Download Analysis of the

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Analysis of the
Inverting Amplifier
Consider an inverting amplifier:
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
v2

ideal
oc
vout
+
Note that we use here the new notation v   v2 and v   v1 .
Now what is the open-circuit voltage gain of this inverting
amplifier?
Let’s start the analysis by writing down all that we know. First,
the op-amp equation:
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oc
vout
 Aop v   v  
Since the non-inverting terminal is grounded (i.e., v+ =0):
oc
vout
 Aop v 
Now let’s apply our circuit knowledge to the remainder of the
amplifier circuit. For example, we can use KCL to determine
that:
i1  i  i2
However, we know that the input current i- of an ideal op-amp
is zero, as the input resistance is infinitely large.
Thus, we reach the conclusion that:
i1  i2
Likewise, we know from Ohm’s Law:
i1 
v1
R1
i2 
v2
R2
and also that:
And so combining:
v1 v 2

R1 R2
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Finally, from KCL we can conclude:
vin  v1  v 
v1  vin  v 

In other “words”, we start at a potential of vin volts (with
respect to ground), we drop a potential of v1 volts, and now we
are at a potential of v  volts (with respect to ground).
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
v2

oc
vout
ideal
+
Likewise, we start at a potential of of v  volts (with respect to
ground), we drop a potential of v 2 volts, and now we are at a
oc
potential of vout
volts (with respect to ground).
oc
v   v2  vout

oc
v2  v   vout
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i2
R2

i1
vin
R1

v1
v-

i- =0
v+
4/7
v2

ideal
oc
vout
+
Combining these last three equations, we find:
oc
vin  v  v   vout

R1
R2
Now rearranging, we get what is known as the feed-back
equation:
oc
R2 vin  R1 vout
v 
R1  R2
oc
Note the feed-back equation relates v  in terms of output vout
.
We can combine this feed-back equation with the op-amp
equation:
oc
vout
 v  Aop
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This op-amp equation is likewise referred to as the feedoc
forward equation. Note this equation relates the output vout
in
terms of v  .
We can combine the feed-back and feed-forward equations to
determine an expression involving only input voltage vin and
oc
output voltage vout
:
oc
oc
R2 vin  R1 vout
vout

R1  R2
Aop
Rearranging this expression, we can determine the output
oc
voltage vout
in terms of input voltage vin .


Aop R2
oc
vout

v
 R1  R2   Aop R1  in


and thus the open-circuit voltage gain of the inverting
amplifier is:
oc


Aop R2
vout
Avo 


vin  R1  R2   Aop R1 
Recall that the voltage gain A of an ideal op-amp is very large—
approaching infinity. Thus the open-circuit voltage gain of the
inverting amplifier is:
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

Aop R2
Avo  lim 

Aop    R  R   A R 
2
op 1 
 1
R
 2
R1
Summarizing, we find that for the inverting amplifier:
Avo 
R2
R1
 R 
oc
vout
  2 vin
 R1 
One last thing. Let’s use this final result to determine the value
of v-, the voltage at the inverting terminal of the op-amp.
Recall:
oc
R2 vin  R1 vout
v 
R1  R2
Inserting the equation:
 R 
oc
vout
  2 vin
 R1 
we find:
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R
R2 vi  R1  2 vin
R1 

v 
R1  R2
R v  R2 vin
 2 in
R1  R2
0
The voltage at the inverting terminal of the op-amp is zero!
Thus, since the non-inverting terminal is grounded (v2 =0), we
find that:
v  v
and 
v  v  0
Recall that this should not surprise us, as we know that if opoc
amp gain Aop is infinitely large, its output vout
will also be
infinitely large (can you say saturation?), unless v+ - v- is
infinitely small.
We find that the actual value of v+ - v- to be:
oc
vout
R2
v  v  

vin
Aop Aop R1
a number which approaches zero as Aop   !