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2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter 14Feedback, Stability and Oscillators
Microelectronic Circuit Design
Richard C. Jaeger
Travis N. Blalock
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Chapter Goals• Review concepts of negative and positive feedback.• Develop 2-port approach to analysis of negative feedback amplifiers.• Understand topologies and characteristics of series-shunt, shunt-shunt,
shunt-series and series-series feedback configurations.• Discuss common errors that occur in applying 2-port feedback theory.• Discuss effects of feedback on frequency response and feedback amplifier
stability and interpret stability in in terms of Nyquist and Bode plots.• Use SPICE ac and transfer function analyses on feedback amplifiers.• Determine loop-gain of closed-loop amplifiers using SPICE simulation or
measurement.• Discuss Barkhausen criteria for oscillation and amplitude stabilization • Understand basic RC, LC and crystal oscillator circuits and present LCR
model of quartz crystal.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Feedback Effects
• Gain Stability: Feedback reduces sensitivity of gain to variations in values of transistor parameters and circuit elements.
• Input and Output Impedances: Feedback can increase or decrease input and output resistances of an amplifier.
• Bandwidth: Bandwidth of amplifier can be extended using feedback.
• Nonlinear Distortion: Feedback reduces effects of nonlinear distortion.eg: removal of dead zone in class-B amplifiers
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Classic Feedback Systems
• A(s) = transfer function of open-loop amplifier or open-loop gain.
• (s) = transfer function of feedback network.
)(f
V)(i
V)(d
V sss
)()(d
V)(oV sAss
)()(oV)(f
V sss
sT
sA
ssA
sA
s
ssvA
11)(i
V
)(oV
T(s) = loop gain
For negative feedback: T(s) > 0For positive feedback: T(s) < 0
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Voltage Gain Calculation)
and ,
2vA
221iA
21iA2
2vA
121iA
11vA1
hh
hh
2vF
221iF
21iF2
2vF
121iF
11vF1
hh
hh
hijhijhijFAT hh F
21A21 hh A
12F12
2v)T
22(1iA
2102
vT221
iA212
i2
vF121
)iT11(
iv
2vF
121iT
111v
LGhhhh
hhIRhh
AA
LGhhI
Rhh
hvA
1)T22
()T11
(F12
A21
A21
iv2
v
hF12
)T22
()T11
(
A21
LGhhI
R
hA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , RI and RL.
• Required h-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown
hF11 hF
22
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Input and Output Resistances)
1i
)T22
(
A21
-F121
)iT11(
iv
LGh
hhhIR
)1(Ainin
))(1T11(
1ii
v
in
ARR
AhIRR
Series feedback at a port increases input resistance at that port.
For output resistance:
1i
1v IR xv
2v
2vxi2
i LG
xvF121
)iT11(0
xv)T22(
1iA
21xi
hhIRLGhh
A
LGh
R
1
T22
1
xixv
out A
RR
1
Aout
out
Shunt feedback at a port reduces resistance at that port.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Example)
• Problem: Find A, , closed-loop gain, input and output resistances.
• Given data: R1=10 k, R2=91 kRid=25 kRo=1 k
• Analysis:0990.0
21
1
01i2
v1
vF12
kΩ1011
21
1
01i2
v2
iF22
kΩ01.921
02
v1i1
vF11
RR
Rh
RRh
RRh
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Voltage Amplifiers: Series-Shunt Feedback (Example contd.)
4730kΩ00.1kΩ96.1
kΩ96.1)410(kΩ01.9kΩ25kΩ1
kΩ25
ivov
A
1.10)0990.0(47301
47301
A
AvA
16.4M)1(Ainin
ARR
41.11
Aout
out A
RR
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresistance Amplifiers: Shunt-Shunt Feedback (Voltage Gain Calculation)
and ,
2vA
221vA
21iA2
2vA
121vA
11iA1yy
yy
2vF
221vF
21iF2
2vF
121vF
11iF1yy
yy
yijyijyijFAT yy F
21A21 yy A
12F12
2v)T
22(1
vA210
2vT
221vA
212i
2vF
121)vT
11(i
i2
vF121
vT111
i
LGyyyy
yyIGyy
AA
LGyy
IGyy
ytrA
1)T22
()T11
(A12
F21
F21
ii2
v
yF12
)T22
()T11
(
A21
LGyy
IG
yA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresitance Amplifiers: Shunt-Shunt Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , RI and RL.
