Chap18-Feedback, Stability and Oscillators

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Jaeger/Blalock 3/10/04

Microelectronic Circuit DesignMcGraw-Hill

Chapter 18

Feedback, Stability and Oscillators

Microelectronic Circuit Design

Richard C. Jaeger

Travis N. Blalock

Chap 18 - 1





Chapter Goals

• Review concepts of negative and positive feedback.

• Develop two-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 two-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.

Chap 18 - 2





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

Chap 18 - 3





Classic Feedback Systems

• A(s) = transfer function of open-

loop amplifier or open-loop gain.

• b (s) = transfer function of

feedback network.

V d ( s)V

i( s)V

f ( s)

V o( s)V

d ( s) A( s)

V f ( s)V

o( s) b ( s)

Av

s V

o( s)

V i( s)

A s

1 A s b s

A s 1T s

T(s) = A b = loop gain

For negative feedback: T(s) > 0

For positive feedback: T(s) < 0

Chap 18 - 4





Voltage Amplifiers: Series-Shunt

Feedback (Voltage Gain Calculation)

1

A

v

11

A

hi1

12

A

hv

2

1

F

v

11

F

hi1

12

F

hv

2

2 Ai 21

Ah i1 22

Ah v2 2

F i 21 F h i

1 22

F h v2

ijT h ij

Ah ij F h and 21

Ah 21 F h , 12

F h 12 Ah

v1 11

T h i1 12

F h v2

vi( R

I 11

T h )i1 12

F h v2

i2 21

Ah i1 22

T h v2 0 21

Ah i1( 22

T h G L

)v2

Av v

2

vi

21 Ah

21 Ah 12

F h ( R I

11T h )( 22

T h G L

) A

1 A b

A 21 Ah

( R I

11T h )( 22

T h G L

) b 12

F h

Chap 18 - 5






Feedback (Two-Port Representation)

• Gain of amplifier should include

effects of , , R I and R L.• Required h-parameters are found

from their individual definitions.

• Two-port representation of the

amplifier is as shown

11 F

h

22 F

h

Chap 18 - 6






Feedback (Input and Output Resistances)

vi( R

I 11

T h )i1

12 F h

- 21 Ah

( 22T h G

L)

i1

Rin

v

i

i1

( R I

11T h )(1 A b )

Rin in A R (1 A b )

Series feedback at a port increases

input resistance at that port.

For output resistance:

v1 R I i1v2 v x i2 i x G Lv2

i x

21 Ah i

1( 22

T h G L

)v x

0 ( R I

11T h )i

1 12

F h v x

Rout

v x

i x

1

22T h G

L

1 A b Rout out A R

1 A b

Shunt feedback at a port reduces

resistance at that port.

Chap 18 - 7





Voltage Amplifiers

Series-Shunt Feedback (Example cont.)

Av

o

vi

25k W1k W 25k W9.01k W

(104)1.96k W

1.96k W1.00k W 4730

Av A

1 A b 4730

1 4730(0.0990)10.1

Rin

in A R (1 A b )16.4 M W

Rout

out A R

1 A b 1.41W

Chap 18 - 9





Transresistance Amplifiers: Shunt-Shunt


1 Ai 11

A y v1 12

A y v2 1

F i 11 F y v

1 12

F y v2

2 Ai 21

A y v1 22

A y v2 2

F i 21 F y v

1 22 F y v

2

ijT y ij

A y ij F y and 21

A y 21 F y , 12

F y 12 A y

i1 11

T y v1 12

F y v2 i

i (G

I 11

T y )v1 12

F y v2

i2 21 A y v1 22T y v2 0 21 A y v1( 22T y G L)v2

Atr

v2

ii

21 A y

21 A y

12 F y (G

I 11

T y )( 22T y G

L)

A

1 A b

A 21 A y

(G I

11T y )( 22

T y G L) b

12

F y

Chap 18 - 10







• Gain of amplifier should includeeffects of , , R I and R L.

• Required y-parameters are foundfrom their individual definitions.

• Two-port representation of theamplifier is as shown.

