129
Feedback Amplifiers And Oscillators
for
Analogue Circuit Fundamentals
by
Prof. Michael Tse
September 2014
Contents
Feedback
Basic feedback configuration
Advantages
The price to pay
Feedback Amplifier Configurations
Series-shunt, shunt-series, series-series, shunt-shunt
Input and output impedances
Practical Circuits with loading effects
Compensation
Op-amp internal compensation
Oscillation
Oscillation criteria
Sustained oscillation
Wein bridge, phase shift, Colpitts, Hartley, etc.
C.K. Tse: Feedback amplifiers and oscillators
2
Basic feedback configuration
The basic feedback amplifier consists of a basic amplifier and a feedback network.
si
input
e
+
–
A
so
output
basic amplifier
Careful!!
sf
f
feedback network
A = basic amplifier gain
f = feedback gain
C.K. Tse: Feedback amplifiers and oscillators
3
Characteristics
si
input
+
e
–
A
so
output
basic amplifier
sf
The input is subtracted by a
feedback signal which is part of the
output, before it is amplified by the
basic amplifier. 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
si
€
Ao
so
Ao =
so
A
=
si 1+ Af
If Af >> 1,
€
C.K. Tse: Feedback amplifiers and oscillators
€
Ao ≈
1
f
4
Simple viewpoint
si
input
+
e
–
A
basic amplifier
sf
so
output
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 si
f
C.K. Tse: Feedback amplifiers and oscillators
€
5
Obvious advantage
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 + δA. From the formula of Ao, we have
δAo # δA &# 1 &
= % (%
(
Ao $ A '$1+ Af '
Obviously, if Af is large, then δAo/Ao will be reduced drastically. 1
€
Feedback reduces gain sensitivity! In fact, the gain is just Ao ≈ .
f
C.K. Tse: Feedback amplifiers and oscillators
€
6
so
Another advantage
Suppose the basic amplifier is
distortive. So, the output does
not give a sine wave for a sine
wave input.
But, with feedback, we see that
the gain is about 1/f anyway,
regardless of what A is (or as
long as Af is large enough).
This gives a very good property
of feedback amplifier in terms
of eliminating distortion.
A
output
e
input
so
1/f
C.K. Tse: Feedback amplifiers and oscillators
si
7
Other advantages
•
•
•
Improve input and output resistances (to be discussed later).
Widening of bandwidth of amplifier (to be discussed later).
Enhance noise rejection capability.
ni
si
ni
A
so
+
si
input
Signal-to-noise ratio:
" so % " si %
$ '=$ '
# no & # ni &
e
–
sf
+
A
’
A
+
basic amplifier
output
f
feedback network
Signal-to-noise ratio improves!
€
so
C.K. Tse: Feedback amplifiers and oscillators
€
" so %
" si %
$ ' = A($ '
# no &
# ni &
8
The price to pay
Of course, nothing is free! Feedback comes with reduced gain, and hence you may need to add a preamplifier 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 oscillators
9
Terminologies
Basic amplifier gain = A
Feedback gain = f
A
1
≈
Overall gain (closed-loop gain) = 1+ Af
f
Some books use T to denote Af.
Loop gain (roundtrip gain) = Af
€
C.K. Tse: Feedback amplifiers and oscillators
10
Feedback amplifiers
What is an amplifier?
si
so
A
Signals can be voltage or current.
General model for voltage amplifier:
+
vin
–
Ro
Rin
+
–
Avin
+
vo
–
voltage amplifier
C.K. Tse: Feedback amplifiers and oscillators
11
Models of amplifiers
Ro
+
vin
–
Rin
+
–
Avin
iin
+
vo
–
Ro
iin
Rin
+
–
Aiin
transresistance amplifier
Aiin
Rin
voltage amplifier
io
Ro
current amplifier
+
vo
–
+
vin
–
io
Avin
Rin
Ro
transconductance amplifier
C.K. Tse: Feedback amplifiers and oscillators
12
Feedback amplifier configurations
Voltage amplifier
si
input
+
e
A
–
so
output
basic amplifier
sf
f
voltage
voltage
feedback network
To subtract voltage from
voltage, we should use series
connection
+
vi
–
To copy voltage, we should use
parallel (shunt) connection
A
– vf +
•
•
+
vo
–
Hence, series-shunt feedback
C.K. Tse: Feedback amplifiers and oscillators
13
Series-shunt feedback (for voltage amplifier)
Ro
+
v
i
–
+
ve
–
+
–
Ri
+
–
Ave
+
v
o
–
fvo
Overall gain (closed-loop gain) :
Ao =
vo
A
=
v i 1+ Af
C.K. €
Tse: Feedback amplifiers and oscillators
14
Series-shunt feedback (for voltage amplifier)
To find the input resistance, we consider the ratio of vi and ii, with output opened.
