Feedback

FEEDBACK AMPLIFIERS AND OSCILLATORS

forElectronic Circuits

by

Prof. Michael Tse

September 2004

C.K. Tse: Feedback amplifiers andoscillators

2

ContentsFeedback

Basic feedback configurationAdvantagesThe price to pay

Feedback Amplifier ConfigurationsSeries-shunt, shunt-series, series-series, shunt-shuntInput and output impedancesPractical Circuits with loading effectsCompensationOp-amp internal compensation

OscillationOscillation criteriaSustained oscillationWein bridge, phase shift, Colpitts, Hartley, etc.


3

Basic feedback configuration

The basic feedback amplifier consists of a basic amplifier and a feedback network.

f

+

–si

e

sf

soA

basic amplifier

feedback network

input output

A = basic amplifier gainf = feedback gain

Careful!!


4

Characteristics

f

+

–

sie

sf

soA

basic amplifier

feedback network

input output

The input is subtracted by afeedback signal which is part of theoutput, before it is amplified by thebasic amplifier.

so = Ae = A(si - sf )

But, since sf = f so, we get

so = A(si - fso)

Ao =sosi=

A

1+ Af

Hence, the overall gain is

si soAo

If Af >>1, Ao ª1

f


5

Simple viewpoint

f

+

–

sie

sf

soA

basic amplifier

feedback network

input output

If A is large, then e must be very small in order to give a finite output.So, the input si must be very close to the feedback signal sf .That means sf ≈ si .

But, sf is simply a scaled-down copy of the output so.

Hence, f so = si orsosiª

1

f


6

Obvious advantage

If the feedback network is constructed from passive elements havingstable characteristics, the overall gain becomes very steady andunaffected by variation of the basic amplifier gain.

Quantitatively, we wish to know how much the overall gain Aochanges if there is a small change in A.

Let assume A becomes A + dA. From the formula of Ao, we have

Obviously, if Af is large, then dAo/Ao will be reduced drastically.

Feedback reduces gain sensitivity! In fact, the gain is just .

dAo

Ao

=dA

A

Ê

ËÁ

ˆ

¯˜

1

1+ Af

Ê

ËÁ

ˆ

¯˜

Ao ª1

f


7

Another advantage

Suppose the basic amplifier isdistortive. So, the output doesnot give a sine wave for a sinewave input.

But, with feedback, we see thatthe gain is about 1/f anyway,regardless of what A is (or aslong as Af is large enough).

This gives a very good propertyof feedback amplifier in termsof eliminating distortion.

e

so

A

so

si

1/f

input

output


8

Other advantages

• Improve input and output resistances (to be discussed later).• Widening of bandwidth of amplifier (to be discussed later).• Enhance noise rejection capability.

f

+

–

sie

sf

soA’

basic amplifier

feedback network

input outputA

ni

+

+A

ni

sono

È

ÎÍ

˘

˚˙ =

sini

È

ÎÍ

˘

˚˙

Signal-to-noise ratio:

si so

sono

È

ÎÍ

˘

˚˙ = ¢A

sini

È

ÎÍ

˘

˚˙Signal-to-noise ratio improves!


9

The price to pay

Of course, nothing is free!

Feedback comes with reduced gain, and hence you may need to add a pre-amplifier to boost the gain.

Also, wherever you have a loop, there is hazard of oscillation, if you don’twant it.

Later, we will also see how we can use feedback to create oscillationdeliberately.


10

Terminologies

Basic amplifier gain = A

Feedback gain = f

Overall gain (closed-loop gain) =

Loop gain (roundtrip gain) = Af

A

1+ Afª

1

f

Some books use T to denote Af.


11

Feedback amplifiers

What is an amplifier? Asi so

Signals can be voltage or current.