11 F y

22 F y

Chap 18 - 11







ii (G

I 11

T y )v1 12

F y-

21 A y

(22T y G

L)

v1

Rin

v

1

ii

1

G I

11T h

(1 A b ) in

A R

(1 A b )

Shunt feedback at a port reduces resistance

at that port.


i1 G I v1 i2 i x G Lv2

i x

21 A y v

1( 22

T y G L

)v x

0(G I

11T y )v

1 12

F y v x

Rout

v x

i x

1

22T y G

L

1 A b Rout out A R

1 A b

Resistance at output port is reduced due to

shunt feedback.

Chap 18 - 12





Transresistance Amplifiers

Shunt-Shunt Feedback (Example)

• Problem: Find A, b , closed-loop

gain, input and output resistances.

• Given data: V A = 50 V, b F = 150

• Analysis: For the dc equivalent

circuit,

11 F

y

i1

v1 v

2 0

1

R F

10-5

S

22 F y

i2

v2 i

2 0

1

R F

10-5 S

12 F y

i1

v2 v1

0

1

R F

10-5 S

I C

V

CC V

BE

RC

R

C R

F

b F

0.970 mA

V CE

V CC

( I C

I B

) RC

1.35 V

g m

40(0.977mA)39.1 mS

r

1

g m

3.84 k W r o

50V 1.35V

0.977mA52.6 k W

Chap 18 - 13





Transresistance Amplifiers

Shunt-Shunt Feedback (Example cont.)

Av

o

ii

4.76k W

4.76k W 3.84k W(150)1.41k W 52.6k W

114 k W

Atr

A

1 A b 114k W

1114k W(0.01mS )53.3 k W

Rin in A R (1 A b ) R

I R

F r

(1 A b ) 995 W

Rout

out A R

1 A b

R L

R F

RC

r o

1 A b 640 W

ib ii

4.76k W

4.76k Wr

vo b oib(1.41k Wr

o)

Chap 18 - 14





Current Amplifiers: Shunt-Series


i1 1

Ai 1 F i

v2 2 Av 2 F v

ijT g ij

A g ij F g and 21

A g 21 F g ,

12

F g 12 A g

i1 11

T g v1 12

F g i2 i

i (G

I 11

T g )v1 12

F g i2

v2 21

A g v1 22

T g i2 0 21

A g v1( 22

T g R L

)i2

Ai

i2

ii

21 A g

21 A g

12 F g (G

I

11T g )(

22T g R

L) A

1 A b

A 21

A g

(G I

11T g )(

22T g R

L) b 12

F

g

Chap 18 - 15







• Gain of amplifier should include

effects of , , R I and R L.• Required g -parameters are found

from their individual definitions.

• Two-port representation of theamplifier is as shown

11 F g

22 F g

Chap 18 - 16







ii(G

I 11

T g )v1 12

F g - 21

A g

(22T g R

L)

v1

Rin

v1

ii

1

G I

11T g

(1 A b ) in A R

(1 A b )

Shunt feedback at a port decreases



Rout

v

x

i2

22T g R

L

1 A b

Rout out

A R 1 A b

Series feedback at output port increases


i1 G I v1 v2 v x R Li2

v x

21 A g v

1( 22

T g R L

)i2

0(G I

11T g )v

1 12

F g i2

Chap 18 - 17





Transconductance Amplifiers: Series-Series


v1 1 Av 1 F vv

2

2 Av

2 F v

ijT z ij

A z ij F z and

21 A z 21

F z ,12 F z 12

A z

v1 11

T z i1 12

T z i2 v

i( R

S 11

T z )i1 12

F z i2

v2 21

T z i1 22

T z i2 0 21

T z i1(

22T z R

L)i

2

Atc

i2

vi

21 A z

21 A z 12

F z ( R I

11T z )( 22

T z R L

) A

1 A b

A 21 A z ( R

I

11T z )( 22

T z R L

) b z

12 F

Chap 18 - 18





Transconductance Amplifiers: Series-Series


Rout

v

x

i2

22T z R

L

1 A b

Rout

out A R 1 A b

Rin

v

i

i1

11T z R

I

1 A b

Rin

in A R 1 A b

Series feedback at input and output

port increases resistance at both ports.

Chap 18 - 19

• Gain of amplifier should includeeffects of , , R I and R L.

• Required y-parameters are foundfrom their individual definitions.

• Two-port representation of theamplifier is as shown.

