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```									FEEDBACK AMPLIFIERS AND OSCILLATORS

for
Electronic Circuits

by
Prof. Michael Tse

September 2004
Contents
Feedback
Basic feedback conﬁguration
The price to pay
Feedback Ampliﬁer Conﬁgurations
Series-shunt, shunt-series, series-series, shunt-shunt
Input and output impedances
Compensation
Op-amp internal compensation
Oscillation
Oscillation criteria
Sustained oscillation
Wein bridge, phase shift, Colpitts, Hartley, etc.

C.K. Tse: Feedback ampliﬁers and           2
oscillators
Basic feedback conﬁguration
The basic feedback ampliﬁer consists of a basic ampliﬁer and a feedback network.

+           e
si                               A                       so
input       –                                           output
basic ampliﬁer

Careful!!             sf
f

feedback network

A = basic ampliﬁer gain
f = feedback gain

C.K. Tse: Feedback ampliﬁers and                    3
oscillators
Characteristics
The input is subtracted by a
+                                                     feedback signal which is part of the
si                     e       A                       so       output, before it is ampliﬁed by the
input          –                                     output      basic ampliﬁer.
basic ampliﬁer
sf                                                  so = Ae = A(si - s f )
f
But, since sf = f so, we get
feedback network
so = A(si - fso )

Hence, the overall gain is
so   A
si                         Ao                   so              Ao =      =
si 1+ Af

1
If Af >> 1,     Ao ª
f
C.K. Tse: Feedback ampliﬁers and                          4
oscillators
Simple viewpoint

+            e
si                          A                   so
input        –                                  output
basic ampliﬁer
sf
f

feedback network

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.
so 1
Hence, f so = si or          ª
si   f

C.K. Tse: Feedback ampliﬁers and             5
oscillators
If the feedback network is constructed from passive elements having
stable characteristics, the overall gain becomes very steady and
unaffected by variation of the basic ampliﬁer gain.

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

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

dAo Ê dA ˆÊ 1 ˆ
= Á ˜Á       ˜
Ao Ë A ¯Ë1+ Af ¯

Obviously, if Af is large, then dAo/Ao will be reduced drastically.
1
Feedback reduces gain sensitivity! In fact, the gain is just Ao ª   .
f

C.K. Tse: Feedback ampliﬁers and               6
oscillators
so
Suppose the basic ampliﬁer is
distortive. So, the output does
output
A
not give a sine wave for a sine
e
wave input.

But, with feedback, we see that
the gain is about 1/f anyway,                                         input
regardless of what A is (or as
long as Af is large enough).
so
This gives a very good property
1/f
of feedback ampliﬁer in terms
si
of eliminating distortion.

C.K. Tse: Feedback ampliﬁers and                      7
oscillators
•    Improve input and output resistances (to be discussed later).
•    Widening of bandwidth of ampliﬁer (to be discussed later).
•    Enhance noise rejection capability.

ni
ni
si              A             so           +                          +
si              e
A’                     A            so
input                              +
–                                           output
basic ampliﬁer
Signal-to-noise ratio:                        sf
È so ˘ È si ˘                                                     f
Í ˙=Í ˙                                               feedback network
Î no ˚ Î ni ˚                                                             È so ˘  È si ˘
Signal-to-noise ratio improves!           Í ˙ = A¢Í ˙
Î no ˚  Î ni ˚

C.K. Tse: Feedback ampliﬁers and                              8
oscillators
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’t
want it.

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

C.K. Tse: Feedback ampliﬁers and                 9
oscillators
Terminologies

Basic ampliﬁer gain = A

Feedback gain = f
A     1
Overall gain (closed-loop gain) =          ª
1+ Af   f

Loop gain (roundtrip gain) = Af                Some books use T to denote Af.

C.K. Tse: Feedback ampliﬁers and                         10
oscillators
Feedback ampliﬁers

What is an ampliﬁer?              si         A             so

Signals can be voltage or current.