General model for voltage amplifier:

+–

Ro

Rin Avinvin

+

–

vo

+

–

voltage amplifier


12

Models of amplifiers

+–

Ro

Rin Avinvin

+

–

vo

+

–

voltage amplifier

RoRin

Aiinio

current amplifier

+–

Ro

Rin Aiin vo

+

–

transresistance amplifier

iin

iin

RoRin

Aiinio

transconductance amplifier

vin

+

–


13

Feedback amplifier configurations

f

+

–

sie

sf

soA

basic amplifier

feedback network

input output

voltage voltage

To copy voltage, we should useparallel (shunt) connection

To subtract voltage fromvoltage, we should use seriesconnection

vi

– vf +

+

–A

+

–vo

•

•

Hence, series-shunt feedback

Voltage amplifier


14

Series-shunt feedback (for voltage amplifier)

ve+

–

+–

+–

+

–

vo+

–

vi Ave

Ro

Ri

fvo

Overall gain (closed-loop gain) : Ao =vovi=

A

1+ Af


15


To find the input resistance, we consider the ratio of vi and ii, with output opened.

ve+

–

+–

+–

+

–

vo+

–

vi Ave

Ro

Ri

fvo

RIN

ii

RIN =viii=

vive /Ri

= Ri

ve + fvove

= Ri(1+ Af )

The input resistance has been enlarged by (1+Af). This is a desirablefeature for voltage amplifier as a large input resistance minimizes loadingeffect to the previous stage.


16


To find the output resistance, we consider shorting the input source and calculatethe ratio of vo and io.

The output resistance has been reduced by (1+Af). This is a desirablefeature for voltage amplifier as a small output resistance emulates a bettervoltage source for the load.

ve+

–

+–

+–

+

–

vo+

–

vi Ave

Ro

Ri

fvo

io

ROUT

First, we have ve = – fvo.Also,

io =vo - AveRo

=vo + Afvo

Ro

Hence,

ROUT =voio=

Ro

1+ Af


17

Series-shunt feedback (for voltage amplifier)Summary of features Equivalent model

Closed-loop gain =

Input resistance = Ri ( 1 + Af )

Output resistance =

A

1+ Afª

1

f

Ro

1+ Af

NOTE: We did not consider loading effect of thefeedback network, i.e., we assume that the feedbacknetwork is an ideal amplifier which feeds a scaled-downcopy of the output to the input.

∞+–

feedback network

Ri ( 1 + Af )

Ro

1+ Af

Avi1+ Af

+– vo

+

–vi

+

–


18

Feedback amplifier configurations

f

+

–

sie

sf

soA

basic amplifier

feedback network

input output

current voltage

To copy voltage, we should useparallel (shunt) connection

To subtract current fromcurrent, we should use shunt(connection) connection

ii A+

–vo

•

•

Hence, shunt-shunt feedback

Transresistance amplifier


19

Shunt-shunt feedback (for transresistance amplifier)

Overall gain (closed-loop gain) : Ao =voii=

A

1+ Af

ie

+–

+

–

vo

ii

Aie

Ro

Ri

fvo


20


To find the input resistance, we consider the ratio of vi and ii, with output opened.

RIN

vi

RIN =viii=Riieii

= Ri

ieie + fvo

=Ri

1+ Af

The input resistance has been reduced by (1+Af). This is a desirablefeature for transresistance amplifier as a small input resistance ensuresbetter current sensing from the previous stage.

+

–

ie

+–

+

–

vo

ii

Aie

Ro

Ri

fvo


21


To find the output resistance, we consider opening the input source (putting ii = 0)and calculate the ratio of vo and io.

The output resistance has been reduced by (1+Af). This is a desirablefeature for transresistance amplifier as a large small resistance emulates abetter voltage source for the load.

ROUT

First, we have ie = – fvo.Also,

io =vo - AieRo

=vo + Afvo

Ro

Hence,

ROUT =voio=

Ro

1+ Af

ioie

+–

+

–

vo

ii = 0

Aie

Ro

Ri

fvo


22

Shunt-shunt feedback (for transresistance amplifier)Summary of features Equivalent model

Closed-loop gain =

Input resistance =

Output resistance =

A

1+ Afª

1

f

Ro

1+ Af

Ro

1+ Af

Aii1+ Af

+–

Ri

1+ Af

Ri

1+ Af

ii

vo

+

–

Similar, we can develop the feedback configurations fortransconductance amplifier and current amplifier.

Transconductance amplifier: series-series feedbackCurrent amplifier: shunt-series feedback


23

Series-series feedback (for transconductance amplifier)

Overall gain (closed-loop gain) :

Input resistance:

Output resistance:

Ao =iovi=

A

1+ Af

+–

Ave

RoRi

f io

+

–

vo ve+

–

io

io

RIN = Ri(1+ Af )

ROUT = Ro(1+ Af )

Desirable!

Desirable!


