Download Problem 4.5 For the op-amp circuit shown in Fig. P4.5: (a) Use the

Problem 4.5 For the op-amp circuit shown in Fig. P4.5:
(a) Use the model given in Fig. 4-4 to develop an expression for the current gain
Gi = iL /is .
(b) Simplify the expression by applying the ideal op-amp model by (taking A → ∞,
Ri → ∞, and Ro → 0).
+
_
υp
υn
R1
Rs
is
iL
RL
Figure P4.5: Circuit for Problem 4.5.
Solution: (a) We start by replacing the op amp in Fig. P4.5 with its equivalent model,
and to simplify the analysis, we will convert the parallel combination of (is , Rs ) into
a voltage source υs = is Rs and a series resistor Rs .
Rs
υp
+
+
isRs _
(υp - υn)
Ro
Ri
-
+
_
−
υn
+
υo
A(υp - υn)
in
R1
υn
i1
iL
RL
At node υn ,
i1 + iL + in = 0
(1)
υn − A(υp − υn )
R1 + Ro
υn
iL =
RL
υn − is Rs
in =
Rs + Ri
i1 =
Hence,
υn − A(υp − υn ) υn υn − is Rs
+
+
=0
R1 + Ro
RL
Rs + Ri
Additionally,
υp − υn = −Ri in = −Ri
υn − is Rs
Rs + Ri
(2)
(3)
c
All rights reserved. Do not reproduce or distribute. 2013
National Technology and Science Press
Using Eq. (3) in Eq. (2) and then solving for υn leads to:
(ARi + R1 + Ro )is Rs RL
.
(Rs + Ri )(RL + R1 + Ro ) + RL (ARi + R1 + Ro )
υn
,
iL =
RL
υn =
and
Gi =
υn
iL
(ARi + R1 + Ro )Rs
=
=
.
is
RL is (Rs + Ri )(RL + R1 + Ro ) + RL (ARi + R1 + Ro )
(b) For the ideal op amp,
A ≈ 106
and Ri ≈ 106 Ω,
so the product of the two is many orders of magnitude larger than all other products.
Hence,
ARi Rs
Rs
Gi ≃
=
.
ARi RL RL
c
All rights reserved. Do not reproduce or distribute. 2013
National Technology and Science Press