chap17 frequency response
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Microelectronic Circuit DesignMcGraw-Hill
Chapter 17
Frequency Response
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
Chap 17- 1
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Chapter Goals
Review transfer function analysis and dominant-pole approximations
of amplifier transfer functions.
Learn partition of ac circuits into low and high-frequency equivalents.
Learn short-circuit and open-circuit time constant methods to estimate
upper and lower cutoff frequencies. Develop bipolar and MOS small-signal models with device
capacitances.
Study unity-gain bandwidth product limitations of BJTs and
MOSFETs.
Develop expressions for upper cutoff frequency of inverting, non-inverting and follower configurations.
Explore gain-bandwidth product limitations of single and multiple
transistor circuits.
Chap 17- 2
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Chapter Goals (cont.)
Understand Miller effect and design of op amp frequency
compensation.
Develop relationship between op amp unity-gain frequency and slew
rate.
Understand use of tuned circuits to design high-Q band-pass
amplifiers.
Understand concept of mixing and explore basic mixer circuits.
Study application of Gilbert multiplier as balanced modulator and
mixer.
Chap 17- 3
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Transfer Function Analysis
Av(s) N(s)
D(s)
a0
a1sa
2s2 ...a
msm
b0
b1sb
2s2 ...b
nsn
Av(s) AmidFL(s)FH(s)
Amid is midband gain between upper
and lower cutoff frequencies.
FL
(s)s Z1L
s Z2L
...s ZkL
s P1L
s P2L
...s PkL
FH(s)
1 s
Z1
H
1 s
Z2
H
...1 s
Zl
H
1 sP1H
1 s
P2H
...1 s
PlH
FH
(j) 1 for ZjL ,
PjL , j =1,k
AH
(s)Amid
FH
(s)
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Low-Frequency Response
FL
(s) ss
P2
L
P2
Pole P2 is called the dominant low-
frequency pole (> all other poles) and
zeros are at frequencies low enough tonot affect L.
If there is no dominant pole at low
frequencies, poles and zeros interact to
determine L.
AL
(s)Amid
FL
(s)Amid
sZ1 sZ2 s
P1 sP2
For L
s jL
, A(j
L)
Amid
2
1
2
L
2 Z1
2 L2 Z22 L
2 P1
2 L2 P22
12
1Z1
2 Z2
2 L
2Z1
2 Z2
2 L
4
1P1
2 P2
2 L
2P1
2 P2
2 L
4
Pole L > all other pole and zero frequencies
In general, forn poles and n zeros,
L P12 P22 2Z12 2Z22
L
Pn
2
n
2 Zn
2
n
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Transfer Function Analysis and
Dominant Pole Approximation Example Problem: Find midband gain,FL(s) andfL for
Analysis: Rearranging the given transfer function to get it in standard form,
Now,
Zeros are at s = 0 and s = -100. Poles are at s = -10, s =-1000
All pole and zero frequencies are low and separated by at least a decade. Dominant pole is
at = 1000 andfL = 1000/2p= 159 Hz. For frequencies > a few rad/s:
AL
(s) 2000s
s
1001
0.1s1 s1000
AL
(s) 200s s100
s10 s1000 F
L(s) s(s100)
(s10)(s1000)
AL
(s)Amid
FL
(s) Amid
200
fL
1
2p
102 10002 2 02 1002
158 Hz
AL
(s) 200 ss1000
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High-Frequency Response
FL
(s) s
1 sP3
H
P3
Pole P3 is called the dominant high-
frequency pole (< all other poles).
If there is no dominant pole at low
frequencies, poles and zeros interact to
determine H.
