Amplifier:two-portpowergains,stability

WatcharapanSuwansan;suk

#12EIE/ENE450AppliedCommunica;onsandTransmissionLines

KingMongkut’sUniversityofTechnologyThonburi

Wewillcoverthesetopics•  Two-portpowergains•  Stability– Stabilitycircles– K-Δtest

ARerthislecture,youwillbeableto•  Compute thepower gain, the available power gain, and thetransducer power gain of a two-port amplifier, given S-parameters

•  Iden;fycomponentsinasingle-stagetransistoramplifier

•  Compute a transducer gain of a single-stage transistoramplifier

•  Defineuncondi(onallystableandcondi(onallystable•  UsetheK-Δtesttocheckfortheuncondi;onalstability•  Interprettheinputandoutputstabilitycircles•  DrawtheinputandoutputstabilitycirclesinaSmithchart

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Signalamplifica;onisoneofthemostbasicfunc;onsinmoderncommunica;on

•  Anamplifierincreasesthesignalpoweratcertainfrequencies

•  Anamplifierisusuallyconstructedfromtransistordevices– BJT,HBT,MESFET,MOSFET,HEMT,...

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www.wenteqmicrowave.com www.bhe-mw.eu www.teseq.com

9kHz...4GHz,24.5dB,max.-10dBm

Amplifiesthesignalinthefrequencybandof9kHz–4GHzby24.5dB.Acceptstheinputpowerupto-10dBm.

Example1:Originalsignal

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AudioSource vs(t)

+

_

speaker(aload)

200 250 300 350 400 450 500 550 600−50

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−40

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−25

−20

−15

−10

Frequency(Hz)

Power(dBm)

d r m D R

Spectrumoftheoriginalsignal

Example1(con;nued):Amplifiedsignal

6

AudioSource

+

_

Amplifierby10dB430-600Hz

200 250 300 350 400 450 500 550 600−50

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−40

−35

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−25

−20

−15

−10

d r m

D RPower(dBm)

10dB

Frequency(Hz)

430 600

vL(t)

Spectrumoftheamplifiedsignal

Atransistorcanbemodeledasa2-portnetwork

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EmiierBase

Collector

Base Collector

Emiier

Scaieringmatrix[S]

Transistor:

2-portnetwork:

Manufacturersusuallyprovidethescaieringparametersinthedatasheets

8

(con;nued)

9

(con;nued)

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6.1TWO-PORTPOWERGAINS

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Recall:themaximumpowerisdeliveredtotherightnetworkwhen.

•  Thecondi;onisthesameas–  becausedoubleconjuga;onsgivesbackthecomplexnumber

12

LeRnetwork

Rightnetwork

Zleft

Zright

Zright = Z⇤left

Z⇤right = Zleft

(z⇤)⇤ = z

VS

Recall:Weprovedtheconjugatematchingcondi;oninsec;on2.6

•  First, reduce the networks to the the source and loadimpedances,asshown

•  Then,applytheconjugatematchingcondi;onofsec;on2.6

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Zleft

ZrightVS

Weconsideratransistorconnec;ngtoasourceandaloadofanyimpedance

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transistor(2-portnetwork)

sourceimpedance

loadACvoltagesource

thereferenceimpedancewhenmeasuringtheSparameters

Zin

[S]

(Z0)ZL

Zout

loadimpedance

VS

ZS

Wewillcover3defini;onsoftwo-portpowergains

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1.  Powergain

powerdissipatedintheloadZLpowerdeliveredtotheinputofthe2-portnetwork

G =PL

Pin

[S]

(Z0)ZL

VS

ZS

(con;nued)

2.  Availablepowergain

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poweravailablefromthe2-portnetwork

poweravailablefromthethesource

GA =Pavn

Pavs

[S]

(Z0)

Zout

ZL = Z⇤out

Zin

Zs = Z⇤in

conjugatematchingformaximumpower

=PL

���ZL=Z⇤

out

Pin

���ZS=Z⇤

in

VS

(con;nued)

3.  Transducerpowergain(ourfocus)

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poweravailablefromthethesource

[S]

(Z0)

Zin

GT =PL

Pavs

powerdissipatedintheloadZL

ZL

conjugatematchingformaximumpower

=PL

Pin

���ZS=Z⇤

in

VS

ZS = Z⇤in

Intermsofthescaieringparameters,thesegainsare...