24

Shunt-series feedback (for current amplifier)

Overall gain (closed-loop gain) :

Input resistance:

Output resistance:

Ao =ioii=

A

1+ Af

ieii

Aie

RoRi

f io

io

io

RIN =Ri

1+ Af

ROUT =Ro

1+ Af

Desirable!

Desirable!


25

Practical feedback circuits (with loading effects)

ie

+–

+

–

voii Aie

Ro

Ri

fvo

In practice, the input source has resistance and the feedback network hasresistance.

What are the effects on the gain, input and output resistances?

Example: shunt-shunt feedback


26

Systematic analysis using 2-port networks

+

–

voiiy21vi

y22y11

y21fvo

The best way to analyze feedback circuits with loading effects is to use two-portmodels.

For shunt-shunt feedback, input and output sides are both parallel connected.Thus, the loading can be combined by summing the conductances. Also, voltageis common at both sides. So, y-parameter is best.

+

–vi

y11f

y22f

1 2

yi

The first step is to put everything in y-parameter:


27

Systematic analysis of shunt-shunt feedback usingy-parameter

+

–

voiiy21vi

y22y11

y12fvo

+

–vi

y11f

y22f

1 2

yi

In order to use the standard results, we have to convertthis model to the standard form (slide 19).


28


+

–

voiiy21vi

y22y11

y12fvo

+

–vi

y11f y22f

1 2

yi

One step closer…


29


+

–

voiiy21vi

y22y11

y12fvo

+

–vi

y11f y22f

1 2

yi

One more step closer…


30


+

–

voiiy21vi

y11+y11f+yi

y12fvo

+

–vi

y22f+y22

1 2

Yet another step closer…


31


+

–

voii

y11+y11f+yi

y12fvo

+

–vi

1/(y22f+y22)

1 2

Yet another step closer…

+–

in resistance (Ω)in conductance (S)

-y21viy22f + y22

Use Thevenin


32


+

–

voii

y11+y11f+yi

y12fvo

+

–vi

1/(y22f+y22)

1 2

Finally, we get the same standard form.

+–

in resistance (Ω)in conductance (S)

-y21iey22f + y22( ) y11 + y11f + yi( )

ie


33


Basic amplifier gain

Feedback gain f = y21f

Overall (closed-loop) gain

Input resistance

Output resistance

A =-y21

y22f + y22( ) y11 + y11f + yi( )=

-y21

yoT yiT

We can simply apply the standard results:

Ao =A

1+ Afª

1

f=

1

y12f

RIN =1

(y11 + y11f + yi)(1+ Af )

ROUT =1

(y22f + y22)(1+ Af )


34

Appropriate 2-port networks for analyzingfeedback circuits

For shunt-shunt feedback, use y-parameter.For shunt-series feedback, use g-parameter.For series-series feedback, use z-parameter.For series-shunt feedback, use h-parameter.

The procedure is essentially the same as in the previous shunt-shunt case.

WHY?


35

General procedure of analysis

1. Identify the type of feedback.

2. Use appropriate 2-port representation.

3. Lump all loading effects in the basic amplifier, giving a modifiedbasic amplifier.

4. Apply Thevenin or Norton to cast the model back to the standardform (without loading).

5. Apply standard formulae to find A, f, RIN and ROUT.


36

Example

RL+vo–

–

+a

Rf

is

Type of feedback: shunt-shuntAppropriate 2-port type: y-parameter

So, the first step is to represent the circuit in y-parameter networks.


37

Example

RL+vo–

–

+a

Rf

is


38

ExampleConverting to y-parameter

RL+vo–

Rf

is+–avi

–vi+

Ri

Ro

y12fvoy11f y22f

y11f =1

Rf

y12f =-1

Rf

y22f =1

Rf

Note: thisgoes to the–ve inputof A.


39

ExampleConverting to y-parameter

RL+vo–

is+–avi

–vi+

Ri

Ro

y12fvoy11f y22f

y11f =1

Rf

y12f =-1

Rf

y22f =1

Rf

REMEMBER:y11f and y22f are conductance!