• Required y-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown.
yF11
yF22
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresistance Amplifiers: Shunt-Shunt Feedback (Input and Output Resistances)
1v
)T22
(
A21
-F121
)vT11(
ii
LGy
yyy
IG
)1(
Ain
)(1
T11
1
ii1
v
in A
R
A
hIG
R
Shunt feedback at a port reduces resistance at that port.
For output resistance:
1v
1i IG
2vxi2
i LG
xvF121
)vT11(0
xv)T22(
1vA
21xi
yyIG
LGyy
A
LGy
R
1
T22
1
xixv
out A
RR
1
Aout
out
Resistance at output port is reduced due to shunt feedback.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresistance Amplifiers: Shunt-Shunt Feedback (Example)
• Problem: Find A, , closed-loop gain, input and output resistances.
• Given data: VA= 50 V, F= 150• Analysis: From dc equivalent
circuit,
S5-101
01
v2v1i
F12
S5-101
02
i2v2
iF22
S5-101
02
v1v1i
F11
FR
y
FR
y
FR
y
V35.1)(
mA970.0
CRBI
CI
CCV
CEV
F
FR
CR
CR
BEV
CCV
CI
39.1mSmA)977.0(40 mg
k6.52mA977.0
V35.1V50or
k84.31
mgr
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transresistance Amplifiers: Shunt-Shunt Feedback (Example contd.)
kΩ114kΩ6.52kΩ41.1)150(kΩ84.3kΩ76.4
kΩ76.4
iiov
A
kΩ3.53)mS01.0(kΩ1141
kΩ1141
A
AtrA
995)1()1(Ainin
ArF
RI
RARR
64011
Aout
out A
orCR
FR
LR
A
RR
)kΩ41.1(
kΩ76.4kΩ76.4
orbioov
riibi
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Current Amplifiers: Shunt-Series Feedback (Voltage Gain Calculation)
and ,
vF2vA
22v
iF1iA11i
gijgijgijFAT gg F
21A21 gg A
12F12
2i)T
22(1
vA210
2iT
221vA
212v
2iF
121)vT
11(i
i2
iF121
vT111
i
LRgggg
ggIGgg
AA
LRgg
IGgg
giA
1)T22
()T11
(F12
A21
A21
ii2
i
gF12
)T22
()T11
(
A21
LRgg
IG
gA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
CurrentAmplifiers: Shunt-Series Feedback (Two-Port Representation)
• Gain of amplifier should include effects of , , RI and RL.
• Required g-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown
gF11
gF22
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Current Amplifiers: Shunt-Series Feedback (Input and Output Resistances)
1v
)T22
(
A21
-F121
)vT11(
ii
LRg
ggg
IG
Shunt feedback at a port decreases resistance at that port.For output resistance:
2iF
121)vT
11(02
i)T22(
1vA
21xv
ggIG
LRgg
AR
AL
RgR
1Aout
1T22
2ixv
out
Series feedback at output port increases resistance at that port.
)1(
Ain
)(1
T11
1
ii1
v
in A
R
A
gI
GR
1v
1i IG
2ixv
2v LR
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transconductance Amplifiers: Series-Series Feedback (Voltage Gain Calculation)
and ,
vF2vA
22v
vF1vA
11v
zijzijzijFAT zz F
21A21 zz A
12F12
2i)T
22(1iT
2102
iT221
iT212
v2
iF121
)iT11(
iv
2iT
121iT
111v
LRzzzz
zzSRzz
AA
LRzz
IRzz
ztcA
1)T22
()T11
(F12
A21
A21
iv2
i
zF12
)T22
()T11
(
A21
LRzz
IR
zA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Transconductance Amplifiers: Series-Series Feedback (Input and Output Resistances)
• Gain of amplifier should include effects of , , RI and RL.
