Freqqypuency Response (Ch. 11)bandi.chungbuk.ac.kr/~ysk/ckt11.pdf · 11.2 High-Frequency Models of Transistors 11.3 Analysis Procedure 114 F R fCE dCSSt11.4 Frequency Response of

Frequency Response (Ch. 11)q y p

김 영 석김 영 석

충북대학교 전자정보대학

2012.9.1.9.

Email: [email protected]

전자정보대학 김영석 11-1

Contents11.1 Fundamental Concepts

11.2 High-Frequency Models of Transistors

11.3 Analysis Procedure

11 4 F R f CE d CS St11.4 Frequency Response of CE and CS Stages

11.5 Frequency Response of CB and CG Stages

11.6 Frequency Response of Followers11.6 Frequency Response of Followers

11.7 Frequency Response of Cascode Stage

11.8 Frequency Response of Differential Pairs

11.9 Additional Examples


11.1 Fundamental Concepts

High Frequency Roll-off of Amplifier: As frequency of operation increases, the gain of amplifier decreases. This chapter analyzes this problem.


ExamplesExample1: Human Voice

Natural Voice Telephone System

Example2: Video Signal

High Bandwidth Low Bandwidth


High Bandwidth Low Bandwidth

Gain Roll-off: Simple Low-pass Filter

2

1||

, ,0

11

1

↓=>↓==>↑

==

outC

inout

VfC

Zf

VVopenCf

π0 ,short , 1 =∞= outVCf

In this simple example, as frequency increases the impedance of p p , q y p C1 decreases and the voltage divider consists of C1 and R1

attenuates Vin to a greater extent at the output.


Gain Roll-off: Common Source

⎛ ⎞1||out m in DL

V g V RC s

⎛ ⎞= − ⎜ ⎟

⎝ ⎠

1222 +=

ωLD

Dm

in

out

CRRg

VV


Example: Figure of Merit

MOFnConsumptioPower

BandwidthGain... •=

LCC

T

C

VICR

RVI 1

p

×=

LCCT

CCC

CVV

VI1 =

This metric quantifies a circuit’s gain bandwidth andThis metric quantifies a circuit s gain, bandwidth, and power dissipation. In the bipolar case, low temperature, supply, and load capacitance mark a superior figure of

itmerit.


Example: Relationship between Frequency Response and Step Responsep p p

( )2 2 2

1H s jω= = ( ) ( )0 1 expouttV t V u t

R C⎛ ⎞−

= −⎜ ⎟⎝ ⎠2 2 2

1 1 1R C ω + 1 1R C⎝ ⎠

The relationship is such that as R1C1 increases, the bandwidth drops and the step response becomes slower.bandwidth drops and the step response becomes slower.


11.1.3 Bode Plot

When we hit a zero, ωzj, the Bode magnitude rises with a slope of +20dB/dec.

When we hit a pole ω the Bode magnitude falls with a slope ofWhen we hit a pole, ωpj, the Bode magnitude falls with a slope of -20dB/dec

⎟⎟⎞

⎜⎜⎛+⎟⎟

⎞⎜⎜⎛+ 11 ss

L

⎞⎛⎞⎛

⎟⎟⎠

⎜⎜⎝+⎟⎟

⎠⎜⎜⎝+

= 210

11)( zzAsH

ωω

L⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛+

21

11pp

ssωω ⎠⎝⎠⎝


Example: Bode Plot

The circuit only has one pole (no zero) at 1/(RDCL), so the slope drops from 0 to -20dB/dec as we pass ωp1.

1|| oDmout

srRg

VV −

=

LDp CR1 =ω

1

1p

in

wsV +


Pole Identification Example I(CS)

1

inSp CR

11 =ω LD

p CR2 =ω

inS

|| oDmout

ssrRg

VV −

=)1)(1(

21 pp

in

ws

wsV ++


Pole Identification Example II(CG)

=1ω LD

p CR1

2 =ω

inm

S

p

Cg

R ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1||1ω

)||)(/1

/1( oDmmS

m

out

rRggR

gV

−+

)1)(1(21 pp

mS

in

out

ws

wsg

V ++=


Circuit with Floating Capacitor

The pole of a circuit is computed by finding the effective resistance and capacitance from a node to GROUND.

