微波工程講義

Chapter 10 Transistor amplifier design10.1 Stability considerations

unconditionally stable, conditionally stable, stability factor, source stability circle, load stability circle

10.2 Amplifier design for maximum gainunilateral case, bilateral case, simultaneously conjugate match,unilateral figure of merit

10.3 Constant gain circlesunilateral case, bilateral case

10.4 Constant noise figure circles10.5 Broadband amplifiers

negative feedback amplifier, balanced amplifier, traveling wave amplifier

10.6 Small signal equivalent circuit models of transistorsBJT, MESFET

10.7 DC bias circuits for transistors

10-1

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10.1 Stability considerationsBasics1. Unconditionally stable

10-2

ΓL

two-portnetwork ZL

Zs

Vs

Γs Γin,

Zin

Γout,

Zout

<−=∆

>∆+−−

=⇔

<Γ−

Γ+=Γ

<Γ−

Γ+=Γ

<Γ<Γ

1

12

1

11

11

1,1 allfor

21122211

2112

2222

211

11

122122

22

122111

SSSSSSSS

k

SSSS

SSSS

s

sout

L

Lin

Ls

2. Conditionally stable: there exists some Γs and ΓL, such that one or both of these conditions violated.

微波工程講義10-3

Discussion1. Stability circles

2211

211222

11

**2211

s11

211222

2222

211222

22

**1122

L22

211211

,)(

with plane-in

circlestability source11

,)(

with plane-in

circlestability load11

∆−=

∆−∆−

=Γ

=−Γ→=Γ−Γ

+=Γ

∆−=

∆−∆−

=Γ

=−Γ→=Γ−Γ

+=Γ

SSSr

SSSC

rCSSSS

SSSr

SSSC

rCSSSS

sss

sss

sout

LLL

LLL

Lin

微波工程講義10-4

2. If ΓL=0→Γ in=S11, the center of Smith chart represent a stable point if ∣S11∣<1

|S11|<1 |Γin|=1 |S11|>1ΓL-plane ΓL-plane

CL RL

3. If Γs=0→Γout=S22, the center of Smith chart represent a stable point if ∣S22∣<1

|S22|<1 |Γout|=1 |S22|>1Γs-plane Γs-plane

Cs Rs

|Γin|<1

|Γin|<1

|Γout|<1

|Γout|<1

微波工程講義10-5

4. Ex.10.1 A transistor has S11=0.894∠-60.6°, S12=0.02 ∠62.4, S21=3.122∠123.6°, S22=0.781∠-27.6° at 2GHz

2.0,5.6813.15.0,7.4636.1

1607.02

1,16964.0

2112

2222

211

21122211

=°∠==°∠=

<=∆+−−

=<=−=∆

ss

LL

rCrC

SSSS

kSSSS

CL rL

rs

Cs

stable region

微波工程講義10-6

10.2 Amplifier design for maximum gainBasics

1.Unilateral case: S12=0stability condition: ∣S11∣<1, ∣S22∣<1 maximum gain: Γs=S11*, ΓL=S22*unilateral transducer gain

[ S ]Go

Zo

Γs Γin Γout ΓL

Outputmatchingcircuit GL

Inputmatchingcircuit Gs

222

2212

11

max222

22

21211

2

11

11

1

1

1

1

SS

SG

SS

SG TU

L

L

s

sTU

−−=→

Γ−

Γ−

Γ−

Γ−=

微波工程講義10-7

2. Bilateral case (simultaneously conjugate match)

*11222

*22111

2211

2222

2222

2111

1

21

211

11

211222

*

2

22

222

22

211211

*

12

212

12

21

211

22

212222

22

212max

**

,

1,1

24

1

24

1

1 asgain stable maximum:),1(

1

1

11

1

1

11

,match conjugateusly simultaneooutput andinput

1 ,1 if

SSCSSC

SSBSSB

CCBB

SSSS

CCBB

SSSS

kSSGKK

SS

SS

SSGG

K

Mss

soutL

MLL

Lins

MSG

s

s

LL

L

sTT

Loutsin

∆−=∆−=

∆−−+=∆−−+=

−−=Γ→

Γ−Γ

+=Γ=Γ

−−=Γ→

Γ−Γ+=Γ=Γ

==−−=

Γ−

Γ−

Γ−=

Γ−

Γ−

Γ−==→

Γ=ΓΓ=Γ→

><∆

微波工程講義10-8

3. Unilateral figure of merit

)1)(1(

)1(1

)1(1

222

211

22122111

22

SS

SSSSU

UGG

U TU

T

−−=

−<<

+

Discussion1. Linear amplifier design procedure(1) if |∆|<1, K>1, use input and output simultaneously conjugate

matches for GTmax

(2) if K<1, plot source and load stability circles to see if input and output simultaneously conjugate matches possible, otherwise select the proper Γs and ΓL for gain or noise figure considerations (to be discussed in the following sections).

