Freqqypuency Response (Ch. 11)bandi.chungbuk.ac.kr/~ysk/ckt11.pdf · 11.2 High-Frequency Models of...
Transcript of Freqqypuency Response (Ch. 11)bandi.chungbuk.ac.kr/~ysk/ckt11.pdf · 11.2 High-Frequency Models 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
전자정보대학 김영석 11-2
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.
전자정보대학 김영석 11-3
ExamplesExample1: Human Voice
Natural Voice Telephone System
Example2: Video Signal
High Bandwidth Low Bandwidth
전자정보대학 김영석 11-4
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.
전자정보대학 김영석 11-5
Gain Roll-off: Common Source
⎛ ⎞1||out m in DL
V g V RC s
⎛ ⎞= − ⎜ ⎟
⎝ ⎠
1222 +=
ωLD
Dm
in
out
CRRg
VV
전자정보대학 김영석 11-6
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.
전자정보대학 김영석 11-7
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.
전자정보대학 김영석 11-8
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 ⎠⎝⎠⎝
전자정보대학 김영석 11-9
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 +
전자정보대학 김영석 11-10
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 ++
전자정보대학 김영석 11-11
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 ++=
전자정보대학 김영석 11-12
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.
전자정보대학 김영석 11-13
11.1.5 Miller’s Theorem
F
AZZ =∴
11111121 VVVAVVVI v ==
−=
−=
vA−111)1/( ZAZZZ vFFF −
v
F
AZZ
/112 −=
전자정보대학 김영석 11-14
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.
전자정보대학 김영석 11-15
Example: Miller Theorem
( )in CRgR +=
11ω out
CR ⎟⎞
⎜⎛
=11
1ω( ) FDmS CRgR +1
FDm
D CRg
R ⎟⎟⎠
⎞⎜⎜⎝
⎛+
11
전자정보대학 김영석 11-16
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.
전자정보대학 김영석 11-17
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.
전자정보대학 김영석 11-18
Typical Frequency Response
전자정보대학 김영석 11-19
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.
전자정보대학 김영석 11-20
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.
전자정보대학 김영석 11-21
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.
전자정보대학 김영석 11-22
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.
전자정보대학 김영석 11-23
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
전자정보대학 김영석 11-24
Example: Transit Frequency Calculation
( )THGSn
T VVL
f −= 2232 μπ
L2
mVVVnmL
THGS 10065
2
=−=
GHzfsVcm
T
n
226)./(400 2
==μ
전자정보대학 김영석 11-25
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.
전자정보대학 김영석 11-26
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.
전자정보대학 김영석 11-27
11.4 (High) Frequency Response of CE/CS StageUnified Model for CE and CS Stages
전자정보대학 김영석 11-28
11.4 (High) Frequency Response of CE/CS StageUnified Model Using Miller’s Theorem
전자정보대학 김영석 11-29
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
전자정보대학 김영석 11-30
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
전자정보대학 김영석 11-31
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
전자정보대학 김영석 11-32
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
전자정보대학 김영석 11-33
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).
전자정보대학 김영석 11-34
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
전자정보대학 김영석 11-35
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
전자정보대학 김영석 11-36
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 ≈= −
전자정보대학 김영석 11-37
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
전자정보대학 김영석 11-38
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
전자정보대학 김영석 11-39
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
≈=
≈=
−
−
)18.11(250),13(22 _3_ ExMHzfdBA CSdBCSv ≈= −
전자정보대학 김영석 11-40
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
전자정보대학 김영석 11-41
Ex 11.23 Source Follower Input Capacitance
( )]||1[ 21 OO rrCCC +
( )( )]||/1||1[
211
2111
OOm
OOGSGDin rrg
CCC+
−+=
전자정보대학 김영석 11-42
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
전자정보대학 김영석 11-43
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
=+
+=
전자정보대학 김영석 11-44
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.
전자정보대학 김영석 11-45
Poles of Bipolar Cascode
( )1ω = 1( )( )111
, 2|| μππ
ωCCrRS
Xp +=
( )22,
1μ
ωCCR CSL
outp +=
1
( )1212
,
21μπ
ωCCC
g CSm
Yp
++=
2gm
전자정보대학 김영석 11-46
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
전자정보대학 김영석 11-47
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
)18.11(250),13(22 _3_ ExMHzfdBA CSdBCSv ≈= −
전자정보대학 김영석 11-48
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||+
=
전자정보대학 김영석 11-49
Comparisons
전자정보대학 김영석 11-50
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.
전자정보대학 김영석 11-51
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.
전자정보대학 김영석 11-52
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
전자정보대학 김영석 11-53
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.
전자정보대학 김영석 11-54
Tail Node Capacitance Contribution
S d f dSource-Body Capacitance of M1, M2 and M3
Gate-Drain Capacitance of M3
전자정보대학 김영석 11-55
Ex11.28: Capacitive Coupling
( )[ ]EBin RrRR 1|| 222 ++= βπ
( ) ( )HzCRr B
L 5422||
1
1111 ×== πω
π ( ) ( )HzCRR inC
L 9.221
222 ×=
+= πω
전자정보대학 김영석 11-56
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
전자정보대학 김영석 11-57
Ex11.29: IC Amplifier – Midband Design
( ) 773||X RRv ( )211 77.3||
out
inDmin
X
Rv
RRgvv
−=−=
22 DmX
out Rgv
−=
전자정보대학 김영석 11-58
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
전자정보대학 김영석 11-59