Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Dynamic Presentation of Key
Concepts
Module 5 – Part 2
Op Amp Circuits with Feedback
Filename: DPKC_Mod05_Part02.ppt
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Overview of this Part
Op Amp Circuits with Feedback
In this part of Module 5, we will cover the
following topics:
• Identifying Negative Feedback
• The Inverting Configuration
• Comparison of Virtual Short Approach with
Equivalent Circuit Approach
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Textbook Coverage
This material is introduced in different ways in different
textbooks. Approximately this same material is covered in
your textbook in the following sections:
• Circuits by Carlson: Section 3.3
• Electric Circuits 6th Ed. by Nilsson and Riedel: Section 5.3
• Basic Engineering Circuit Analysis 6th Ed. by Irwin and
Wu: Section 3.3
• Fundamentals of Electric Circuits by Alexander and
Sadiku: Section 5.4
• Introduction to Electric Circuits 2nd Ed. by Dorf: Section
6.6
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Solving Op Amp Circuits
We have seen that we
can solve op amp
circuits by using two
assumptions:
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+.
If not, the output
saturates.
The key to using these assumptions is being
able to determine whether the op amp has
negative feedback. Remember that we have
negative feedback when a portion of the output
is returned to the input, and subtracted from it.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Negative Feedback Identification
We have seen that
we can solve op amp
circuits by using two
assumptions:
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+.
If not, the output
saturates.
For ideal op amps, we can assume
that the op amp has negative
feedback if there is a signal path
from the output to the inverting
input of the op amp.
Feedback Path
Inverting Input
Output
Noninverting Input
+
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Negative Feedback Identification
Most of the time,
this feedback path is
provided by using a
resistor between the
output and the
inverting input.
For ideal op amps, we can assume
that the op amp has negative
feedback if there is a signal path
from the output to the inverting
input of the op amp.
Feedback Path
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+.
If not, the output
saturates.
Inverting Input
Output
Noninverting Input
+
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Negative Feedback Identification
In general the rule is
this: If, when the output
voltage increases, the
voltage at the inverting
input also increases
immediately, then we
have negative feedback.
For ideal op amps, we can assume
that the op amp has negative
feedback if there is a signal path
from the output to the inverting
input of the op amp.
Feedback Path
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+.
If not, the output
saturates.
Inverting Input
Output
Noninverting Input
+
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Negative Feedback Identification
In general the rule is
this: If, when the output
voltage increases, the
voltage at the inverting
input also increases
immediately, then we
have negative feedback.
For ideal op amps, we can assume
that the op amp has negative
feedback if there is a signal path
from the output to the inverting
input of the op amp.
These are two different ways of saying the same thing. However,
for most students this becomes clearer once we see some
examples. We will look at one example in detail in this part, and
then more examples in Part 3.
Go back to
Overview
slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Inverting Configuration of the Op Amp
One of the simplest op amp amplifiers is
called the inverting configuration of the op
amp.
R
f
Ri
+
+
vi
+
vo
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Inverting Configuration of the Op Amp
The inverting configuration is distinguished by
the feedback resistor, Rf, between the output and the
inverting input, and the input resistor, Ri, between
the input voltage
and the inverting input.
R
The noninverting input
R
is grounded.
f
i
+
+
vi
+
vo
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Inverting Configuration of the Op Amp
Note that the feedback resistor, Rf, between
the output and the inverting input, means that
we have negative feedback.
Rf
Ri
+
+
vi
+
vo
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Inverting Configuration of the Op Amp
Note that the feedback resistor, Rf, between the
output and the inverting input, means that we have
negative feedback. Thus, we will have a virtual
short at the input of the op amp, and we can apply
our virtual-short rule, and get
Rf
v  v  0.
Ri
+
+
vi
+
vo
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
Let’s find the voltage gain, which is the
ratio of the output voltage vo to the input
voltage vi. To get this, let’s define two
currents, ii and if.
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
Next, since we know that the voltage v- is
zero, we can write that the current ii is
vi  v
ii 

Ri
vi  0 vi
ii 
 .
Ri
Ri
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
Following a similar approach, since we
know that the voltage v- is zero, we can write
that the current if is
v  vo
if 

Rf
0  vo vo
if 

.
Rf
Rf
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
Next, by applying KCL at the inverting
input terminal, we can write
ii  i  i f . Knowing from our op amp assumptions that
i = 0, we can write
Rf
ii  i f , or
vi
vo
 .
Ri
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
Finally, we solve for vo/vi, by dividing both
sides by vi, and then by multiplying both sides
by -Rf, and we get
vo
1