微波工程講義10-9

2. Ex.10.2 A transistor has (Zo=50Ω), S11=0.606∠-155°, S12=0, S21=6∠180°, S22=0.48∠-20° at 1GHz → design an amplifier to give GTUmax

dBS

SS

G

S

TU

69.189.733.13658.148.01

16606.01

11

11

1020.48 S,1550.606 S

148.0,1606.0S

22

2222

2212

11max

22L11s

2211

==××=−−

=−−

=

°∠==Γ°∠==Γ

<=<=∗∗

ΓsΓL

Γs ΓL1

1: y=j1.7, or z=-j0.452: z=j0.23: z=-j1.38, or y=j0.484: y=-j0.16

2

3

4

微波工程講義10-10

3. Ex.10.2 A transistor has (Zo=50Ω), S11=0.614∠-167.4°, S12=0.046∠65°, S21=2.187∠32.4°, S22=0.716∠-83° at 6GHz →design an amplifier to give GTmax

dBKKSSG

CCBB

CCBB

K

T

MsLout

MLsin

58.14728.28)1(

48.849011.02

4

76.1698673.02

4

11296.1 ,13419.0

2

12

21max

1

21

211*

2

22

222*

==−−=→

°∠=−−

=Γ→Γ=Γ

°∠=−−

=Γ→Γ=Γ→

>=<=∆

微波工程講義10-11

ΓMS

ΓML

21: ΓMS=0.8673∠169.76°2: y=-j2.73: ΓML=0.9011∠84.48°4: z=j3.4

1

3

4

Γs ΓL

0.093λ

0.065λ

0.194λ

Ζo Ζo

G

微波工程講義10-12

4. Ex.10.4 transistor A: S11=0.45∠150°, S12=0.01∠-10°, S21=2.05∠10°, S22=0.4∠-10°transistor B: S11=0.641∠-171.3°, S12=0.057∠16.3°, S21=2.058∠28.5°, S22=0.572∠-95.7°→compare their Us