ii
+
v
i
–
vi
vi
=
ii v e /Ri
v e + fv o
= Ri
ve
= Ri (1+ Af )
RIN =
€
Ro
+
ve
–
+
–
Ri
Ave
+
v
o
–
RIN
+
–
fvo
The input resistance has been enlarged by (1+Af). This is a desirable
feature for voltage amplifier as a large input resistance minimizes loading
effect to the previous stage.
C.K. Tse: Feedback amplifiers and oscillators
15
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
+
v
i
–
+
ve
–
+
–
Ri
io
+
v
o
–
Ave
First, we have ve = – fvo.
Also,
io =
v o − Av e v o + Afv o
=
Ro
Ro
Hence,
ROUT
+
–
€
fvo
ROUT =
vo
Ro
=
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
voltage source for the load.
C.K. Tse: Feedback amplifiers and oscillators
16
Series-shunt feedback (for voltage amplifier)
Summary of features
Equivalent model
A
1
≈
Closed-loop gain =
1+ Af
f
Input resistance = Ri ( 1 + Af )
€
Ro
Output resistance = Ro
1+ Af
+
v
i
–
Ri ( 1 + Af )
€
1+ Af
€
+
–
Av i
1+ Af
+
v
o
–
€
NOTE: We did not consider loading effect of the
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 oscillators
17
Feedback amplifier configurations
Transresistance amplifier
si
input
+
e
–
A
so
output
basic amplifier
sf
f
current
voltage
feedback network
To subtract current from current,
we should use shunt
(connection) connection
ii
A
To copy voltage, we should use
parallel (shunt) connection
•
•
+
vo
–
Hence, shunt-shunt feedback
C.K. Tse: Feedback amplifiers and oscillators
18
Shunt-shunt feedback (for transresistance amplifier)
ii
ie
Ro
+
–
Ri
Aie
+
v
o
–
fvo
Overall gain (closed-loop gain) :
Ao =
vo
A
=
ii 1+ Af
C.K. €
Tse: Feedback amplifiers and oscillators
19
Shunt-shunt feedback (for transresistance amplifier)
To find the input resistance, we consider the ratio of vi and ii, with output opened.
ii
v i Riie
=
ii
ii
ie
= Ri
ie + fv o
Ri
=
1+ Af
RIN =
€
+
v
i
–
ie
Ro
+
–
Ri
+
v
o
–
Aie
RIN
fvo
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 oscillators
20
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.
ii = 0
ie
Ro
+
–
Ri
io
+
v
o
–
Aie
First, we have ie = – fvo.
Also,
io =
v o − Aie v o + Afv o
=
Ro
Ro
Hence,
ROUT
€
fvo
ROUT =
vo
Ro
=
io 1+ Af
€
The output resistance has been reduced by (1+Af). This is a desirable
feature for transresistance amplifier as a small resistance emulates a better
voltage source for the load.
C.K. Tse: Feedback amplifiers and oscillators
21
Shunt-shunt feedback (for transresistance amplifier)
Summary of features
Equivalent model
A
1
≈
Closed-loop gain =
1+ Af
f
Ri
Input resistance =
1+ Af
€
Ro
Output resistance = €
1+ Af
Ro
1+ Af
ii
Ri
1+ Af
€
€
+
–
Aii
1+ Af
+
v
o
–
€
€
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 oscillators
22
Series-series feedback (for transconductance amplifier)
io
+
v
o
–
+
ve
–
Ro
Ave
Ri
io
+
–
f io
io
A
=
v i 1+ Af
RIN = Ri (1+ Af )
Overall gain (closed-loop gain) :
Ao =
Input resistance:
Output resistance:
€
€
ROUT = Ro (1+ Af )
C.K. Tse: Feedback amplifiers and oscillators
€
Desirable!
Desirable!
