8 2 4
1
CHAPTER 8
THE S H U N T - S H U N T A N D SHUNT-SERIES FEEDBACK AMPLIFIERS
8.6
FEEDBACK
A circuit
(a)
The A circuit is
nr
O
+
R, <
V,
Ru
R
ZJ
22
-WV—0 -
J
I_.
in
•~1
r"
O'
A
y'
Ri
R. of
where R
n
and R
is obtained from
o
Feedback
network
©
22
is obtained from
©
Feedback
network
I
ß circuit
and the gain A is defined as
A = j-
FIGURE 8 . 2 2 Ideal structure for the shunt-series feedback amplifier.
(b)
ß is obtained from
V,
li;.vc
Feedback
network
= 0 ®
¡¡¡¡¡ÍIIM
v¡ =
©
Feedback
network
0
FIGURE 8 . 2 4 Finding the A circuit and B for the current-mixing current-sampling (shunt-series) feedback
amplifier of Fig. 8.23.
and that m o s t of the r e v e r s e transmission takes place in the feedback network,
FIGURE 8 . 2 3 Block diagram for a practical shunt-series feedback amplifier.
^12! basic
amplifier
A g a i n w e note that the shunt connectio n at the input r e d u c e s the input resistance. T h e output
resistance w i t h feedback is the resistance seen by breaking the output circuit, such as between
O and O ' , and looking b e t w e e n the t w o terminals thus generate d (i.e., b e t w e e n O a n d O')T h i s resistance, R , is given b y
(8.46)
Ic?12| feedback
^
network
Finally, w e note that o n c e R and R h a v e b e e n d e t e r m i n e d using the feedback equations
(Eqs. 8.43 a n d 8.44), the input and output resistances of the amplifier proper, R and R
(Fig. 8.23), can b e found as
if
of
m
out
of
R
of
= R (l+AB)
1
(8.44)
a
if
R
w h e r e w e n o t e that the increase in output resistance is
G i v e n a practical s h u n t - s e r i e s feedback amplifier,
d i a g r a m of Fig. 8.23, w e follow the m e t h o d given in
H e r e agai n the analysis m e t h o d is predicated on the
t r a n s m i s s i o n occurs in the basic amplifier,
d u e to the current (series) sampling.
such as that represented b y the block
Fig. 8.24 in order to obtain A and B.
assumption that m o s t of the forward
^out
=
1
R
(8.47)
s
(8.48)
Rof~RL
5
IS21I feedback
network
^
\82l\
basic
Figure 8.25 shows a feedback circuit of the shunt-series type. Find I /I__,
the transistors to have B = 100 and V = 75 V.
out
(8.45)
R , and R _ . Assume
m
0
t
A
amplifier
Solution
5
For this amplifier topology, the most convenient set of two-port parameters to use is the set of g
parameters; it is the only set that provides a representation that is composed of a parallel circuit at the
input and a series circuit at the output (see Appendix B).
I
We begin by determining the dc operating points. In this regard we note that the feedback
signal is capacitively coupled; thus the feedback has no effect on dc bias. Neglecting the effect
CHAPTER 8
THE S H U N T - S H U N T A N D SHUNT-SERIES FEEDBACK AMPLIFIERS
8.6
FEEDBACK
The required input resistance R
of finite transistor ¿8 and V , the dc analysis proceeds as follows:
is given by (see Fig. 8.25b).
in
A
y
12——
=
B l
100+15
= 1.57 V
= 1 . 5 7 - 0 . 7 = 0.87 V
I
= 0.87/0.87 = 1 mA
El
= 7~~
\IR -\IR
in
if
V
E1
R
Since R
m
- R
s
W;
y
i 4 r -
=
-
3
f
8
7
A
/
A
= 12-10x1 = 2 V
c l
V
Note that because Ap > 1 the closed-loop gain is approximately equal to 1 / / ? .
Now, the required current gain is given by
= 2 - 0 . 7 = 1.3 V
E2
7
= 29.5 Q
s
it follows from Fig. 8.25(b) that 4 = I . The current gain A is given by
if>
£ 2
V
c2
= 1 . 3 / 3 . 4 ~ 0.4 m A
{out
= 1 2 - 0 . 4 x 8 = 8.8 V
7
Rg
{out _
I
jn
[c
L
C2
Rg
I_o
R +
R I
=
R +R I
s
S
L
C2
S
Thus,
The amplifier equivalent circuit is shown in Fig. 8.25(b), from which w e note that the feedback network is composed of R
f
2
c
gain, 7
0 U t
/ / , will be slightly different than the closed-loop current gain A
i n
f
=
0
xl
= flRJ/iR^
V
b2
= -g iV, {r ///? //[r
m
r
e2
+
1
+
c l
J [ 2
R
of
= R (l+AB)
-
0
An estimate of the required output resistance i ?
140.1 k Q
can be obtained using the technique employed
out
in Example 8.2, namely, by considering that the effect of feedback is to place a resistance
R h\
of
2
+ (j8 + l ) ( / ? / / / ? ) ] }
£2
*out =
/
Substituting, r
results in
(R IIR )
E2
= "3-44 A / A
is given by
the emitter of Q (see Fig. 8.25e). Thus, using Eq. (6.78), w e can write
RfV/RJ/r^]
o l
of
s
feedback network using the rules of Fig. 8,24. For the A circuit we can write
V
The output resistance R
I /I .
T h e A circuit is shown in Fig. 8.25(c), where w e have obtained the loading effects of the
0
W i n
and R . The feedback network samples the emitter current of
E2
Q , I„, which is approximately equal to the collector current I . Also note that the required current
f
o2
u2
o2i^+g i{rJIR )]
m
R
where we have neglected the effect of r . These equations can be combined to obtain the openloop current gain A,
r
= 7 5 / 0 . 4 = 187.5 k Q , g
oat
m2
of
= 16 m A / V , r
= 6.25 k Q , and R
n2
of
= 140.1 k Q ,
= 18.1 M O
Thus, while negative feedback considerably increases R ,
mt
the increase is not by the factor (1 + A/3),
simply because the feedback network samples the emitter current and not the collector current.
A= — — - 2 0 1 . 4 5 A/A
h
Thus, in effect, the feedback network "does not k n o w " about the existence of r .
o2
T h e input resistance R is given by
t
Ri = RJI(R
+ R )IIR llr
E2
f
B
= 1.535 k Q
nl
The output resistance R is that found by looking into the output loop of the A circuit between
0
nodes Y and Y' (see Fig. 8.25c) with the input excitation l
effect of r
o2
t
set to zero. Neglecting the small
it can be shown that
1.7 Use the feedback method to find the voltage gain V /V , the input resistance R- . and the output resistance R
of the inverting o p - a m p configuration of Fig. E 8 . 7 . Let the o p a m p have o p e n - l o o p gain
H = 1 0 V/V, R = 100 k Q , and r„ = 1 k Q . (Hint: T h e feedback is of the shunt-shunt type.)
0
s
m
om
4
u
R
0
= (R f/ )
E2
Rf
+r
e2
+
jj^
R
= 2.69 k Q
R, =
R j l
I
P
0
=
^ ? _ = - M = _
R + Rf
13.4
E2
0.254
1 kil
- W v
T h e circuit for determining ¿8 is shown in Fig. 8.25(d), from which w e find
- v.
O
: A'
:
ki»
Thus,
1 + A / 3 = 52.1
F I G U R E ES.7
The input resistance R is given by
if
Ans. -
\
\ ; | s i i I)
:
o2 11
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