• Required g-parameters are found from their individual definitions.
• Two-port representation of the amplifier is as shown
zF11 zF
22
AR
AL
RzR
1Aout
1T22
2ixv
out
AR
AI
RzR
1Ain
1T11
1ii
v
in
Series feedback at input and output port increases resistance at both ports.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Erroneous Application of 2-port Feedback Theory
• Problem: Find A, , closed-loop gain, input and output resistances.
• Given data: VREF = 5 V, o= 100, VA = 50 V, Ao = 10,000, Rid =25 k, Ro =0
• Analysis: The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
This case seems to use series-series feedback.
ie is sampled by feedback network instead of io. This assumption is made since o is approximately 1.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Erroneous Application of 2-port Feedback Theory (contd.)
Z-parameters are found as shown. From A-circuit, IE=1 mA
k501mA
V50or k5.2
mA1V)025.0(1000
r
Roro
oAR
idR
idR
ivoi )1(
mS200.0)S(5k64.11
S64.11
A
A
ivoi
tcA
MΩ2461)(1Ainin
AR
idRARR GΩ7.271A
inin
ARR
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Erroneous Application of 2-port Feedback Theory (contd.)
SPICE analyses confirm results for Atc and Rin, but results for Rout are in error. For Atc and Rin, amplifier can be properly modeled as a series-shunt feedback amplifier, as collector of Q1 can be directly connected to ground for calculations and a valid 2-port representation exists as shown.
Results for Rout are in error because output of op amp is referenced to ground, base current of BJT is lost from output port and feedback loop and Rout is limited to
3 and 4 are not valid terminals as current entering 3 is not same as that exiting 4. Amplifier can’t be reduced to a 2-port.
oroR out
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Analysis of Shunt-Series Feedback Pair• Problem: Find A, , closed-loop gain, input and output resistances.
• Given data: o= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point for Q2:(1.6 mA, 7.5 V)
• Analysis: The circuit is redrawn to identify amplifier and feedback networks and appropriate 2-port parameters of feedback network are found.
Shunt-shunt transresistance configuration is used.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Analysis of Shunt-Series Feedback Pair (contd.)
Small signal parameters are found from given Q-points.
For Q1, r=3.79 k, r = 155 k.
For Q2, r =1.56 k, r = 64.8 k.
kΩ88.81
kΩ10i
i51094.4
)1
(1
1i
ith
v
or
thR
CR
or
orB
RB
R
)kΩ901.0)(12
(2
kΩ88.8
)kΩ901.0)(12
(
thv
2v
oro
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Analysis of Shunt-Series Feedback Pair (contd.)
51043.4
ii2
vA S
91001F
12 y
89101 A
AtrA
86.1)1(
Aout
out
5.42)1(
1)1(
Ain
in
A
RR
A
rB
R
A
RR
Closed-loop current gain is given by:
79.9901
901
trAo
ii
2v
o
ii
eio
iioi
iA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Direct Calculation of Loop Gain
• Original input source is set to zero.
• Test source is inserted at the point where feedback loop is broken.
xvrv
xv)xv0(ovrv
AT
AA
Example:
is added for proper termination of feedback loop.
123RRR
312
1
xvrv
3xv
12
1
12
1ovrv
Rid
Rid
R
RR
RAT
Rid
Rid
RA
RR
R
RR
R
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Calculation of Loop Gain using Successive Voltage and Current Injection
Voltage injection: Voltage source vX is inserted at arbitrary point P in circuit.
where
xv1
)1(xv
1v
2v
xv11
v
AA
A
BR
AR
AR
AR
BR
AR
BR
TAvT
11
1v2
v
Current injection: Current source iX is inserted again at P.