23
Shunt-series feedback (for current amplifier)
ii
io
ie
Ro
Aie
Ri
io
f io
i
A
Overall gain (closed-loop gain) :
Ao = o =
ii 1+ Af
Ri
Input resistance:
RIN =
1+ Af
Output resistance:
€
€
ROUT = Ro (1+ Af )
C.K. Tse: Feedback amplifiers and oscillators
€
Desirable!
Desirable!
24
Practical feedback circuits (with loading effects)
In practice, the input source has resistance and the feedback network has
resistance.
Example: shunt-shunt feedback
ie
ii
Ro
Ri
+
–
Aie
+
v
o
–
fvo
What are the effects on the gain, input and output resistances?
C.K. Tse: Feedback amplifiers and oscillators
25
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. It is convenient to use admittance or conductance.
ii
yi
+
v
–
i
y11
+
v
o
–
y22
y21vi
1
2
y11f
y22f
y12fvo
C.K. Tse: Feedback amplifiers and oscillators
26
Systematic analysis of shunt-shunt feedback using
admittances or conductances
ii
yi
+
v
–
i
y11
+
v
o
–
y22
y21vi
1
2
y11f
y22f
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 oscillators
27
Systematic analysis of shunt-shunt feedback using
admittances or conductances
y11f
ii
y22f
+
v
–
i
yi
y11
y21vi
1
+
v
o
–
y22
2
y12fvo
One step closer…
C.K. Tse: Feedback amplifiers and oscillators
28
Systematic analysis of shunt-shunt feedback using
admittances or conductances
y11f
ii
+
v
–
i
y22f
y11
yi
y21vi
1
+
v
o
–
y22
2
y12fvo
One more step closer…
C.K. Tse: Feedback amplifiers and oscillators
29
Systematic analysis of shunt-shunt feedback using
admittances or conductances
y11+y11f+yi
ii
+
v
–
i
y22f +y22
y21vi
1
+
v
o
–
2
y12fvo
Yet another step closer…
C.K. Tse: Feedback amplifiers and oscillators
30
Systematic analysis of shunt-shunt feedback using
admittances or conductances
in conductance (S)
in resistance (Ω)
y11+y11f+yi
1/(y22f +y22)
ii
+
v
–
i
−y 21v i
y 22f + y 22
+
v
o
–
+
–
1
2
Use Thévenin
€
y12fvo
Yet another step closer…
C.K. Tse: Feedback amplifiers and oscillators
31
Systematic analysis of shunt-shunt feedback using
admittances or conductances
in conductance (S)
y11+y11f+yi
1/(y22f +y22)
ie
ii
in resistance (Ω)
+
v
–
i
+
v
o
–
+
–
1
2
( y 22f
−y 21ie
+ y 22 )( y11 + y11f + y i )
y12fvo
€
Finally, we get the same standard form.
C.K. Tse: Feedback amplifiers and oscillators
32
Systematic analysis of shunt-shunt feedback using
admittances or conductances
We can simply apply the standard results:
A=
Basic amplifier gain
Feedback gain
( y 22f
−y 21
−y 21
=
+ y 22 )( y11 + y11f + y i ) y oT y iT
f = y21f
A
1
1
Ao =
≈ =
1+ Af
f y12f
€
Overall (closed-loop) gain
1
RIN =
(y11 + y11f + y i )(1+ Af )
Input resistance
€
1
ROUT =
(y 22f + y 22 )(1+ Af )
Output resistance
€
C.K. Tse: Feedback amplifiers and oscillators
€
33
The Secret is…
For shunt-shunt feedback, the feedback network converts voltage to current.
For shunt-series feedback, the feedback network converts current to current.
For series-series feedback, the feedback network converts current to voltage.
For series-shunt feedback, the feedback network converts voltage to voltage.
There are really just four types of simplest practical feedback networks!!
e.g., for series-shunt, obviously, just a voltage divider:
R2
R1
=
C.K. Tse: Feedback amplifiers and oscillators
+
–
34
Similarly, for shunt-series, obviously current divider:
R1
R2
=
NOTE on calculating resistance:
To this 2-port, the input is current source io, output is current if through a shortcircuit load on the left side. Therefore, when we calculate the resistance seen from the left-side, we take the
source io as an open circuit. When we calculate the resistance seen from the
right-side, we take the load at the left port as short circuit.