General model for voltage ampliﬁer:
Ro
+                                        +
+
vin             Rin   –   Avin           vo
–                                        –

voltage ampliﬁer

C.K. Tse: Feedback ampliﬁers and             11
oscillators
Models of ampliﬁers
Ro                              iin                         io
+                                        +                              Aiin
+
vin            Rin   –   Avin            vo                           Rin        Ro
–                                        –

voltage ampliﬁer                                      current ampliﬁer

iin                 Ro                                                          io
+          +                   Aiin
+                              vin               Rin        Ro
Rin   –   Aiin            vo
–          –

transresistance ampliﬁer                                  transconductance ampliﬁer

C.K. Tse: Feedback ampliﬁers and                             12
oscillators
Feedback ampliﬁer conﬁgurations
Voltage ampliﬁer
+            e
si                           A                       so
input        –                                      output
basic ampliﬁer
sf
f
voltage                                                                  voltage
feedback network

To subtract voltage from
To copy voltage, we should use
voltage, we should use series
parallel (shunt) connection
connection
+                                               •         +
vi
–
A                        vo
–
•
– vf +                                      Hence, series-shunt feedback
C.K. Tse: Feedback ampliﬁers and                         13
oscillators
Series-shunt feedback (for voltage ampliﬁer)

Ro
+
+          +                        +                  vo
vi         ve            Ri         –   Ave
–                                           –
–

+
–        fvo

vo    A
Overall gain (closed-loop gain) :             Ao =      =
v i 1+ Af

C.K. Tse: Feedback ampliﬁers and             14
oscillators
Series-shunt feedback (for voltage ampliﬁer)
To ﬁnd the input resistance, we consider the ratio of vi and ii, with output opened.

ii                                Ro
+
+            +                      +               vo
vi           ve          Ri         –   Ave
–                                      –
–
vi    vi               RIN
RIN =      =
ii v e /Ri
v e + fv o                              +
= Ri                                          –        fvo
ve
= Ri (1+ Af )

The input resistance has been enlarged by (1+Af). This is a desirable
effect to the previous stage.
C.K. Tse: Feedback ampliﬁers and                       15
oscillators
Series-shunt feedback (for voltage ampliﬁer)
To ﬁnd the output resistance, we consider shorting the input source and calculate
the ratio of vo and io.
Ro               io          First, we have ve = – fvo.
+                                                             +     Also,
+
ve        Ri
+                            vo            v o - Av e v o + Afv o
vi                               –   Ave                             io =             =
–
–                                                 –                 Ro          Ro
Hence,
ROUT
vo   Ro
ROUT =        =
+
fvo
io 1+ Af
–

The output resistance has been reduced by (1+Af). This is a desirable
feature for voltage ampliﬁer as a small output resistance emulates a better
C.K. Tse: Feedback ampliﬁers and                              16
oscillators
Series-shunt feedback (for voltage ampliﬁer)
Summary of features                                       Equivalent model

A     1                                                      Ro
Closed-loop gain =       ª                                                       1+ Af
1+ Af   f
+                            +         Av i     +
Input resistance = Ri ( 1 + Af )              vi           Ri ( 1 + Af )    –
1+ Af
vo
–                                               –
Ro
Output resistance =
1+ Af
feedback network, i.e., we assume that the feedback
network is an ideal ampliﬁer which feeds a scaled-down
copy of the output to the input.

+
–           ∞

feedback network

C.K. Tse: Feedback ampliﬁers and                                17
oscillators
Feedback ampliﬁer conﬁgurations
Transresistance ampliﬁer
+            e
si                          A                       so
input       –                                      output
basic ampliﬁer
sf
f
current                                                                 voltage
feedback network

To subtract current from
To copy voltage, we should use
current, we should use shunt
parallel (shunt) connection
(connection) connection
ii                                             •         +
A                        vo
–
•

Hence, shunt-shunt feedback
C.K. Tse: Feedback ampliﬁers and                        18
oscillators
Shunt-shunt feedback (for transresistance ampliﬁer)

ii          ie                         Ro
+
+                 vo
Ri         –   Aie
–

fvo

vo   A
Overall gain (closed-loop gain) :         Ao =      =
ii 1+ Af

C.K. Tse: Feedback ampliﬁers and          19
oscillators
Shunt-shunt feedback (for transresistance ampliﬁer)
To ﬁnd the input resistance, we consider the ratio of vi and ii, with output opened.

ii         ie                       Ro
+
+
+                  vo
vi                         Ri         –   Aie
–
–
v i Riie
RIN =      =                  RIN
ii   ii
ie
= Ri
ie + fv o                                           fvo