AH
(s)Amid
FH
(s)
AH
(s)Amid 1(s/Z1) 1(s/Z2) 1(s/P1
) 1(s/P2)
For=H
s = jH
, A(j
H)
Amid
2
1
2
1(H
2 /Z1
2 ) 1(H2 /Z22 ) 1(
H
2 /P1
2 ) 1(H2 /P22 )
1
2
1
H
2
Z1
2
H
2
Z2
2
H
4
Z1
2 Z2
2
1
H
2
P1
2
H
2
P2
2
H
4
P1
2 P2
2
Pole H < all other pole and zero frequencies
In general,
H
11
P12 1
P22 2
Z12 2
Z22
H
11
Pn
2n
21
Zn
2n
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Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier
Vo(s)I
o(s)R
3 g
mVgs(s)
RD
R3
(1/sC2)R
3
R3
gm
(R3RD
) s
s 1C
2(R
DR3)
Vgs(s)
Vg(s) sC1RGsC1(R
IR
G)1
Vi(s)
Vgs(s)V
g-V
s
s(1/C3R
S)
s 1C
3(1/g
m) R
S
Vg(s)
Av(s)Vo(s)
Vi(s)
AmidFL(s)
Amid
gm
(R3RD
)RG
RG
RI
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Direct Determination of Low-Frequency
Poles and Zeros: C-S Amplifier (cont.)
The three zero locations are: s = 0, 0, -1/(RSC3).
The three pole locations are:
Each independent capacitor in the circuit contributes one pole and one
zero. Series capacitors C1
and C2
contribute the two zeros at s = 0 (dc),
blocking propagation of dc signals through the amplifier. Third zero due
to parallel combination ofC3 andRSoccurs at frequency where signal
current propagation through MOSFET is blocked (output voltage is zero).
FL(s)
s2 s(1/C3R
S)
s 1C
1(R
IR
G)
s 1
C3
(1/gm
) RS
s 1C
2(R
DR
3)
s 1C
1(R
IR
G), 1
C3
(1/gm
)RS
, 1C
2(R
DR
3)
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Short-Circuit Time Constant Method to
Determine L
Lower cutoff frequency for a
network with n coupling and
bypass capacitors is given by:
whereRiSis resistance at
terminals ofith capacitorCi with
all other capacitors replaced by
short circuits. ProductRiSCi isthe short-circuit time constant
associated with Ci.
L
1
RiS
Ci
i 1
n
Midband gain and upper and lower
cutoff frequencies that definebandwidth of amplifier are of more
interest than complete transfer function.
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Estimate ofL for C-E Amplifier
Using SCTC method for C1:
R1S
RI
(RB in
CER )R2 (RB rp)
ForC2
,
R2S
R3(R
C outCER )R3 (RC ro)R3 RC
ForC3,
R3S
RE out
CCR RErp
Rth
o 1
RE
rp
(RIRB
)
o 1
L
1R
iSC
ii 1
3
Chap 17- 11
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Estimate ofL for C-S Amplifier
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Estimate ofL for C-S Amplifier
Using the SCTC method for C1,
R1S
RI
(RG in
CSR )RSRG
For C2,
R2S
R3(R
D outCSR )R3 (RD ro)
R2S
R3R
D
For C3,
R3S
RS out
CGR RS1
gm
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Estimate ofL for C-B Amplifier
Apply the SCTC method :
For C1,
R1S
RI
(RE in
CBR )RI(RE1
gm
)
For C2,
R2S
R3(R
C outCBR )R3 RC
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Estimate ofL for C-G Amplifier
Chap 17- 15
Apply the SCTC method :
For C1,
R1S
RI
(RS in
CGR )RI(RS1
gm
)
For C2,
R2S
R3(R
D outCGR )R3 RD
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Estimate ofL for C-C Amplifier
Chap 17- 16
Apply the SCTC method :
For C1,
R1S
RI
(RB in
CCR )
R1S
RI
RB
rp
o
1
R
ER
3
For C2,
R2S
R3(R
E outCCR )R3 RE
rp
Rth
o
1
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Estimate ofL for C-D Amplifier
Chap 17- 17
Apply the SCTC method :
For C1
,
R1S
RI
(RG in
CDR )RIRG
For C2,
R2S
R3R
S outCDR R3 RS
1
gm
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Frequency-dependent Hybrid-Pi Model
for BJT
Capacitance between base andcollector terminals is:
Cmo is total collector-base junctioncapacitance at zero bias, Fjc is thejunctions built-in potential.
Cm
C
mo
1(VCB
/jc
)
Capacitance between base andemitter terminals is:
tFis the forward transit-time of the
BJT. Cpappears in parallel with rp.
As frequency increases, for a given
input signal current, impedance of
Cp reduces vbe and thus reduces thecurrent in the controlled source at
transistor output.
