ece 2c, notes set 7: basic transistor circuits; high ... · basic transistor circuits;...
TRANSCRIPT
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class notes, M. Rodwell, copyrighted 2013
ECE 2C, notes set 7: Basic Transistor Circuits; High-Frequency Response
Mark Rodwell
University of California, Santa Barbara
[email protected] 805-893-3244, 805-893-3262 fax
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Goals
gengineerin electrical of partsmost in important :analysis LaPlace
useful. g,interestin real, are they
:exercises good make circuits Transistor
methods. LaPlacein expert Become
analysis.circuit in expert Become
:Goals
reponse.Transient response.Frequency
rolloff.frequency -High rolloff.frequency -Low
).( functionsansfer circuit tr calculate :notes These sH
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Field-Effect Transistor High-Frequency Model
given. as model thisesimply tak weece2c,For
classes.later in covered be willescapacitanc theseoforigin Physical
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Field-Effect Transistor High-Frequency Model
voltage.bias vary withescapacitanc theseofMany
es.capacitancjunction PNbulk -drain and )(substratebulk -source are and
ct vias.interconne and fields fringing source-gate toduePartly
ecapacitanc channel-gate frompart in arises
ct vias.interconne and fields, fringing source-gate toduePartly
ecapacitanc channel-gate frommostly arises
dbsb
dg
gs
CC
C
C
sbC dbC
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Basic Common-Source Amplifier
s.frequenciehigh at gain circuit thereduce will and
s.frequencie lowat gain circuit thereduce will and
gdgs
outin
CC
CC
biasing RC style-s1950' :circuit This
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class notes, M. Rodwell, copyrighted 2013
Basic Common-Source Amplifier
s.frequenciehigh at gain circuit thereduce will and
s.frequencie lowat gain circuit thereduce will and
gdgs
outin
CC
CC
:circuit equivalent signal-Small
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class notes, M. Rodwell, copyrighted 2013
Basic Common-Source Amplifier
. and neglect ly temporari therefore willWe
s.frequenciehigh at responseconsider first usLet
outin CC
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Nodal Analysis: How to solve circuits
equations.t independenlinearly ofset needed thegives Always
voltage. theknownot doyou which nodeeach at 0currents Write
.*tindependenlinearly * bemust equations the
voltagesnodeunknown of # unknowns of # equations of #
:circuit solve To
(red) ltageunknown vo (blue), ageknown volt
?known voltage theis :eachFor nodes.circuit all Label
node
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Common-Source Amplifier: Simplifying
:modelNorton a oThevenin t a fromgenerator eConvert th
:formulasresistor parallel use then And
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
gengengdoutgdgsiin
gengengdoutgdingsiniin
in
GVsCVsCsCGV
GVsCVsCVsCVGV
V
)()(
)(
:zero is at currents of Sum
0)()(
0)()()(
:zero is at currents of Sum
gdLeqoutgdmin
Leqoutgdoutgdinmin
out
sCGVsCgV
GVsCVsCVgV
V
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
0)()(
)()(
gdLeqoutgdmin
gengengdoutgdgsiin
sCGVsCgV
GVsCVsCsCGV
0
gengen
out
in
gdLeqgdm
gdgdgsi GV
V
V
sCGsCg
sCsCsCG
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Common-Source: Nodal Analysis
0
gengen
out
in
gdLeqgdm
gdgdgsi GV
V
V
sCGsCg
sCsCsCG
gen
gen
out
in
gdLeqgdm
gdgdgsiV
G
V
V
sCGsCg
sCsCsCG
0
0
1 gen
genout
in
gdLeqgdm
gdgdgsi G
VV
V
sCGsCg
sCsCsCG
0/
/gen
genout
genin
gdLeqgdm
gdgdgsi G
VV
VV
sCGsCg
sCsCsCG
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
0)(
and
)(
where
)()(
)()()()(
gdm
gengdgsi
gdLeqgdm
gdgdgsi
gengenout
sCg
GsCsCGsN
sCGsCg
sCsCsCGsD
sVsD
sNsVsHsV
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
gdmgsgdgdLeqgsLeqgdiLeqi
gdmgdgdLeqgdgsi
gdLeqgdm
gdgdgsi
sCgsCsCsCGsCGsCGGGsD
sCgsCsCGsCsCGsD
sCGsCg
sCsCsCGsD
)(
)(
)(
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
gdgsgdmgdLeqgsLeqgdiLeqi CCssCgCGCGCGsGGsD
s
2)(
: of powers into organize
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
gdgsleqigdleqleqmgdigsi
gdgsleqigdleqimgdigsigdleqleqi
gdgsleqigdmgdLeqgsLeqgdileqi
Leqi
CCRRsCRRgCRCRs
CCRRsCRRgCRCRCRsRRsD
CCRRsCgCGCGCGRsR
GGsD
2
2
2
)1(1
1)(
1
/)(
:responsefrequency unitless aby multiplied
units carryingconstant a into Separate
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
mgdmgengdgenmgen
gdmgen
gdm
gengdgsi
gsCgGsCGgG
sCgGsCg
GsCsCGsN
/1
0)(
:numerator for the rules same theFollow
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class notes, M. Rodwell, copyrighted 2013
Common-Source: Nodal Analysis
gdgsleqigdleqleqmgdigsi
mgdmleqigen
gen
out
gdgsleqigdleqleqmgdigsi