Ri
=
1+ Af
The input resistance has been reduced by (1+Af). This is a desirable
feature for transresistance ampliﬁer as a small input resistance ensures
better current sensing from the previous stage.
C.K. Tse: Feedback ampliﬁers and                       20
oscillators
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.
First, we have ie = – fvo.
ii = 0     ie                        Ro             io
+    Also,
+                                             v o - Aie v o + Afv o
Ri          –   Aie                          vo    io =            =
–                Ro         Ro
Hence,
ROUT
vo   Ro
ROUT =        =
io 1+ Af
fvo

The output resistance has been reduced by (1+Af). This is a desirable
feature for transresistance ampliﬁer as a large small resistance emulates a
better voltage source for the load.
C.K. Tse: Feedback ampliﬁers and                                    21
oscillators
Shunt-shunt feedback (for transresistance ampliﬁer)
Summary of features                                      Equivalent model

A     1                                            Ro
Closed-loop gain =       ª                                             1+ Af
1+ Af   f                    ii
Aii    +
Ri                                     Ri     +
Input resistance =                                                 –
1+ Af
vo
1+ Af                                  1+ Af                –
Ro
Output resistance =
1+ Af

Similar, we can develop the feedback conﬁgurations for
transconductance ampliﬁer and current ampliﬁer.
Transconductance ampliﬁer: series-series feedback
Current ampliﬁer: shunt-series feedback

C.K. Tse: Feedback ampliﬁers and                 22
oscillators
Series-series feedback (for transconductance ampliﬁer)
io
+               +                       Ro
vo              ve         Ri
–                      Ave
–

io

+
–        f io

io      A
Overall gain (closed-loop gain) :        Ao =  =
v i 1+ Af
Input resistance:                    RIN = Ri (1+ Af )           Desirable!

Output resistance:                  ROUT = Ro (1+ Af )           Desirable!

C.K. Tse: Feedback ampliﬁers and                      23
oscillators
Shunt-series feedback (for current ampliﬁer)
ii          ie                                io
Ro
Ri
Aie

io

f io

Overall gain (closed-loop gain) :              io   A
Ao =       =
ii 1+ Af
Input resistance:                                Ri
RIN =
1+ Af       Desirable!
Output resistance:                              Ro
ROUT     =             Desirable!
1+ Af
C.K. Tse: Feedback ampliﬁers and                24
oscillators
In practice, the input source has resistance and the feedback network has
resistance.

Example: shunt-shunt feedback
ie                        Ro
+
ii                                        +               vo
Ri      –   Aie
–

fvo

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

C.K. Tse: Feedback ampliﬁers and                25
oscillators
Systematic analysis using 2-port networks
The best way to analyze feedback circuits with loading effects is to use two-port
models.
For shunt-shunt feedback, input and output sides are both parallel connected.
Thus, the loading can be combined by summing the conductances. Also, voltage
is common at both sides. So, y-parameter is best.

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

+
+
ii          yi         vi           y11                y22       vo
–                  y21vi                  –

1                               2

y22f
y11f       y21fvo

C.K. Tse: Feedback ampliﬁers and                      26
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter

+
+
ii       yi     vi          y11               y22       vo
–                 y21vi                 –

1                              2

y22f
y11f      y12fvo

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

C.K. Tse: Feedback ampliﬁers and             27
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter
y11f                                   y22f
+
+
ii      yi      vi              y11               y22              vo
–                     y21vi                        –

1                           2

y12fvo

One step closer…

C.K. Tse: Feedback ampliﬁers and                        28
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter
y11f                                   y22f
+
+
ii    vi       yi         y11               y22              vo
–                         y21vi                        –

1                           2

y12fvo

One more step closer…

C.K. Tse: Feedback ampliﬁers and                    29
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter

y11+y11f+yi    y22f +y22
+
+
ii     vi                                              vo
–                       y21vi                   –

1                              2

y12fvo

Yet another step closer…

C.K. Tse: Feedback ampliﬁers and             30
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter
in conductance (S)             in resistance (Ω)

y11+y11f+yi 1/(y22f +y22)
+
+
ii                                         +                       vo
vi                                  –
–                      -y 21v i                             –
y 22f + y 22
1                               2
Use Thevenin

y12fvo

Yet another step closer…

C.K. Tse: Feedback ampliﬁers and                            31
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter
in conductance (S)          in resistance (Ω)

ie       y11+y11f+yi 1/(y22f +y22)
+
+
ii                                     +                              vo
vi                              –
–                                                              –

1                               2                           -y 21ie
( y 22f   + y 22 )( y11 + y11f + y i )

y12fvo

Finally, we get the same standard form.

