each of Rf V1 R+R For source -A, V, only (V set to zero), RL (-A,V) %3D R+ R, The total voltage V, is then V, = V + V R1 (-A,V) %3D R +R R + Rf which can be solved for V, as Rf (10.7) Vi Rf + (1 + A)R1 If A, >1 and A, R > R, as is usually true, then Vi = -V1 A,R1 Solving for V/V, we get -A,V-A, RV1 V; A,R Ry V R Vi %3D %3D Vi Vi so that Rf (10.8) R1

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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proof the equation 10.8 which represents the gain of inversting amplifer ... the proof is already exist in page 621 but there are some expressions were missed before the equation 10.7, you should start this proof from the begining and adding that missed expressions.
each of the sources. PU
V
R+ R
For source -A, V, only (V set to zero),
RL
(-A,V)
R+ R
The total voltage V; is then
V; = V + Vig
R1
-V+
(-A,V)
%3D
R + R;
R + Rj
which can be solved for V, as
(10.7)
-V1
Rf + (1 + A,)R1
V, =
If A, >1 and A, R > R as is usually true, then
Rf
V =
A,R1
Solving for V/V, we get
V
Ry Vi
-A, RV1
V A,R1
-A,Vi
%3D
%3D
R1 Vị
so that
Vo
Rf
(10.8)
VI
R1
Transcribed Image Text:each of the sources. PU V R+ R For source -A, V, only (V set to zero), RL (-A,V) R+ R The total voltage V; is then V; = V + Vig R1 -V+ (-A,V) %3D R + R; R + Rj which can be solved for V, as (10.7) -V1 Rf + (1 + A,)R1 V, = If A, >1 and A, R > R as is usually true, then Rf V = A,R1 Solving for V/V, we get V Ry Vi -A, RV1 V A,R1 -A,Vi %3D %3D R1 Vị so that Vo Rf (10.8) VI R1
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