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Chapter 14
Single-Transistors Amplifiers
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
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Chapter Goals
Detailed study of three broad classes of amplifiers
Inverting amplifiers- that provide high voltage gain with a 1800
phase shift, common-emitter and common-source configurations,
Followers- that provide nearly unity gain similar to op amp voltage
follower, common-collector and common-drain configurations, Noninverting amplifiers- that provide high voltage gain with no
phase shift, common-base and common-gate configurations.
Detailed design of voltage gain, input voltage range, current gain, input
and output resistances, coupling and bypass capacitor design and lower
cutoff frequency for each type of amplifier. Understand differences between SPICE ac (small-signal), transient
(large-signal) and transfer function analysis modes.
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Signal Injection and Extraction: BJT
In forward-active region,
To cause change in current, vBE= vB - vEmust be changed.Base or emitter terminals are used to inject signal because
even if Early voltage is considered, collector voltage has
negligible effect on terminal currents.
Substantial changes in collector or emitter currents cancreate large voltage drops across collector and emitter
resistors and collector or emitter can be used to extract
output. Since iB is a factor ofbFsmaller than iCoriE
currents, base terminal is not used to extract output.
TVBEv
F
SI
Ei
TVBEv
SI
Ci
exp
exp
T
VBEv
FO
SI
Bi exp
b
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Signal Injection and Extraction: FET
In pinch-off region,
To cause change in current, vGS= vG - vSmust be changed.Gate or source terminals are used to inject signal because
even with channel-length modulation, drain voltage has
negligible effect on terminal currents.
Substantial changes in drain or source currents can createlarge voltage drops across drain and source resistors and
drain or source can be used to extract output. Since iG is
always zero, gate terminal is not used to extract output.
2
2
TN
VGS
vnK
Di
Si
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Amplifier Families
Constraints for signal injection and extraction yield three families of
amplifiers
Common-Emitter (C-E)/Common- Source (C-S)
Common-Base (C-B)/Common- Gate (C-G) Common-Collector (C-C)/Common- Drain (C-D)
All circuit examples here use the four-resistor bias circuits to establish
Q-point of the various amplifiers
Coupling and bypass capacitors are used to change the ac equivalent
circuits.
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Inverting Amplifiers: Common-Emitter
and Common-Source Circuits
AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier
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Inverting Amplifiers: Terminal Voltage
Gain
Using test source vb to drive the base
terminal of the transistor, neglecting
ro,
ER
mg
LR
mg
ACEvt
ERomgo
LR
o
bv
ov
vtA
1
)1(/ bb
b
1obAssuming
For C-S Amplifier, take limit of voltagegain of C-E amplifier as
and
IfREandRSare set to zero, the voltagegain has an upper bound of
IfREandRSare large, the voltage gainhas a lower bound of
r
b rmgo
SR
mg
LR
mg
ACSvt
1
CCVA
CEvt
LRmgA
CSvtA
CEvt
10
DDVA
CSvt
R
LR
ACSvtA
CEvt
R is the unbypassed
resistor in the
emitter or source.
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Inverting Amplifiers: Input Signal Range
For small-signal operation, magnitude ofvbe developed across r in small-
signal model must be less than 5 mV.
If , vb can be increased beyond 5 mV limit.
In case of FET, magnitude ofvgs must be less than 0.2(VGS- VTN).
Presence ofRSincreases permissible value ofvg
r
ER
ERmg
r
1
bv
ibev
V)1(005.01005.0E
Rmg
o
ER
ERmgb
v
b
1E
Rmg
)(2.01 TN
VGS
V
SRmg
gvgsv
)1)((2.0 SRmgTN
VGS
Vgv
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Inverting Amplifiers: Condition forgmR
>>1
This condition simplifies the gain expression and is used to stabilize
voltage gain, achieve high input and output resistances and increase
input signal range.
