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Chapter 10
Analog Systems
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
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Chapter Goals
Develop understanding of linear amplification concepts such as:
Voltage gain, current gain, and power gain,
Gain conversion to decibel representation,
Input and output resistances,
Transfer functions and Bode plots,
Cutoff frequencies and bandwidth,
Low-pass, high-pass, band-pass, and band-reject amplifiers,
Biasing for linear amplification,
Distortion in amplifiers, Two-port representations of amplifiers,
g-, h-, y-, and z-parameters,
Use of transfer function analysis in SPICE.
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Example of Analog Electronic System:
FM Stereo Receiver
Linear functions: Radio and audio frequency amplification, frequency
selection (tuning), impedance matching(75-W input, tailoring audio
frequency response, local oscillator
Nonlinear functions: DC power supply(rectification), frequency
conversion (mixing), detection/demodulation
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Amplification: Introduction
A complex periodic signal can be represented as the sum of many
individual sine waves. We consider only one component with
amplitude VS=1 mV and frequency wSwith 0 phase (signal is used
as reference):
Amplifier output is sinusoidal with same frequency but different
amplitude VO and phase :
tssVsv wsin
)(sin w tsoVov
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Amplification: Introduction (contd.)
Amplifier output power is:
Here,PO = 100 W andRL=8 W
Output power also requires output current which is:
Input current is given by
phase is zero because circuit is purely resistive.
LR
oVoP
12
2
V40810022 L
RoPoV
)(sin w tsoIoi
A58V40
W
LR
oVoI
A81082.1k50k5
V3-10 WW
in
RsRsV
sI
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Amplification: Gain
Voltage Gain:
Magnitude and phase of voltage gain are given by
and
For our example,
Current Gain:
Magnitude of current gain is given by
sVoV
sVoV
svov
vA 0
sVoV
vA vA
4104V310
V40
sVoV
vA
sIoI
sIoI
sioi
iA
0
81075.2A8-101.82
5A
sIoI
iA
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Amplification: Gain (contd.)
Power Gain:
For our example,
On decibel scale, i.e. in dB
iAvA
sIoI
sVoV
sIsV
oIoV
sPoP
PA
22
22
131010.181082.1310
540
P
A
PA
PdBA log10
vAvdBA log20
iA
idBA log20
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Amplifier Biasing for Linear Operation
iv
IV
Iv
VI= dc value ofvI,
vi = time-varying component
For linear amplification- vImust be biased in desired region of outputcharacteristic by VI.
ovOV
Ov
If slope of output characteristic is positive, input and output are in
phase (amplifier is non-inverting).
If slope of output characteristic is negative, input and output signals
are 1800 out of phase (amplifier is inverting).
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Amplifier Biasing for Linear Operation
(contd.)
IV
IvI
vov
vA
Voltage gain depends on bias
point.Eg: if amplifier is biased at VI=
0.5 V, voltage gain will be +40
for input signals satisfying
If input exceeds this
value, output is distorted due tochange in amplifier slope.
V1.0i
v
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Amplifier Biasing for Linear Operation
(contd.)
Output signals for 1 kHZ sinusoidal
input signal of amplitude 50 mV
biased at VI= 0.3 V and 0.5V:V)2000sin05.03.0( t
IAv
ForVI=0.3V:
V)2000sin14( tOA
v
ForVI=0.5V:
Gain is 20, output variesabout dc level of 4 V.
V)2000sin05.05.0( t
IB
v
V)2000sin210( tOB
v
Gain is 40, output varies
about dc level of 10 V.
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Distortion in Amplifiers
Different gains for positive and negative values of input cause
distortion in output.
Total Harmonic Distortion (THD) is a measure of signal distortion
that compares undesired harmonic content of a signal to the desired
component.
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Total Harmonic Distortion
...)3
sin3(3
)2
sin2(2
)1
(sin1
)( www toVtoVtoVoVtv
dc desired
output
2nd harmonic
distortion
3rd harmonic
distortion
1
2
%100 2
V
Vn
THD
Numerator= sum of rms amplitudes of distortion terms,
Denominator= desired component
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Two-port Models for Amplifiers
Simplifies amplifier-behavior modeling in complex systems.
Two-port models are linear network models, valid only under
small-signal conditions.
Represented byg-, h-, y- andz-parameters.
(v1, i1) and (v2, i2) represent signal components of voltages and
currents at the network ports.
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g-parameters
Using open-circuit (i=0) and short-
circuit (v=0) termination
conditions,
2
i
221
v
212
v2i
121v
111i
gg
gg
01v2
i2
v
22
02i1v
2v
21
01v2
i1i
12
02i1
v1i
11
g
g
g
g Open-circuit input
conductance
Reverse short-circuit
current gain
Forward open-circuit
voltage gain
Short-circuit output
resistance
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g-parameters:Example
Problem:Findg-parameters.
Approach: Apply specified boundary
conditions for eachg-parameter, use
circuit analysis.
Forg11 andg21: apply voltage v1 toinput port and open circuit output port.
