulse response.


CB Stage: Elementary Bandwidth Analysis

Rgen Rcc

RL

CL

bmjecbin gCC τ+=,

RinRgen Rcc

RL

CLCin,cb CcbVin

V

kTqIjg ω =/)(

Rin

RL

Rin gm*Vin

Vgen

cm j

jgωτ

ω+

=1

)(

Here we have a problem. To the extent that the CB stage is modeled by a very very simple hybrid-pi model (explicitly, with zero Rbb), we find (by very simple analysis) very high bandwidth with poles having time constants equal to tau banalysis) very high bandwidth, with poles having time constants equal to tau_b, to tau_c, and to the product of the load resistance times (Ccb+CL).

Note that1) I i i i d d d D i l d ff f1) Input capacitance is indeed as noted. Does not include effect of tau_c

2) Ignoring Rbb in CB stage analysis, while appealing for simplicity (e.g. undergrad classes) is quite unreasonable, as CcbRbb often dominates g ) q ,high frequency rolloff. More regarding this later.


M th d fMethod ofTi C t tTime Constants


make revision for 2009----first before MOTC, give by summary without derivation the standard stage expressions

M th d f

without derivation the standard stage expressions.

then define MOTC, first and second order

Method ofTi C t t

then show a 1-stage Darlington diff amp, and say caps to ground, caps between inputs and outputs.

Time ConstantsGive expression for caps to groundGive expression for caps between in and out of general block

then use this for CD stage Cgs onlythen use this for CC stage Cbe onlythen do for CE stage Ccb only

then work the full Darlington diff ampthen work the full Darlington diff amp

then show how CE (with degen) CB CC are same problem

then re-show stage relationship


Finding Bandwidth: Method of Time Constants

ebmVg '


Ω= 6500cR

ebV '

B C

=πRebmg

Ω= 3.4exR

eb'

E

=diffCfF182

=jeCfF38

Ω350


Ω= 6500cR

B C

Q2Q1

Vgen

Ω= 3.4exR

ebmVg 'ebV '

E

=diffCfF182

=jeCfF38

=πRΩ350

take a general RC network (no inductors or delays tau), and separate into 2 parts, network without capacitors, and the capacitors:


MOTC: Separation into Capacitors & Resistive N-port

The internal capacitor-free network is now frequency-independent The MOTC method (not proven here) relies onindependent. The MOTC method (not proven here) relies on results from n-port network theory


MOTC: Open-Circuit Resistances

i hdii lllhi0

by determined is This circuited.-open portsother allwithoneport at measuredresistancesignal small theis 0

11R

e)(or voltagcurrent resulting the thisfromcomputingandport at thecurrent)(or age test volta applying


MOTC: the Dominant Time Constant

theus givedirectly constants order time-first MOTC The

dB-3 the,negligible is constant timesecondary theif)only (andIf circuit.theofconstant imedominant t

2

1

aa

. determine toconstants timecircuit)-(shortorder -second theusemust We.2/1 isbandwidth

2

1

aaπ


MOTC: Are We Saving Any Work ?

Are we saving work relative to brute-force nodal analysis:MOTC would be of only moderate value if we had to calculateMOTC would be of only moderate value if we had to calculateall the Ri's each time. Fortunately, most terms involve quantities already found in midband stage analysis: input and output impedances, load impedances, etc.


MOTC: Short-Circuit Resistances

circuited,-openportsotherall withoneport at measured resistance signal small theis 2

11R

shorted is which 2,port for except circuited,open portsother all


MOTC: The Second-Order Time-Constant


MOTC: Working these Efficiently

inchoices2havealwaysweBecause 00 RRRR xy =

work the tois trick The MOTC. in the each term finding

in choices2have alwayswe, Because RRRR yyxxyyyxx =

impedancesloadouput,input,torelated are terms1):possible asmuch as that so problem

willIoften thatsoarisewhichcasesfunny""2areThereanalysis. in earlier, found ones are terms2)

ppp)

1a

)illkllhl d intimately are these(note pages 2next on the themgive

willIoften thatso arisewhich casesfunny 2 are There

effect)Miller known - well the torelated


MOTC and the Miller Effect

outviyxx RARR ++= )1( outvixx )(

outRvARiR

C

RoutRvA

iR[ ] CRARa •++== )1(τ [ ] CRARa outvi •++== )1(1 τ


MOTC: port impedances between collector and base

xR parallel thedenote tois that explicitly decide weIf

Leq

be

RR

ofeffect combined thedenotessimilarly that and , and sresistancecircuit externalany ofn combinatio

LeqLeqmxyy

ce

RRgRR

R

++= )1(

then , and resistors external0

LeqLeqmxyy g )(


MOTC: Port Impedances Between Emitter & Base


MOTC: Multistage Example

work on the board...


E dEnd

ECE145A / 218A Notes : Basic Analysis of Analog Circuits...Basic Analysis of Analog Circuits Mark...

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