For BJT:
For MOSFET:
V025.0
1
ER
EI
TV
ER
EI
TV
ER
CI
ERmg
2
12
TNV
GSV
SR
DI
TNV
GSV
SR
DI
SR
mg
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Inverting Amplifiers: Input Resistance
and Overall Voltage GainInput resistance looking into the base
terminal is given by
For C-S Amplifier,
)1(
)1(
ERmgrRCEin
ERor
i
bv
RCEin
b
r
RCSin
RCEinBR
IR
RCEinBRA
CEvt
ivb
vA
CEvt
ivb
v
bvov
ivovA
CEv
Overall voltage gain is
For C-S Amplifier,
GR
IR
GR
ACSvtA
CSv
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Inverting Amplifiers: Voltage Gain
Calculations (Example)
Problem: Find overall voltage gain.
Given data: Q-point values and values forRI,R1,R2,R3,R7 ,for both
BJT and FET as well as values forREandRS.
Assumptions: Small-signal operating conditions.
Analysis: For C-E Amplifier,
k18k100k22
73
k313k3(101)k160)1(
k104k300k16021
RR
L
R
ERmgrR
CEin
RRB
R
75.5k313
)k18(100
RCEin
LR
oA
CEvt
b61.5
RCEinB
RI
R
RCEinBR
ACEvtA
CEv
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Inverting Amplifiers: Voltage Gain
Calculations (Example contd.)
Analysis: For C-S Amplifier,
k18k100k2273
k892M2.2.5M121
RRL
R
RRG
R
46.4)k2(mS)491.0(1
)k18(mS)491.0(
1
S
Rm
g
LRmgACSvt
45.4
GRIR
GR
ACS
vtA
CS
v
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Inverting Amplifiers: Output Resistance
0ixi0i
0ev0)1(
1ev
i)1(ev
o
rth
RE
Ro
ERo
b
b
b
rthR
ev-
i
ButRout = ro whenRE= 0, not infinite.
Now, we also include ro in our analysis.
xixvoutR
b
b
rth
RE
RE
RoorR
rthRER
ER
ERr
thR
oro
1out
xii
xiev
ev)ixi(evrvxv
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Inverting Amplifiers: Output Resistance
(contd.)
b rmgo Assuming and , with
.for
Finite current gain of BJT places an upper limit on size of output
resistance. rappears in parallel withREifRth is neglected. If we letREbe
infinite, maximum value of output resistance is
Output resistance of C-S amplifier is given by,
thR
ERr )( E
Ror
)(out
)()(1out
ERr
fR
ERr
for
ERrmgorR
1)(
E
Rrmg
oroR )1(out b
S
RmgorR 1out
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Inverting Amplifiers: Output Resistance
(Example) Problem: Find output resistance.
Given data: Q-point values and values forRI,R1,R2,R3,R7 ,for both
BJT and FET as well as values forREandRS.
Assumptions: Small-signal operating conditions. Small -signal values
are known.
Analysis: For C-E Amplifier,
For C-S Amplifier,
4.55Mk3k2.10k96.1
k3100(1k2191
out
b
rth
RE
RE
RoorR
k442k20.491mS)(1k2231out
SRmgorR
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Inverting Amplifiers: Current Gain
Terminal current gain is the ratio of the current delivered to the load
resistor to the current being supplied to the base terminal.
oACEit b
ACSit
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Inverting Amplifiers: Summary
C-E and C-S amplifiers have similar voltage gains.
C-S amplifier provides extremely high input resistance but that of C-E is also
substantial due to the fREterm.
Output resistance of C-E amplifier is much higher than that of C-S amplifier as
f is much larger for BJT than for FET.
Input signal range of C-E amplifier is also higher than that of C-S amplifier.
Current gains of both are identical to those of individual transistors.
Following transformation is used to simplify circuit analysis by absorbingRE
(orRS) into the transistor (For FET, current gain and input resistance are
infinite).