C.K. Tse: Feedback amplifiers and oscillators
35
Similarly, for shunt-shunt, we have:
=
NOTE on calculating resistance:
To this 2-port, the input is voltage source vo, output is current if through a shortcircuit load on the left side. Therefore, when we calculate the resistance seen from the left-side, we take the
source vo as short circuit. When we calculate the resistance seen from the rightside, we take the load at the left port as short circuit too.
C.K. Tse: Feedback amplifiers and oscillators
36
Similarly, for series-shunt, we have:
=
+
–
NOTE on calculating resistance:
To this 2-port, the input is current source io, output is voltage vf through a opencircuit load on the left side. Therefore, when we calculate the resistance seen from the left-side, we take the
source io as open circuit. When we calculate the resistance seen from the rightside, we take the load at the left port as open circuit.
C.K. Tse: Feedback amplifiers and oscillators
37
General procedure of analysis
1.  Identify the type of feedback.
2.  Find the appropriate simplest feedback network.
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
form (without loading).
5.  Apply standard formulae to find A, f, RIN and ROUT.
C.K. Tse: Feedback amplifiers and oscillators
38
Example
Rf
is
–
a
+
+
vo
–
RL
Type of feedback:
shunt-shunt
Appropriate 2-port type: see page 36
So, the first step is to represent the feedback network as in page 36.
C.K. Tse: Feedback amplifiers and oscillators
39
Example
is
–
a
+
+
vo
–
RL
Rf
C.K. Tse: Feedback amplifiers and oscillators
40
Example
Converting to the appropriate 2-port (page 36)
is
Note: this
goes to the
–ve input
of A.
–
vi
+
Ri
+
–
Ro
avi
+
vo
–
RL
Rf
y12fvo
y11f =
€ and oscillators
C.K. Tse: Feedback amplifiers
1
Rf
y11f
y22f
−1
Rf
y 22f =
y12f =
1
Rf
41
Example
Converting to y-parameter
–
vi
+
is
Ri
y12fvo
y11f =
€
1
Rf
+
–
Ro
+
vo
–
avi
y11f
y12f =
−1
Rf
RL
y22f
y 22f =
1
Rf
REMEMBER:
y11f and y22f are conductance!
C.K. Tse: Feedback amplifiers and oscillators
42
Example
Casting it to standard form
–
vi
+
is
Ri || R f
+
–
Ro
avi
Rf||RL
+
vo
–
€
y12fvo
y11f =
€
1
Rf
y12f =
−1
Rf
y 22f =
1
Rf
C.K. Tse: Feedback amplifiers and oscillators
43
Example
Casting it to standard form
Ro || R f || RL
–
vi
+
is
Ri || R f
+
a(R f || RL )
–
vi
Ro + R f || RL
+
vo
–
€
€
€
−v o
Rf
€
C.K. Tse: Feedback amplifiers and oscillators
44
Example
Finally, we get the standard form
ie
–
vi
+
is
Ri || R f
+
–
Ro || R f || RL
Aie
+
vo
–
€
€
Using Thévenin theorem,
€
Aie = av i
−v o
Rf
R f || RL
(R
f
|| RL ) + Ro
R || R )( R || R )
(
A = −a
(R || R ) + R
i
f
f
f
L
L
o
Ri R 2f RL
1
A = −a
(Ri + R f ) ( R f RL + Ro R f + Ro RL )
€
C.K. Tse: Feedback amplifiers and oscillators
€
45
Example
Apply standard results:
Ri R 2f RL
1
A
=
−a
Basic amplifier gain (transresistance)
(Ri + R f ) ( R f RL + Ro R f + Ro RL )
−1
Feedback gain:
f =
Rf
€
A
1
if Af >> 1
A
=
≈
= −R f
o
Overall (closed-loop) gain
1+ Af
f
€
R || R f
RIN = i
Input resistance
1+ Af
€
R || R f || RL
Output resistance
ROUT = o
1+ Af
€
€
C.K. Tse: Feedback amplifiers and oscillators
46
Frequency response
Gain and bandwidth
si
input
+
e
–
A(jω)
basic amplifier
sf
f
feedback network
so
output
Suppose the basic amplifier
has a pole at p1, i.e., ALF
A( jω ) =
jω
1+
p1
20log10|A| (dB)
€
ALF
slope = –20dB/dec
p1
C.K. Tse: Feedback amplifiers and oscillators
ω
47
Frequency response
Gain and bandwidth
Hence, we see that the overall gain has a
pole at pc = p1(1 + fALF) and the low-frequency gain is lowered to
ALF
Ao,LF =
1+ fALF
The overall (closed-loop) gain is
A( jω )
1+ A( jω ) f
ALF
=
#
jω &
%1+
( + fALF
p
$
1 '
Ao ( jω ) =
)
,
.
