EEE421 ANALOGUE ELECTRONICS II
M. Lupupa, 2022/2023
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1. FEEDBACK AND ITS EFFECTS IN AMPLIFIERS
-
Feedback plays a great role especially in op-amp circuits.
-
Depending on the relative polarity of the signal being fed back into the circuit, one may
have negative (degenerative) or positive (regenerative) feedback.
-
Negative feedback results in decreased voltage gain, for which a number of circuit features
are improved.
-
Positive feedback drives a circuit into oscillation as in various types of oscillator circuits.
-
Although negative feedback results in reduced overall voltage gain, a number of
improvements are obtained, among them being:
1. Better gain stability or gain desensitivity: that is, it makes the value of the gain less
sensitive to variations in the values of circuit components, such might be caused by
changes in temperature.
2. Reduces nonlinear distortion: that is, it makes the output proportional to the input (it
makes the gain constant, independent of signal level).
3. Reduces the effect of noise or interference reduction: that is, it minimises the
contribution to the output of the unwanted electric signals generated, either by the
circuit components themselves or by unrelated interference.
4. Higher input impedance
5. Lower output impedance
6. Improved frequency response or bandwidth extension
-
A typical feedback amplifier connection is shown below.
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-
The open-loop amplifier has a gain A ; thus its output xo is related to the input xi by:
xo = Axi
-
(1.1)
The output xo is fed to the load as well as to the feedback network, which produces a
sample of the output. This sample or feedback signal x f is related to xo by the feedback
factor,  .
x f =  xo
-
(1.2)
The input signal xi is obtained by subtracting the feedback signal x f from the source signal
xs .
xi = xs − x f
(1.3)
-
We note that it is the subtraction that makes the feedback negative. In essence negative
feedback reduces the signal that appears at the input of the basic amplifier.
-
The gain of the feedback amplifier can be obtained by combining the above equations, and
can be expressed as:
Af =
xo
A
=
xs 1 + A
(1.4)
-
The quantity A is called the loop gain.
-
For the feedback to be negative, the loop gain must be positive, i.e. the feedback signal x f
should have the same sign as xs .
-
The quantity 1 + A is called the amount of feedback.
-
If the loop gain A is large, i.e. A
expressed as:
Af 
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
1, then the gain of the feedback amplifier can be
(1.5)
3
-
The feedback signal x f can also be expressed as:
xf =
-
A
xs
1 + A
(1.6)
For A >>1 , x f  xs , which implies that the signal xi at the input of the basic amplifier is
reduced to almost zero. Thus if a large amount of negative feedback is employed, the
feedback signal x f becomes an almost identical replica of the source signal xs .
-
The difference between xs and x f , which is xi is sometimes referred to as the error signal.
-
The input differencing circuit is often called a comparison circuit or a mixer.
-
An expression for xi can be determined as:
xi =
1
xs
1 + A
(1.7)
Example
(a) If the open-loop gain, A= 104 V / V and  =
R
R1
find 2 to obtain a closed-loop gain
R1
R1 + R2
of Af = 10V / V .
(b) What is the amount of feedback in decibels?
(c) If Vs = 1V , find V0 , V f and Vi .
(d) If A decreases by 20%, what is the corresponding decrease in Af ?
Effects of Negative Feedback
The following are some of the effects of negative feedback on the performance of feedback
amplifiers.
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(a) Better gain stability or gain desensitivity
-
An important advantage of negative feedback is that the resultant gain of the amplifier can
be made independent of transistor parameters or the supply voltage variations. We have,
Af =
-
A
1 + A
(1.4)
For negative feedback in an amplifier to be effective, the designer deliberately makes
A >>1. Therefore, in the above expression, 1 can be neglected as compared to  A . Hence,
Af =
A
1

A 
(1.5)
-
It may be observed that the gain now depends only upon feedback fraction,  . As the
feedback network is usually made up of resistive components, it is unaffected by changes
in temperature, variations in transistor parameters and frequency. Hence, the gain of the
amplifier is extremely stable.
-
If the feedback transfer function  is a constant, then taking the derivative of Af with
respect to A in Af =
A
we get:
1 + A
dAf
dA
=
1
A
1
−
=
2
(1 + A ) (1 + A ) (1 + A )2
dAf =
-
Dividing both sides by Af =
dAf
Af
=
dA
(1 +  A)
2
A
yields:
1+  A
1
dA
(1 +  A) A
(1.8)
which says that the percentage change in Af is less than the corresponding percentage
change in the open loop gain A by the amount of feedback, 1 + A , which is also called
the desensitivity factor.
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Example
Calculate the percent change in the closed-loop gain, Af , given a change in the open-loop gain,
A . We have A = 105 , A f = 50 and  = 0.01999 . Assume that the change in the open-loop
gain is dA = 104 (a 10 percent change).
(b) Improved frequency response or bandwidth extension
-
Consider an amplifier whose frequency response is characterized by a single pole. Its gain
at mid and high frequencies can be expressed as:
A( s ) =
AM
s
1+
(1.9)
H
where AM is the midband gain and H is the upper 3-dB frequency.
-
The closed-loop gain of the feedback amplifier can be expressed as:
Af ( s ) =
A( s )
1 +  A( s)
(1.10)
where we assume that the feedback transfer function  is independent of frequency.
-
Substituting for A(s) into A f ( s ) yields:
AM
s
1+
Af ( s ) =
-