Cp
gm
tF
Chap 17- 18
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Unity-gain Frequency of BJT
Ic(s) (g
msC
m)V
be(s)
Ic(s) (g
msC
m)I
b(s)
rp
s(Cp
Cm)rp
1
(s)I
c(s)
Ib(s)
o
1 sCmg
m
s(Cp
Cm)rp
1
The right-half plane transmission zero Z= +
gm/Cmoccurring at high frequency can be
neglected.
= 1/ rp(Cm+ Cp ) is the beta-cutoff frequency
where
andfT = T /2p is the unity gain-bandwidth
product. Above fT, the BJT has no current gain.
(s)
o
s(Cp
Cm
)rp
1
o
(s/
)1
(s)
o
s
T
s
T o
o(C
pC
m)rp
g
mCp
Cm
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Unity-gain Frequency of BJT (cont.)
Current gain is o =gmrpat low
frequencies and has single pole roll-
off at frequencies >f, crossing
through unity gain at T. Magnitude
of current gain is 3 dB below itslow-frequency value atf.
Cp
g
m
T
Cm
40I
C
T
Cm
Chap 17- 20
Cp is calculated using
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High-frequency Model of MOSFET
Id(s) (g
msC
GD)V
gs(s)
Id(s)I
b(s)
(gm
sCGD
)
s(CGS
CGD
)
(s)I
d(s)
Ig(s)
T
s1 s
T
1(CGS
/CGD
)
T
g
m
CGS
CGD
fT
mnCox
" W
LVGSVTN
(2/3)Cox
"WL
3
2
mn
VGSVTN L
2
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Limitations of High-frequency Models
Above 0.3fT, behavior of simple pi-models begins to deviatesignificantly from the actual device.
Also, Tdepends on operating current as shown and is not constant asassumed.
For given BJT, a collector currentICMexists that yields maximum
fTmax. For the FET in saturation, CGSand CGD are independent of Q-point
current, so
T
gm
ID
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Effect of Base Resistance
on Midband Amplifiers
Base current enters the BJT through
external base contact and traverses a
high resistance region before entering
active area. rx models voltage drop
between base contact and active areaof the BJT.
To account for base resistance rx is absorbed
into equivalent pi model and can be used to
transform expressions for C-E, C-C and C-B
amplifiers.
igm
vgm
rp
rprxvbe
gm
' vbe
gm' g
m
rp
rp
rx
o
rp
rx
rp' r
pr
x
o'
o
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Direct High-Frequency Analysis
C-E Amplifier
The small-signal model can be simplified by
using a Norton source transformation.
RL
R3
RC
100k 4.3k RB
R1R
2 30k10k
vth viR
BR
IR
B
Rth
RIR
BR
IR
B
is
v
th
Rth
rx
rpo
rp
(Rth
rx
)
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Direct High-Frequency Analysis
C-E Amplifier (Pole Determination)
From nodal equations for the circuit in
frequency domain,
High-frequency response is given by 2
poles, one finite zero and one zero at
infinity. Finite right-half plane zero, Z=+gm/Cm> T can easily be neglected.
For a polynomials2 +sA1 +A0 with roots
a and b, a A1 and b A0/A1.
V2(s)I
s(s)
(sCm-g
m)
s2
Cp CmCL
CpCL
s Cpg
LC
mg
Lg
mg
p
C
Lgpo
g
Lgpo
CT
Cp
Cm
1gm
RL
R
L
rpo
C
L
RL
rpo
P1
A
0
A1
1rpo
CT
P2
g
m
Cp
1(CL
/Cm)
C
L
Smallest root that gives first pole limits
frequency response and determines H.
Second pole is important in frequencycompensation as it can degrade phase
margin of feedback amplifiers.
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Direct High-Frequency Analysis
C-E Amplifier (Overall Transfer Function)
Vo(s)
Vth(s)
Rth
rx
(sCm-g
m)
gLgpo
1(s/P1
)
1(s/
P2)
Vo(s)
Vth(s)
Rth
rx
(gmRLrpo
)1(s/
Z)
gLgpo
1(s/P1
)
1(s/
P2)
Vo(s) -
Vth(s)
Rth
rx
gmRLrpo
1(s/P1
)
Avth
(s)Vo(s)
Vth(s)
Amid
1(s/P1
)
Amid
oR
L
Rth
rx
rp
P1
1rpo
CT
Chap 17- 26
Dominant pole model at high
frequencies for C-E amplifier is as
shown.
