# 6. Negative Feedback in Single

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Lecture 8: Introduction to electronic analog circuits 361-1-3661
6. Negative Feedback in SingleTransistor Circuits
&copy; Eugene Paperno, 2008
Our aim is to study the effect of negative feedback on the
small-signal gain and the small-signal input and output
impedances of the single-transistor circuits. Our study will be
based on generic (functional) models of the circuits (see Fig.
1).
ss
ss
sε
AOL
&lt;1
so
Example
rs
vo
6.1. Single-transistor circuits with no feedback
vs
Let us start from analyzing the small-signal gain of a circuit
with no feedback [see Fig. 1(a)]
A = G AOL ,
vbe ib
hie
hfeib
gmvbe
RC io
ro
Rin
(1)
G=
where G is the small-signal input transmission
Rin
rs+Rin
AOL= −gm(RC||ro)
(a)
s
G≡ ε ,
ss
(2)
ss
ss
ssG
ss is the signal source value, sε is the signal at the control port
of the dependent source of the transistor model, and AOL is the
small-signal open-loop gain
Σ
sε
sε AOLβ
AOL
&lt;1
sε AOL
Σ
AOL
βfwd
&lt;1
(3)
Equation (1) shows that the small-signal gain A is directly
proportional to the small-signal open-lop gain AOL. This may
be a serious disadvantage because AOL depends on the
transistor small-signal parameters, which are very sensitive to
the transistor technology and temperature. On the other hand,
this can be an advantage if the maximum gain is required and
its exact value is not important. We will also see later when
studying positive feedback, that the circuit with no feedback is
always stable provided its G and AOL gains are stable. In the
next course on electronic analog circuit, you will see as well
that adding negative feedback can limit the frequency range of
the circuit.
6.2. Single-transistor circuits with feedback
Let us now find the closed-loop gain of a single-transistor
circuit with feedback [see Fig. 1(b)]
s
≡ o .
sε
so
ssGβfwd
(b)
Fig. 1. Functional (block) diagrams of electronic circuits (a) without and (b)
with feedback.
ACL ≡
s s Gβ fwd
so
sε AOL
=
+
+
β
s
s
A
ss
ss
ε
ε OL
G
=G
AOL
A
+ Gβ fwd = G OL + D ,
1 + AOL β
1 + RR
=G
AOL
+D
DSF
(4)
where β is the small-signal feedback transmission of the
feedback network, βfwd is the small-signal feedforward
transmission of the feedback network,
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Lecture 8: Introduction to electronic analog circuits 361-1-3661
RR≡AOLβ
(5)
D≡Gβfwd
(6)
is the return ratio,
is the small-signal direct transmission, and
DSF≡1+AOLβ
(7)
is the desensitivity factor or the amount of feedback.
We will define a feedback as negative or positive if it
decreases or increases, correspondingly, the closed loop gain
ACL relative to GAOL. It is obvious from (4) that |DSF|&gt;1
corresponds to a negative feedback, |DSF|&lt;1 corresponds to a
positive feedback, and DSF=1 corresponds to no feedback.
Advantages of negative feedback
From mathematical point of view, both the advantages and
disadvantages of negative feedback are related to the
denominator, 1+AOLβ or DSF, in (4).
For AOLβ&gt;&gt;1, the closed loop gain ACL becomes insensitive
to the open-loop gain AOL:
ACL = G
1
β
+ Gβ fwd .
(8)
ACL mainly depends on the small-signal input transmission G
and the transmissions β and βfwd of the feedback circuit.