<<−

<<−→

<<

<<→

−<<

+

==

→−−

=

dBGGdB

dBGGdB

GGGG

UGG

U

UU

SS

SSSSU

BTU

T

ATU

T

BTU

T

ATU

T

TU

T

B

A

9976.08948.0

0238.00476.0

2582.18238.0

0055.19891.0

)1(1

)1(1

1085.000551.0

)1)(1(

22

222

211

22122111

微波工程講義10-13

10.3 Constant gain circlesBasics1. Unilateral case (S12=0)

222

222

222

*22

LLmax

211

211

211

*11

SSmax

222

max222

2

211

max211

2

222

22

21211

2

)1(1

)1(1,

)1(1

plane in circlesgain constant

)1(1

)1(1,

)1(1

plane in circlesgain constant

11,

1

1,

11,

1

1

1

1

1

1

Sg

SgR

SgSgd

RdGGg

Sg

SgR

SgSgd

RdGG

g

SG

SG

SG

SG

GGGS

SS

G

L

LL

L

LL

LLL

LL

s

ss

s

ss

sss

ss

LL

LLs

s

ss

LosL

L

s

sTU

−−

−−=

−−=

=−ΓΓ→≡

−−

−−=

−−=

=−ΓΓ→≡

−=

Γ−

Γ−=

−=

Γ−

Γ−=

=Γ−

Γ−

Γ−

Γ−=

微波工程講義10-14

2. Bilateral case (S12≠0)(1) Unconditionally stable case, use GP or GA

max2

12

21max

2

2112max

0

112222222

2221122112

2222

*2

LL211

222

2

221

22

11

22

2112112

22

22

212

)1(

)1(1

,)(1

21,

)(1

plane in circlesgain constant 1

1

11,

1

1

11

TPP

PP

R

P

PPP

P

PP

PPLL

LP

PP

L

L

L

Lin

L

L

inP

GKKSS

GG

KKSS

gg

SSCSg

gSSgSSKR

SgCgc

RcSS

g

gSG

SS

SSSS

SSG

P

=−−==

−−==→

∆−=∆−+

+−=

∆−+=

=−ΓΓ→∆Γ−−Γ−

Γ−≡

=

Γ−∆Γ−

=Γ−Γ

+=ΓΓ−

Γ−

Γ−=

=

∗

微波工程講義10-15

max2

12

21max

2

2112max

0

221112211

2221122112

2211

*1

ss222

211

2

221

11

22

11

2112222

2212

11

2

)1(

)1(1

,)(1

21,

)(1

plane in circlesgain constant 1

1

11,

11

1

1

TAA

AA

R

A

AAs

A

As

ssss

sA

AP

s

s

s

sout

outs

sA

GKKSS

GG

KKSS

gg

SSCSg

gSSgSSKR

SgCgc

RcSS

g

gSG

SS

SSSSS

SG

s

=−−==

−−==→

∆−=∆−+

+−=

∆−+=

=−ΓΓ→∆Γ−−Γ−

Γ−≡

=

Γ−∆Γ−

=Γ−Γ

+=ΓΓ−Γ−

Γ−=

=

∗

微波工程講義10-16

(2) Potentially unstable caseuse GP to plot constant gain circles in ΓL-plane and plot load stability

circles → properly select ΓL, calculate Γs= Γ∗in

use GA to plot constant gain circles in Γs-plane and plot source stability circles → properly select Γs, calculate ΓL= Γ∗

outDiscussion1. Ex. 10.5 A MESFET with S-parameters (Zo=50Ω)

S11 S21 S12 S223GHz 0.8∠-90° 2.8∠100° 0 0.66∠-50°4GHz 0.75∠-120° 2.5∠80° 0 0.6∠-70°5GHz 0.71∠-140° 2.3∠60° 0 0.68∠-85°→ plot constant gain circles @4GHz for GL=0, 1dB, and Gs=2, 3dBdesign an amplifier of 11dB gain

dBGdBSG

dBS

GdBS

GGHz

GGGS

SS

G

TUo

Ls

LosL

L

s

sTU

47.1392.196.759.3,96.725.6

92.15625.11

1,59.3288.21

1,4@

1

1

1

1

max2

21

222

max211

max

222

22

21211

2

=++====

==−

===−

=

=Γ−

Γ−

Γ−

Γ−=

微波工程講義10-17

=

°∠°∠

=→

=→

==

=

=

°∠°∠

=→

=→

==

=

44.0303.0

,7044.07052.0

64.08064.0

1026.11

294.0166.0

,120627.0120706.0

691.0875.0

58.1223

LLLL

SSSS

RdgdBdB

G

RdgdBdB

G

A

Gs=2dBGs=3dB

B

GL=1dBGL=0dB

chose GTU(4GHz) 11dB=2dB+8dB+1dB to design the circuit → GTU(3GHz)=7.33dB, GTU(5GHz)=6.96dB

0.044λ

0.431λ0.183λ

0.09λ

微波工程講義10-18

2. Ex. 10.6 A GaAs FET with S-parameters (Zo=50Ω, 1GHz) S11=2.27∠-120°, S21=4∠50°, S12=0, S22=0.6∠-80°

→ (1) calculate Zin, plot unstable region in Γs-plane(2) plot constant gain circles for Gs=3, 5dB(3) find Zs for Gs=3dB with maximum degree of stable, determine ZL to

give maximum GL, then design the amplifier (4) calculate GTU

unstable be to5.24

235.2412027.2)1( 11

Ω<→

Ω−−=→+−

=Γ=°−∠=

s

inoin

oinin

R

jZZZZZS

Γs-plane

rs=0.49

微波工程講義10-19

=

°∠°∠

=→

−−

=→

==

=2698.02174.0

,1204.012045.0

3.813.13

2316.35

)2( SSSS RdgdBdB

G

AGs=3dBGs=5dB

(3) Select A for Zs to give a maximum possible Rs

25507624.0: jZA ss +=→°∠=ΓFor maximum GL, select C: 5.5127806.022 jZS LL +=→°∠==Γ ∗

C

dBGS

G TUL 175056.142,56.11

1)4( 22

22max ==××==

−=

0.067λ

0.342λ0.036λ

0.436λ

微波工程講義10-20

3. Ex. 10.7 A GaAs FET with S-parameters (Zo=50Ω, 6GHz) S11=0.641∠-171.3°, S21=2.058∠28.5°, S12=0.057∠16.3°, S22=0.572∠-95.7°→ design an amplifier to give GP=9dB

°∠=Γ→°−∠=Γ→°∠==°∠=

=−ΓΓ→=

=−−==→=∆=

6.17563.0:6.17563.05036.0:Cselect 43.0,1045.0

, plane in circlesgain constant 9

38.11)1(3.0,5.1

LL

2

12

21max

sinL

PP

PPP

PP

AΓRc

RcdBG

dBKKSS

GGK

A

GP=9dBC

微波工程講義10-21

4. Ex. 10.8 A GaAs FET with S-parameters (Zo=50Ω, 8GHz) S11=0.5∠-180°, S21=2.5∠70°, S12=0.08∠30°, S22=0.8∠-100°→ design an amplifier to give GP=10dB