The circuit above creates a problem since neither terminal of CF is groundedgrounded.


11.1.5 Miller’s Theorem

F

AZZ =∴

11111121 VVVAVVVI v ==

−=

−=

vA−111)1/( ZAZZZ vFFF −

v

F

AZZ

/112 −=


Miller Multiplication

With Miller’s theorem, we can separate the floating capacitor.With Miller s theorem, we can separate the floating capacitor. However, the input capacitor is larger than the original floating capacitor. We call this Miller multiplication.


Example: Miller Theorem

( )in CRgR +=

11ω out

CR ⎟⎞

⎜⎛

=11

1ω( ) FDmS CRgR +1

FDm

D CRg

R ⎟⎟⎠

⎞⎜⎜⎝

⎛+

11


11.1.6 General Freq Response:High-Pass Filter Response

11=ωCRVout

121

21

21 +ωCRVin

The voltage division between a resistor and a capacitor can be configured such that the gain at low frequency is reducedreduced.


Example: Audio Amplifier

Ω=100i KRAssume

Ω= 200/1mg

FC

HzCR

fii

lowin

679

202

1, ==

π1nFCi 6.79≥

nFC

kHzCg

f

L

Lmhighout

8.39

20)/1(2

1,

≤

==π

In order to successfully pass audio band frequencies (20 y p q (Hz-20 KHz), large input and output capacitances are needed.


Typical Frequency Response


11.2 High Freq Model of TransistorsHigh-Frequency Bipolar Model

b jeC C Cπ = +

At high frequency, capacitive effects come into play. Cb

represents the base diffusion charge, whereas Cμ and Cje

are the junction capacitancesare the junction capacitances.


High-Frequency Model of Integrated Bipolar Transistor

Since an integrated bipolar circuit is fabricated on top of aSince an integrated bipolar circuit is fabricated on top of a substrate, another junction capacitance exists between the collector and substrate, namely CCS.


MOS Intrinsic Capacitances

For a MOS there exist oxide capacitance from gate to channelFor a MOS, there exist oxide capacitance from gate to channel, junction capacitances from source/drain to substrate, and overlap capacitance from gate to source/drain.


Gate Oxide Capacitance Partition and Full Model

The gate oxide capacitance is often partitioned betweenThe gate oxide capacitance is often partitioned between source and drain. In saturation, C2 ~ Cgate, and C1 ~ 0. They are in parallel with the overlap capacitance to form C and CCGS and CGD.


11.2.3 Transit Frequency

Transit frequency, fT, is defined as the frequency where the current gain from input to output drops to 1.

)(1||

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=−= π CC

rIgVgI inminmout

1)(21

)(

=++

−=

⎥⎦⎢⎣ +

π

μπ

π CCfjrg

II

CCs

Tm

i

out

πCgf m

T ≈2GS

mT C

gf ≈π2

)(21 ++ μππ CCfjI Tin

πCT GS


Example: Transit Frequency Calculation

( )THGSn

T VVL

f −= 2232 μπ

L2

mVVVnmL

THGS 10065

2

=−=

GHzfsVcm

T

n

226)./(400 2

==μ


11.3 High Frequency Circuit Analysis Procedure

Determine which capacitor impact the low-frequency region of the response and calculate the low-frequency pole (neglect transistor capacitance).pole (neglect transistor capacitance).

Calculate the midband gain by replacing the capacitors with short circuits (neglect transistor capacitance).

Include transistor capacitances.

Merge capacitors connected to AC grounds and omit those that play no role in the circuit.those that play no role in the circuit.

Determine the high-frequency poles and zeros.

Plot the frequency response using Bode’s rules or exact analysis.


11.4 (Low) Frequency Response of CE/CS StageDegenerated CS stage

( ) ( )pz

zDmout

CgRw

CRwf

wswsRgs

VV

)/1||(1 ,1 low, @

/1/1

===+

+−=

bmSbSpX CgRCRwsV )/1||(/1+

In order to increase the midband gain a capacitor C isIn order to increase the midband gain, a capacitor Cb is placed in parallel with Rs. The pole frequency must be well below the lowest signal frequency to avoid the effect of degeneration.