ERmg
mgmg1
' )1(' ERmgrr )1(' ERmgoror
ormgo bb ''' formgf
'''
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Follower Circuits: Common-Collector
and Common-Drain Amplifiers
AC equivalent for C-C Amplifier AC equivalent for C-D Amplifier
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Follower Circuits: Terminal Voltage
Gain
Neglecting ro,
LR
mg
LR
mg
ACCvt
L
Ro
r
LR
o
b
v
ov
vtA
1
)1(
)1(
b
b
1obAssuming
For C-S Amplifier, take limit of voltagegain of C-E amplifier as
and
In most C-C and C-D amplifiers,
Output voltage follows input voltage,
hence theses circuits are called
followers. BJT gain is closer to unity
than FET. Mostly,
ro can be neglected as gain
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Follower Circuits: Input Signal Range
For small-signal operation, magnitude ofvbe developed across r in small-
signal model must be less than 5 mV.
If , vb can be increased beyond 5 mV limit.Since only small
portion of input signal appears across base-emitter or gate-source terminals,
followers can be used with relatively large input signals without violating
small-signal limits.
In case of FET, magnitude ofvgs must be less than 0.2(VGS- VTN).
rL
R
LRmg
r
1
bv
ibev
V)1(005.01005.0L
Rmg
o
LR
LRmgb
v
b
1L
Rmg
)(2.01 TN
VGS
V
LRmg
gv
gsv )1)((2.0 L
RmgTNV
GSVgv
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Follower Circuits: Input Resistance and
Overall Voltage GainInput resistance looking into the base
terminal is given by
For C-S Amplifier,
LRor
i
bv
RCCin
)1( b
r
RCDin
RCCinBR
IR
RCCinBRA
CEvt
ivb
vA
CCvt
ivb
v
bvov
ivovA
CCv
Overall voltage gain is
For C-S Amplifier,
GR
IR
GR
ACDvtA
CCv
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Follower Circuits: Voltage Gain
Calculations (Example)
Problem: Find overall voltage gain.
Given data: Q-point values and values forRI,R1,R2,R4,R7 ,for both
BJT and FET.
Assumptions: Small-signal operating conditions.
Analysis: For C-C Amplifier,
k117)k2.10(101k2.10)1(
k5.1174
k10421
L
RorRCE
in
RRL
R
RRB
R
b
991.0k117
)k5.11(101)1(
RCCin
LR
oA
CCvt
b956.0
RCCinB
RI
R
RCCinBR
ACCvtA
CCv
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Follower Circuits: Voltage Gain
Calculations (Example contd.)
Analysis: For C-D Amplifier,
k7.1074
892k21
RRL
R
RRG
R
840.0)k7.10(mS)491.0(1
)k7.10(mS)491.0(
1
L
Rm
g
LRmgACDvt
838.0
GRIR
GR
ACD
vtA
CDv
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Follower Circuits: Output Resistance
111
xvxviixi
o
thR
o
r
o
rth
R
RCCout
rth
Ro
rth
Ro
bb
b
b
b
1
1
1
o
thR
mgo
thR
mgoR
CCout bb
In case of FET,
Thus equivalent resistance looking into
emitter or source of a transistor is
approximately 1/gm.
mgR
CD
out
1
Current is injected
into emitter of BJT.
1
iiev
o
thR
mgo
b
Current oi coming out of collector mustbe supported by veb = oi/gm, given by
first term. ib =-i/bo+1creates voltage
drop inRth given by second term
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Follower Circuits: Output Resistance
(Example) Problem: Find output resistance. Given data: Q-point values and values forRI,R1,R2,R4,R7 ,for both
BJT and FET.
Assumptions: Small-signal operating conditions. Small -signal values
are known.
Analysis: For C-C Amplifier,
For C-D Amplifier,
120101
k96.1
mS80.9
990.0
1
1
o
thR
mgR
CCout b
k04.2mS91.4
11
mgR
CDout
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Follower Circuits: Current Gain
Terminal current gain is the ratio of the current delivered to the load
resistor to the current being supplied from the Thevenin source.