, and then
Ri
vi R f
Rf
vo
  .
Ri vi
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Gain for the Inverting Configuration
This is the result that we were looking for.
As implied by our analysis of negative
feedback, the gain is not a function of the op
amp gain at all. The gain is the ratio of two
R
resistor values,
f
Rf
vo
 .
vi
Ri
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Input Resistance for the Inverting
Configuration
Let’s find the quantity called the input
resistance of this amplifier, which is defined as
the Thevenin resistance seen by the source.
Here, the source is the voltage source vi.
Rf
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Input Resistance for the Inverting
Configuration
The Thevenin resistance seen by the source
will be the ratio of vi/ii. We have already
solved for ii, and found that
Rf
vi
ii  . Thus,
Ri
vi
Ri  .
ii
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
Let’s find the output resistance of this
amplifier, which is defined as the Thevenin
resistance seen by the load. The load is the
resistor RLOAD.
R
f
Ri
+
ii
+
vi
-
i+
+
if
+
vo
vv+
-
-
-
RLOAD
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
The Thevenin
resistance seen by the
load can be found by
setting all independent
sources equal to zero.
The voltage source
thus becomes a short
circuit. We then
applying a test source
at the output, in place
of the load. We do
this here, applying a
current source.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
Now, we are
solving for vo/it,
which is the output
resistance, Rout.
We know that
v- = 0, what we call
a virtual short, due
to the presence of
negative feedback.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
Now, we are
solving for vo/it, which
is the output
resistance, Rout.
We know that
v- = 0, due to the
presence of negative
feedback. Thus,
Rf
Ri
+
ii
vi  v
00
ii 
, or ii 
 0.
Ri
Ri
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
Now, we are
solving for vo/it,
which is the output
resistance, Rout.
We know that
i- = 0, due to our
first assumption.
Thus,
i f  ii  i  0.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the
Inverting Configuration
Now, we are solving
for vo/it, which is the
output resistance, Rout.
Next, we write KVL
around the loop marked
with a dashed green
line. We get,
v  i f R f  vo  0, or
0  0 R f  vo  0, or
vo  0.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Go back to
Overview
slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Output Resistance for the Inverting
Configuration
Now, we are
solving for vo/it,
which is the output
resistance, Rout.
Since vo = 0, we
have
Rout
vo 0
   0.
it it
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Testing the Virtual Short Assumption
Let’s test the results we have obtained, so test the virtual
short assumption. We found the gain,
R
input resistance,
and output resistance
R
+
i
for this configuration.
+
i
Let’s check our
v
i
+
A(v -v ) +
v
approach, by going
v
+
R
back to the original
v
equivalent circuit for
the op amp. That is,
we replace the op amp
with a dependent source.
f
i
f
-
o
i
+
i
-
-
LOAD
+
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Testing the Virtual-Short Assumption
We can solve this circuit for the gain, vo/vi. This
requires some circuit analysis, but when we do so, we get
AR f
vo

.
vi
ARi  R f  Ri
If we take the limit as A
goes to infinity, we get
the same answer we had
before,
Rf
vo
 .
vi
Ri
Rf
Ri
+
ii
+
vi
-
v-
+
if
i-
vo
A(v+-v-) +
+
-
RLOAD
v+
-
-
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Testing the Virtual-Short Assumption
We can solve this circuit for the input resistance, vi/ii.
This requires some analysis, but when we do so, we get
vi ARi  R f  Ri    Ri  R f 

.
ii
A  R f  Ri   R f  Ri
2
If we take the limit as A
goes to infinity, we get
the same answer we had
before,
vi
 Ri .
ii
Rf
Ri
+
ii
+
vi
-
v-
if
+
i-
vo
A(v+-v-) +
+
-
RLOAD
v+
-
-
-
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Testing the Virtual-Short Assumption
We can solve this circuit for the output resistance, vt/it,
for the circuit below. This requires some analysis, but
when we do so, we get
vt
 0.
it
In this case, we do
not have to take
the limit as A goes
to infinity to get
the same answer
we had before.
Rf
Ri
+
ii
v-
if
+
iA(v+-v-) +
+
vt
v+
-
-
-
it
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Is the Virtual-Short Assumption Really Valid?
• This is a good question.
• You can check this by performing the
solutions with actual values for real op amps.
Try an open loop gain A of 106, and resistor
values such as Rf = 10[kW] and Ri = 1[kW],
and see how close your answers are.
• You can also check this by building an op
amp circuit, and measuring the actual gain,
and other parameters. You might
be surprised by how accurate
Go back to
the virtual-short assumption is.
Overview
slide.
Dave Shattuck
University of Houston
© Brooks/Cole Publishing Co.
Is the Virtual-Short Assumption
Really Valid?
• Strictly speaking, the answer is no. It is only
an approximation.
• However, it is such a good approximation,
we are going to use it for the rest of this
module, in all cases where negative feedback
is present.
Go back to
Overview
slide.