ALF +
1
+
.
=
jω
1+ fALF +1+
.
+* p1 (1+ fALF ) .-
€
20log10|A| (dB)
basic amplifier
ALF
feedback amplifier
Ao,LF
p1
€
C.K. Tse: Feedback amplifiers and oscillators
ω
pc
48
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 oscillators
49
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
input
+
–
so
A
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 oscillators
50
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:
Af = 1
1.
This is known as the Barkhausen criteria.
2.
Af = ±180o
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 oscillators
51
The loop gain T
An important parameter to test stability is the loop gain, usually denoted
by T.
T = Af
|T| (dB)
€
0dB
φ
crossover frequency
(where the gain is 1)
ωo
ω
ω
φT
If φΤ = –180o, OSCILLATES!
C.K. Tse: Feedback amplifiers and oscillators
52
Phase margin
Phase margin is an important parameter to evaluate how stable the
system is.
Phase margin φPM = –180o – φT
|T| (dB)
crossover frequency
(where the gain is 1)
0dB
φ
ωo
ω
ω
φT
–180o
phase margin φPM (the larger the better)
C.K. Tse: Feedback amplifiers and oscillators
53
Compensation
Compensation is to make the amplifier more stable, i.e., to increase φPM.
REMEMBER: We should always look at T, not A or Ao. |T| (dB)
crossover frequency
(where the gain is 1)
0dB
φ
p1
p2
ωo
ω
ω
φT
–180o
phase margin φPM (how to increase it?)
C.K. Tse: Feedback amplifiers and oscillators
54
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
function Gc is
1
Gc ( jω ) =
jω
1+
pa
crossover frequency
before compensation
0dB
pa
φ
€
crossover frequency
after compensation
|T| (dB)
p1
p2
ω
ω
before compensation
after compensation
–180o
phase margin φPM
after compensation
C.K. Tse: Feedback amplifiers and oscillators
phase margin φPM
before compensation
55
Method 2: Lead compensation
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
function Gc is
jω
1+
za
Gc ( jω ) =
jω
1+
pa
|T| (dB)
crossover frequency
before compensation
0dB
φ
za
p1
p2
crossover frequency
after compensation
ω
ω
before compensation
after compensation
€
–180o
phase margin φPM
before compensation
C.K. Tse: Feedback amplifiers and oscillators
phase margin φPM after
compensation
56
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 oscillators
57
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 oscillators
58
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.
RL
Equivalent model:
RL
next stage
to next stage
C
rπ
ro//RL
2C
RIN
The dominant pole is
1
pa =
= 2π (5)
2(ro || RL || RIN )C
We can find C from this equation.
C.K. Tse: Feedback amplifiers
€ and oscillators
59
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
pa =
€
1
2(ro || RL || RIN )C
Using this method, C can be as large as hundreds
of pF, which is too large to be implemented on
chip. NOT practical!
= 2π (5)
Better solution: Use
Miller effect.
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 gain
stage in order to use
Miller effect. In this way,
C can be much smaller,
say a few pF.
to active load
C
RL
RL
output stage
main gain stage
CE stage
C.K. Tse: Feedback amplifiers and oscillators
60
Example: Op-amp 741 internal compensation
+Vcc
Q1
differential
input stage
+
Q2
Q13B
–
Q13A
Q14
Q3
Q4
output
+Vcc
Q20
Q16
–VEE
Q5
Q6
Data:
DC gain = 70 dB
Poles:
30 kHz
500 kHz
10 MHz
Q23
Q17
main gain stage
CE stage
–VEE
C.K. Tse: Feedback amplifiers and oscillators
61
Example: Op-amp 741 internal compensation
Unity-gain feedback (worst case stability problem): T = A
p1 = 30 kHz
Bad stability because
of the substantial
phase shift!