Forg12 andg22: apply current i2 to
output port and short circuit input port.
998.0)k200)(51(
1102i1v
2v
21
S81079.9
)k200(514102
1
02i1
v1i
11
W
WW
gg
g
0196.0k20
391
01
v2i
2v
22
391
k20
51
k200
11
01
v2i1i
12
W
W
W
W
W
g
g
2i21091.3
1v998.0
2v
2i21096.1
1v81079.9
1i
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Hybrid orh-parameters
Using open-circuit (i=0) and short-
circuit (v=0) termination
conditions,
2
v
221
i
212
i2
v121
i111
v
hh
hh
01i2
v2i
22
02
v1i
2i
21
01i2
v1v
12
02
v1i1v
11
h
h
h
h Short-circuit input
resistance
Reverse open-circuit
voltage gain
Forward short-circuit
current gain
Open-circuit output
conductance
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h-parameters:Example
Problem:Find h-parameters for the
same network (used ing-parameters
example).
Approach: Apply specified boundary
conditions for each h-parameter, usecircuit analysis.
Forh11 and h21: apply current i1 to input
port and short circuit output port.
Forh12 and h22: apply voltage v2 to
output port and open circuit input port.
51
02
v1i2i
21
4102
02
v1i1
v
11
W
h
h
S6105k200
1
01i2
v2i
22
10
1i2
v1
v
12
W
h
h
2v6105
1i51
2i
2v
1i4102
1v
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Admittance ory-parameters
Using open-circuit (i=0) and short-
circuit (v=0) termination
conditions,
2
v
221
v
212
i2
v121
v111
i
yy
yy
01v2
v2i
22
02
v1v
2i
21
01v2
v1i
12
0
2
v1v1i
11
y
y
y
y Short-circuit input
conductance
Reverse short-circuit
transconductance
Forward short-circuit
transconductance
Short-circuit output
conductance
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y-parameters:Example
Problem:Findy-parameters for the
same network (used ing-parameters
example).
Approach: Apply specified boundary
conditions for eachy-parameter, use
circuit analysis.
Fory11 andy21: apply voltage v1 to input
port and short circuit output port.
Fory12 andy22: apply voltage v2 to
output port and short circuit input port.
S31055.2k20
51
02
v1v2i
21
S5105k20
1
02
v1v1i
11
W
W
y
y
S31056.2391
1
01i2
v1i
22
S5105k20
1
01
v2v1i
12
W
W
y
y
2v31056.2
1v31055.2
2i
2v5105
1v5105
1i
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Impedance orz-parameters
Using open-circuit (i=0) and short-
circuit (v=0) termination
conditions,
2
i
221
i
212
v2i
121i
111v
zz
zz
01i2
i2
v
22
02i1i
2v
21
01i2
i1v
12
0
2
i1i1v
11
z
z
z
z Open-circuit input
resistance
Reverse open-circuit
transresistance
Forward open-circuit
transresistance
Open-circuit output
resistance
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z-parameters:Example
Problem:Findz-parameters for the
same network (used ing-parameters
example).
Approach: Apply specified
boundary conditions for eachz-
parameter, use circuit analysis.
Forz11 andz21: apply current i1 to
input port and open circuit output
port. Forz12 andz22: apply current i2
to output port and open circuit inputport.
W
WWW
M2.10
0
2
i1i2
v
21
M2.10)k200(51k20
02i1
i1v
11
z
z
W
W
k200
01i2
i2
v
22
k200
01i2
i1
v
12
z
z
2i51000.2
1i71002.1
2v
2i51000.2
1i71002.1
1v
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Mismatched Source and Load
Resistances: Voltage Amplifier
LR
outR
LR
inRsR
inR
A
sVoV
vA
inRsR
inR
LR
outR
LR
A
sv1v
1vov
IfRin >>Rs andRout
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Mismatched Source and Load
Resistances: Current Amplifier
h-parameter representation (h12=0) with Norton equivalent of input source:
LR
outR
outR
inRsRsR
sIoI
iA
inRsRsR
LR
outR
outR
si1i
1ioi
IfRs >>Rin andRout>> RL,
i
A
In an ideal current amplifier,andRin=0
outR
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Amplifier Transfer Functions
ssV
soVsvA
Av(s)=Frequency-dependent voltage gain
Vo(s) and Vs(s) = Laplace Transforms of input and output voltages of
amplifier, w js
3...
21
...21
pspsps
mzszszsKsvA
(-z1, -z2,-zm)=zeros (frequencies for which transfer function is zero)
(-p1, -p2,-pm)=poles (frequencies for which transfer function is infinite)
www jvAjvAjvA
Bode plots display magnitude of the transfer function in dB and the
phase in degrees (or radians) on a logarithmic frequency scale..
(In factorized form)
(In polar form)
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Low-pass Amplifier: Description
Amplifies signals over a range of frequencies including dc.
Most operational amplifiers are designed as low pass amplifiers.