C.K. Tse: Feedback ampliﬁers and                                          32
oscillators
Systematic analysis of shunt-shunt feedback using
y-parameter

We can simply apply the standard results:
-y 21                 -y 21
Basic ampliﬁer gain            A=                                           =
( y 22f   + y 22 )( y11 + y11f + y i ) y oT y iT
Feedback gain                  f = y21f

A    1  1
Overall (closed-loop) gain     Ao =         ª =
1+ Af  f y12f

1
Input resistance              RIN =
(y11 + y11f      + y i )(1+ Af )

1
Output resistance            ROUT =
(y 22f + y 22 )(1+ Af )
C.K. Tse: Feedback ampliﬁers and                                 33
oscillators
Appropriate 2-port networks for analyzing
feedback circuits

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

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

C.K. Tse: Feedback ampliﬁers and                34
oscillators
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 modiﬁed
basic ampliﬁer.

4.   Apply Thevenin or Norton to cast the model back to the standard

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

C.K. Tse: Feedback ampliﬁers and               35
oscillators
Example
Rf

–
a
is                 +                 +
vo     RL
–

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

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

C.K. Tse: Feedback ampliﬁers and                  36
oscillators
Example

–
a
is             +                 +
vo     RL
–

Rf

C.K. Tse: Feedback ampliﬁers and        37
oscillators
Example
Converting to y-parameter

–                  Ro            +
+
is        vi    Ri    –                    vo           RL
+               avi              –

Note: this
goes to the              Rf
–ve input
of A.                                                                           y11f        y22f
y12fvo

1             -1             1
y11f =        y12f =        y 22f =
Rf            Rf             Rf

C.K. Tse: Feedback ampliﬁers and                                 38
oscillators
Example
Converting to y-parameter

–                                  Ro                  +
+
is        vi                 Ri        –                         vo       RL
+                                avi                   –

y11f         y22f
y12fvo

1                  -1               1          REMEMBER:
y11f =             y12f =          y 22f =
Rf                 Rf               Rf         y11f and y22f are conductance!

C.K. Tse: Feedback ampliﬁers and                            39
oscillators
Example
Casting it to standard form

–                              Ro                  +
+
is       vi       Ri || R f       –
Rf||RL     vo
+                            avi                   –

y12fvo

1               -1              1
y11f =          y12f =         y 22f =
Rf              Rf              Rf

C.K. Tse: Feedback ampliﬁers and   40
oscillators
Example
Casting it to standard form

–                          Ro || R f || RL     +
+
is       vi     Ri || R f     –     a(R f || RL )       vo
+                                        vi    –
Ro + R f || RL

-v o
Rf

C.K. Tse: Feedback ampliﬁers and   41
oscillators
Example
Finally, we get the standard form
ie

–                        Ro || R f || RL   +
+
is        vi    Ri || R f     –                      vo
+                        Aie               –

Using Thévenin theorem,
R f || RL
Aie = av i
-v o
(R   f   || RL ) + Ro

Rf
A = -a
(R || R )(R || R )
i         f       f       L

(R || R ) + R
f         L       o

Ri R 2 RL
f                 1
A = -a
(Ri + R f ) ( R f RL + Ro R f + Ro RL )

C.K. Tse: Feedback ampliﬁers and                                        42
oscillators
Example
Apply standard results:
Ri R 2 RL
f                1
Basic ampliﬁer gain (transresistance)          A = -a
(Ri + R f ) ( R f RL + Ro R f + Ro RL )

Feedback gain:                                        -1
f =
Rf

A    1
Overall (closed-loop) gain                     Ao =         ª = -R f                 if Af >> 1
1+ Af  f

Ri || R f
Input resistance                             RIN =
1+ Af
Ro || R f || RL
Output resistance                          ROUT =
1+ Af