C.K. Tse: Feedback amplifiers and oscillators
62
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 oscillators
63
Example: Op-amp 741 internal compensation
Graphical construction method
p1 = 30 kHz
pa
Bad stability because
of the substantial
phase shift!
C.K. Tse: Feedback amplifiers and oscillators
64
Example: Op-amp 741 internal compensation
Exact calculation of pa :
slope = –20 dB/dec
70 dB
pa
€
=
0 − 70
−70
=
log p1 − log pa 4.477 − log pa
Hence,
pa = 9.5 Hz
p1 = 30 kHz
C.K. Tse: Feedback amplifiers and oscillators
65
Example: Op-amp 741 internal compensation
After compensation, the
phase margin is 45º.
C.K. Tse: Feedback amplifiers and oscillators
66
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
input stage
+
Q2
Q13B
–
Q13A
Cc
Q3
Q4
Q14
output
+Vcc
Q20
Q16
–VEE
Q5
Q6
Q23
Q17
main gain stage
CE stage
–VEE
C.K. Tse: Feedback amplifiers and oscillators
67
Example: Op-amp 741 internal compensation
+Vcc
Q1
differential
input stage
+
Q3
Q2
–
Q4
Q13B
Given:
Ro17 = 5 MΩ
Ro13 = 720 kΩ
Ri23 = 100 kΩ
Q13A
Cc
Q14
output
+Vcc
Q20
Q16
–VEE
Q5
Q6
Q23
Q17
main gain stage
CE stage
–VEE
Gm = 6 mA/V
C.K. Tse: Feedback amplifiers and oscillators
Gain of CE stage:
ACE = Gm[Ro17||Ro13||Ri23]
Miller-effect capacitor
CM = Cc (ACE + 1)
= 518.74 Cc
68
Example: Op-amp 741 internal compensation
+Vcc
Q1
differential
input stage
+
Q3
Q2
–
Q4
Q13B
Cc
Equivalent ckt:
Ro4||Ro6
+Vcc
Q16
–VEE
Q5
Q13A
Given:
Ri16 = 2.9 MΩ
Ro4 = 10 MΩ
Ro6 = 20 MΩ
Q6
Ri16
CM
Q17
main gain stage
CE stage
–VEE
pa = 1 / 2π CM [Ro4||Ro6|| Ri16]
and CM = 518.74 Cc
Hence, Cc = 15 pF
C.K. Tse: Feedback amplifiers and oscillators
69
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 φT = ±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 oscillators
70
Oscillator principle
As T = Af, we can deliberately create phase shift in A or f.
si
input
+
–
A(jω)
basic amplifier
f(jω)
so
output
NOTE: Since this model is a
negative feedback, we need the total
phase shift of A(jω) and f(jω) to be
180o at the frequency of oscillation.
If a positive feedback is used, we
need the total phase shift to be 360o.
|A(jω) f(jω)| (dB)
feedback network
ωo
ω
ω
–180o
C.K. Tse: Feedback amplifiers and oscillators
71
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 oscillate 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 oscillators
72
The Wien bridge oscillator
Model:
R2
R1
–
+
+
A
Zp
C
C
R
R
Zs
Zp
Zs
R2
A
=
1+
Basic amplifier gain
R1
−Z p
Feedback gain
f =
Zs + Z p
where Z p =
R
1+ jωCR
and Z s =
1+ jωCR
jωC
C.K. Tse: Feedback amplifiers and oscillators
€
€
73
The Wien bridge oscillator
Oscillation frequency
Note that we define the standard feedback structure with negative feedback. So,
the loop gain is
$ R2 '
−&1+ )
% R1 (
T( jω ) =
$
1 '
3 + j&ωCR −
)
%
(
ω
CR
Applying the oscillation criteria, we can find the oscillation frequency and the
€ as follows:
resistor values
R2
T(
j
ω
)
=
1
⇒
1+
= 3 ⇒ R2 = 2R1
R1
1
1
φT = ±180 o ⇒ ω oCR =
⇒ ωo =
ω 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 oscillators
74
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
+
| f |
1/3
φf
freq
fo
+90
freq
–90
Clearly, the roundtrip gain will be 1 for
f = fo if the basic amplifier has a gain of
3.