mgdmgen
leqi
gdgsleqigdleqleqmgdigsileqi
mgdmgen
CCRRsCRRgCRCRs
gsCgRRGsH
sV
sV
CCRRsCRRgCRCRs
gsCgG
RRsD
sN
CCRRsCRRgCRCRsRRsD
gsCgGsN
2
2
2
)1(1
/1)(
)(
)(
So
)1(1
/1
)(
)(
)1(1)(
/1)(
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Common-Source: Nodal Analysis
Ampingen
Ampin
leqm
Ampingengen
Ampingen
leqm
gen
Ampingen
leqm
gen
ileqmmleqigenbandmid
gdgsleqigdleqleqmgdigsi
mgd
normalized
normalizedbandmid
gen
out
RR
RRg
RRR
RRRg
R
RRRg
R
RRggRRGH
CCRRsCRRgCRCRs
gsCsH
sHHsHV
V
,
,
,
,
,
2
)(
||
)1(1
/1)(
)()(
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Common-Source Nodal Analysis: the answer
mgd
gdgsleqi
gdleqgdleqimgdigsi
normalized
Ampingen
Ampin
leqmbandmid
normalizedbandmid
gen
out
gCb
CCRRa
CRCRRgCRCRa
sasa
sbsH
RR
RRgH
sHHV
V
/ and
:constant (time)order -Second
:constant order time-First
response.frequency amplifier theis 1
1)(
gain. band-mid theis
where
)(
1
2
2
1
2
21
1
,
,
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Comments on the analysis
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Finding Poles: Separated Pole Approximation
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Finding Poles: Separated Pole Approximation
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Amplifier Frequency Response
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Example
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Frequency Response Example (1): DC
V10/1
V 3.0
mA/V1)2/(
:FET
2
th
ggox
V
LWc
V 3.3
A 50
V 7.0
conditionsDesign
DD
D
DS
V
I
V
V. 52.0
V22.0mA/V1A 50)V 3.0(
)2/()()(
A 50)ignored term())(2/(
:Analysis
2/12
2/12/1
2
gs
gs
ggoxDthgs
thgsggoxD
V
V
LWcIVV
VVLWcI
k 25A 50/)V 7.0V 3.3(
/)(
DDSDDD
DDDDDS
IVVR
RIVV
analysis. signal small
in the term)1( theignorenot Do
.accuratelyfairly yet quickly thiscalculate
to trickssomelearn willECE137A weIn
Hard. formulas. quadratic
solving involves analysis DC thenot, do weIf
error. small some causes thisDoing
analysis. bias
in the term)1( theignoring are we*
DS
DS
V
V
A 1rough current th
k 50 ,k 100
1
g
Lgen
R
RR
,M 782.A 10V)/52.03.3(
,k 520A 1V/52.0
2
1
g
g
R
R
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Frequency Response Example (2): Component Values
V 3.3
k 25
k 100
k 520 k 50
M 782.
V 52.0 V 7.0
A 50 A 1
k 200
1S5
V10
A 50
)1(
mS 471.0)V/10V7.01)(V3.0V52.0)(mA/V2(
)1)()(/()1())(2/(
:FET
2
2
D
DS
D
ds
Dds
m
DSthgsggox
gs
DmDSthgsggoxD
IV
I
V
IG
g
VVVLWcV
IgVVVLWcI
S5
mS 47.0
ds
m
G
g
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Frequency Response Example (3): Component Values
k5.25
k50||k 25k 100
k438
2.78M|| k 520
mS 47.0mg
00k2
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Frequency Response Example (4): Component Values
k82
mS 47.0mg
k6.22
pF 1 pF .250
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Frequency Response Example (5): (time)n constants
ns 0.532mS47.0pF/25.0/
ion works.approximat pole separated so, ,/
ns42.1s326.0pF/25.01pFk6.22k82/
s326.0
pF25.0k6.22pF25.0k6.22k82mS47.0pF25.0k821pFk82
. 1
1)(
65.8814.066.10||
||
)(
1
12
122
1
2
21
1
21
21
mgd
gdgsleqi
gdleqgdleqimgdigsi
normalized
gg
gg
leqmbandmid
normalizedbandmid
gen
out
gCb
aaa
aaCCRRa
CRCRRgCRCRa
sasa
sbsH
RR
RRRgH
sHHV
V
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Frequency Response Example (6): Transfer Function
ns 0.532 ns,42.1/ s,326.0
))/(1)(1(
165.8
)(
)(
1121
121
1
baaa
saasa
sb
sV
sV
gen
out
plane)-s theof halfright in the (zero MHz 29932ps5
159.0
MHz 112ns42.1
159.0 kHz, 488
s326.0
159.0
)/1)(/1(
/165.8
)2(
)2(
21
21
z
pp
pp
z
gen
out
f
ff
fjffjf
fjf
fjV
fjV
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The Miller Effect
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The Miller Approximation
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The Miller Approximation
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The Miller Approximation
ivgd
iLeqmgd
RAC
RRgC
)1( i.e.
)1( see You will
analysis source-common theback toRefer
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Using Miller Approximation to Understand Response
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Using Miller Approximation to Understand Response
millergd CC with replace :eApproximat
gdLeqmmiller CRgC )1(
gsCiR
imVgLeqR
outV
Leqmouti RgVV is to fromgain frequency -Low
gdLeqmigsimillergsipole CRgRCRCCRf )1()(2/1
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Miller and Exact Solutions
solutionExact
2
1
gdgdleqi
gdleqgdleqimgdigsi
CCRRa
CRCRRgCRCRa
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Further Comments Regarding Miller Approximation
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class notes, M. Rodwell, copyrighted 2013
Frequency/Transient Reponse: Real, Important