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Direct High-Frequency Analysis
C-E Amplifier (Example) Problem: Find midband gain, poles, zeros andfL.
Given data: Q-point = ( 1.60 mA, 3.00 V),fT= 500 MHz, o = 100, Cm= 0.5
pF, rx = 250,CL0
Analysis:gm= 40IC= 40(0.0016) = 64 mS, rp= o/gm =1.56 k
Cp g
m
2pfT
Cm19.9 pF
RL
R3
RC
100k 4.3k 4.12k
Rth
RB
RI
7.5k1k 882
rpo
rp
(Rth
rx
) 656
CT
Cp
Cm
1gmR
L
RL
rpo
C
L
RL
rpo
156 pF
fP1
1
2prpoCT
1.56 MHz
P2
1RLCm
gm
Cp
1 1gmrpo
1gmRL
fP2
P2
2p
603MHz fZ
gm
2pCm
20.4 GHz
Avth
oRL
Rth
rx
rp
153
Overall gain is reduced to -135 as vth = 0.882vs.
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Gain-Bandwidth Product Limitations of
C-E Amplifier
IfRth is reduced to zero in order to increase bandwidth, then rpo would
not be zero but would be limited to approximately rx.
IfRth = 0, rx
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High-Frequency Analysis: C-S Amplifier
Rth
RI
RG
RL
RD
R3
vth
vi
RG
RI
RG
CT CGSCGD 1gmRL
RLR
th
Z
g
m
CGD
P1
1R
thC
T
P2
1R
LC
GD
gmC
GS
1 1g
mR
th
1g
mR
L
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Miller Multiplication
Vo(s)AV
1(s) I
s(s)sC V
1(s)V
o(s)
Y(s)I
s(s)
V1(s)
sC(1A)
Total input capacitance = C(1+A) because
total voltage across Cis vc = vi(1+A) due to
inverting voltage gain of amplifier.
For the C-E amplifier,
CT
Cp
Cm(1A) C
pC
m(1g
mR
L)
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Miller Integrator
Assuming zero current in input
terminal of amplifier,
V1 VinR
sC(Vin
Vo)
Vo
AVin
Av(s)
Vo
V1
1
RC
A
1A
s 1RC(1A)
A
o
so
o
1
RC(1A)where
For frequencies >> o, assumingA >> 1,
which is the transfer function of an
integrator.
Av(s)
Ao
s
1
sRC
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Open-Circuit Time Constant Method to
Determine HAt high frequencies, impedances of
coupling and bypass capacitors are small
enough to be considered short circuits.
Open-circuit time constants associated
with impedances of device capacitances
are considered instead.
H
1
Rio
Ci
i 1
m
whereRio is resistance at terminals of
ith capacitorCi
with all other
capacitors open-circuited.
For a C-E amplifier, assuming CL = 0
Rpo
rpo
Rmo
vx
ix
rpo
(1
gmRL
R
L
rpo
)
H
1R
poCp
Rmo
Cm
1rpo
CT
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Gain-Bandwidth Trade-off Using
Emitter Resistor
Amid
oR
L
Rth
rx
rp
(o
1)RE
R
L
RE
rp
Rth
rxfor and
gain decreases as emitter resistanceincreases and bandwidth of stage will
correspondingly increase.
To find bandwidth using OCTC method:
gmR
E1
Req
vx
ix
Rth
rx
RE
1gm
RE
Rpo
rp
Req
R
thr
xR
E
1gmR
E
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Gain-Bandwidth Trade-off Using
Emitter Resistor (cont.)
Test source ix is first split into two
equivalent sources and then
superposition is used to find vx = (vb - vc).