1.  Powergain

2.  Availablepowergain

3.  Transducerpowergain

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G =|S21|2 (1� |�L|2)

(1� |�in|2) |1� S22�L|2

GA =|S

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|2 (1� |�S |2)|1� S

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�S |2 (1� |�out

|2)

�in = S11 +S12S21�L

1� S22�L

�out

= S22

+S12

S21

�S

1� S11

�S

�L =ZL � Z0

ZL + Z0

�S =ZS � Z0

ZS + Z0

GT =1� |�S |2

|1� �in�S |2|S21|2

1� |�L|2

|1� S22�L|2

Hereareremarksaboutthesegains• Whenand,thegainismaximizedandthethreedefini;onsareequal:

• When,wegetandthetransducerpowergainreducesto

• Whenorisnegligiblysmall(manytransistorshavethisproperty),thetransducerpowergainbecomes

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ZS = Z⇤in ZL = Z⇤

out

G = GA = GT

ZS = ZL = Z0 �S = �L = 0

GT = |S21|2

S12 = 0

GTU =1� |�S |2

|1� S11�S |2|S21|2

1� |�L|2

|1� S22�L|2

Thisgainiscalledtheunilateraltransducerpowergain

Example1:Comparisonofpowergaindefini;ons

•  A silicon bipolar junc;on transistor has the followingscaiering parameters at 1 GHz, with a 50 Ω referenceimpedance(i.e.,Z0=50Ω)

•  Compute the power gain, available power gain, andtransducerpowergain

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S11 = 0.38\� 158� S12 = 0.11\54�

S21 = 3.50\80� S22 = 0.40\� 43�

ZS = 25 ⌦

ZL = 40 ⌦transistorVS

Example1:Solu;on1.  Powergain

2.  Availablepowergain

3.  Transducerpowergain

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G =|S21|2 (1� |�L|2)

(1� |�in|2) |1� S22�L|2

GA =|S

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|2 (1� |�S |2)|1� S

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�S |2 (1� |�out

|2)

�in = S11 +S12S21�L

1� S22�L

�out

= S22

+S12

S21

�S

1� S11

�S�S =

ZS � Z0

ZS + Z0

GT =1� |�S |2

|1� �in�S |2|S21|2

1� |�L|2

|1� S22�L|2

�L =ZL � Z0

ZL + Z0Instudent’sversions:Removeanswerattheunderlines

Example1(Solu;on):WecanalsowritethesegainsintheunitofdB

1.  Powergain=

2.  Availablepowergain=

3.  Transducerpowergain=

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Instudent’sversions:Removeanswerattheunderlines

Wewillcoveronedesignmethod:asingle-stagetransistoramplifica;on

•  The amplifier circuit has two matching networks, whichcontroltheimpedancestobeZSandZL

•  ThevaluesofZSandZLarechosenaspartofthedesign

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Z0

Transistor[S]

(Z0)

Outputmatchingcircuit

Inputmatchingcircuit

Z0

ZLZS

VS

Thesingle-stagetransistoramplifica;oncanbereducedtothepreviousmodel

•  So we can also compute various gains of the single-stagetransistoramplifier– usingthesameexpressionsforG,GA,GT

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Transistor[S]

(Z0)

ZL

ZS

VS

Thegaincommonly-usedinthesingle-stagedesignisthetransducerpowergain

•  Sameexpressionasbefore,butwecaninterprettheterms:

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Z0

Transistor[S]

(Z0)

Outputmatchingcircuit

Inputmatchingcircuit

Z0

ZLZS

VS

GT =1� |�S |2

|1� �in�S |2|S21|2

1� |�L|2

|1� S22�L|2

Gainduetheinputmatchingnetwork

Gainduetheoutputmatchingnetwork

12.2STABILITY

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Oscilla;onoccursifwedidnotdesignanamplifierproperly

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AudioSource

+

_Unstableamplifier

vun(t)

Power(dBm)

Frequency(Hz)

Spectrumofanunstable,amplifiedsignal

200 250 300 350 400 450 500 550 600−50

−45

−40

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−20

−15

−10

(TheamplifiercannotamplifythefrequenciesofDandR)

d r m

l t

(con;nued)

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AudioSource

+

_

Anotherunstableamplifier

200 250 300 350 400 450 500 550 600−50

−45

−40

−35

−30

−25

−20

−15

−10

Power(dBm)

Frequency(Hz)

Spectrumofanunstable,amplifiedsignal

(Theamplifiercannotamplifythefrequenciesofl,t,DandR)

d r m

vuns(t)

Acondi;onforoscilla;ondependsonthereflec;oncoefficientsatinputandoutput

•  Oscilla;onoccursbecauseofanega;vereal-partofaninputoroutputimpedance:or

•  Inotherwords,becauseofor

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Zin Zout

Re{Zin} < 0 Re{Zout

} < 0

|�out

| > 1|�in| > 1

Z0

Transistor[S]