The world is noisy. Signals of all
frequencies exist everywhere! But signals at all frequencies except fo
will be reduced after a round trip. Only
signals at fo will have a roundtrip
gain of 1. 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 oscillators
75
The Wien bridge oscillator
+15V
Amplitude control
3k
If we choose R2/R1 = 2.03, then amplitude may
grow. We have to stabilize the amplitude. The
following is an amplitude limiter circuit.
Diode D1 (D2) conducts when vo reaches its
positive (negative) peak.
D2
20.3k
10k
A
–
+
vo
Just when D1 conducts, we have vA = vB.
10k
A
20.3k
1k
16n 10k
vo
1k
16n
10k
B
3k
3
1
(v o + 15) = v o
4
3
3k
–15V
⇒
v o = 9 V.
Similar procedure applies for the negative peak.
So, the amplitude is 9 V.
C.K. Tse: Feedback amplifiers and oscillators
€
B
D1
–15V
−15 +
1k
76
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 exactly
180o. At this frequency, if the roundtrip gain is
1, oscillation occurs. Note that the negative
feedback already gives 180o phase shift. R2
R1
–
+
| f |
C
C
1
29
C
freq
R'
R1 || R'= R
€
R
R
€
f
φf
fo
180o
C.K. Tse: Feedback amplifiers and oscillators
€
freq
77
The phase shift oscillator
C
From the filter characteristic of the feedback network, we
know that the phase shift is 180o at fo, where its gain is
1/29.
So, oscillation starts at fo if A ≥ 29. This means we need
to have R2 ≥ 29R1.
We can prove that
1
fo =
2π 6CR
R
C
R
C
R
| f |
1
29
freq
€
φf
fo
180o
freq
€
Note that the leftmost resistor in the feedback filter is R’ (not R). But R’//R1 is
exactly R. This will adjust the loading effect of the basic amplifier and make the
overall filter circuit easier to analyze since it is then simply composed of three
identical RC sections.
C.K. Tse: Feedback amplifiers and oscillators
78
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
jX1
jX2
+
vi
–
Ri
+
–
Ro
Avi
jX3
pure reactive π-feedback network
jX3
–
vi
+
jX1
C.K. Tse: Feedback amplifiers and oscillators
jX2
79
Resonant circuit oscillators
Analysis:
−A( jX1 )( jX 2 )
The loop gain is
T( jω ) =
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.
+
vi
Thus, the
€ oscillation criteria become
–
X
+
X
+
X
=
0
1
2
3
AX1
=1
X1 + X 3
In practice, we may have
–
(a) X1€and X2 are capacitors and X3 is inductor.
vi
+
OR
(b) X1 and X2 are inductors and X3 is capacitor.
Ri
+
–
Ro
Avi
jX3
jX1
C.K. Tse: Feedback amplifiers and oscillators
jX2
80
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 =
jωC1
jωC2
jX 3 = jωL3
Ro
+
+
From X1 + X 2 + X 3 = 0
vi
Ri
–
€ Avi
–
the oscillation frequency can be found:
€
ωo =
€
1
# C1C2 &
L3 %
(
$ C1 + C2 '
–
vi
+
L3
C1
C.K. Tse: Feedback amplifiers and oscillators
C2
81
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
=
sC
+
+
€
2
1
Z
ro || Rc
T
sL +
sC1
Putting s = jω, and applying the Barkhausen criterion:
ωo =
€
1
# CC &
L3 % 1 2 (
$ C1 + C2 '
G m C 2 (ro || Rc )
>1 ⇒
C1
€
€
ZT
C1 G m Rc ro
<
C2
Rc + ro
for oscillation to start.
C.K. Tse: Feedback amplifiers and oscillators
82
Hartley oscillator
When X1 and X2 are inductors and X3 is capacitor, we have the Hartley oscillator.
In this case, we have
jX1 = jωL1 and jX 2 = jωL2
1
jX 3 =
jωC3
Ro
+
+
From X1 + X 2 + X 3 = 0
vi
Ri
–
€ Avi
–
the oscillation frequency can be found:
€
1
ωo =
C3 ( L1 + L2 )
C3
For both the Colpitts and Hartley oscillators, the
–
gain
vi
L1
L2
€ of the amplifier has to be large enough to ensure that the loop gain magnitude is larger than
+
1.
C.K. Tse: Feedback amplifiers and oscillators
83