For an arbitrary return ratio AOLβ and ACL&gt;&gt;D, we can find
the sensitivity of ACL to AOL as follows:
⎡
⎤
dACL
AOL β
1
= G⎢
−
2⎥
dAOL
⎣ 1 + AOL β (1 + AOL β ) ⎦
⎡ 1 + AOL β
⎤
AOL β
= G⎢
−
2
2⎥
(1 + AOL β ) ⎦
⎣ (1 + AOL β )
=G
AOL
1
1
1
. (9)
=G
2
(
1
)
(
1
)
β
β
+
+
A
A
A
(1 + AOL β )
OL4
OL
OL
14
424
3
ACL if ACL &gt;&gt; D
= ACL
⇒
1
1
1 + AOL β AOL
dACL dAOL
1
1
=
, or δACL =
δAOL
ACL
AOL 1 + AOL β
DSF
This means that the relative change in ACL is by a factor of
DSF lower than the relative change in AOL. Note that with no
feedback, like the case of Fig. 1(a), δACL = δAOL. Due to the
negative feedback, ACL becomes by a factor of DSF less
sensitive to AOL.
The main disadvantage of the negative feedback is related
to the frequency dependence of the return ratio AOLβ. If there
is a frequency where AOLβ =−1, then in (4), ACL=∞; the
feedback turns out to be positive, and the circuit (amplifier)
becomes unstable: it can produce a sustained output, for
example, sustained oscillations, with no input. We will study
such a behavior of transistor circuits in the lectures dedicated
to positive feedback oscillators.
Finding partial gains
To define the partial gains in (4)–(7) in terms of the signals
at the circuit input, ss, output, so, and the control terminals of
the dependent source, sε, we first assume that in a generic
single-transistor circuit (see Fig. 2) there is only a single
dependent source and, then, we will solve this circuit by
applying superposition. Yes, we will apply superposition
despite the fact that one in the sources in Fig. 2 depends on
the other. However, we will do it carefully.
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Lecture 8: Introduction to electronic analog circuits 361-1-3661
We have no difficulty to find the contribution of the ss
source to all the other signals in Fig. 2(a). To do this, we
simply suppress the dependent source as shown in Fig. 2(b).
However, when we have to find the contribution of the
dependent source aOLsε, we cannot simply suppress the
independent source ss. This is so because zeroing the ss source
also zeroes the aOLsε source and it contributes nothing. To let
the dependent source aOLsε to contribute the same way it does
in Fig. 2(a), before applying superposition, we have to
&quot;remind&quot; the aOLsε source in Fig. 2(c) its original value aOLsε,
which it had in Fig. 2(a). [If you do not like the word to
&quot;remind&quot; something to a dependent source, you can replace it
in Fig. 2(d) with an equivalent independent source having the
value aOLsε. Or you can assume that the dependent source in
Fig. 2(c) still depends on its original value in Fig. 2(a).]
We now can define
G≡
s' ε
,
ss
(11)
AOL ≡
so′′
,
sε
(12)
so′′
i =1
D
Single-transistor circuit
(a)
aOL=0
s'ε
s'o
ss
G
βfwd
RL
D
aOLsε
s&quot;ε
RR
β
AOL
s&quot;o
RL
Single-transistor circuit
(c)
sε′′
.
sε
n
∑
RL
(13)
(14)
In a general case, the number of independent signal sources
in a single-transistor circuit can be greater than one, but by
applying superposition, the circuit can always be solved for
each of them separately. Each independent source si will have
different Gi, βfwd, and Di gains, where i is the dependent
source number. The total closed-loop gain can be found in this
case as
AOL
DSF
β
so
βfwd
ss=0
,
RR ≡ AOL β ≡ −
ACL ≡
ss
AOL
Single-transistor circuit
(b)
so′
,
sε′
s ′ε′
RR
G
(10)
β fwd ≡
β ≡−
aOLsε
sε
aOLsε
s&quot;ε
RR
ss=0
β
AOL
s&quot;o
RL
Single-transistor circuit
(d)
n
Gi +
∑D .
i
(15)
i =1
It important to note that the solution for the dependent
source remains the same and AOL and DSF in (13) should be
found only once. Thus, the very important advantages of the
new generic approach are that it allows:
 to recognize the feedback in a single-transistor circuit,
 to find the partial transmissions and gains of the circuit,
 to distinguish between negative an positive feedback,
Fig. 2. Finding partial small-signal transitions and gains of a single-transistor
circuit.
 to solve the circuit in the simplest way: each time it is
solved for a single source only (this approach is most
powerful and insightful in noise analysis, when quite a
few independent equivalent noise sources are
connected to the circuit and contributions of all of them
should be found separately to see the dominant noise
source).