 A
1+   M
 1+ s
 

H






After some manipulation A f ( s ) becomes:
Af ( s ) =
-
H
AM / (1 +  AM )
1 +  s / H (1 +  AM ) 
 Af (0) =
AM
(1 +  AM )
(1.11)
Thus the feedback amplifier will have a midband gain of AM / (1 +  AM ) and an upper 3dB frequency Hf given by:
Hf = H (1 +  AM )
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(1.12)
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-
If the open-loop gain is characterised by a dominant low frequency pole giving rise to a
lower 3-dB frequency, L , then the feedback amplifier will have a lower 3-dB frequency,
Lf given as:
Lf =
L
(1 +  AM )
(1.13)
Example
Determine the bandwidth of a feedback amplifier. Consider a feedback amplifier with an openloop midband-frequency gain of AM = 10 4 , an open-loop bandwidth of H = 200 rad/s , and
a closed-loop midband-frequency of Af (0) = 50 .
c) Reduces the effect of noise or interference reduction
-
In any electronic system, unwanted random signals may be present in addition to the desired
signal. These random signals are called noise.
-
The noise level in amplifiers can be reduced considerably by the use of negative feedback.
-
The input signal-to-noise ratio is defined as:
( SNR )i =
Si Vs
=
Ni Vn
(1.14)
where Si = Vs is the input source signal and N i = Vn is the input noise signal.
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-
The output signal-to-noise ratio is:
( SNR )o =
So Si  Ai 
=  
N o Ni  An 
(1.15)
where the desired output signal is S o = Si Ai , and the output noise signal is No = Ni Ai .
-
The parameter Ai is the amplification factor that multiplies the source signal, and the
parameter An is the amplification factor that multiplies the noise signal.
-
A large signal-to-noise ratio allows the signal to be detected without any loss of
information. This is a desirable characteristic.
d) Reduced non-linear distortion
-
Large signal amplifiers suffer from non-linear distortions, such as amplitude distortion,
harmonic distortion and intermodulation distortions. Those distortions can be effectively
reduced by employing negative feedback in large signal amplifiers. It can be proved that
Df =
where
D
1 + A
(1.16)
D f = distortion in amplifiers with feedback
D = distortion in amplifier without feedback
-
It is clear that by applying negative feedback to an amplifier, distortion is reduced by a
factor (1 + A ) .
Stability of Feedback Amplifiers
-
We know that an amplifier gain will change with change in frequency, in addition to this,
the phase shift of the amplifier will also change with frequency.
-
Due to these changes the amplifier can break into oscillations, and in this case it is no longer
useful as an amplifier.
-
Thus a proper feedback amplifier design requires that the circuit be stable at all frequencies.
-
In judging the stability of a feedback amplifier as a function of frequency, the loop gain,
A and the phase shift between the input and output are the determining factors.
The Nyquist Plot
-
One of the techniques used to investigate stability is the Nyquist plot. This plots the loop
gain and phase shift as a function of frequency on a complex plane. For example,
T ( j) = A( j) ( j) =| T ( j) | T ( j)
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Consider the figure below:
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Note that the radial distance is | A | and the angle is the phase angle  .
-
The solid line plot is for positive frequencies.
-
The plot for negative frequencies is the mirror image of the positive frequency plot about
the real axis, these are shown by the broken line plots.
-
The Nyquist plot intersects the negative real axis at the frequency 180 . Thus, if this
intersection occurs to the left of the point (-1,0), we know that the magnitude of loop gain
at this frequency is greater than unity and amplifier is unstable.
-
On the other hand, if the intersection occurs to the right of the point (-1,0), the amplifier
will be stable.
-
If the Nyquist plot encircles the point (-1,0) then the amplifier will be unstable.
-
Another way of stating this is to say, a feedback amplifier is stable if its loop gain, A is
less than 1 or 0-dB when its phase is 180 .
Effect of Feedback on the Amplifier Poles/ Root locus of Feedback Amplifier
-
The amplifier frequency response and stability are determined directly by its poles.
-
For an amplifier or any other system to be stable, its poles should lie in the left half of the
s -plane, (a).
-
A pair of complex-conjugate poles on the j -axis gives rise to sustained sinusoidal
oscillations, (c).
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-
From the closed loop gain, Af ( s ) =
A( s )
, we see that the poles of the feedback
1 +  A( s)
amplifier are the zeros of 1 +  A(s) .
-
That is, the feedback amplifier poles are obtained by solving the equation:
1 +  A(s) = 0
(1.29)
which is the characteristic equation of the feedback loop.
Amplifier with a Single-Pole Response
-
Consider an amplifier whose open-loop gain is characterised by a single pole:
A( s) =
A0
1+
s
(1.30)
p
where A0 is the low-frequency gain, and  p is the frequency where the pole is located.
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-
The closed-loop gain is given by:
Af ( s ) =
-
A0 / (1 + A0  )
1 +  s /  p (1 + A0  ) 
 Af ( j ) =
A0 / (1 + A0  )
(1.31)
1 +  j /  p (1 + A0  ) 
The feedback moves the pole along the negative real axis to a frequency  pf , where
 pf =  p (1 + A0  )
(1.32)
-
Since the pole of the closed-loop amplifier never enters the right half to the s-plane, the
single pole amplifier is stable for any value of  . This amplifier is said to be
unconditionally stable.
-
The second figure shows that applying negative feedback to an amplifier results in
extending its bandwidth at the expense of a reduction in gain.
Amplifier with Two-Pole Response
-
Consider an amplifier whose open-loop gain is characterised by two real poles:
A( s) =
-
A0

