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Analysis of the
Inverting Amplifier
Consider an inverting amplifier:
i2
R2

i1
vin
ide+
i- =0
D
Equ
atio
n.D
SM
T4
vin

R1
v1
v-

i- =0
v+
v2

ideal
oc
vout
+
l
Note that we use here the new notation v   v2 and v   v1 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
2/12
Pay attention to your TA!
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:
oc
vout
 Aop v   v  
Since the non-inverting terminal is grounded (i.e., v+ =0):
oc
vout
 Aop v 
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
3/12
First some KCL…
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.
i1  i2
Thus, we reach the conclusion that:
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
4/12
And then some Ohm’s law…
Likewise, we know from Ohm’s Law:
i1 
v1
R1
i2 
v2
R2
and also that:
And so combining:
v1 v 2

R1 R2
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
5/12
Followed by KVL…
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+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
6/12
And yet another KVL…
Likewise, we start at a potential of of v  volts (with respect to ground), we drop
oc
a potential of v 2 volts, and now we are at a potential of vout
volts (with respect
to ground).
oc
v   v2  vout

i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
oc
v2  v   vout
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
7/12
The feed-back equation
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
.
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
8/12
The feed-forward equation
We can combine this feed-back equation with the op-amp equation:
oc
vout
 v  Aop
This op-amp equation is likewise referred to as the feed-forward equation.
oc
Note this equation relates the output vout
in terms of v  .
We can combine the feed-back and feed-forward equations to determine an
oc
expression involving only input voltage vin and output voltage vout
:
oc
oc
R2 vin  R1 vout
vout

R1  R2
Aop
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
9/12
…and the open-circuit voltage gain appears!
oc
Rearranging this expression, we can determine the output 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:


Aop R2
Avo  lim 

Aop    R  R   A R 
2
op 1 
 1
R
 2
R1
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
10/12
Summarizing
Summarizing, we find that for the inverting amplifier:
Avo 
R2
R1
 R 
oc
vout
  2 vin
 R1 
i2
R2

i1
vin
R1

v1
v-

i- =0
v+
Jim Stiles
v2

ideal
oc
vout
+
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
11/12
The non-inverting terminal
is at ground potential
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:
Inserting the equation:
oc
R2 vin  R1 vout
v 
R1  R2
 R 
oc
vout
  2 vin
 R1 
we find:
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!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
5/12/2017
582759121
12/12
The logic behind the virtual short
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.
oc
We know that if op-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   !
Jim Stiles
The Univ. of Kansas
Dept. of EECS