Assuming that o >> 1 and
Rth rx rp(o 1)RE
Rmo
vxix
(Rth
rx
)1 gmRL1gmRE
RLRthrx
H
1
(Rthrx)Cp
1gmRE
1RE
Rthrx
C
m1gmRL
1gmRE
RL
Rthrx
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Dominant Pole for C-B Amplifier
Rth RE RI RL RC R3
Rpo
rp
vx
ix
rp
Rth
rx
1gmR
th
R
thr
x
1gmR
th
Using split-source transformation assuming
that o >>1 and rx
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Dominant Pole for C-G Amplifier
Rth
R4
RI R
LR
DR
3
RGSo
R
th
1gmRth 1
Gth gm
RGDo
RL
H
1
CGS
Gth
gm
CGDRL
1C
GD
RL
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Dominant Pole for C-C Amplifier
Rth RB RI RL RE R3
Rpo
rp
vx
ix
rp
Rth
rx
RL
1gmR
L
R
thr
xR
L
1gm
RL
Rmo
(Rth
rx
) inCCR (Rth rx ) rp(o 1)RL
Rmo (Rth rx )
H
1
(Rth
rxR
L)
Cp
1gmR
L
(Rth
rx
)Cm
A better estimate is obtained if we setRL = 0
in the expression forRpo.
H
1
(Rthrx)Cp
1gmRL
Cm
GBW (1)H
1
Cmrx
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Dominant Pole for C-D Amplifier
Substituting rpas infinite and rx as
zero in the expression for the emitter
follower,
Rth
RG
RI R
LR
SR
3
H
1R
th
1gmRL
CGS
CGDR
th
1
Rth
CGS
1gmRL
CGD
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Differential Amplifier
Frequency Response
CEE is total capacitance at emitter node
of the differential pair.
Differential mode half-circuit is similar
to a C-E stage. Bandwidth is determined
by the product. As emitter is avirtual ground, CEEhas no effect on
differential-mode signals.
For common-mode signals, at very low
frequencies,
Transmission zero due to CEEis
T
C
o
r
p
Acc
(0) R
C
2REE
1
s Z
1C
EER
EE
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Differential Amplifier
Frequency Response (cont.)
Common-mode half-circuit is similar to a
C-E stage with emitter resistor 2REE.OCTC forCpand Cm is similar to the C-E
stage. OCTC forCEE/2 is:
REE O
2REE
rprx
o1
1g
m
P
1
rx
Cp
12gm
REE
12R
EE
rx
C
m1
gm
RC
12gm
REE
R
C
rx
CEE
2gm
AsREEis usually designed to be large,
P
1CpCEE
2gm
Cm(RCrx)
1
Cm(RCrx)
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Frequency Response
Common-Collector/ Common-Base Cascade
REE is assumed to be large and
neglected.
Sum of the OCTC ofQ1 is:
outCC1R
rp1
rx1
o1
1 1
gm1
inCB2
R
rp2
rx2
o2 1
1
gm2
rx1
Cp1
1gm1
1
gm2
Cm1
rx1C
p1
2 Cm1
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Frequency Response
Common-Collector/ Common-Base Cascade (cont.)
Sum of the OCTC ofQ2 is:
Combining the OCTC forQ1 and Q2, and assuming that transistors
are matched,
rx2
Cp2
1gm2
1
gm1
Cm2
1gmR
C
1gm2
1
gm1
Cm2R
Cr
x2
Cp2
2C
m21
gmR
C
2R
C
rx2
H
1
rx Cp Cm 2gm
RC
2 RCrx
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Frequency Response
Cascode Amplifier
OCTC ofQ1 with load resistor 1/gm2 :
AsIC2 =IC1,gm2 =gm1, gain of first stage is
unity. Assuminggm2
rpo1
>>1,
OCTC ofQ1, a C-B stage forro1 >>RL and
mf>>1:
Assuming matched devices,
Rpo1C
p1R
mo1C
m1r
po1CT1
rpo1
Cp1
Cm1
1gm1
gm2
1
gm2rpo1
rpo1
CT1
rpo1
Cp1
2Cm1
Rpo2
Cp2
Rmo2
Cm2
Cp2
gm2
(rx2 RL)Cm2
H
1
rpo1
Cp
2Cm
r
xR
L
Cm
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Frequency Response
MOS Current Mirror
rpo
1g
m1
RL
ro2
Cp
CGS1
CGS2
Cm
CGD2
P1
1
2CGS
gm1
2CGD2ro2
1
2CGD2
ro2
For matched transistors,
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Frequency Response
Multistage Amplifier
Problem:Use open-circuit and short-circuit time constant methods to
estimate upper and lower cutoff frequencies and bandwidth.