C.K. Tse: Feedback ampliﬁers and                                   43
oscillators
Frequency response
Gain and bandwidth                                             Suppose the basic ampliﬁer
has a pole at p1, i.e.,
+            e                                                              ALF
si                         A(jw)                    so                A( jw ) =
jw
input       –                                     output                           1+
basic ampliﬁer                                            p1

sf                                          20log10|A| (dB)
f

feedback network                   ALF
slope = –20dB/dec

w
p1

C.K. Tse: Feedback ampliﬁers and                              44
oscillators
Frequency response
Gain and bandwidth                              Hence, we see that the overall gain has a
pole at
The overall (closed-loop) gain is                          pc = p1(1 + fALF)
and the low-frequency gain is lowered
A( jw )                        to                ALF
Ao ( jw ) =                                         Ao,LF =
1+ A( jw ) f
1+ fALF
ALF
=
Ê   jw ˆ
Á1+    ˜ + fALF
Ë   p1 ¯                                20log10|A| (dB)
È               ˘
ALF Í        1        ˙
ALF               basic ampliﬁer
=         Í               ˙
1+ fALF Í1+     jw      ˙                                        feedback ampliﬁer
Í               ˙
Î p1 (1+ fALF ) ˚             Ao,LF

w
p1       pc

C.K. Tse: Feedback ampliﬁers and                              45
oscillators
Stability of feedback ampliﬁer
Deﬁnition: A feedback system is said to be stable if it does not
oscillate by itself at any frequency under a given circuit condition.

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

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

feedback circuit
sustained oscillation at certain frequency

C.K. Tse: Feedback ampliﬁers and                             46
oscillators
Why and how does it oscillate?
The feedback system oscillates because of the simple fact that it has a
closed loop in which signals can combine constructively.

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

+
si                     A                            so
input       –                                       output

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

C.K. Tse: Feedback ampliﬁers and                  47
oscillators
Why and how does it oscillate?
Clearly, if SB’ and SB are same in magnitude and have a 360o phase
difference, then the closed loop will oscillate by itself.

Oscillation criteria:

1.           Af = 1

2.           Af = ±180o           This is known as the Barkhausen criteria.

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

C.K. Tse: Feedback ampliﬁers and                    48
oscillators
The loop gain T
An important parameter to test stability is the loop gain, usually
denoted by T.
T = Af

|T| (dB)

crossover frequency
(where the gain is 1)
wo
0dB                                    w
f
w

fT

If fT = –180o, OSCILLATES!

C.K. Tse: Feedback ampliﬁers and                 49
oscillators
Phase margin
Phase margin is an important parameter to evaluate how stable the
system is.
Phase margin fPM = –180o – fT

|T| (dB)

crossover frequency
(where the gain is 1)
wo
0dB                                  w
f
w

fT
–180o
phase margin fPM (the larger the better)

C.K. Tse: Feedback ampliﬁers and             50
oscillators
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.

|T| (dB)

crossover frequency
(where the gain is 1)
wo
0dB                                                         w
p1        p2
f
w

fT
–180o
phase margin fPM (how to increase it?)

C.K. Tse: Feedback ampliﬁers and                   51
oscillators
Method 1: Lag compensation
Add a pole at a low frequency point. The aim is to make the crossover point
appear at a much lower frequency. The drawback is the reduced bandwidth.

Compensation                      |T| (dB)                  crossover frequency
after compensation
function Gc is
crossover frequency
1
Gc ( jw ) =                                                        before compensation
jw
1+
pa            0dB                                                          w
pa       p1   p2
f
w
before compensation
after compensation

–180o
phase margin fPM
phase margin fPM             before compensation
after compensation
C.K. Tse: Feedback ampliﬁers and                           52
oscillators
Add a zero near the ﬁrst pole. The aim is to reduce the phase shift and hence
increase the phase margin and keep a wide bandwidth. But the drawback is the
more difﬁcult design.
Compensation               |T| (dB)
function Gc is                                        crossover frequency   crossover frequency
jw                                     before compensation
1+                                                              after compensation
za
Gc ( jw ) =
jw    0dB               za
1+                              p1   p2
w
pa
f
w
before compensation
after compensation

–180o
phase margin fPM
phase margin fPM after
before compensation
compensation

C.K. Tse: Feedback ampliﬁers and                          53
oscillators
Op-amp stability problem
The op-amp has a high DC gain, and hence at crossover it is likely that the
phase shift is signiﬁcant. The worst-case scenario is when the feedback gain is
1 (maximum for passive feedback). We call this unity feedback condition, and
use 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)

op-amp frequency response

p1          p2

–90o                                             phase margin too small
–180o

C.K. Tse: Feedback ampliﬁers and                      54
oscillators
Op-amp internal compensation
Usually, op-amps are internally compensated. The technique is by lag
compensation, i.e., adding a pole at low frequency such that the phase margin
can reach at least 45o.