11.4 (High) Frequency Response of CE/CS StageUnified Model for CE and CS Stages


11.4 (High) Frequency Response of CE/CS StageUnified Model Using Miller’s Theorem


11.4 (High) Frequency Response of CE/CS StageEx 11.15: CE Stage

mAIR

C

S

1001200

===

β

Ω

fFCfFC

20100

100

===

μ

π

β

fFCCS 30=

VkVI tTC /

MHzCKCR

f

VVRgVVKk

IVrmS

VIg

in

Lmb

out

B

T

T

Cm

462])1()[||(2

1

/80,5.2,40

==

−=−====== π Ω

GHzCKCR

f

CKCrRf

CSLout

Sin

59.1])/11([2

1])1()[||(2

=−+

=

−+

μ

μππ

π

π

The input pole is the bottleneck for speed.

CSL )( μ

p p s sp


11.4.4 Direct Analysis of CE and CS Stages(High Freq Response)

mg=||ω

XYz C

1

||ω

( ) ( )( ) ( )

outXYLinThevThevXYLmp CCRCRRCRg ++++=

11|| 1ω

( ) ( )( )tiXYtXYiLTh

outXYLinThevThevXYLmp CCCCCCRR

CCRCRRCRg++

++++=

1|| 2ω

Direct analysis yields different pole locations and an extra zero.

( )outinXYoutXYinLThev CCCCCCRR ++

y y p


Ex 11.18: Comparison Between Different Methods

=Ω=

fFCR

GS

S

250200

( )Ω=

==

−g

fFCfFC

m

DB

GD

GS

150

10080

1( )

Ω==

KRL

m

20λ

( )MHzinp 5712, ×= πω ( )MHzinp 2642, ×= πω ( )MHzinp 2492, ×= πω

Miller’s Exact Dominant Pole

( )MHzoutp 4282, ×= πω ( )GHzoutp 53.42, ×= πω ( )GHzoutp 79.42, ×= πω

Not Accurate for High Gain Accurate


11.4.5 Input Impedance of CE and CS Stages

1( )[ ] π

μπ

rsCRgC

ZCm

in ||1

1++

≈( )[ ]sCRgC

ZGDDGS

in ++≈

11

( )[ ]μπ Cm ( )[ ]sCRgC GDDmGS ++ 1


11.5 Frequency Response of CB/CG StagesLow Frequency Responseq y p

Cmm sRgg 1))(/

/1(( )

imSp

p

pCm

mS

in

out

CgRw

wsw

ggR

sVV

)/1(1 ,

/1

))(/1

(

+=

+

+=

As with CE and CS stages, the use of capacitive coupling leads to low-frequency roll-off in CB and CG stages (although a CB stage is shown above, a CG stage is similar).


11.5 Frequency Response of CB/CG StageHigh Frequency Response: No Miller Cap., Good Freq. Response

)(1 ,

11

,,CSL

YpXp CCR +=

⎟⎞

⎜⎛

= ωω∞=Or

CB

)(1|| CSL

mS

CCRC

gR

+⎟⎟⎠

⎞⎜⎜⎝

⎛ μπ

CG

)(1 ,

)(1||

1,,

DBGDLYpXp CCR

CCR+

=+⎟⎟

⎞⎜⎜⎛

= ωω)(|| SBGS

mS CC

gR +⎟⎟

⎠⎜⎜⎝

The input pole is on the order of fT so rarely a speed bottleneck

∞=Or


The input pole is on the order of fT, so rarely a speed bottleneck.