°∠=Γ→°−∠=Γ→°∠==°∠=

=−ΓΓ→=

=°∠==°∠=

→=∆=

3.17952.0:3.17952.0971.0:Cselect 4733.0,2.9757.0

, plane in circlesgain constant 10

3388.0,9717.1 circlestability load998.0,6.17067.1 circlestability source

2228.0,399.0

LL

sinL

PP

PPP

LL

ss

AΓRc

RcdBG

rCrC

K

AGP=10dB

C

load stability circle

source stability circle

微波工程講義10-22

5. Ex. 10.9 A BJT with S-parameters (Zo=50Ω, 750MHz) S11=0.277∠-59°, S21=1.92∠64°, S12=0.078∠93°, S22=0.848∠-31°→ design an amplifier to give GTmax

°∠=°∠=→=∆=

85.339511.0,1357298.0 match conjugately simutaneou3242.0,0325.1

MLMS ΓΓK

6. Ex. 10.10 use the BJT in Ex.10.9 → design an amplifier to give GP=10dB, determine Gs, GL for GPmax

MSs

MLLP

s

inL

PP

PPP

ΓΓdBGA

ΓRc

RcdBG

Γ=°∠=Γ→=°∠=→=

°∠=Γ→°−∠=Γ→°∠=

=°∠==−ΓΓ→=

1357298.084.339511.08.12

45.563455.0:45.563455.03456.0:Cselect

2142.0,8.3378.0, plane in circlesgain constant 10

max

LL

A

GP=10dB

C

微波工程講義10-23

10.4 Constant noise figure circlesBasics

1

)1(,

1

circle figure noiseconstant

parameter figure noise:1/41

r transistoof resistance noise equivalent ,,:parameter noise

1)1(

4

amplifierport - twoaFor

2

s

2min2

2

min

22

2

min

2

min

+

Γ−+=

+Γ

=

=−Γ→

Γ+−

=Γ−

Γ−Γ≡

Γ+Γ−

Γ−Γ+=−+=

N

NNNR

NC

RC

ZRFFN

RYF

ZR

FYYGR

FF

optNF

optNF

NFNF

optons

opts

nopt

opts

opts

o

nopts

s

n

微波工程講義10-24

Discussion1. Ex. 10.11 A BJT with S-parameters (Zo=50Ω, 1GHz) S11=0.707∠-

155°, S21=5∠180°, S12=0, S22=0.51∠-20°, Fmin=3dB, Rn=4Ω, Γopt= =0.45∠180° → design an amplifier to give GT=16dB, F<3.5dB

Ls

NFNF

LLL

sssTU

TU

DΓRCdBF

RddBGRddBG

dBG

dBG

Γ=°∠=→=

=°∠=→==°∠=→=

→=

=++=

:,:Bselect 3953.0,1803658.05.3

258.0,2046.078.0345.0,15556.022.1

16

29.1831.198.133max

B

GL=0.78dB

D

Gs=1.22dB

F=3.5dBΓs ΓL

微波工程講義10-25

2. Ex. 10.12 A GaAs FET with S-parameters (Zo=50Ω, 4GHz) S11=0.6∠-60°, S21=1.9∠81°, S12=0.05∠26°, S22=0.5∠-60°, Fmin=1.6dB, Rn=20Ω, Γopt= =0.62∠100°→ (1) assume S12=0, design an amplifier to give GTUmax, F<2dB(2) repeat for S12≠0

s

ss

sss

NFNF

L

TU

T

ΓRddBRddB

G

RCdBFSΓ

dBG

dBGGdB.UK

:Aselect 15.0,60586.07.12.0,6056.05.1

245.0,10056.02605.0:Cselect

76.825.157.5938.1

59.05.005940,3713.0,778.2)1(

22

TUmax

→

=°∠=→=°∠=→

=

=°∠=→=°∠==

=++=

<<−→==∆=

∗

B

C

Gs=1.5dB

F=2dBGs=1.7dB

微波工程講義10-26

°∠=Γ→°−∠=Γ→°∠==°∠=→=

=−ΓΓ→

8.67504.08.67504.090535.0select 234.0,7.64527.08

plane in circlesgain constant use )2( ss

Louts

ssA

ssA

ΓRcdBG

RcG

Γs

F=2dB GA=8dB

ΓL

3. Low noise amplifier design procedure: if unconditionally stable, use GA constant gain circles and constant noise figure circles to determine Γs , and ΓL= Γ*outif potentionally unstable, consider source and load stability circles in addition