(Z0)

Outputmatchingcircuit

Inputmatchingcircuit

Z0VS

1

�in

1

�out

Thedefini;onsofstabilitywillbestatedintermsofthereflec;oncoefficients

•  Recallthat

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Transistor[S]

(Z0)

Outputmatchingcircuit

Inputmatchingcircuit

ZL

VS

�L =ZL � Z0

ZL + Z0�S =

ZS � Z0

ZS + Z0

�out

=Zout

� Z0

Zout

+ Z0

= S22

+S12

S21

�S

1� S11

�S�in =

Zin � Z0

Zin + Z0= S11 +

S12S21�L

1� S22�L

Zin ZoutZS

Thereare2maindefini;onsofstability• Wesaysthatanamplifierisuncondi(onallystableifforallandsuchthatand

• Wesaysthatanamplifieriscondi(onallystableorpoten(allyunstableifforsomeandsuchthatand

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�S �L |�L| < 1|�S | < 1

|�in| < 1 |�out

| < 1and

�S �L |�L| < 1|�S | < 1

|�in| < 1 |�out

| < 1and

K-ΔTESTFORUNCONDITIONALSTABILITY

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Hereisasimpletestofuncondi;onalstability

•  Theorem (K-Δ test):An amplifier is uncondi;onally stable ifandonlyifthetwocondi;onsbelowarebothsa;sfied:

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and

(ThisiscalledRollet’scondi;on)

Δisthedeterminantofthescaieringmatrix[S]

Example2:K-Δtest•  A Gallium Arsenide (GaAs) MESFET transistor has thefollowingscaieringparametersat4GHz(Z0=50Ω)

•  Ques;on:Isthetransistoruncondi;onallystableat4GHz?•  Answer:YesThescaieringparametersgivesusand

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S11 = 0.72\� 116� S12 = 0.03\57�

S21 = 2.60\76� S22 = 0.73\� 54�

|�| = 0.487 < 1K = 1.19 > 1

Example2:Matlabclear all;! !S_abs = [0.72 0.03 ; 2.60 0.73]; % magnitude of Sij!S_ang = [-116 57 ; 76 -54]; % angle of Sij (degree)! !S = S_abs .* exp( j*S_ang*pi/180 ); % scattering matrix! !abs_Delta = abs( det( S ) ) % magnitude of Delta!K = (1 - abs( S(1,1) )^2 - abs( S(2,2) )^2 + ...! abs_Delta^2 ) / ( 2*abs( S(1,2)*S(2,1) ) ) % parameter K! !% print out the result!if ( K > 1 && abs_Delta < 1 )! fprintf('Unconditionally stable\n');!else! fprintf('Conditionally stable\n'); !end !

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Example3:Isthetransistoruncondi;onallystableat1.9GHz?

•  The Triquint T1G6000528-Q3 Gallium nitride (GaN) HEMTtransistorhasthefollowingscaieringparametersat1.9GHz(Z0=50)

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S11 = �0.810� j0.315

S22 = �0.231� j0.451

S12 = 0.031� j0.005

S21 = 2.049 + j3.724

(A) Yes (B) No

STABILITYCIRCLEFORCONDITIONALSTABILITY

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Stabilitycirclesdefinetherangesofloadandsourceimpedances...

•  thatanamplifierisstable

•  The output stability circle define the set of the loadimpedances’s(orequivalently’s)thatyieldstability

•  The input stability circle define the stable set of the sourceimpedances’s(orequivalently’s)thatyieldstability

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ZL �L

�SZS

Transistor[S]

(Z0)

Outputmatchingcircuit

Inputmatchingcircuit

ZL

VS

ZS

Theoutputstabilityregionisdefinedtobeasetinthecomplexplane

•  Eachelementofisthereflec;oncoefficient(attheload)thatyieldsastablesignalattheinputportofthetransistor

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Sout

=

⇢�L is a complex number : |�L| < 1 and

����S11

+

S12

S21

�L1� S

22

�L

���� < 1

�

Thisisthereflec;oncoefficientwhen

�in

�L = �L

Sout

Sout

i.e.,|�in| < 1

Todeterminegraphically,usetheSmithchart

1.  Drawtheoutputstabilitycirclewhosecenterandradiusare

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Sout

Smithchart

outputstabilitycircleCL

RL

CL =(S22 ��S⇤

11)⇤

|S22|2 � |�|2 (center)

