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c
°
Copyright 2009. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of
Electrical and Computer Engineering.
Feedback Ampli
f
ers
Co
l
lect
iono
fSo
lvedProb
lems
A collection of solved feedback ampli
f
er problems can be found at the below link. The solutions
are based on the use of the Mason Flow Graph described below.
http://users.ece.gatech.edu/~mleach/ece3050/notes/feedback/FBExamples.pdf
Basic Description of Feedback
A feedback ampli
f
er is one in which the output signal is sampled and fed back to the input to form
an error signal that drives the ampli
f
er. The basic block diagrams of noninverting and inverting
feedback ampli
f
ers are shown in Fig. 1. Depending on the type of feedback, the variables
x
,
y
,
and
z
are voltages or currents. The diagram in Fig. 1(a) represents a noninverting ampli
f
er. The
summing junction at its input subtracts the feedback signal from the input signal to form the error
signal
z
=
x
−
by
which drives the ampli
f
er. If the ampli
f
er has an inverting gain, the feedback
signal must be added to the input signal in order for the feedback to be negative. This is illustrated
in Fig. 1(b). The summing junction at the input adds the feedback signal to the input signal to
form the error signal
z
=
x
+
by
. In both diagrams, the gain around the loop is negative and equal
to
−
bA
,whe
rebo
th
A
and
b
are positive real constants. Because the loopgain is negative, the
feedback is said to be negative. If the gain around the loop is positive, the ampli
f
er is said to have
positive feedback which causes it to be unstable.
Figure 1: Feedback ampli
f
er block diagrams. (a) Noninverting. (b) Inverting.
In the noninverting ampli
f
er of Fig. 1(a), the error signal is given by
z
=
x
−
by
. The output
signal can be written.
y
=
Az
=
A
(
x
−
by
)
(1)
This can be solved for the gain to obtain
y
x
=
A
1+
bA
(2)
We see that the e
f
ect of the feedback is to reduce the gain by the factor
(1 +
bA
)
. This factor is
called the “amount of feedback”. It is often speci
f
ed in dB by the relation
20 log

bA

.
In the inverting ampli
f
er of Fig. 1(b), the error signal is given by
z
=
x
+
by
.W
h
e
n
x
goes
positive,
y
goes negative, so that the error signal represents a di
f
erence signal. The output signal
can be written
y
=
−
Az
=
−
A
(
x
+
by
)
(3)
1
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View Full DocumentThis can be solved for the gain to obtain
y
x
=
−
A
1+
bA
(4)
We see that the amount of feedback for the inverting ampli
f
er is the same as for the noninverting
ampli
f
er.
If
A
is large enough so that
bA >>
1
, the gain of the non inverting ampli
f
er given by Eq. (2)
can be approximated by
y
x
'
A
bA
=
1
b
(5)
The gain of the inverting ampli
f
er given by Eq. (4) can be approximated by
y
x
'
−
A
bA
=
−
1
b
(6)
These are important results, for they show that the gain is set by the feedback network and not by
the ampli
f
er. In practice, this means that an ampli
f
er without feedback can be designed without
too much consideration of what its gain will be as long as the gain is high enough. When feedback
is added, the gain can be reduced to any desired value by the feedback network.
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 Summer '08
 HOLLIS

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