Ex 11.19: CG Stage Pole Identification

1

( )11

,1||

1

GSSBS

Xp

CCR +⎟⎟⎠

⎞⎜⎜⎝

⎛=ω

( )2211

, 11

DBGSGDDB

Yp

CCCCg

+++=ω

( )111

|| GSSBm

S g ⎟⎠

⎜⎝ 2mg


Ex 11.20: Frequency Response of CG Stage

==

Ω=

fFCfFC

R

GD

GS

S

80250

200

( )Ω=

=−g

fFCfFC

m

DB

GD

150

10080

1( )( )

GHzXp 31.52, ×= πω

Ω==

KRd 20λ ( )MHzYp 4422, ×= πω

)1811(250)13(22

450),7.5(15 _3_

EMHfdBA

MHzfdBA CGdBCGv ≈= −

)18.11(250),13(22 _3_ ExMHzfdBA CSdBCSv ≈= −


11.6 Emitter and Source Followers

Emitter Follower: Direct Analysisy

1)0( fVout

( )SR

1)0( ==fVin

1+ sgC

Vπ

( )LLm

S CCCCCCgRa

⎞⎛

++= πμπμ

12 ++=

bsasg

VV m

in

out

m

LS

mS g

CrR

gCCRb ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++=

π

πμ 1


11.6 Emitter and Source Followers

Source Follower: Direct Analysisy

Follower)(Emitter ∞=πr

1)0( fVout 1)0( ==fVin

1+ sgC

VGS ( )SBGSSBGDGSGD

m

S CCCCCCgRa ++=

12 ++=

bsasg

VV m

in

outSBGD

GDS gCCCRb +

+=mg


Ex 11.21: Frequency Response of Source Follower

250100200

=Ω=

L

S

fFCfFC

R

( )[ ]GHzjGHz 57.279.121 +−= πω

10080250

===

DB

GD

GS

fFCfFCfFC ( )[ ]

( )[ ]GHzjGHz

GHzjGHz

p

p

57.279.12

57.279.12

2

1

−−=

+

πω

πω

54)1(0 GHzfdBA ≈( )0

150 1

=Ω= −

λmg

)1811(250)13(22

)20.11(450),7.5(15

5.4),1(0

_3_

_3_

EMHfdBA

ExMHzfdBA

GHzfdBA

CGdBCGv

CDdBCDv

≈=

≈=

−

−



11.6.1 Input Capacitance of Emitter/Source Follower

∞=OrO

R

mL

Lv

CACCgR

RA+

=

)1(/1

i

mL

Lv

CACCgR

RA

−+=+

=

)1(/1

πμ

L

GSGD

GSvGDin

RgCC

CACC

++=

−+=

1

)1(

Lm

vin

RgCC

CACC

++=

+

1

)1(

πμ

πμ

LmRg+1


Ex 11.23 Source Follower Input Capacitance

( )]||1[ 21 OO rrCCC +

( )( )]||/1||1[

211

2111

OOm

OOGSGDin rrg

CCC+

−+=


Output Impedance of Emitter/Source Follower

RCRV R1

1++++

==βππ

πππ

sCrRrsCrR

IVZ SS

X

Xout

Sout

S

mout

RZshortCf

Rg

ZopenCf

=∞=+

+==

, ,1

1 , ,0

π

π β

mGS

GSS

X

Xout gsC

sCRIVZ

++

==1

SoutGS

moutGS

RZshortCfg

ZopenCf

=∞=

==

, ,

1 , ,0

Usually, RS>1/gm


Ex 11.24: Output Impedance

∞=OrO

||1 rrV

)1(1

||1

33

21

sCg

rrgI

VZ

GSm

OO

mX

Xout

++==

),1

1(

3

πββ

zgRg

Z mB

m

GS

=+

+=


11.7 Frequency Response of Cascode Stage

12

1, −≈

−=

m

mXYv g

gAXYx CC 2≈

For cascode stages, there are three poles and Miller multiplication is smaller than in the CE/CS stage

2mg

multiplication is smaller than in the CE/CS stage.