Suppose we add a low-frequency dominant pole at pa. If we can put pa such
that p1 (original dominant pole) is at crossover, then the phase margin is about
45o.               |A| (dB)
op-amp frequency response
before compensation

op-amp frequency response
after compensation

pa                 p1          p2

–90o                                                  phase margin too small
–180o
phase margin ≈ 45o
C.K. Tse: Feedback ampliﬁers and                           55
oscillators
Op-amp internal compensation
Typically, pa is about a few Hz, say 5 Hz. Then, we have to create a pole at
such a low frequency.

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

Equivalent model:
RL                RL                                                                    next stage
to next stage
rπ                ro//RL        2C           RIN
C

The dominant pole is
1
pa =                      = 2p (5)
2(ro || RL || RIN )C

We can ﬁnd C from this equation.

C.K. Tse: Feedback ampliﬁers and                                  56
oscillators
Op-amp internal compensation
If we use the previous method of inserting a C between collectors of the
differential stage, the size of C required is very large, as can be found from
1                             Using this method, C can be as large as hundreds
pa =                          = 2p (5)        of pF, which is too large to be implemented on
2(ro || RL || RIN )C
chip. NOT practical!

Better solution: Use                                                           to active load
Miller effect.                                                             C
RL                 RL
output stage
Miller effect can expand
capacitor size by a factor
of the gain magnitude.
So, we may put the
capacitor across the input
and output of the main
main gain stage
gain stage in order to use
CE stage
Miller effect. In this way,
C can be much smaller,
say a few pF.
C.K. Tse: Feedback ampliﬁers and                                57
oscillators
Example: Op-amp 741 internal compensation
+Vcc
Data:
Q1 differential Q2                   Q13B         Q13A
input stage                                                                   DC gain = 70 dB
+                             –
Poles:
Q14               30 kHz
Q3                 Q4                                                                500 kHz
+Vcc                                     output
10 MHz
Q20

–VEE                        Q16                  Q23
Q17
Q5                 Q6
main gain stage
CE stage

–VEE

C.K. Tse: Feedback ampliﬁers and                           58
oscillators
Example: Op-amp 741 internal compensation
Unity-gain feedback (worst case stability problem): T = A

p1 = 30 kHz

because of the
substantial phase
shift!

C.K. Tse: Feedback ampliﬁers and                 59
oscillators
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)

op-amp frequency response
after compensation

pa                p1

–90o
–180o
phase margin ≈ 45o

C.K. Tse: Feedback ampliﬁers and                           60
oscillators
Example: Op-amp 741 internal compensation
Graphical construction method

p1 = 30 kHz

pa

because of the
substantial phase
shift!

C.K. Tse: Feedback ampliﬁers and                 61
oscillators
Example: Op-amp 741 internal compensation

Exact calculation of pa :

0 - 70          -70
slope = –20 dB/dec      =                  =
70 dB                                      log p1 - log pa 4.477 - log pa

Hence, pa = 9.5 Hz
pa                      p1 = 30 kHz

C.K. Tse: Feedback ampliﬁers and                62
oscillators
Example: Op-amp 741 internal compensation

After compensation, the
phase margin is 45º.

C.K. Tse: Feedback ampliﬁers and   63
oscillators
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.
+Vcc

Q1 differential Q2                   Q13B         Q13A
+        input stage          –
Cc                                  Q14
Q3                 Q4
+Vcc                                output
Q20

–VEE                        Q16                 Q23
Q17
Q5                 Q6
main gain stage
CE stage

–VEE

C.K. Tse: Feedback ampliﬁers and                          64
oscillators
Example: Op-amp 741 internal compensation
+Vcc
Given:
Q1 differential Q2                                                                 Ro17 = 5 MΩ
Q13B         Q13A
+         input stage          –
Ro13 = 720 kΩ
Cc                                                   Ri23 = 100 kΩ
Q14
Q3                 Q4
+Vcc                                   output
Q20

–VEE                        Q16                   Q23
Q17
Q5                 Q6                                                    Gain of CE stage:
main gain stage
CE stage                                   ACE = Gm[Ro17||Ro13||Ri23]

–VEE                                       Miller-effect capacitor
Gm = 6 mA/V                   CM = Cc (ACE + 1)
= 518.74 Cc
C.K. Tse: Feedback ampliﬁers and                             65
oscillators
Example: Op-amp 741 internal compensation
+Vcc
Given:
Q1 differential Q2                                            Ri16 = 2.9 MΩ
Q13B         Q13A
+         input stage          –
Ro4 = 10 MΩ
Cc                              Ro6 = 20 MΩ

Q3                 Q4                                        Equivalent ckt:
+Vcc
Ro4||Ro6

–VEE                        Q16                                    Ri16           CM
Q17
Q5                 Q6
main gain stage
CE stage                        pa = 1 / 2π CM [Ro4||Ro6|| Ri16]
and CM = 518.74 Cc
–VEE
Hence, Cc = 15 pF

C.K. Tse: Feedback ampliﬁers and                            66
oscillators
Oscillation
In designing feedback ampliﬁers, we want to make sure that oscillation does
not occur, that is, we want stable operation.