AR
xv1i
BR
A
BR
A
1xvxvxv
2i
iTvT
iTvT
T
AR
BR
TA
RBR
AR
BRA
iT
2
1
1/1
1
1i2
i
As A = T
iTvT
AR
BR
1
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Simplifications to Successive Voltage and Current Injection Method
• Technique is valid even if source resistances with vX and iX are included in analysis.• If at P, RB is zero or RA is infinite, T can be found by only one measurement and T =
Tv . In ideal op amp, such point exists at op amp input.• If at P, RB is zero, T = Tv . In ideal op amp, such point exists at op amp output.• If RA = 0 or RB is infinite, T = TI .• In practice, if RB >> RA or RA >> RB, the simplified expressions can be used.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem• First we select ports where resistance is to be calculated.• Next we select one controlled source in the amplifier’s equivalent
circuit and use it to disable the feedback loop and also as reference to find TSC and TOC.
RCL = resistance of closed-loop amplifier looking into one of its ports (any terminal pair)
RD = resistance looking into same pair of terminals with feedback loop disabled.
TSC = Loop gain with a short-circuit applied to selected port
TOC = Loop gain with same port open-circuited.
OCT
SCT
DRCL
R
1
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 1)
For output resistance:MΩ18.3
)(
)(1
idRRrid
RRoorDR
9940))(1(
))(1(
11
oridRRor
oridRRo
oAvoA
SCT
6350)(
)(
1
idRRr
idRR
oAvoAOC
T
Problem: Find input and output resistances.Given data:VREF =5 V, R =5 k o=100, VA=50 V, Ao=10,000, Rid=25 k, Ro=0Assumptions: Q-point is known, gm = 0.04 S, r =25 k, ro =25 k.
MΩ06.563501/99401MΩ18.3out
R
For input resistance: Ωk25)1( mgR
idRDR
MΩ2491/99401kΩ25in
R
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 2)Problem: Find input and output resistances.Given data:o= 100, VA = 100 V, Q-point for Q1:(0.66 mA, 2.3 V), Q-point for Q2:(1.6 mA, 7.5 V). For Q1, r=3.79 k, r = 155 k, For Q2, r =1.56 k, r = 64.8 k.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 2 contd.)
kΩ321)
22/1()
1(
2
)22
/1(1
2
y for
CRr
y fo
orDR
7.4812kΩ901.0)1(kΩ56.1
kΩ901.0)1(.
kΩ901.0)1(kΩ56.1kΩ10kΩ79)1
)(kΩ79.3kΩ10kΩ1.9(
y fo
o
omg
SCT
MΩ99.233.41/7.481kΩ321out
R
For input resistance:kΩ11.2
1rkΩ1.9kΩ10 DR
Ω5.42)7.481/(01kΩ11.2in
R
For output resistance:
33.49.1kΩ
10.901kΩ1.56kΩ
0.901kΩ93kΩ30.66mA)(1.kΩΩ)(411.2(
OCT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Blackman’s Theorem (Example 3)
• Problem: Find expression for output resistance.• Analysis: Feedback loop is
disabled by setting reference source i
to zero.
Next, i is set to 1
3)1
/1(32
)1
/1(31
3 or
mgr
or
mg
oo
rDR
Assuming gm1= gm2 = gm3 and f >> o >>1.
21111
out
1
)1(12
1
)1()1
3(
orooorR
oSCT
o
f
oo
bi
oeii
11
1 f
oiei
OCT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Feedback to Control Frequency Response
ssA
sAsvA
1
))((
Hs
Ls
sHoA
sA
where
HLsoA
HLs
sHoA
svA
)1(2
LoAH )1(Assuming ,
)1(BW
1
oAHF
oALF
L
)1( oAHFH
Upper and lower cutoff frequencies as well as bandwidth of amplifier are improved, gain is stabilized at
1
1mid
oAoA
A
HoAFA BWmid
GBW
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Nyquist Plot to Determine Stability • If gain of amplifier is greater than or equal to
1 at the frequency where feedback is positive, instability can arise.