11 F z

22 F z





Erroneous Application of Two-Port

Feedback Theory (cont.)

z -parameters are found

as shown. From A-

circuit, I E = 1 mA

r

1000(0.025V)

1mA2.5k W r

o 50V

1mA50k W

iovi

Rid

Rid R Ao

b o

r ( b o1) R

Atc io

vi

A1 A b

1.64S 11.64S (5k W)

0.200 mS

Rin

in A R 1 A b

( R

id R)1 A b

246 M W R

in in

A R 1 A b

27.7 GW

Chap 18 - 21





Erroneous Application of Two-Port

Feedback Theory (cont.)

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.

Rout

b or

o

Chap 18 - 22





Analysis of Shunt-Series Feedback Pair

• Problem: Find A, b , closed-loop gain, input and output resistances.

• Given data: b o= 100, V A = 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.

Chap 18 - 23






(cont.)

Small signal parameters are found

from given Q-points.

For Q1, r = 3.79 k W, r o = 155 k W.

For Q2, r

= 1.56 k W, r

o = 64.8 k W.

vthii

R B

R Br

1

b o1

(r o1 RC )

vth4.94105i

i

Rth 10k W r o18.88k W

v2vth

( b o21)(0.901k W)

8.88k W r 2( b o21)(0.901k W)

Chap 18 - 24






(cont.)

Av

2

ii

4.43105 W

b 12

F y 1

9100S

Atr

A

1 A b 8910 W

Rin

in

A R

(1 A b )

R B

r 1

(1 A b ) 42.5 W

Rout

out A R

(1 A b )1.86 W

Closed-loop current gain is given by:

Ai

io

ii

a

oi

e

ii

a o

v2

901Wi

i

a

o

901W A

tr 9.79

Chap 18 - 25





Direct Calculation of Loop Gain

• Original input source is set to zero.

• Test source is inserted at the point

where feedback loop is broken.

vr b vo b A(0v x) b Av x

T A

b vr

v x

Example:

is added for proper

termination of feedback loop.

R3 R

2 R

1

vr v

o

R1

R1 R2

R

1

R1 R2

Av x

Rid

Rid R3

T v

r

v x

A R

1

R1 R

2

Rid

Rid

R3

Chap 18 - 26





Calculation of Loop Gain using

Successive Voltage and Current Injection

Voltage injection: Voltage source v X is

inserted at arbitrary point P in circuit.

v1

b

1 A b v x

where b R A

R A R B

v2v1v x b (1 A b )

1 A b v x

T vv

2

v1

1 A b b

b T 1 R B

R A

R B

R A

for T = A b

Current injection: Current

source i X is inserted again at P.

i1

v x

R A

i2

v x

Av x

R B

v x1 A R B

T ii2

i1

1 A

R B

1/ R A

R B

R A

T 1 R B

R A

T

T v

T i

1

2T vT i

R B R A

1T v

1T i

Chap 18 - 27





Simplifications to Successive Voltage

and Current Injection Method

• Technique is valid even if source resistances with v X and i X are included in analysis.

• If at P, R B is zero or R A is infinite, T can be found by only one measurement and T =

T v . In ideal op amp, such point exists at op amp input.• If at P, R B is zero, T = T v . In ideal op amp, such point exists at op amp output.

• If R A = 0 or R B is infinite, T = T I .

• In practice, if R B >> R A or R A >> R B, the simplified expressions can be used.

Chap 18 - 28





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 T SC and T OC .

RCL = resistance of closed-loop amplifier looking into one of its ports (any

terminal pair)

R D = resistance looking into same pair of terminals with feedback loop

disabled.

T SC = Loop gain with a short-circuit applied to selected port

T OC = Loop gain with same port open-circuited.

RCL R D

1 T SC

1 T OC

Chap 18 - 29





Blackman’s Theorem (Example 1)

For output

resistance:

R D

r o1

b o( R R

id )

r

( R Rid

)

3.18 M W

T SC

A

ov

1

1 A

o

( b o1)( R R

id r

o)

r

( b o1)( R R

id r

o)

9940

T OC

Aov

1 A

o

( R Rid

)

r

( R Rid

) 6350

Problem: Find input and output resistances.