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Lecture 8: Introduction to electronic analog circuits 361-1-3661
VCC
Finding closed-loop gain: example circuit 1
Let us now to solve the elementary CC amplifier (see Fig.
3) for the closed-loop gain:
CC
vO
VBB
G≡
s ′ε
ss
β fwd ≡
=
i ′b
vs
s ′o
=
sε′
v s =1
= i ′b =
v ′o
ib′
i ' b =1
vs
1
,
hie + RE || ro
RE
(1)
B
= v ′o = RE || ro ,
C
(2)
ib
hfeib
hie
vs
AOL
s ′′ v ′′
≡ o= o
sε ib
RR ≡ AOL β ≡ −
= vo′′ = h fe ( hie || RE || ro ) ,
(3)
RE
i b =1
sε′′
i ′′
=− b
sε
ib i
= h fe
b =1
RE || ro
.
RE || ro + hie
(4)
B
h fe ( hie || RE || ro )
so
v
1
= o =
RE || ro
ss
v s hie + RE || ro 1 + h
fe
RE || ro + hie
RE
h ( R || r )
h fe ie E o
hie + ( RE || ro )
RE || ro
1
=
+
RE || ro
hie + RE || ro 1 + h
hie + RE || ro
fe
RE || ro + hie
=
⎡
⎤
h fe hie
RE || ro
&times;⎢
+ 1⎥
hie + RE || ro ⎢⎣ hie + RE || ro + h fe ( RE || ro ) ⎥⎦
=
⎡
⎤
h fe (1 + h fe ) re
RE || ro
&times;⎢
+ 1⎥ . (6)
hie + RE || ro ⎣⎢ (1 + h fe ) re + (1 + h fe )( RE || ro ) ⎥⎦
=
RE || ro
hie + RE || ro
=
RE || ro
re + RE || ro
re + ( RE || ro )
ro
(5)
1
( RE || ro )
hie + RE || ro
hie
647
4
8
h fe re + re + ( RE || ro )
v'o
E
In the exam, you do not have to simplify any solution for
ACL but we will do it here to be sure that the new solution is
identical to that obtained in Lecture 3.
ACL
hfe=0
hie
vs
.
+
ro
C
i'b
ACL ≡
vo
E
B
C
i&quot;b
hfeib
hie
vs=0
v&quot;o
E
RE
ro
Fig. 3. Example circuit 1: the elementary CC amplifier, solved with the π
model of the transistor.
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Lecture 8: Introduction to electronic analog circuits 361-1-3661
VCC
Finding closed-loop gain: example circuit 2
Let us now find the closed-loop gain for the same circuit
but by using the T model for the transistor.
CC
vO
VBB
G≡
s ′ε
ss
β fwd ≡
=
s ′o
sε′
i ′b
vs
=
v s =1
= i ′b =
v ′o
ib′
i ' b =1
1
,
re + RE || ro
vs
RE
(1)
C
= v ′o = RE || ro ,
(2)
hfeib
B
AOL
s ′′ v ′′
≡ o= o
sε ib
= vo′′ = 0 ,
ib
(3)
i b =1
re
vs
vo
E
RR ≡ AOL β ≡ −
ACL ≡
sε′′
i ′′
=− b
sε
ib i
= h fe .
RE
(4)
b =1
C
so
1
0
=
ss
re + RE || ro 1 + h fe
hfe=0
B
1
+
( RE || ro )
re + RE || ro
=
ro
.
i'b
(5)
re
vs
v'o
E
RE || ro
re + RE || ro
RE
ro
C
hfeib
B
vs=0
i&quot;b
re
v&quot;o
E
RE
ro
Fig. 4. Example circuit 2: the elementary CE amplifier, solved with the T
model of the transistor.
REFERENCES
[1]
P. R. Gray, P. J. Hurst, S. H. Lewis, and R. G. Meyer., Analysis and
Design of Analog Integrated Circuits. (4th Edition.)