Approach: Coupling and bypass capacitors determine low-frequency
response, device capacitances affect high-frequency response.
At high frequencies, ac model for multi-stage
amplifier is as shown.
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Frequency Response
Multistage Amplifier (Estimate ofL)
SCTC for each of the six independent coupling and bypass capacitors has
to be determined.
R1S
RI
(RGRin1
)10k1M
R1S
1.01M
R2SRS1 1
gm1
200 10.01S
66.7
R3S
RD1
RO1
R
B2Rin2
R3S
RD1
ro1
R
B2rp2
2.69 k
Rth2 RB2 RD1 ro1 571
R4S
RE2
Rth2
rp2
o2
119.4
R5S
RC2
RO2
R
B3R
in3
R5S
RC2
ro2
R
B3rp3
(o3
1)(RE3
RL
)
R5S18.4kR
th3R
B3R
C2r
o2 3.99 k
R6S
RL
RE3
Rth3
rp3
o3
1 311
L 1
RiS
Cii
1
n 3300 rad/s
fL
L
2p 530 Hz
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Frequency Response
Multistage Amplifier (Estimate ofH)
OCTC for each of the two capacitors associated with each transistor has to
be determined.
ForM1,
ForQ2,
RL1
RI12
rp2
478
RthCT1Rth CGS1CGD1 1gm1RL1
RL1
Rth
Rth2
RI12
ro1
570
rpo2
rp2
(Rth2
rx2
) 610
RL2 RI23 Rin3
RL2
RI23
rp3
(o3
1)(RE3
RL
)
RL2
3.54 k
rpo2CT2
rpo2
Cp2
Cm2
1gm2RL2
RL2
rpo2
rpo2CT2
1.74107s
ForQ3,
Rth3 RI23 ro2 3.99 k
Rp3OCp3R
m3OCm3(Rth3
rx3
)
1gm3REE
Cp3(Rth3
rx3
)Cm3
Rp3OCp3R
m3OCm31.51108 s
H
1
RioCi
i 1
m
3.3810
6
rad/s
fHH
2p538 kHz
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Single-pole Op Amp Compensation
Frequency compensation forces overall amplifier to have a single-pole
frequency response by connecting compensation capacitor around second gain
stage of the basic op amp.
Av(s)
Ao
B
sB
T
sB
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Three-stage MOS Op Amp Analysis
Input stage is modeled by its Norton
equivalent- current source Gmvdm and output
resistanceRo. Second stage has gain of
gm5ro5= mf5 and follower output stage is a
unity-gain buffer. Vo(s) = Vb(s) = -Av2Va(s)
Av(s)
Vo(s)
Vdm
(s)
-Av2
Va(s)
Vdm
(s)
Gm
RoA
v2
1sRoC
C(1A
v2)
Av(s)
T
sB
A
o
B
sB
B 1R
oC
C(1A
v2) T GmAv2
CC
(1Av2
)
For large Av2
, T
G
m
CC
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Transmission Zeros in FET Op Amps
Incorporating the zero determined bygm5 in the analysis,
This zero cant be neglected due to lowratio of transconductances ofM2 andM5.Zerocan be canceled by addition ofRZ=1/gm5.
Avth
(s) (gm5
ro5
)1(s/
Z)
1(s/P1
)
Z
gm5
CCCGD5
T
gm5
gm2
P1
1
RoCT
CT
CGS5
(CC
CGD5
)1mf5
r
o5
Ro
Z
1(1/g
m5)R
Z
C
C
Chap 17- 50
Note error
in Eq.
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Bipolar Amplifier Compensation
Bipolar op amp can be compensated in
the same manner as a MOS amplifier
Transmission zero occurs at too high a
frequency to affect the response due to
higher transconductance of BJT that
FET for given operating current.