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

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

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

In practice, T is a function of frequency, and the above criteria are
satisﬁed for one particular frequency. This frequency is the oscillation
frequency.

C.K. Tse: Feedback ampliﬁers and                     67
oscillators
Oscillator principle
As T = Af, we can deliberately create phase shift in A or f.
NOTE: Since this model is a
+                                              negative feedback, we need the total
si                  A(jw)                     so      phase shift of A(jw) and f(jw) to be
input       –                               output     180o at the frequency of oscillation.
basic ampliﬁer                    If a positive feedback is used, we
need the total phase shift to be 360o.
f(jw)
|A(jw) f(jw)| (dB)
feedback network

wo           w

w

–180o

C.K. Tse: Feedback ampliﬁers and                              68
oscillators
Sustained oscillation
There are two problems! How does oscillation start? And how can oscillation
be maintained?

First, there is noise everywhere! So, signals of all frequencies exist and go
around 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 the
Barkhausen criteria.

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

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

C.K. Tse: Feedback ampliﬁers and                          69
oscillators
The Wien bridge oscillator

R2                              Model:
R1
–

+                                                         A
+

Zp
C    R                                                    Zs
C        R
Zp
Zs

R2
Basic ampliﬁer gain       A = 1+
R1
Feedback gain                        -Z p                        R              1+ jwCR
f =               where Z p =           and Z s =
Zs + Z p                  1+ jwCR             jwC

C.K. Tse: Feedback ampliﬁers and                            70
oscillators
The Wien bridge oscillator
Oscillation frequency
Note that we deﬁne the standard feedback structure with negative feedback. So,
the loop gain is                  Ê     ˆR
-Á1+ 2 ˜
Ë R1 ¯
T( jw ) =
Ê        1 ˆ
3 + jÁwCR -     ˜
Ë      wCR ¯

Applying the oscillation criteria, we can ﬁnd the oscillation frequency and the
resistor values as follows:
R
T( jw ) = 1 ﬁ 1+ 2 = 3 ﬁ R2 = 2R1
R1
1                1
fT = ±180 o ﬁ w oCR =             ﬁ wo =
w oCR              CR

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

C.K. Tse: Feedback ampliﬁers and                        71
oscillators
The Wien bridge oscillator
Frequency response viewpoint
Suppose the ampliﬁer has a ﬁxed gain of A. The feedback network, however,
has a bandpass frequency response.

WHY OSCILLATE?
A                                Clearly, the roundtrip gain will be 1
+
for f = fo if the basic ampliﬁer has a
gain of 3.
|f|
The world is noisy. Signals of all
1/3                                          frequencies exist everywhere!

freq
But signals at all frequencies except fo
ff                                     will be reduced after a round trip.
fo
Only signals at fo will have a
+ 90
roundtrip gain of 1.
freq
–90                                          Hence, the oscillation frequency is fo.
From the ﬁlter structure, we can ﬁnd
that fo is equal to 1/2πCR.

C.K. Tse: Feedback ampliﬁers and                                      72
oscillators
The Wien bridge oscillator                                                                  +15V

Amplitude control                                                                              3k
If we choose R2/R1 = 2.03, then amplitude may                                          D2
grow. We have to stabilize the amplitude. The
following is an amplitude limiter circuit.
10k           20.3k                 1k
Diode D1 (D2) conducts when vo reaches its                               –
A                        vo
positive (negative) peak.
+
Just when D1 conducts, we have vA = vB.
1k
10k     A 20.3k
vo                                              16n 10k          B
1k                              16n 10k
B                                                                      3k
3k                                                     D1

–15V
–15V
3             1
-15 +     (v o + 15) = v o    ﬁ     v o = 9 V.     Similar procedure applies for the negative peak.
4             3                            So, the amplitude is 9 V.
C.K. Tse: Feedback ampliﬁers and                                 73
oscillators
The phase shift oscillator
This circuit matches exactly our negative
feedback model. The basic ampliﬁer gain is R2/R1,
and the feedback network is frequency dependent.