40

ExampleCasting it to standard form

+vo–

is+–avi

–vi+

Ro

y12fvo

Rf||RL

y11f =1

Rf

y12f =-1

Rf

y22f =1

Rf

Ri ||Rf


41

ExampleCasting it to standard form

+vo–

is+–

–vi+

Ri ||Rf

Ro ||Rf ||RL

a(Rf ||RL )

Ro + Rf ||RL

vi

-voRf


42

ExampleFinally, we get the standard form

+vo–

is+–Ri ||Rf

Ro ||Rf ||RL

Aie

-voRf

ie

Aie = aviRf ||RL

Rf ||RL( ) + Ro

A = -aRi ||Rf( ) Rf ||RL( )Rf ||RL( ) + Ro

A = -aRiRf

2RL

(Ri + Rf )

1

Rf RL + RoRf + RoRL( )

Using Thévenin theorem,

–vi+


43

Example

Apply standard results:

Basic amplifier gain (transresistance)

Feedback gain:

Overall (closed-loop) gain

Input resistance

Output resistance

A = -aRiRf

2RL

(Ri + Rf )

1

Rf RL + RoRf + RoRL( )

f =-1

Rf

Ao =A

1+ Afª

1

f= -Rf

RIN =Ri ||Rf

1+ Af

ROUT =Ro ||Rf ||RL

1+ Af

if Af >> 1


44

Frequency response

Gain and bandwidth

f

+

–

sie

sf

soA(jw)

basic amplifier

feedback network

input output

Suppose the basic amplifierhas a pole at p1, i.e.,

A( jw) =ALF

1+jw

p1

p1w

20log10|A| (dB)

slope = –20dB/decALF


45

Frequency response

Gain and bandwidth

The overall (closed-loop) gain is

Ao( jw) =A( jw)

1+ A( jw) f

=ALF

1+jw

p1

Ê

ËÁ

ˆ

¯˜ + fALF

=ALF

1+ fALF

1

1+jw

p1(1+ fALF )

È

Î

ÍÍÍÍ

˘

˚

˙˙˙˙

p1w

20log10|A| (dB)

basic amplifier

Hence, we see that the overall gain has apole at

pc = p1(1 + fALF)and the low-frequency gain is loweredto

pc

Ao,LF =ALF

1+ fALF

Ao,LF

ALF

feedback amplifier


46

Stability of feedback amplifier

Definition: A feedback system is said to be stable if it does notoscillate by itself at any frequency under a given circuit condition.

Note that this is a very restrictive definition of stability, but isappropriate for our purpose.

Therefore, the issue of stability can be investigated in terms of thepossibility of sustained oscillation.

feedback circuitsustained oscillation at certain frequency


47

Why and how does it oscillate?

The feedback system oscillates because of the simple fact that it has aclosed loop in which signals can combine constructively.

Let us break the loop at an arbitrary point along the loop.

f

+

–

si soAinput output

B B’

Signal at B, as it goes around the loop, will be multiplied by f and A, andalso –1.

SB’ = – A f SB


48

Why and how does it oscillate?

Clearly, if SB’ and SB are same in magnitude and have a 360o phasedifference, then the closed loop will oscillate by itself.

Oscillation criteria:

1.

2. Af = ±180o

The idea isIf the signal, after making a round trip through A and f, has a gain of 1and a phase shift of exactly 360o, then it oscillates. But, in the negativefeedback system, there is already a 180o phase shift. Therefore, thephase shift caused by A and f together will only need to be 180o tocause oscillation.

Af =1

This is known as the Barkhausen criteria.


49

The loop gain T

An important parameter to test stability is the loop gain, usuallydenoted by T.

T = Af

w

w

wo

|T| (dB)

f

fT

crossover frequency(where the gain is 1)

If fT = –180o, OSCILLATES!

0dB


50

Phase margin

Phase margin is an important parameter to evaluate how stable thesystem is.

w

w

wo

|T| (dB)

f

fT

–180o


phase margin fPM (the larger the better)

0dB

Phase margin fPM = –180o – fT


51

Compensation

Compensation is to make the amplifier more stable, i.e., to increase fPM.REMEMBER: We should always look at T, not A or Ao.

w

w

wo

|T| (dB)

f

fT

–180o


phase margin fPM (how to increase it?)