Unity gain frequency is given by:
Z
g
m5
CC
T
IC5
IC2
T
g
m2
CC
40I
C2
CC
20I
C1
CC
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Slew rate of Op Amp
Slew-rate limiting is caused by limitedcurrent available to charge/discharge
internal capacitors. For very largeAv2,
amplifier behaves like an integrator:
For CMOS amplifier,
For bipolar amplifier,
IC1
CC
dvB
(t)
dt C
C
dvo(t)
dt
SRdv
o(t)
dtmax
I1
CC
T
Gm
/I1
SR TG
m/I
1
T I1K
n2
SR
T
Gm
/I1
T
20
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Tuned Amplifiers
Amplifiers with narrow bandwidth are often required in RF
applications to be able to select one signal from a large number of
signals.
Frequencies of interest > unity gain frequency of op amps, so active
RC filters cant be used.
These amplifiers have high Q (fHandfL close together relative to center
frequency)
These applications use resonant RLC circuits to form frequency
selective tuned amplifiers.
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Single-Tuned Amplifiers
RLC network selects thefrequency, parallel combination
ofRD,R3 and ro set the Q and
bandwidth.
Neglecting right-half plane
zero,
Av
(s)V
o(s)
Vi(s)
sCGD
gm
GPs(CCGD)(1/sL)G
Pg
oG
DG
3
Av(s) Amid
s
o
Q
s2 soQ
o
2
o
1L(CC
GD)
QoR
P(CC
GD)
RP
oL
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Single-Tuned Amplifiers (contd.)
At center frequency,s =jo,
Av =Amid.
AmidgmRPgm(ro RD R3)
BW
o
Q 1
RP
(CCGD
)
o
2L
RP
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Use of tapped Inductor- Auto
Transformer
CGD and ro can often be small enough todegrade characteristics of the tuned
amplifier. Inductor can be made to work
as an auto transformer to solve this
problem.
These results can be used to transform the
resonant circuit and higherQ can be
obtained and center frequency doesnt
shift significantly due to changes in CGD.
Similar solution can be used if tuned
circuit is placed at amplifier input instead
of output
Vo(s)I2(s) nV1(s)Is(s)/n n
2 V1(s)Is(s) Zs(s) n2Zp(s)
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Multiple Tuned Circuits
Tuned circuits can be placed at both inputand output to tailor frequency response.
Radio-frequency choke (an open circuit atthe operating frequency) is used for biasing.
Synchronous tuning uses two circuits tunedto same center frequency for high Q.
Stagger tuning uses two circuits tuned toslightly different center frequencies torealize broader band amplifiers.
Cascode stage is used to provide isolationbetween the two tuned circuits andeliminate feedback path between them dueto Miller multiplication.
BWn
BW1
21/ n 1
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Mixers: Conversion Gain
Amplifiers discussed so far have always been assumed to be linear and gainexpressions involve input and output signals at same frequency.
Mixers are nonlinear devices whose output signal frequency is different from
the input signal frequency.
A mixers conversion gain is the ratio of phasor representation of output signalto that of input signals, the fact that the two signals are at different frequenciesis ignored.
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Single-Balanced Mixer
Eliminates one of the two inputsignals from the output.
No signal energy appears at1 , but
2 appears in output spectrum, so
circuit is single-balanced.
Up-conversion uses component (2-1) and down-conversion uses
(2+1) component.
iEE
IEE
I1sin
1t
v2(t)
4
npnodd sinn2t
Vo(t) 4
npnodd
IEE
RC
sinn2t
I
1R
C
2cos(n
2
1)t
I1R
C
2cos(n
2
1)t
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Double-Balanced Mixer/ Modulator
The Gilbert Multiplier
Double-balanced mixers dontcontain spectral components ateither of the two input frequencies.
Modulator applications give doublesideband suppressed carrier outputsignal. Amplitude-modulated signal
can also be obtained if
iC1
IBB
V
m
2R1
sinm
t iC2
IBB
V
m
2R1
sinm
t
vo(t)V
mRCR
1
4npnodd
cos(nc
m
)t cos(nc
m
)t
vo(t)Vm
RC
R1
4
npnodd sinn
ctM
2cos(ncm)tM
2cos(ncm)t
v1 Vm(1Msin
mt)
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End of Chapter 17
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