We will cover these topics Amplifier: two-port power gains...
Transcript of We will cover these topics Amplifier: two-port power gains...
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)
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speaker(aload)
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Frequency(Hz)
Power(dBm)
d r m D R
Spectrumoftheoriginalsignal
Example1(con;nued):Amplifiedsignal
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AudioSource
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Amplifierby10dB430-600Hz
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d r m
D RPower(dBm)
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Frequency(Hz)
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vL(t)
Spectrumoftheamplifiedsignal
Atransistorcanbemodeledasa2-portnetwork
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EmiierBase
Collector
Base Collector
Emiier
Scaieringmatrix[S]
Transistor:
2-portnetwork:
Manufacturersusuallyprovidethescaieringparametersinthedatasheets
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(con;nued)
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(con;nued)
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6.1TWO-PORTPOWERGAINS
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Recall:themaximumpowerisdeliveredtotherightnetworkwhen.
• Thecondi;onisthesameas– becausedoubleconjuga;onsgivesbackthecomplexnumber
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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
21
|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
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vun(t)
Power(dBm)
Frequency(Hz)
Spectrumofanunstable,amplifiedsignal
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(TheamplifiercannotamplifythefrequenciesofDandR)
d r m
l t
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AudioSource
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Anotherunstableamplifier
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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
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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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