(radius)RL =

����S12S21

|S22|2 � |�|2

����

(con;nued)

2.  Outputstabilityregionistheshadedarea,dependingonthevalueof

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Sout

|S11|

CLRL

CLRL

(a)For|S11| < 1 |S11| > 1(b)For

Sout

Sout

Example4:Outputstabilitycircle•  Recall the scaiering parameters of a poten;ally unstabletransistorinexample3(Z0=50Ω):

•  Determinetheoutputstabilityregion

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S11 = �0.810� j0.315

S22 = �0.231� j0.451

S12 = 0.031� j0.005

S21 = 2.049 + j3.724

Example4:Solu;on

•  Answer:TheoutputstabilityregionistheshadedareaoftheSmithchart

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0.2

0.5

1.0

2.0

5.0

+j0.2

−j0.2

+j0.5

−j0.5

+j1.0

−j1.0

+j2.0

−j2.0

+j5.0

−j5.0

0.0 '

Outputstabilityregion

1.60\133�0.92

Example4:Solu;on

•  Drawtheoutputstabilitycircle:

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CL =(S22 ��S⇤

11)⇤

|S22|2 � |�|2 = 1.60\133�

(center) (radius)

CL

RL

Smithchart

Outputstabilitycircle

length=1.60

133°

length=0.92

RL =

����S12S21

|S22|2 � |�|2

���� = 0.92

Example4:Solu;on•  ThemagnitudeofS11islessthanone:sothestabilityregionisoutsidetheoutputstabilitycircle

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|S11| = |� 0.810� j0.315|

=p(�0.810)2 + (�0.315)2

= 0.87 < 1

YoumayusescalesontheSmithcharttoobtaintheangleandlength

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Forangle,usethescalemarked“Angleofreflec;oncoefficientindegrees”

Forlength,usetheboiom-mostscalemarked“Origin”

Theinputstabilityregionisdefinedtobeasetinthecomplexplane

•  Eachelementofisthereflec;oncoefficient(atthesource)thatyieldsastablesignalattheoutputportofthetransistor

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Thisisthereflec;oncoefficientwhen

i.e.,

Sin

Sin =

⇢�S is a complex number : |�S | < 1 and

����S22 +S12S21�S1� S11�S

���� < 1

�

�out

�S = �S

Sin

|�out

| < 1

Todeterminegraphically,usetheSmithchartinasimilarwayto’s

1.  Drawtheinputstabilitycirclewhosecenterandradiusare

48

Smithchart

inputstabilitycircle

(center)

(radius)

Sin

CS =(S11 ��S⇤

22)⇤

|S11|2 � |�|2

RS =

����S12S21

|S11|2 � |�|2

����

RS

CS

Sout

(con;nued)

2.  Outputstabilityregionistheshadedarea,dependingonthevalueof

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|S11|

(a)For|S11| < 1 |S11| > 1(b)For

RS

CS

RS

CS

Sin

Sin

Sin

Example5:Inputstabilitycircle•  Recall the scaiering parameters of a poten;ally unstabletransistorinexample3(Z0=50Ω):

•  Determinetheinputstabilityregion

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S11 = �0.810� j0.315

S22 = �0.231� j0.451

S12 = 0.031� j0.005

S21 = 2.049 + j3.724

0.2

0.5

1.0

2.0

5.0

+j0.2

−j0.2

+j0.5

−j0.5

+j1.0

−j1.0

+j2.0

−j2.0

+j5.0

−j5.0

0.0 '

Example5:Solu;on

•  Answer: The input stability region is the shaded area of theSmithchart

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1.09\162�

0.21

center Inputstabilityregion

Example5:Solu;on

•  Drawtheinputstabilitycircle:

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(center) (radius)

Smithchart

Inputstabilitycircle

length=1.09162°radius=0.21

CS =(S11 ��S⇤

22)⇤

|S11|2 � |�|2 = 1.09\162� RS =

����S12S21

|S11|2 � |�|2

���� = 0.21

CS

RS

Example5:Solu;on•  ThemagnitudeofS11islessthanone:sothestabilityregionisoutsidetheoutputstabilitycircle

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|S11| = |� 0.810� j0.315|

=p(�0.810)2 + (�0.315)2

= 0.87 < 1

Summary•  Examplesofamplifiedsignals

•  Transistorasa2-portnetwork•  Two-portpowergains– Powergain– Availablepowergain– Transducerpowergain

•  Stability– Uncondi;onalstability– Condi;onalstability– Testofuncondi;onalstability:K-Δtest– Stabilitycircles

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