微波工程講義10-27

4. Ex. 10.13 A BJT with S-parameters (Zo=50Ω, 1GHz) S11=0.6∠-130°, S21=7.12∠86°, S12=0.039∠35°, S22=0.5∠-38°, Fmin=1.3dB, Rn=8Ω, Γopt= =0.06∠49° → design an amplifier to give 16dB power gain, F<2.5dB

°∠=Γ→°−∠=Γ→

°∠==°∠=→=

=°∠==−ΓΓ→=

=°∠==°∠=

→=∆=

45.36469.0:45.36469.0

013.0:Bselect 65.0,400346.05.2

6731.0,1723542.0, plane in circlesgain constant 16

296.1,8.5016.2 circlestability load8566.0,7.17174.1 circlestability source

189.0,7667.0

ss

L

out

s

NFNF

aa

aaA

LL

ss

D

ΓRCF

RcRcdBG

rCrC

K

Γs

F=2.5dBGA=16dB

ΓL

load stabilitycircle

source stabilitycircle

微波工程講義10-28

10.5 Broadband amplifiersBasics1. Negative feedback amplifier

Vo

β

AV

Vi

coco

o

ooocc

oc

o

o

V

c

oV

V

V

i

oV

fAfAAAAAff

Affj

AA

G

ffj

AA

AA

VVG

''1

'),1('

)1(1

1

1

1

=+

=+=

++

+=→

+=

+==

ββ

β

β

β

微波工程講義

2. Balanced amplifier

input

output

G1

G2

90° hybrid 90° hybrid

Zo

Zo∠0°

∠-90°

∠0°∠-180°

∠-90°∠-90°

Bandwidth is limited by the 90° hybrids and amplifier gain.

10-29

微波工程講義10-30

2. Traveling wave amplifier (distributed amplifier)

gate line

drain line

FET equivalent circuit

TEM line “extreme wide operation bandwidth”

Bandwidth is limited by the FET and line loss.

微波工程講義10-31

10.6 Small-signal equivalent circuit models of transistorBasics1. BJT

B

C

E

CB

E

rbeC be+

Vbe

- gme-jτVbe

Vb(+) Vc(+) Ice

Vce

Ib=+ mA

rbe: due to base dopingCbe: due to p-n junction

微波工程講義10-32

G

D

S

2. MESFET Vg(-) Vd(+)Ids Vgs = 0V

Vgs = - 1V

Vgs = Vp

S G D

gate length

gate width

↑⇒↑

↑⇒↓

power widthgatef length gate

DG

S

Cgd

Ri

Cgs+Vgs

-gmVgs

CdsRds

Ig

微波工程講義10-33

10.7 DC bias circuits for transistorsDiscussion1. Resistive bias with voltage feedback for BJT

DC block

RF block

matchingcircuit GND

+Vcc

Lc

Rc

LB

RB

IB

IC

IE

ββ

β

β

,VI

RRVV

RR

VVI

IIIIIVIRIRV

BEE

BCBEcc

BC

BEccE

CBEEC

BEBBEccc

in changes tosensitive less have to1

,

1

,,

+>>>>→

++

−=→

+==++=

微波工程講義10-34

2. Bias circuit with emitter bypassed resistor for BJT

+Vcc

Rc

VTh

IB

IC

IE

RE

RTh

+Vcc

Lc

Rc

LB1

R1

IBIC

IE

RE CER2

LB2

↓↓→

≈↓↑→

+>>>>→

++

−=→

+++

=++=

+=

+=

inTh

ccTh

CECBTh

BEE

ThEBETh

ThE

BEThE

BEEETh

EEBEBThTh

ThccTh

RRVV

VVV,VI

RRVV

RR

VVI

VIRR

IRVIRV

RRRRRV

RRRV

%20~%15usually swing,output limited ,

in changes tosensitive less have to1

,

1

)1

(

,21

21

21

2

ββ

β

β

微波工程講義10-35

3. Unipolar bias circuits for MESFET

Ids Vgs = 0V

Vgs = - 1V

Vgs = Vp

low-noiselow-power

low-noisehigh-power

high-powerhigh-gain

high-powerhigh-efficiency

DC block

RF block

matchingcircuit

+VD

LD

Cc1

LG

CD

IC

CS

Cc2

RS

DC block

RF block

matchingcircuit

GND

+VD

LD

Cc1

LG

CD

IC

CS

Cc2

+VSLS

Hw #8 (due 2 weeks)5, 11, 15