• Poles are at frequencies where T(s)=-1.• In Nyquist plots, each value of s in s-plane
has corresponding value of T(s).• Values of s on j axis are plotted.• If -1 point is enclosed by boundary, there is
some value of s for which T(s)=-1, pole exists in RHP and amplifier is unstable.
• If -1 point lies in outside interior of Nyquist plot, all poles of closed-loop amplifier are in LHP and amplifier is stable.
sT
sAsvA
1
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
First-Order Systems
For a simple low-pass amplifier,
It can also represent a single-pole op amp with resistive feedback
os
oT
osooA
sT
1)(
joT
jT
At dc, T(0) = To, but for >>1,
As increases, magnitude monotonically approaches zero and phase asymptotically approaches -900.
As changes, value of T(0) = To is scaled but as T(0) changes, radius of circle changes, but it can never enclose the -1 point, so amplifier is stable regardless of value of To.
oT
jjT )(
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Second-Order Systems
2
s1
1
s1
oT
2
s1
1
s1
oAsT
In given example,
T(0) =14, but, for high frequencies
As increases, magnitude monotonically decreases from 14 towards zero and phase asymptotically approaches -1800 The transfer function can never enclose the -1 point but can come arbitrarily close to it.
21)(
j
14jT
222)()(
1414jjT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Phase Margin
Phase Margin is the maximum increase in phase shift that can be tolerated before system becomes unstable.
)1
(180)180()1
( jTjTm
Where
First we determine frequency for which magnitude of loop gain is unity, corresponding to intersection of Nyquist plot with unit circle shown and then determine phase shift at this frequency. Difference between this angle and -1800 is phase margin.
Small phase margin causes excessive peaking in closed-loop frequency response and ringing in step response.
1)1
( jT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Third-Order SystemsIn given example,
T(0) = 7, but, for high frequencies
As increases, polar plot asymptotically approaches zero along positive imaginary axis and plot can enclose the -1 point under any circumstances and system is unstable.
3
s1
2
s1
1
s1
oTsT
2323)(
sss
14sT
333)()(
14j14jjT
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Gain Margin
Gain Margin is the reciprocal of magnitude of T(j) evaluated at frequency for which phase shift is 1800.
)180
(
1GMjT
Where
If magnitude of T(j) is increased by a factor equal to or exceeding gain margin, then closed-loop system becomes unstable, because Nyquist plot then encloses -1 point.
180)180
( jT
GM)log(20dB
GM
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Bode Plots
710610510
19102
sssA
At 1.2e+6 rad/s, magnitude of loop gain is unity and corresponding phase shift is 1450, and phase margin is given by 1800 - 1450 = 350.Amplifier can tolerate additional phase shift of 350 before it becomes unstable.
At 3.2e+6 rad/s, phase shift is exactly 1800 and corresponding magnitude of loop gain is -17 dB, and phase margin is given by 17 dB.Gain of amplifier must increase by 17 dB before amplifier becomes unstable.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Use of Bode plot to Determine Stability
1log20log20log20 AA
Frequency at which curves corresponding to magnitudes of open-loop gain and reciprocal of feedback factor intersect is the point at which loop gain is unity, phase margin is found from phase plot.
Assuming feedback is independent of frequency,For 1/ =80 dB, m=850, amplifier is stable.For 1/ =50 dB, m=150, amplifier is stable, but with significant overshoot and ringing in its step response.For 1/ =0 dB, m= -450, amplifier is unstable
8106103510
24102
sssA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Operational Amplifier Compensation Example• Problem: Find value of compensation capacitor for m=700.
• Given data: RC1=3.3 k , RC2 =12 k ,SPICE parameters-BF=100, VAF=75 V, IS=0.1 fA, RB=250 , TF=0.75 ns, CJC= 2 pF.