1i
1i
oA
CC
itb
ACDit
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Follower Circuits: Summary
Both C-C and C-D amplifiers have voltage gains approaching unity.
C-D amplifier provides extremely high input resistance because of
infinite resistance looking into gate terminal of FET as compared to C-
C amplifier.
Output resistance of C-C amplifier is much lower than the C-D
amplifier due to higher transconductance of BJT than an FET for given
operating current.
Both C-C and C-D amplifiers can handle relatively large input signal
levels..
Current gains of FET is inherently infinite, whereas that of BJT is
limited by its finite value ofbo.
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Noninverting Amplifiers: Common-Base
and Common-Gate Circuits
AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier
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Noninverting Amplifiers: Terminal
Voltage Gain and Input Resistance
Polarities ofvbe and dependentcurrent sourcegmvbe are both
reversed, signal source is
transformed to its Norton
equivalent ro is neglected.
LRmg
ev
ovA
CBvt
For C-S Amplifier, take limit of voltage
gain of C-E amplifier as
r
LRmgA
CGvt
mgo
r
i
ev
RCBin
1
1
b
m
gR
CGin
1
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Noninverting Amplifiers: Input Signal
Range
For small-signal operation,
In case of FET,
Relative size ofgm andRIdetermine signal-handling limits.
4
4
)4
(1
iv
ebv
RI
R
R
RI
Rmg
V)1(005.0I
Rmgbv
)1)((2.0I
RmgTNV
GSV
iv
)1(ebv
iv
IRmg forRI>>R4.
)1(sgviv
IRmg
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Noninverting Amplifiers: Overall
Voltage Gain
44
)4
(1
4
4
RI
R
R
RI
RmgL
R
m
g
RCBinR
IR
RCBinR
ACBvt
ivev
evov
ivovA
CBv
Overall voltage gain is
For C-S Amplifier,
ForRI>>R4,
4
4
)4(1 RIR
R
RIRmg
LRmg
ACGv
IRmgL
RmgA
CGCBv
1
,
For ,
This is the upper bound on gain.
For ,
1IRmg
LRmgA
CGvt
LRmgA
CBvt
1th
Rmg
IR
LR
A
CG
vtA
CB
vt
ro can be neglected as gain
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Noninverting Amplifiers: Voltage Gain
Calculations (Example)
Problem: Find overall voltage gain. Given data: Q-point values and values forR1,R2,R3,R7 ,for both BJT
and FET,RI=2 k,R4 =12 k.
Assumptions: Small-signal operating conditions.
Analysis: For C-E Amplifier,
For C-S Amplifier,
k1873
102/1
RRL
R
mgRCBin176 LRmgACBvt
59.8
4
4
)4
(1
RI
R
R
RI
Rmg
ACBvtA
CBv
k1873 RRLR 84.8
LRmgACSvt
k04.2/1 mgRCGin 11.4
4
4
)4
(1
RI
R
R
RI
Rmg
ACGvtACGv
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Noninverting Amplifiers: Output
Resistance
b
rth
Rth
RoorR
CBout 1
Desired resistance is that looking intocollector with base grounded and resistorRth
in emitter.
The redrawn equivalent circuit is same as
that for C-E amplifier except resistance in
base is zero and resistance in emitter is
relabeled asRth.
Using b rmgo
)()(1th
Rrfth
RrmgorRCGoutR
CBout
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Noninverting Amplifiers: Current Gain
Terminal current gain is the ratio of the current delivered to the load
resistor to the current being supplied to the base terminal.
1ei1i
oA
CB
it
1ACGit
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Noninverting Amplifiers: Summary
C-B and C-G amplifiers have similar voltage and current gains.
Numerical differences occur due to difference in parameter values of
BJT and FET at similar operating points.