Simplest (single-pole) low-pass amplifier is described by
Ao = low-frequency gain or mid-band gain
wH= upper cutoff frequency or upper half-power point of
amplifier.
H
soA
Hs
HoA
svA
ww
w
1
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Low-pass Amplifier: Magnitude Response
ForwwH:
ForwwH:
Gain is unity (0 dB) atwAowH , called gain-bandwidth product
Bandwidth (frequency range with constant amplification )= wH (rad/s)
22log20log20dB
22
HHoAjvA
H
HoA
Hj
HoA
jvA
wwww
ww
w
ww
w
w
dB)log20( oAoAjvA w
dBlog20log20
HoA
HoA
jvAw
w
w
w
w
dB3)log20(2
oA
oAjvA w
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Low-pass Amplifier: Phase Response
IfAo
positive: phase angle = 00
IfAo negative: phase angle = 1800
At wC: phase =450
One decade below wC: phase =5.70
One decade above wC: phase =84.30
Two decades below wC: phase =00
Two decades above wC: phase =900
H
oA
H
j
oAjvA ww
w
ww1tan
1
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RC Low-pass Filter
Problem: Find voltage transfer
function
Approach: Impedance of the where
capacitor is 1/sC, use voltage
division
H
sRR
R
sVoV
sCR
sCRR
sCR
sCR
sVoV
w1
1
21
2
/12
/2
1
/12
/2
CRRH
21
1w
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High-pass Amplifier: Description
True high-pass characteristic impossible to obtain as it requires
infinite bandwidth.
Combines a single pole with a zero at origin.
Simplest high-pass amplifier is described by
wH= lower cutoff frequency or lower half-power point of amplifier.
Ls
soAsvAw
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High-pass Amplifier: Magnitude and
Phase Response
Forw>>wL :
Forw
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RC High-pass Filter
Problem: Find voltage transfer
function
Approach: Impedance of the where
capacitor is 1/sC, use voltage
division
Ls
s
RR
R
sVoV
RsC
R
R
sVoV
w21
2
2
1
1
2
CRRL
21
1w
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Band-pass Amplifier: Description
Band-pass characteristic obtained by combining highpass and
low-pass characteristics.
Transfer function of a band-pass amplifier is given by
Ac-coupled amplifier has a band-pass characteristic:
Capacitors added to circuit cause low frequency roll-off Inherent frequency limitations of solid-state devices cause
high-frequency roll-off.
1
1)())((
H
sLs
soA
Hs
Ls
HsoAsvA
wwww
w
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Band-pass Amplifier: Magnitude and
Phase Response
The frequency response shows a wide band of operation.
Mid-band range of frequencies given by , whereHLwww oAjvA )( w
)22)(22())((
HL
HoA
Hj
Lj
HjoAjvA
wwww
ww
wwww
ww
w
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Band-pass Amplifier: Magnitude and
Phase Response (contd.)
At both wHand wL, assumingwL
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Narrow-band or High-Q Band-pass
Amplifiers
Gain maximum at center frequency woand
decreases rapidly by 3 dB at wHand wL.
Bandwidth defined aswH - wL, is a small
fraction ofwowith width determined by:
For high Q, poles will be complex and
Phase response is given by:
BWo
f
Lf
Hf
of
LHoQ ww
w
22
oQo
ss
Qos
oAsvA
w
w
w
2211tan090
ww
www
o
oQo
AjvA
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Band-Rejection Amplifier or Notch
Filter
Gain maximum at frequencies far from wo
and exhibits a sharp null at wo.
To achieve sharp null, transfer function has
a pair of zeros onjwaxis at notch frequency
wo , and poles are complex.
Phase response is given by:
22
22
oQ
oss
osoAsvA
www
2211tan22
ww
wwwww
o
oQoo
AjvA
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All-pass Function
Uniform magnitude response at all frequencies.
Can be used to tailor phase characteristics of a signal
Transfer function is given by:
For positive Ao,
osos
oAsvA ww
o
jvA
oAjvA
www
w
1tan2
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Complex Transfer Functions
1
5
1
431
2mid
5431
2
wwww
w
wwww
w
ssss
ssA
ssss
sKssvA
Amplifier has 2 frequency ranges withconstant gain. Midband region is
always defined as region of highest
gain and cutoff frequencies are
defined in terms of midband gain.
2
midA
HjvAL
jvA ww
ww
2 3434BW
ff
Since wH= w4and wL = w3,
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J /Bl l k Mi l t i Ci it D i
Bandwidth Shrinkage
If critical frequencies arent widely spaced, the poles and zeros
interact and cutoff frequency determination becomes
complicated.
Example : for which ,Av(0) =Ao
Upper cutoff frequency is defined by or
2)1
(
2
1w
w
s
oA
svA
2oA
HjvA w 222
1
1
Ao
H
Ao
w
w
Solving forwHyields wH=0.644w1.The cutoff frequency of
two-pole function is only 64% that of a single-pole function.
This is known as bandwidth shrinkage.
Ch 10 39