Assumptions: Dominant pole is set by CC and pnp C-E stage. RZ is included to remove zero associated with CC, pnp and npn transistors are identical, quiescent value of Vo =0, VJC=0.75 V, MJC=0.33. Q4 and Q5 are in parallel, small signal resistances of diode-connected Q7 and Q8 can be neglected.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Operational Amplifier Compensation Example (contd.)
• Analysis: IC1 = IC2 =250 A. For Vo =0,voltage across RC2 =12-0.75=11.3 V and IC3 =11.3V/12 k=938 A. Q4 and Q5 mirror currents in Q7 and Q8 , so, IC4 = IC5 =938 A. For Vo =0, VCE4 =12 V, VCE5=12 V, VCE3 =11.3 V. For VI =0, VCE2 =12.8 V, VCE1 =12-3300(0.25 mA)+0.75= 11.9 V
Small signal parameters are found using their respective formulae.
93.7)kΩ07.3kΩ3.3kΩ696(201.0
)311
2(2
11
rCR
orm
g
vA
338
)500)117(2kΩ09.3kΩ12kΩ92(0375.0
)14
(24
23(
22
LR
o
r
CR
or
mg
vA
974.0
500)117(2
3090500)117(
)14
(24
)1(
3
LR
o
rL
Rov
A
2610321
vA
vA
vAvA
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Operational Amplifier Compensation Example (contd.)
Input stage pole:
Emitter Follower pole: Q4 and Q5 are in parallel, composite parameters are-gm =0.02 S, rx =125 , C =56.2 pF, C =1.60 pF, Rth =1/ gm3 =267 .
MHz2.59
22
121
xrC
RC
RmgCCr
Hf
MHz5.82
1)(
121
xrthRC
LRmg
CL
RxrthR
Hf
At fT ,dominant pole due to CC contributes phase shift of 900. For m=700, other 2 poles can contribute more phase margin of 200.
pF652
12
113
MHz2.12
MHz5.821tan
MHz2.591tan20
Tfm
g
Tm
GC
CC
Tf
Tf
Tf
RZ =1/ gm3 =27.5 is included to remove zero associated with CC.
4.67kHz2610
MHz2.12 oAT
f
Bf
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Barkhausen’s Criteria for Oscillation.
• For sinusoidal oscillator, poles of closed-loop amplifier should be at frequency 0 on j axis.
• Use positive feedback through frequency-selective feedback network to ensure sustained oscillation at 0 .
sT
sA
ssA
sAsvA
11
• For sinusoidal oscillations,
• Barkhausen’s criteria state-
• Phase shift around feedback loop should be zero degrees and magnitude of loop gain must be unity.
• Loop gain greater than unity causes distorted oscillations.
101 ojTojT
1
0
ojTojT
Or even multiples of 3600
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Oscillators with Frequency-Selective RC Networks: Wien-Bridge Oscillator
)(2
)(1
)(2)(
1V)(oV
sZsZ
sZss
13)2221()(I
V
)(oV
sRCCR
sRCGs
sT(s)
Phase shift will be zero if = 0,
At 0 =1/RC
This oscillator is used for frequencies upto few MHz, limited primarily by characteristics of amplifier.
)2221( CR
3G)oT(j
3G)oT(j 0)oT(j
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Oscillators with Frequency-Selective RC Networks: Phase-Shift Oscillator
)(2
V
)(1
V
G)(2sC
G)(2sC
0
)('oVs
ssC
sC
ssC
1422231
233
)('oV)(oV
)(
1)(2
V)(oV
sRCCRs
RRCs
ss
sT
sCRss
Phase shift will be zero if = 0,
At 0
)22231( CRo
RCo 31
R
R
4
RRCo)oT(j 11211
22
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization
• Loop gain of oscillator changes due to power supply voltage, component value or temperature changes.
• If loop gain is too small, desired oscillation decays and if it is too large, waveform is distorted.
• Amplitude stabilization or gain control is used to automatically control loop gain and place poles exactly on j axis.