s 
s
1+
1 +


  p1   p2



(1.33)
In this case, the closed-loop poles are obtained from 1 +  A(s) = 0 , which leads to:
(
)
s 2 + s  p1 +  p2 + (1 + A0  )  p1 p2 = 0
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(1.34)
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The closed-loop poles are given by:
s=
-
(
) (
−  p1 +  p2 
p1
+  p2
)
2
− 4 (1 + A0  )  p1 p2
(1.35)
2
From the above equation, as the loop gain (1 + A0  )  p1 p2 is increased from zero, the poles
are brought closer together.
-
A value is then reached at which the poles become coincident.
-
If the loop gain is further increased, the poles become complex conjugate and move along
the vertical line.
-
The figure below shows the locus of the poles for increasing loop gain. This plot is called
a root-locus diagram.
-
From the root-locus diagram, we see that this feedback amplifier is also unconditionally
stable.
-
To be more specific, the characteristic equation of a second-order network can be written
in the standard from:
 
s 2 + s  0  + 02 = 0
Q
where
(1.36)
0 is called the pole frequency
Q is called the pole Q factor
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-
The Q factor of poles of the feedback amplifier is expressed as:
Q=
(1 +  A0 )  p  p
1
2
p + p
1
(1.37)
2
-
The poles are complex if Q is greater than 0.5.
-
When Q = 0.707 (poles at 45 angles) results in the maximally flat response.
-
The figure below shows a number of possible responses obtained for various values of Q .
Stability Study Using Bode Plots
-
Another technique that can be used to investigate stability is the Bode plot.
Gain and Phase Margins
-
Gain margin is an indication of excess gain before instability. Thus 0dB or A = 1 is on
the border of stability, and any negative decibel value is stable. The gain margin can also
be defined as the difference in gain between the value of | A | at 180 and unity or 0-dB.
-
Phase margin is an indication of excess phase before -180° phase shift at unity gain. The
phase margin can also be defined as the difference between the phase angle at which the
value of | A | is 0-dB and 180°. If at the frequency of unity loop gain magnitude, the phase
angle is less (in magnitude) than 180°, then the amplifier is stable. If at the frequency of
unity loop gain magnitude, the phase lag is in excess of 180°, the amplifier is unstable.
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-
The figure below shows both the gain margin and the phase margin.
Example
An amplifier with a low-frequency gain of 100 and poles at 104 rad/s and 106 rad/s is
incorporated in a negative-feedback loop with feedback factor  .
a) For what value of  do the poles of the closed-loop amplifier coincide?
b) What is the corresponding Q of the resulting second-order system?
Effect of Phase Margin on Closed Loop Response
-
Feedback amplifiers are usually designed with a phase margin of 45 .
-
Let us consider a feedback amplifier with a large low-frequency loop gain, A0 
-
Then its closed loop gain at low frequencies is approximately
-
If we write the frequency at which the magnitude of loop gain is unity as 1 , then
A ( j1 )  = 1 e − j
1

1.
.
(1.38)
where  = 180 − phase margin
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-
At 1 the closed loop gain is:
Af ( j1 ) =
-
A( j1 )
1 +  A( j1 )
(1.39)
Substituting A ( j1 )  = 1 e − j into Af ( j1 ) =
1 e− j
Af ( j1 ) =

1+ e
− j
1
=

A( j1 )
gives:
1 +  A( j1 )
e− j
(1.40)
1 + e− j
Note: e− j = cos − j sin 
Thus
1
e − j =   ( cos  − j sin  ) whose magnitude can be written as:

 
1
1

2
e
− j
1

-
2
1
1
1
=   ( cos  − j sin  ) = Re2 + Im 2 =   cos 2  +   sin 2 
 
 
 
2
e
− j
2
1
1
1
=   cos 2  + sin 2   =   =

 
 
The magnitude of the closed-loop gain at 1 can then be expressed as:
1
| Af ( j1 ) |=
-

1 + e− j
(1.41)
For a phase margin of 45 ,  = 135 the magnitude of the closed-loop gain at 1 can
then be expressed as:
| Af ( j1 ) |= 1.3
1

(1.42)
Example
Consider an op amp having a single-pole, open-loop response with A0 = 105 , f p = 10 Hz and
feedback factor  = 0.01 . Find the frequency at which A = 1 . Find the phase margin.
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Example
Find the closed-loop gain at 1 relative to the low-frequency gain when the phase margin is
30 .
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