For the feedback network, we want to ﬁnd
the frequency at which the phase shift is
R2                    exactly 180o. At this frequency, if the
R1
–                         roundtrip gain is 1, oscillation occurs. Note
that the negative feedback already gives 180o
+                         phase shift.
|f|
1
C       C     C                  29
freq

R'   R        R                                      fo
ff

R1 || R'= R                        f               180o
freq

C.K. Tse: Feedback ampliﬁers and                            74
oscillators
The phase shift oscillator                                                  C        C   C

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

So, oscillation starts at fo if A ≥ 29. This means we need        |f|
to have R2 ≥ 29R1.
1
We can prove that                                                 29
freq
1
fo =                                                            fo
2p 6CR                                        ff

180o
freq

Note that the leftmost resistor in the feedback ﬁlter is R’ (not R). But R’//R1 is
overall ﬁlter circuit easier to analyze since it is then simply composed of three
identical RC sections.

C.K. Tse: Feedback ampliﬁers and                               75
oscillators
Resonant circuit oscillators
A general class of oscillators can be constructed by a pure reactive π-feedback network.

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

+                      Ro
jX1               jX2                                           +
vi          Ri   –
–                    Avi

jX3

pure reactive π-feedback network                                     jX3
–
vi     jX1                jX2
+

C.K. Tse: Feedback ampliﬁers and                                     76
oscillators
Resonant circuit oscillators
Analysis:
-A( jX1 )( jX 2 )
The loop gain is T( jw ) =
j ( X1 + X 2 + X 3 ) Ro + jX 2 ( jX1 + jX 3 )
AX1 X 2
=
j ( X1 + X 2 + X 3 ) Ro - X 2 ( X1 + X 3 )

For oscillation to start, we need T = –1.
+                    Ro
+
Thus, the oscillation criteria become                                 vi        Ri   –
–                  Avi
X1 + X 2 + X 3 = 0
AX1
=1
X1 + X 3

In practice, we may have
jX3
(a) X1 and X2 are capacitors and X3 is inductor.                       –
vi      jX1             jX2
OR                                                                     +

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

C.K. Tse: Feedback ampliﬁers and                              77
oscillators
Colpitts oscillator
When X1 and X2 are capacitors and X3 is inductor, we have the Colpitts oscillator.

In this case, we have
1                   1
jX1 =           and jX 2 =
jwC1               jwC2
jX 3 = jwL3

+                           Ro
+
From X1 + X 2 + X 3 = 0                                     vi          Ri        –
–                         Avi
the oscillation frequency can be found:

1
wo =
Ê C1C2 ˆ
L3 Á         ˜
Ë C1 + C2 ¯
L3
–
vi              C1        C2
+

C.K. Tse: Feedback ampliﬁers and                        78
oscillators
A practical form of Colpitts oscillator
The basic ampliﬁer can be realized by a common-emitter ampliﬁer.

The loop gain is
1/sC1
T(s) = Gm ZT
sL + 1/sC1

where      1              1          1                                                   ZT
= sC2 +           +
ZT                1    ro || Rc
sL +
sC1
Putting s = jw, and applying the Barkhausen criterion:
1
wo =
Ê CC ˆ
L3 Á 1 2 ˜
Ë C1 + C2 ¯

Gm C2 (ro || Rc )      C1 Gm Rc ro
>1 ﬁ   >                    for oscillation to start.
C1               C2 Rc + ro

C.K. Tse: Feedback ampliﬁers and                           79
oscillators
Hartley oscillator
When X1 and X2 are inductors and X3 is capacitor, we have the Hartley oscillator.

In this case, we have
jX1 = jwL1 and jX 2 = jwL2
1
jX 3 =
jwC3

+                       Ro
+
From X1 + X 2 + X 3 = 0                                     vi          Ri    –
–                     Avi
the oscillation frequency can be found:

1
wo =
C3 ( L1 + L2 )

For both the Colpitts and Hartley oscillators, the           –                C3
gain of the ampliﬁer has to be large enough to               vi          L1              L2
ensure that the loop gain magnitude is larger than           +
1.

C.K. Tse: Feedback ampliﬁers and                         80
oscillators

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