C-B amplifier can achieve high output resistance due to higher
amplification factor of BJT.
C-B amplifier can more easily reach low levels of output resistance due
to higher transconductance of BJT for a given operating current.
Input signal range of C-G amplifier is inherently larger than that of C-B
amplifier.
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Selecting Amplifier Configuration
A single-transistor amplifier with a gain of 80 dB and input resistance of 100
k.
Av = 1080/20 = 10,000. For even best BJTs, gain< f= 40VA = 40(150) =
6000 and FET typically has much lower intrinsic gain. Hence such large
gain cant be achieved by single-transistor amplifier. A single-transistor amplifier with gain of 52 dB, input resistance of 250 k.
Av = 1052/20 = 400. Since we need large gain and relatively large input
resistance, we can use C-E amplifier.Av = 20VCC, so, VCC=20 V.
which is small but acceptable.
For FET, even with small gate overdrive, VDD =100 V which is too large
A105105.2
)V025.0(100k250 b
C
I
CI
TVor
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Selecting Amplifier Configuration
(contd.)
A single-transistor amplifier with gain of 30dB and input resistance of 5M.
Av = 1030/20 = 31.6. Since we need large input resistance and moderate
gain, we can use C-S amplifier.
Input resistance of can be set by our choice of gate-bias resistors. For C-
E amplifier, required high input resistance could be attained but values of
base bias resistor could be a limiting factor.
A single-transistor amplifier with gain of 0dB and input resistance of 20M
with load resistor of 10k. Gain of 0 dB implies a source follower,Rin=boRL=100(10k)=1M, so
BJT cant meet input resistance requirement, A source follower can be
used easily.
30V5.0
DD
V
TNV
GSV
DDV
vA
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Coupling and Bypass Capacitor Design
Since impedance of a capacitor increases with decreasing frequency,
coupling and bypass capacitors reduce amplifier gain at low
frequencies.
To choose capacitor values, short-circuit time constant method is used:each capacitor is considered separately with all other capacitors
replaced by short circuits.
To neglect a capacitor, the magnitude of capacitive impedance must be
much smaller than the equivalent resistance appearing at its terminals.
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Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers
RCEinB
Rin
R For C-E amplifier,
RCEoutRout
R3
For C-S amplifier,
RCSinG
Rin
R RCSoutRout
R3
For coupling capacitorC1,
For coupling capacitorC3,
w is chosen to be lowest frequency
for which midband operation is
needed in given application.
inR
IR
Cw
11
outRR
C
7
13
w
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Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers (contd.)
In this case, we can neglect impedancesof capacitors C1 and C3 , the find the
equivalent resistance looking up into
emitter or source of amplifier.
mgE
RR
C
16
1
2w
mgS
RR
C
16
12
w
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Coupling and Bypass Capacitor Design:
C-E and C-S Amplifiers (Example)
Problem: Choose values of coupling and bypass capacitors. Given data:f= 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
Analysis:For C-E amplifier:
k1.78 RCEinBRinR
F02.01
1.99nF1
1
w
C
inR
IR
C
F0.682
nF2.67)/1(
6
12
w
Cm
gE
RRC
F015.03
nF31.1
7
13
w
C
outRR
C
For C-S amplifier:
k
892
GRinRpF1800
1178pF
11
C
inR
IR
Cw
F0.562
nF3.55)/1(
6
12
w
Cm
g
S
RRC
F015.03
nF31.1
7
13
w
C
outRR
C
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Coupling and Bypass Capacitor Design:
C-C and C-D Amplifiers
RCCinB
Rin
R
For C-E amplifier,
RCCoutRout
R4
For C-S amplifier,
R
CS
inGR
inR
R
CD
outR
outR
4
For coupling capacitorC1,
For coupling capacitorC3,
inR
IR
Cw
11
outRR
C
7
13 w
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Coupling and Bypass Capacitor Design:
C-C and C-D Amplifiers (Example)
Problem: Choose values of coupling and bypass capacitors. Given data:f= 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
Analysis:For C-C amplifier:
k5.95 RCCinBRinR
pF82001
816pF1
1
C
inR
IR
Cw
pF82003
pF795
7
13
C
outRR
Cw
For C-D amplifier:
k892G
R
in
R
pF10001
89pF1
1
Cin
RI
RC
w
pF82003
pF782
7
13
C
outRR
Cw
1204
RCCoutRout
R k74.1
4R
CDoutRout
R
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Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers
RCBin
Rin
R4
For C-E amplifier,
RCBoutRout
R3
For C-S amplifier,
RCGinRinR 4 R
CGoutRoutR 3
For coupling capacitorC1,
For coupling capacitorC3,
inR
IR
Cw
11
outRR
C
7
13 w
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Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers (contd.)