Given data:V REF = 5 V, R = 5 k W, b o= 100, V A= 50 V, Ao= 10,000, Rid = 25 k W, Ro= 0

Assumptions: Q-point is known, g m = 0.04 S, r = 25 k W, r o = 25 k W.

Rout 3.18 M W 1 9940

/1 6350

5.06 M W

For input

resistance:

R D

Rid

( R 1 g

m

) 25 k W

Rin

25k W 1 9940

/1

249 M W

Chap 18 - 30



Jaeger/Blalock 3/10/04 Microelectronic Circuit DesignMcGraw-Hill


Problem: Find input and output resistances.

Given data: b o = 100, V A = 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 W, r o = 155 k W, For Q2, r = 1.56

k W, r o = 64.8 k W.

Chap 18 - 31




Blackman’s Theorem (Example 2 cont.)

R Dr o2

1b o(1/

22 F y )

r 2( RC r o1

)(1/22 F y )

321 k W

T SC

(9.1k W10k W 3.79k W)( g m1

) 79k W 10k W 1.56k W ( b o

1)0.901k W

.

( b o1)0.901k W

1.56k W( b o 1)0.901k W 12

F

y 48.7

T OC

(2.11k W)(400.66mA)(1.93k W)0.901k W

1.56k W 0.901k W

1

9.1k W

4.33

Rout

321k W 1 48.7

/1 4.33

2.99 M W

For input resistance:

R D10k W9.1k W r

12.11 k W

Rin2.11k W 10

147.8 42.5 W


Chap 18 - 32





• Problem: Find expression for output resistance of the Wilson source.

Analysis: Feedback loop is disabled by

setting reference source i to zero.

R Dr o3 1

b o3

(1/ g m1

)

r o2r

3(1/ g

m1)

r o3

Assuming g m1= g m2 = g m3 and m f >> b o >> 1.

Next, current source i is set to 1:

i ie ( b

o31)i

b

( b o

1)

12 b

o1

m f

( b o1)

T SC

b o1 T

OC

ie

i1 b o 1

m f

1

Rout

r o

1 b o1

11

b

or

o

2

Chap 18 - 33




Use of Feedback to Control Frequency

Response

Av

s A s

1 A s b s

A s A

o

H s

( s L

)( s H

)where

Av

s A

o

H s

s2 L

H

(1 Ao b )

s

L

H

Assuming H

1 Ao b

L,

L F

L

1 Ao b H

F

H

(1 Ao

b )

BW F

H

(1 Ao b )

Upper and lower cutoff frequencies as well as

bandwidth of amplifier are improved, gain is

stabilized at

Amid Ao

1 Ao b

1 b

GBW Amid

BW F

Ao

H

Chap 18 - 34




Use of Nyquist Plot to

Determine Stability

• If gain of amplifier is greater than or equal to 1 at the frequency wherefeedback 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, a pole exists in RHP and the amplifier is

unstable.• If -1 point lies outside the interior of the Nyquist plot, all poles of the closed-loopamplifier are in the LHP and theamplifier is stable.

Av

s A s

1T s

Chap 18 - 35




First-Order Systems

For a simple low-pass amplifier,

It can also represent a single-pole

op amp with resistive feedback

T s Ao o s

o

b T o s

o

T ( j )T o

j 1

At dc, T (0) = T o, but for >>1,

As increases, magnitude

monotonically approaches zero and

phase asymptotically approaches -900.

As b changes, value of T (0) = T o 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 T o.

T ( j ) jT o

Chap 18 - 36




Second-Order Systems

T s Ao

1 s

1

1 s

2

b T o

1 s

1

1 s

2

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 it

can come arbitrarily close to it.

T ( j ) 14

j 1

2

T ( j )( j)2 14

2

14

2

Chap 18 - 37




Phase Margin

Phase Margin is the maximumincrease in phase shift that can be

tolerated before system becomes

unstable.

m T ( j 1)(180)180T ( j

1)

Where

First we determine frequency for which

magnitude of loop gain is unity,corresponding to intersection of the

Nyquist plot with unit circle, 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.

T ( j 1) 1

Chap 18 - 38




Third-Order Systems

In 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 manycircumstances and system is unstable.

T s T

o

1 s

1

1 s

2

1 s

3

T ( s) 14

s3 s2 3 s2

T ( j )( j)3 14 3 j14

3

Chap 18 - 39




Gain Margin

Gain Margin is the reciprocal of magnitude of T ( j ) evaluated at

frequency for which phase shift is

1800.