• At power on, loop gain is larger than that required for oscillation.As oscillation builds up, gain is reduced o minimum required to sustain oscillations.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in RC Oscillators: Method 1
R1 is replaced by a lamp. Small-signal resistance of lamp depends on temperature of bulb filament.If amplitude is large, current is large, resistance of lamp increases, gain is reduced. If amplitude is small, lamp cools, resistance decreases, loop gain increases. Thermal time constant of bulb averages signal current and amplitude is stabilized.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in RC Oscillators: Method 2
For positive signal at vo, D1 turns on as voltage across R3 exceeds diode turn-on voltage. R4 is in parallel with R3, loop gain is reduced. D2 functions similarly at negative signal peak.
2
1
32
R
RR2
1
432
R
RRR
Thus, when diodes are off, op amp gain is slightly >3 ensuring oscillation, but, when one diode is on, gain is reduced to slightly<3.
Same method can also be used in phase shift oscillators.
14
3
41
1
22
3ov
41
vov
31
vovi
RR
R
R
R
RDV
RDV
R
1
213ov
1v
R
R
2
1
2 R
R
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
LC Oscillators: Colpitts Oscillator
)/(1 orSRG GS
CCC 23
L
CC
sL
Gmg
GCGCGD
CsCCGD
CCCs
s
s
GmgCCs
sC
mgCs
sLGD
CCs
)31
(3
)3
()31
(31
2
)(sV
)(gV
)31
(3
)3
(
/1)3
(
0
0
=0, collect real and imaginary parts and set them to zero.
TCLCo
131
31CC
CCGD
CTC
C
At 0
13
CC
Rmg
Generally more gain is used to ensure oscillation with amplitude stabilization.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
LC Oscillators: Hartley Oscillator
2
1
1
1
212
1
2
)(sV
)(gV
)2
/1()1
/1(2
/1
)2
/1(2
/1
0
0
LLC
LLssLmg
ogmgsC
s
s
ogmgsLsL
sL
mgsL
sLsC
=0, collect real and imaginary parts and set them to zero.
)21
(1
LLCo
At 0
21
LL
f
Generally more gain is used to ensure oscillation with amplitude stabilization.
G-S and G-D capacitances are neglected, assume no mutual coupling between inductors.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Amplitude Stabilization in LC Oscillators
• Inherent nonlinear characteristics of transistors are used to limit oscillation amplitude. Eg: rectification by JFET gate diode or BJT base-emitter diode.
• In MOS version, diode and RG form rectifier to establish negative bias on gate, capacitors act as rectifier filter.
• Practically, onset of oscillation is accompanied by slight shift in Q-point values as oscillator adjusts to limit amplitude.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Crystal Oscillators
Crystal: A piezoelectric device that vibrates is response to electrical stimulus, can be modeled electrically by a very high Q (>10,000) resonant circuit.
L, CS, R represent intrinsic series resonance path through crystal. CP is package capacitance. Equivalent impedance has series resonance where CS resonates with L and parallel resonance where L resonates with series combination of CS and CP.
SC
PC
SC
PC
TC
TLCL
RssS
LCLRss
PsC
SZ
PZ
SZ
PZ
CZ
12
12
1
Below S and above P, crystal appears capacitive, between S and P it exhibits inductive reactance.
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Crystal Oscillators: Example• Problem: Find equivalent circuit elements for crystal with given parameters.
• Given data: fS=5 MHz, Q=20,000 R =50 , CP =5 pF
Analysis:
5.02MHz
fF)6.31mH)(8.31(21
2
1
fF8.31
)0318.0(2710
12
1
mH8.31)6105(2
)000,20(50
SC
PC
SC
PC
LPf
LS
SC
S
RQL
2 Microelettronica – Circuiti integrati analogici 2/edRichard C. Jaeger, Travis N. Blalock
Copyright © 2005 – The McGraw-Hill Companies srl
Crystal Oscillators: TopologiesColpitts Crystal Oscillator Crystal Oscillator using BJT
Crystal Oscillator using JFET
Crystal Oscillator using CMOS inverter as gain element.