In this case, we can neglect impedancesof capacitors C1 and C3 , the find the
equivalent resistance looking up into
emitter or source of amplifier.
21
)4)(1(21
RRRCGeq
IRRorRRRCBeq
b
RCGCB
eq
C,
12
w
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Microelectronic Circuit Design
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Coupling and Bypass Capacitor Design:
C-B and C-G Amplifiers (Example)
Problem: Choose values of coupling and bypass capacitors. Given data:f= 1000Hz, values of all resistors and input and output
resistances for both C-E and C-S amplifiers.
Analysis: For C-B amplifier:
100k
102k
134
R
CB
inRinR
F82.01
75.8nF1
1
w
C
inR
IR
C
F015.03
nF31.1
7
13
w
C
outRR
C
k9.21M93.3k223
RCBoutRoutR
F0.027nF38.2
)4
)(1(21
1
2
bw
IRR
orRR
C
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Coupling and Bypass Capacitor Design: C-
B and C-G Amplifiers (Example contd.)
For C-G amplifier:
k74.1k04.2k124
RCGinR
inR
F42.01
42.6nF1
1
w
C
inR
IR
C
F015.03
nF31.1
7
13
w
C
outRR
C
k9.20k410k224
RCGoutRoutR
pF1800pF178
21
12
RR
Cw
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Lower Cutoff Frequency of an Amplifier
We can choose capacitor values to set the lower cutoff frequency of the
amplifier at desired value.
Pole associated with a capacitor occurs at the frequency at which
capacitive reactance is equal to resistance at the capacitor terminals.
In discussed amplifiers, there are several poles and a bandwidth
shrinkage occurs at low frequencies.
A transfer function with n identical poles at wo is given by
Lower cutoff frequency is higher than frequency corresponding to
individual poles.
n
o
nAjT
22
mid)(
ww
ww
1/122mid)(
no
L
A
LjT
www
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Microelectronic Circuit Design
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Dominant Pole Design
Instead of having the lower cutoff frequency set by the interaction of
several poles, it can be set by the pole associated with just one of the
capacitors. The other capacitors can be chosen to have their pole
frequencies much belowfL.
Capacitor associated with emitter or source part of the circuit tends to
be the largest due to low resistance presented by emitter or source
terminal of transistor and is commonly used to setfL.
Values of other capacitors are increased by a factor of 10 to push their
corresponding poles to much lower frequencies.
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Follower Design Example
Problem: Design an amplifier with given specifications.
Given data:Av>0.95,Rin>20M,Rout 6600 (beyond range of normal
BJTs). So we choose source-follower
configuration.
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Follower Design Example (contd.)
M22G
Rin
R
Tolerance of 5% is included.
Now,
LRD
InK
LRmgL
RmgL
Rmg
2
19
1995.01
k3k3 SRLR
We chooseRL=1.5 k
mA96.8D
InK
GSV
SR
DI
SSV
TNV
GSVn
K
DI
2
2
We can chooseID = 5
mA withKn =20mA/V2
to give VSS=16.7+ VTN.If we choose a MOSFET
with VTN= 1.5 V,V21.2
GSV
k3S
R
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Follower Design Example (contd.)