GM 1

T ( j 180

)

where

If magnitude of T ( j ) is increased by a

factor equal to or exceeding gain margin,

then closed-loop system becomes

unstable, because the Nyquist plot thenencloses -1 point.

T ( j 180

)180

GM)log(20dB

GM

Chap 18 - 40




Bode Plots

A b 21019

s105

s106

s107

At 1.2 Mrad/s, magnitude of the loop gain is

unity and corresponding phase shift is 145o.

Phase margin is given by 180o - 145o = 35o.

Amplifier can tolerate additional phase shift

of 35o before it becomes unstable.

At 3.2 Mrad/s, phase shift is exactly 180o and

corresponding magnitude of the loop gain is -17 dB. Phase margin is given by 17 dB.

Gain of amplifier must increase by 17 dB

before amplifier becomes unstable.

Chap 18 - 41




Use of Bode plot to Determine Stability

b

1log20log20log20 A A

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/ b =80 dB, m=850, amplifier is stable.

For 1/ b =50 dB, m=150, amplifier is stable, butwith significant overshoot and ringing in its step

response.

For 1/ b =0 dB, m= -450, amplifier is unstable

A b 210 24

s 10 5

s 310 6

s 10 8

Chap 18 - 42




Operational Amplifier

Compensation Example

• Problem: Find value of compensation capacitor for m = 700.

• Given data: RC 1 = 3.3 k W , RC 2 = 12 k W ,SPICE parameters: BF =

100, VAF = 75 V, IS = 0.1 fA, RB = 250 W, TF = 0.75 ns, CJC = 2 pF.

Assumptions: Dominant pole is set by C C and pnp C-E stage. R Z is

included to remove zero associated

with C C . pnp and npn transistors are

identical; quiescent value of V o = 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.

Chap 18 - 43



Jaeger/Blalock

3/10/04


McGraw-Hill


Compensation Example (cont.)

• Analysis: I C 1 = I C 2 = 250 mA. For V o = 0,voltage across RC 2 = 12-0.75 = 11.3 V and I C 3 = 11.3V/12k W= 938 mA. Q4 and Q5 mirror

currents in Q7 and Q8 , so, I C 4 = I C 5 = 938 mA. For V o = 0, V CE4 =12 V, V CE5 = 12 V, V CE3 = 11.3V. For V I = 0, V CE2 = 12.8 V,V CE1 = 12-3300(0.25 mA)+0.75 =11.9 V

Small-signal parameters arefound using their respectiveformulae.

Av1

g m1

2(2r

o1 R

C 1r 3

)

Av1

0.01

2(696k W 3.3k W 3.07k W) 7.93

Av2

g m2

(r o3

RC 2

r 4

2

( b o4

1) R L

Av2

0.0375 92k W12k W 3.09k W2

(117)500W

Av2

338

Av3

( b

o1) R

L

r 4

2 ( b o4 1) R L

(117)500

30902 (117)500

Av3

0.974

Av A

v1 A

v2 A

v3 2610

Chap 18 - 44



Jaeger/Blalock

3/10/04


McGraw-Hill


Compensation Example (cont.)

Input stage pole:

Emitter Follower pole: Q4 and Q5 are

in parallel, composite parameters are-

g m = 0.02 S, r x = 125 W, C = 56.2 pF,

C m = 1.60 pF, Rth = 1/ g m3 = 267 W.

f H

1

2

1

r

C

C m

2g

m R

C

2

RC

r x

f H

59.2 MHz

f H 12

1

( Rth

r x

R L

)C

1 g m R

L

C m

Rth

r x

f H

82.5 MHz

At f T ,dominant pole due to C C

contributes phase shift of 900. For m= 700, other 2 poles can contribute

more phase margin of 200.

20 tan1f

T

59.2 MHz

tan1

f T

82.5 MHz

f T

12.2 MHz

C C

C m 3

G

m1

T

g m1

21

2 f T

65 pF

R Z = 1/ g m3 = 27.5 W is included toremove zero associated with C C.

f B

f

T

Ao

12.2 MHz

2610 4.67 kHz

Chap 18 - 45



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Barkhausen’s Criteria for Oscillation.