0dBp1 p2


52

Method 1: Lag compensationAdd a pole at a low frequency point. The aim is to make the crossover pointappear at a much lower frequency. The drawback is the reduced bandwidth.

w

w

|T| (dB)

f

–180o

crossover frequencybefore compensation

phase margin fPM

before compensation

0dBp1 p2pa

phase margin fPM

after compensation

crossover frequencyafter compensation

Compensationfunction Gc is

Gc ( jw) =1

1+jw

pa

before compensationafter compensation


53

Method 2: Lead compensationAdd a zero near the first pole. The aim is to reduce the phase shift and henceincrease the phase margin and keep a wide bandwidth. But the drawback is themore difficult design.

w

w

|T| (dB)

f

–180o

crossover frequencybefore compensation

phase margin fPM

before compensation

0dBp1 p2

za

phase margin fPM aftercompensation

crossover frequencyafter compensation

before compensationafter compensation

Compensationfunction Gc is

Gc ( jw) =1+

jw

za

1+jw

pa


54

Op-amp stability problemThe op-amp has a high DC gain, and hence at crossover it is likely that thephase shift is significant. The worst-case scenario is when the feedback gain is1 (maximum for passive feedback). We call this unity feedback condition, anduse this to test the stability of an op-amp.

Under unity-gain feedback condition, the loop gain T = Af = A, because f = 1.

|A| (dB)

–90o

–180o

phase margin too small

p1 p2

op-amp frequency response


55

Op-amp internal compensationUsually, op-amps are internally compensated. The technique is by lagcompensation, i.e., adding a pole at low frequency such that the phase margincan reach at least 45o.

Suppose we add a low-frequency dominant pole at pa. If we can put pa suchthat p1 (original dominant pole) is at crossover, then the phase margin is about45o. |A| (dB)

–90o

–180o

phase margin too small

p1 p2pa

phase margin ≈ 45o

op-amp frequency responsebefore compensation

op-amp frequency responseafter compensation


56

Op-amp internal compensationTypically, pa is about a few Hz, say 5 Hz. Then, we have to create a pole atsuch a low frequency.

First, consider the input differential stage of an op-amp. One way to add thepole is to put a capacitor between the two collectors of the differential stage.

RL RL

to next stage

C

Equivalent model:

rπ ro//RL 2C

next stage

RIN

pa =1

2(ro ||RL ||RIN)C= 2p (5)

The dominant pole is

We can find C from this equation.


57

Op-amp internal compensationIf we use the previous method of inserting a C between collectors of thedifferential stage, the size of C required is very large, as can be found from

pa =1

2(ro ||RL ||RIN)C= 2p (5)

Using this method, C can be as large as hundredsof pF, which is too large to be implemented onchip. NOT practical!

Better solution: UseMiller effect.

Miller effect can expandcapacitor size by a factorof the gain magnitude.So, we may put thecapacitor across the inputand output of the maingain stage in order to useMiller effect. In this way,C can be much smaller,say a few pF.

RL RL

Cto active load

main gain stageCE stage

output stage


58

Example: Op-amp 741 internal compensation

Q4


output

differentialinput stage

+Vcc

+Vcc

–VEE

–VEE

Q3

Q2Q1

Q5 Q6

Q16

Q17

Q13B Q13A

Q23

Q20

Q14

Data:

DC gain = 70 dBPoles:

30 kHz500 kHz10 MHz

+ –


59

Example: Op-amp 741 internal compensationUnity-gain feedback (worst case stability problem): T = A

Bad stabilitybecause of thesubstantial phaseshift!

p1 = 30 kHz


60


Compensation trick (based on lag compensation approach):

• Introduce a low-frequency pole at pa such that p1 is at crossover.• This ensures the phase angle at crossover = –135º. Hence, PM = 45º.

|A| (dB)

–90o

–180o

p1pa

phase margin ≈ 45o

op-amp frequency responseafter compensation


61

Example: Op-amp 741 internal compensationGraphical construction method

Bad stabilitybecause of thesubstantial phaseshift!

p1 = 30 kHz

pa


62


Exact calculation of pa :

70 dB

pa p1 = 30 kHz

slope = –20 dB/dec =0 - 70

log p1 - log pa=

-70

4.477 - log pa

pa = 9.5 HzHence,


63


After compensation, thephase margin is 45º.


64


Question: How to create the 9.5 Hz pole with a reasonably small C ?Solution: Take advantage of Miller effect to boost capacitance.

Q4


output


+Vcc

+Vcc

–VEE

–VEE

Q3

Q2Q1

Q5 Q6

Q16

Q17

Q13B Q13A

Q23

Q20

Q14

+ –

Cc


65


Q4


output


+Vcc

+Vcc

–VEE

–VEE

Q3

Q2Q1

Q5 Q6

Q16

Q17

Q13B Q13A

Q23

Q20

Q14

+ –Cc

Gm = 6 mA/V

Gain of CE stage:ACE = Gm[Ro17||Ro13||Ri23]

Miller-effect capacitorCM = Cc (ACE + 1)

= 518.74 Cc

Given:Ro17 = 5 MΩRo13 = 720 kΩRi23 = 100 kΩ


66


Q4



+Vcc

+Vcc

–VEE

–VEE

Q3

Q2Q1

Q5 Q6

Q16

Q17

Q13B Q13A+ –

Cc

Given:Ri16 = 2.9 MΩRo4 = 10 MΩRo6 = 20 MΩ

CM

Ro4||Ro6

Ri16

Equivalent ckt:

pa = 1 / 2π CM [Ro4||Ro6|| Ri16]and CM = 518.74 Cc

Hence, Cc = 15 pF


67

Oscillation

In designing feedback amplifiers, we want to make sure that oscillation doesnot occur, that is, we want stable operation.

However, oscillation is needed to make an oscillator. As shown before, thecriteria for oscillation in a feedback amplifier are

1. Loop gain magnitude | T | = 12. Roundtrip phase shift fT = ±180o

Thus, the same feedback structure can be used to make an oscillator. In otherwords, we construct a feedback amplifier, but try to make it satisfy the abovetwo criteria.

In practice, T is a function of frequency, and the above criteria aresatisfied for one particular frequency. This frequency is the oscillationfrequency.


68

Oscillator principle

As T = Af, we can deliberately create phase shift in A or f.

f(jw)

+

–

si soA(jw)

basic amplifier

feedback network

input output

NOTE: Since this model is anegative feedback, we need the totalphase shift of A(jw) and f(jw) to be180o at the frequency of oscillation.If a positive feedback is used, weneed the total phase shift to be 360o.

–180o

|A(jw) f(jw)| (dB)

wo w

w


69

Sustained oscillation

There are two problems! How does oscillation start? And how can oscillationbe maintained?

First, there is noise everywhere! So, signals of all frequencies exist and goaround the loop. Most of them get reduced and do not show up as oscillation.But the one at the oscillation frequency starts to oscillation as it satisfies theBarkhausen criteria.

If | T | is slightly bigger than 1, oscillation amplitude will grow and go toinfinity. But if | T | is slightly less than 1, oscillation subsides. The question ishow to maintain oscillation with a constant magnitude.

We need a control that changes | T | continuously. Typically, this is done by anonlinear amplitude stabilizing circuit, for example, an amplifier whose gaindrops when its output increases, and rises when its output decreases.


70

The Wien bridge oscillator

C RC R

–

+

R1R2

Zp

Zs

A

Zp

Zs

+

Model:

A =1+R2

R1

f =-Zp

Zs + Zp

Basic amplifier gain

Feedback gain where Zp =R

1+ jwCR and Zs =

1+ jwCR

jwC


71


Note that we define the standard feedback structure with negative feedback. So,the loop gain is

Applying the oscillation criteria, we can find the oscillation frequency and theresistor values as follows:

We can choose R2/R1 to be slightly larger than 2, say 2.03, to start oscillation.

T( jw) =

- 1+R2

R1

Ê

ËÁ

ˆ

¯˜

3+ j wCR -1

wCR

Ê

ËÁ

ˆ

¯˜

T( jw) =1 fi 1+R2

R1

= 3 fi R2 = 2R1

fT = ±180o fi woCR =1

woCRfi wo =

1CR

Oscillation frequency


72


Suppose the amplifier has a fixed gain of A. The feedback network, however,has a bandpass frequency response.

Frequency response viewpoint

1/3

| f |

fffreq

freq

fo

A+

+90

–90

Clearly, the roundtrip gain will be 1for f = fo if the basic amplifier has again of 3.

The world is noisy. Signals of allfrequencies exist everywhere!

But signals at all frequencies except fowill be reduced after a round trip.Only signals at fo will have aroundtrip gain of 1.

Hence, the oscillation frequency is fo.From the filter structure, we can findthat fo is equal to 1/2πCR.

WHY OSCILLATE?


73


If we choose R2/R1 = 2.03, then amplitude maygrow. We have to stabilize the amplitude. Thefollowing is an amplitude limiter circuit.

Diode D1 (D2) conducts when vo reaches itspositive (negative) peak.

Amplitude control

16n 10k

16n 10k

–

+

10k 20.3k

D2

D1

3k

1k

1k

3k

+15V

–15V

vo

Just when D1 conducts, we have vA = vB.

–15V

A

B

vo10k 20.3k

1k

3k

-15 +3

4vo +15( ) =

1

3vo fi vo = 9 V. Similar procedure applies for the negative peak.

So, the amplitude is 9 V.

A

B


74

The phase shift oscillator

This circuit matches exactly our negativefeedback model. The basic amplifier gain is R2/R1,and the feedback network is frequency dependent.

–

+

R1R2

R R

C C C

R'

R1 ||R'= R

For the feedback network, we want to findthe frequency at which the phase shift isexactly 180o. At this frequency, if theroundtrip gain is 1, oscillation occurs. Notethat the negative feedback already gives 180o

phase shift.

1

29

| f |

ff

180o

freq

fo

freqf


75

The phase shift oscillator

From the filter characteristic of the feedback network,we know that the phase shift is 180o at fo, where its gainis 1/29.

So, oscillation starts at fo if A ≥ 29. This means we needto have R2 ≥ 29R1.

We can prove that1

29

| f |

ff

180o

freq

fo

freq

fo =1

2p 6CR

Note that the leftmost resistor in the feedback filter is R’ (not R). But R’//R1 isexactly R. This will adjust the loading effect of the basic amplifier and make theoverall filter circuit easier to analyze since it is then simply composed of threeidentical RC sections.

R R

C C C

R


76

Resonant circuit oscillators

A general class of oscillators can be constructed by a pure reactive π-feedback network.

A

jX1 jX2

jX3

For a voltage amplifier implementation, this structurecan be modelled as a series-shunt feedback circuit:

+–Avi

Ri

Ro+vi–

jX3

jX1 jX2

pure reactive π-feedback network

–vi+


77

Resonant circuit oscillators

Analysis:

The loop gain is

For oscillation to start, we need T = –1.

Thus, the oscillation criteria become

In practice, we may have

(a) X1 and X2 are capacitors and X3 is inductor.

OR

(b) X1 and X2 are inductors and X3 is capacitor.

+–Avi

Ri

Ro+vi–

jX3

jX1 jX2

–vi+

T( jw) =-A jX1( ) jX2( )

j X1 + X2 + X3( )Ro + jX2 jX1 + jX3( )

=AX1X2

j X1 + X2 + X3( )Ro - X2 X1 + X3( )

X1 + X2 + X3 = 0

AX1

X1 + X3

=1


78

Colpitts oscillator

When X1 and X2 are capacitors and X3 is inductor, we have the Colpitts oscillator.

In this case, we have

From

the oscillation frequency can be found:

+–Avi

Ri

Ro+vi–

–vi+

L3

C2C1

jX1 =1

jwC1

and jX2 =1

jwC2

jX3 = jwL3

wo =1

L3

C1C2

C1 + C2

Ê

ËÁ

ˆ

¯˜

X1 + X2 + X3 = 0


79

A practical form of Colpitts oscillator

The basic amplifier can be realized by a common-emitter amplifier.

The loop gain is

where

Putting s = jw, and applying the Barkhausen criterion:

wo =1

L3

C1C2

C1 + C2

Ê

ËÁ

ˆ

¯˜

T(s) =GmZT1/sC1

sL +1/sC1

1

ZT= sC2 +

1

sL +1sC1

+1

ro ||Rc

GmC2(ro ||Rc )

C1

>1 fiC1

C2

>GmRcroRc + ro

for oscillation to start.

ZT


80

Hartley oscillator

When X1 and X2 are inductors and X3 is capacitor, we have the Hartley oscillator.

In this case, we have

From

the oscillation frequency can be found:

For both the Colpitts and Hartley oscillators, thegain of the amplifier has to be large enough toensure that the loop gain magnitude is larger than1.

+–Avi

Ri

Ro+vi–

–vi+

C3

L1

jX1 = jwL1 and jX2 = jwL2

jX3 =1

jwC3

wo =1

C3 L1 + L2( )

X1 + X2 + X3 = 0

L2

Documents

Feedback