VSS=16.7+ VTN. gives a lower bound. Ifwe choose VSS= 20 V,RS=3.56 k .
To choose VDD :
For small-signal operation,
If we choose VDD =1.2 V, small-signal
criteria are fulfilled and MOSFET stays
saturated for all signals.
V5.1
SvTNVSvDDV
TNV
GSV
Sv
GSV
DDv
V7.2)19)(71.0(2.0
1)1(2.0
LRmgL
Rmg
LRmgTN
VGS
Vggv
Coupling capacitors mustnt affectcircuit at frequencies > 50 Hz.
Choose C1 =1500 pF and C2 =10 F.
pF1451
M22
1)Hz50(2
1 C
C
F03.12
k097.3
2)Hz50(2
1
C
C
4.691
mg
SRoutR
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Common-Source Amplifier Design
Example
Problem: Design an amplifier with given specifications.
Given data:Av> 50,Rin= 75,f >500 kHz.
Analysis:Moderate input resistance of C-E and C-S amplifiers can be
limited by reducing sizes of bias resistors (called swamping of
impedance level). JFET is chosen over BJT due to its better signalhandling capability and simple bias circuit design.
To deliver overall gain of 50, amplifier
must deliver a gain of 100.
We can choose VDD =20 V
and VGS- VP= 0.2 V
27575
75 iv
iv
GR
IR
GR
ivgsv
PV
GSV
DDV
vA
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Common-Source Amplifier Design
Example (contd.)
k100
k100
DR
DR
LR
ID of 0.2 mA inRD = 100 k needs avoltage drop equal to total power
supply of 20 V. So power supply
voltage must be increased.
For pinch-off region operation,
Choose VDD = 25 V
k9
DIGS
V
SR
VGS- VPmust be small for high gain,
so JFET with largeIDSShas to be
chosen for a reasonableID. Choose
JFET withIDSS= 20 mA and VP=-2 V
mA2.0
2
1
PVGS
V
DSSIDI
k50mg
v2A
LR
V22)2()8.1(8.120
DDV
DDV
PVGSV
SRDIDRDIDDV
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Common-Source Amplifier Design
Example (contd.)
Coupling capacitors mustnt affectcircuit at frequencies > 500 kHz.
Choose C1 =0.022 F, C2 =0.0068 F
and C3 =20 pF
nF12.21
150
1)kHz500(2
1 C
C
pF6442
494
2)kHz500(2
1 C
C
4741k1.9 mg
eqR
pF59.13k
2003
)kHz500(2
1
CC
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Common-Source Amplifier with Body
Effect
SR
mbgmgS
R )bsvgsv(oisv
svbsv svgvgsv
thv
)1(1sv
SR
mg
SRmg
mgmbg
gv)1(1
oi
SR
mg
mg
SRmg
LRmgL
R
vtA )1(1gv
oi-
gvov
gsv)1(
gsv)(bsvgsvi
mgmb
gmgmbgmg
S
RmgorRCSout )1(1
RCSin
ACSit
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Common-Drain Amplifier with Body
Effect
gv)1(1
oi
L
Rmgmg
LRmg
LRmgL
R
vtA
)1(1gv
oi
gvov
)1(
1
mgR
CDout
RCDin
ACDit
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Common-Gate Amplifier with Body
Effect
LRmgA
CGvt )1(
)1(
1
mgR
CGin
))(1(1
SR
IRmgorR
CGout
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J /Bl l k Mi l i Ci i D i
Results of Body Effect
Gain of C-G amplifier more closely approaches theRL/Rth limit.
Gain of source follower is degraded.
Input resistance of C-G and output resistance of C-D amplifier is
lowered. Output resistance of both C-S and C-G amplifiers is raised.
Body effect increases input signal range.
Ch 14 59
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