• For sinusoidal oscillator, poles of closed-loop amplifier should be at

frequency o on j axis.

• Use positive feedback through

frequency-selective feedback

network to ensure sustained

oscillation at 0 .

Av

s A s

1 A s b s

A s 1T s

• For sinusoidal oscillations,

• Barkhausen’s criteria state

• Phase shift around feedback loop

should be zero degrees and

magnitude of loop gain must beunity.

• Loop gain greater than unity causes

distorted oscillations.

1T j o 0T j

o 1

T j o 0

T j o 1

Or even multiples

of 3600

Chap 18 - 46



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Oscillators with Frequency-Selective RC

Networks: Wien-Bridge Oscillator

V o( s)V

1( s) Z

2( s)

Z 1( s) Z

2( s)

T ( s )V o( s)

V I ( s)

sRCG

(1 2 R2C 2)3 sRC 1

Phase shift will be zero if ,

This oscillator is used for frequencies up to a

few MHz, limited primarily by characteristics

of amplifier .

(1 2 R2C 2)0

At o

= 1/ RC T ( j o )G

3

T ( j o )G

3 T ( j

o )0o

Chap 18 - 47



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Oscillators with Frequency-Selective RC

Networks: Phase-Shift Oscillator

sCV o'( s)

0

(2 sC G)

sC

sC

(2 sC G)

V 1( s)

V 2( s)

V o( s)

V 2( s)

sCR1

T ( s) V o( s)

V o'( s)

s3C 3 R2 R

13 s2 R2C 2 4 sRC 1

Phase shift will be zero if (1 3 o2 R2C 2) 0,

RC o

3

1

At o , T ( j

o )

o2C 2 RR1

4

1

12

R1 R

Chap 18 - 48



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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 toolarge, waveform is distorted.

• Amplitude stabilization or gain control is used to automaticallycontrol 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 tosustain oscillations.

Chap 18 - 49



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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.

Chap 18 - 50



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LC Oscillators: Colpitts Oscillator

G1/( RS

r o) C

3C

2C

GS

0

0

s(C 3 C GD)1/ sL

s(C 3 g

m)

sC 3

s(C 1C

3) g

mG

V g ( s)

V s( s)

s2 C 1C

3C

GD(C

1C

3)

s (C

GDC

3)GGC

3

g

m

G

sL (C

1

C 3

)

L

= 0, collect real and imaginary parts and set them

to zero.

o 1

LC TC

C TC

C GD

C 1C

3

C 1C

3

At o, g

m R C 3

C 1

Generally more gain is used to ensure oscillation

with amplitude stabilization.

Chap 18 - 52



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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 biason 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.

Chap 18 - 54



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Crystal Oscillators

Crystal: A piezoelectric device that vibrates

is response to electrical stimulus, can bemodeled electrically by a very high Q (

> 10,000) resonant circuit.

L, C S , R represent intrinsic series resonance

path through crystal. C P is package

capacitance. Equivalent impedance has seriesresonance where C S resonates with L and

parallel resonance where L resonates with

series combination of C S and C P .

Z C

Z

P Z

S

Z P

Z S

1

sC P

s2 s R

L 1

LC S

s2 s R

L 1

LC T

C T

C P C S

C P

C S

Below S and above P ,

crystal appears capacitive,

between S and P it exhibits

inductive reactance.

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Crystal Oscillators: Example

• Problem: Find equivalent circuit elements for crystal with given parameters.

• Given data: f S = 5 MHz, Q = 20,000 R = 50 W, C P = 5 pF

Analysis:

L RQ

S

50(20,000)

2 (5106

)

31.8 mH

C S

1

S

2 L 1

107

2

(0.0318)

31.8 fF

f P

1

2 L C P C S

C P

C S

1

2 (31.8mH )(31.6 fF )

f P

5.02 MHz

Chap 18 - 56



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Crystal Oscillators: Topologies

Colpitts Crystal Oscillator Crystal Oscillator using BJT

Crystal Oscillator using JFET

Crystal Oscillator using CMOS

inverter as gain element.

Chap 18 - 57



J /Bl l k Mi l t i Ci it D i

End of Chpater 18

Chap18-Feedback, Stability and Oscillators

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