Poles of Bipolar Cascode

( )1ω = 1( )( )111

, 2|| μππ

ωCCrRS

Xp +=

( )22,

1μ

ωCCR CSL

outp +=

1

( )1212

,

21μπ

ωCCC

g CSm

Yp

++=

2gm


Poles of MOS Cascode

1

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

=

12

11

,

1

1

GDm

GSS

Xp

CggCR

ω

( )1

outp CCR=ω⎦⎣ ⎠⎝ 2mg ( )22

,GDDBL

outp CCR +

⎤⎡ ⎞⎛=

1Ypω

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++ 1

1

221

2

,

11GD

m

mGSDB

m

Yp

CggCC

g


Ex 11.25: Frequency Response of Cascode

=Ω=

fFCR

GS

S

250200

( )Ω=

==

−g

fFCfFC

DB

GD

150

10080

1442),13(22 _3_ MHzfdBA CascodedBCascodev ≈= −( )GHzXp 95.12, ×= πω( )

Ω==

Ω

KR

g

L

m

20

150λ

)20.11(450),7.5(15

)21.11(5.4),1(0

_3_

_3_

ExMHzfdBA

ExGHzfdBA

CGdBCGv

CDdBCDv

≈=

≈=

−

−( )( )MHz

GHz

outp

Yp

4422

73.12

,

,

×=

×=

πω

πω



I/O Impedance of Bipolar/MOS Cascode

1( )sCC

rZin11

1 21||

μππ +

=

( )sCCRZ

CSLout

22

1||+

=μ

CgCZ

min

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

=11

1

sCggC GD

m

mGS ⎥

⎦⎢⎣

⎟⎟⎠

⎜⎜⎝++ 1

2

11 1

1( )sCC

RZDBGD

Lout22

1||+

=


Comparisons


11.8 Bipolar Differential Pair Frequency Response

Half Circuit

Since bipolar differential pair can be analyzed using half-circuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s.


MOS Differential Pair Frequency Response

Half Circuit

Since MOS differential pair can be analyzed using half-

Half Circuit

p y gcircuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of the half circuit’s.


Ex 11.27: MOS Differential Pair 1

Xp =ω1311

,

1])/1([

Yp

GDmmGSSXp CggCR

⎤⎡ ⎞⎛=

++

ω

ω

11

331

3

,

11GD

m

mGSDB

m

Yp

CggCC

g ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++

( )33,

1GDDBL

outp CCR +=ω


11.8.1 Common Mode Frequency Response

SS

m

SSSSmp C

gCRg

2)2)(2||/1(

1≈=ω

( )12

11 ++

+Δ=

Δ=

ΔΔ SSSSDmDout

RCRsCRRgR

VV

12)1||(2/1 +++Δ SSmSSSS

SSSSm

CM RgsCRsC

RgV

Css will lower the total impedance between point P to ground at high frequency, leading to higher CM gain which degrades the CM rejection ratio.degrades the CM rejection ratio.


Tail Node Capacitance Contribution

S d f dSource-Body Capacitance of M1, M2 and M3

Gate-Drain Capacitance of M3


Ex11.28: Capacitive Coupling

( )[ ]EBin RrRR 1|| 222 ++= βπ

( ) ( )HzCRr B

L 5422||

1

1111 ×== πω

π ( ) ( )HzCRR inC

L 9.221

222 ×=

+= πω


Ex11.29: IC Amplifier – Low Frequency Design

( )Rg S 44221CS)dC1(blA 11 +

( ) ( )2 92.621:C2by poleA L MHz×== πω

( )MHzCR

Rg

S

SmL 4.4221:CS)edC1(Degeratby poleA

11

111 ×=

+= πω

( ) ( )

2222

2

2212

,1

yp

Dmvv

Fi

inDL

RgAA

RR

CRR

−=−

=

+

2v


Ex11.29: IC Amplifier – Midband Design

( ) 773||X RRv ( )211 77.3||

out

inDmin

X

Rv

RRgvv

−=−=

22 DmX

out Rgv

−=


Ex11.29: IC Amplifier – High Frequency Design

:stage1st

)15.2(2

)308(2

2

1

GHz

MHz

p

p

×=

×=

πω

πω

1 :stage 2d

=ω

)21.1(2)15.1( 222

3

GHzCCR DBGDL

p

×=+

=

π

ω


Documents

Freqqypuency Response (Ch. 11)bandi.chungbuk.ac.kr/~ysk/ckt11.pdf · 11.2 High-Frequency Models of Transistors 11.3 Analysis Procedure 114 F R fCE dCSSt11.4 Frequency Response of