2 Microelettronica – Circuiti integrati analogici 2/ed 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 14Feedback, Stability and Oscillators

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock



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.



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



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



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



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



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.



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



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



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



Transresitance Amplifiers: Shunt-Shunt Feedback (Two-Port Representation)


• Required y-parameters are found from their individual definitions.

• Two-port representation of the amplifier is as shown.

yF11

yF22



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.


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.



Transresistance Amplifiers: Shunt-Shunt Feedback (Example)


• 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



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



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



CurrentAmplifiers: Shunt-Series Feedback (Two-Port Representation)


• Required g-parameters are found from their individual definitions.


gF11

gF22



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



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



Transconductance Amplifiers: Series-Series Feedback (Input and Output Resistances)


• Required g-parameters are found from their individual definitions.


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.



Erroneous Application of 2-port Feedback Theory


• 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.



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



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



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.



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



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



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



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



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.



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



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



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.



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


33.49.1kΩ

10.901kΩ1.56kΩ

0.901kΩ93kΩ30.66mA)(1.kΩΩ)(411.2(

OCT



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



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



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



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 )(



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



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



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



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



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.



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



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.



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



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



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



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



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



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.



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.



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



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.



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.



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.



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.



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



Crystal Oscillators: TopologiesColpitts Crystal Oscillator Crystal Oscillator using BJT

Crystal Oscillator using JFET

Crystal Oscillator using CMOS inverter as gain element.

2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock...

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Transcript of 2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock...