FEEDBACK AMPLIFIERS AND OSCILLATORS
forElectronic Circuits
by
Prof. Michael Tse
September 2004
C.K. Tse: Feedback amplifiers andoscillators
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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.
C.K. Tse: Feedback amplifiers andoscillators
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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!!
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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
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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
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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
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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
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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!
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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.
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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.
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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
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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
+
–
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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
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Series-shunt feedback (for voltage amplifier)
ve+
–
+–
+–
+
–
vo+
–
vi Ave
Ro
Ri
fvo
Overall gain (closed-loop gain) : Ao =vovi=
A
1+ Af
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Series-shunt feedback (for voltage amplifier)
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.
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Series-shunt feedback (for voltage amplifier)
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
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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
+
–
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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
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Shunt-shunt feedback (for transresistance amplifier)
Overall gain (closed-loop gain) : Ao =voii=
A
1+ Af
ie
+–
+
–
vo
ii
Aie
Ro
Ri
fvo
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Shunt-shunt feedback (for transresistance amplifier)
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
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Shunt-shunt feedback (for transresistance amplifier)
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
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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
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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!
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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!
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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
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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:
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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).
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Systematic analysis of shunt-shunt feedback usingy-parameter
+
–
voiiy21vi
y22y11
y12fvo
+
–vi
y11f y22f
1 2
yi
One step closer…
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Systematic analysis of shunt-shunt feedback usingy-parameter
+
–
voiiy21vi
y22y11
y12fvo
+
–vi
y11f y22f
1 2
yi
One more step closer…
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Systematic analysis of shunt-shunt feedback usingy-parameter
+
–
voiiy21vi
y11+y11f+yi
y12fvo
+
–vi
y22f+y22
1 2
Yet another step closer…
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Systematic analysis of shunt-shunt feedback usingy-parameter
+
–
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
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Systematic analysis of shunt-shunt feedback usingy-parameter
+
–
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
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Systematic analysis of shunt-shunt feedback usingy-parameter
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 )
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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?
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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.
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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.
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Example
RL+vo–
–
+a
Rf
is
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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.
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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!
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ExampleCasting it to standard form
+vo–
is+–avi
–vi+
Ro
y12fvo
Rf||RL
y11f =1
Rf
y12f =-1
Rf
y22f =1
Rf
Ri ||Rf
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ExampleCasting it to standard form
+vo–
is+–
–vi+
Ri ||Rf
Ro ||Rf ||RL
a(Rf ||RL )
Ro + Rf ||RL
vi
-voRf
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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+
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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
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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
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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
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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
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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
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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.
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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
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Phase margin
Phase margin is an important parameter to evaluate how stable thesystem is.
w
w
wo
|T| (dB)
f
fT
–180o
crossover frequency(where the gain is 1)
phase margin fPM (the larger the better)
0dB
Phase margin fPM = –180o – fT
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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
crossover frequency(where the gain is 1)
phase margin fPM (how to increase it?)
0dBp1 p2
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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
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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
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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
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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
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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.
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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
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Example: Op-amp 741 internal compensation
Q4
main gain stageCE stage
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
+ –
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Example: Op-amp 741 internal compensationUnity-gain feedback (worst case stability problem): T = A
Bad stabilitybecause of thesubstantial phaseshift!
p1 = 30 kHz
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Example: Op-amp 741 internal compensation
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
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Example: Op-amp 741 internal compensationGraphical construction method
Bad stabilitybecause of thesubstantial phaseshift!
p1 = 30 kHz
pa
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Example: Op-amp 741 internal compensation
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,
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Example: Op-amp 741 internal compensation
After compensation, thephase margin is 45º.
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Example: Op-amp 741 internal compensation
Question: How to create the 9.5 Hz pole with a reasonably small C ?Solution: Take advantage of Miller effect to boost capacitance.
Q4
main gain stageCE stage
output
differentialinput stage
+Vcc
+Vcc
–VEE
–VEE
Q3
Q2Q1
Q5 Q6
Q16
Q17
Q13B Q13A
Q23
Q20
Q14
+ –
Cc
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Example: Op-amp 741 internal compensation
Q4
main gain stageCE stage
output
differentialinput stage
+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
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Example: Op-amp 741 internal compensation
Q4
main gain stageCE stage
differentialinput stage
+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
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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.
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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
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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.
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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
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The Wien bridge oscillator
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
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The Wien bridge oscillator
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?
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The Wien bridge oscillator
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
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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
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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
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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+
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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
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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
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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
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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