Download 3355LectureSet04v24

ECE 3355 Electronics
Lecture Notes
Set 4 -- Version 24
Operational Amplifiers
Dr. Dave Shattuck
Dept. of ECE, Univ. of Houston
Dave Shattuck
University of Houston
© University of Houston
Overview of this Part
Fundamentals of Operational Amplifiers
In this part, we will cover the following
topics:
• Basic Operational Amplifier
Requirements
• Equivalent Circuit for Operational
Amplifiers
• Negative Feedback and What it Does
Dave Shattuck
University of Houston
© University of Houston
Operational Amplifiers
(Op Amps)
Operational Amplifiers
(op amps) are devices
that amplify voltages.
Because of the way the
op amps are built, they
facilitate the application
of negative feedback,
which in turn allows
• easy design of special
applications, and
• for the op amps to
behave very ideally.
Operational amplifiers are most useful
because of their ability to process signals
easily. One example: An audio mixer
provides the functions of providing variable
gain (amplification), and adding different
signals together with separate gains for
each signal. Op amps can perform these
functions, and do so in an easy to use and
design form.
Op Amps:
A Structural Definition
Dave Shattuck
University of Houston
© University of Houston
An op amp is a differential input, single-ended
output, amplifier. The schematic symbol for the op
amp has three terminals that are always shown:
1. Inverting input (which should not be called the
negative input)
2. Noninverting input (which should not be called
the positive input)
3. Output (which can be called the output)
Inverting Input
Output
Noninverting Input
+
Schematic Symbol for the Op Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Structural Definition
This means
that at the
input, the
difference in
voltage
between the
input terminals
is used.
An op amp is a differential
input, single-ended output,
amplifier.
Inverting Input
Output
Noninverting Input
+
Schematic Symbol for the Op Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Structural Definition
This means
that at the
output, the
voltage is
obtained with
respect to a
reference,
usually called
ground.
An op amp is a differential
input, single-ended output,
amplifier.
Inverting Input
Output
Noninverting Input
+
Schematic Symbol for the Op Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Structural Definition
There are actually five terminals which are always present in an op
amp. These are shown in the more complete schematic shown below.
The dc power supplies must be connected for the op amp to work. They
may not be shown, since they do not affect the signal behavior in many
cases. However, the connections must be present, whether they are
shown in the schematic or not.
Positive dc
power supply
Inverting Input
Output
Noninverting Input
+
Negative dc
power supply
Complete Schematic Symbol for the Op Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Structural Definition
It is important to note that, in the positive and negative dc power
supplies, the “positive” and “negative” here are relative. The voltages for
these terminals may have any relationship to ground. For example, the
voltages could be:
Positive dc
• +15[V] and ground,
power supply
• ground and -15[V],
• +15[V] and -15[V], or
• +15[V] and +5[V].
Inverting Input
Output
Noninverting Input
+
Negative dc
power supply
Complete Schematic Symbol for the Op
Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Structural Definition
Actually, most op amps have at least two more terminals for use in
correcting for some of the non-ideal characteristics of the op amp.
However, for this course, we will assume that our op amps are ideal, and
will not use these additional terminals.
Positive dc
power supply
Inverting Input
Output
Noninverting Input
+
Negative dc
power supply
Complete Schematic Symbol for the Op Amp
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Functional Definition
Op amps take the difference between the voltages at the two inputs,
and amplify it by a large amount, and provide that voltage at the output
with respect to ground. This can be shown with an equivalent circuit.
Equivalent Circuit for
the Op Amp
+
vi
+
+
Avi
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Functional Definition
The effective resistance between the input terminals, and the resistances
between the input terminals and ground, are typically large compared to
other resistances in the op-amp circuits, and can often be large enough to
be considered effectively infinite. The output resistance is small enough to
be ignored.
Equivalent Circuit for the Op Amp
Ignoring all these
resistances gives us the
relatively simple equivalent
vcircuit at right. In some
situations, these resistances
can be estimated and
included in the equivalent
+
vo
circuit to make it more
Av
i
accurate. For the purposes
of this course, this model will
vi
be accurate enough for all
+
the problems we consider.
v+
+
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Functional Definition
Note that the output is a function of only the difference
between the inputs. This means that if
v- = v+ = 500,000[V],
the output should be zero volts.
This is hard to obtain
in practice, and is
called common
mode rejection. The
part of the two
voltages, v- and v+,
that are common, is
called the common
mode.
Equivalent Circuit for the Op Amp
v-
-
+
-
v+
vi
+
+
Avi
vo
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Functional Definition
The coefficient in the dependent source is the
gain.
The gain A is called
the differential gain,
and also called the
open loop gain, for
reasons that will
become obvious
soon.
Equivalent Circuit for the Op Amp
v-
-
+
-
v+
vi
+
+
Avi
vo
Dave Shattuck
University of Houston
© University of Houston
Op Amps:
A Functional Definition
Conceptually, the gain A is a function of
frequency.
The response is
good for all
frequencies, even at
dc, in the ideal case.
The value for the
gain, A, is very large,
in general.
Equivalent Circuit for the Op Amp
v-
-
+
-
v+
vi
+
+
Avi
vo
Op Amps:
A Functional Definition
Dave Shattuck
University of Houston
© University of Houston
The output voltage is limited. The output voltage cannot be
higher than the positive dc power supply voltage (+VDC), and
cannot be lower than the negative dc power supply
voltage (-VDC).
vO
Equivalent Circuit for the Op Amp,
for Region Marked in Red
Saturation
+VDC
+VDC
iv-
-
vI
-VDC
Saturation
In this region,
the op amp
behaves like a
dependent
source
+
i+
v+
Avi
vi
+
+
-VDC
vo
Dave Shattuck
University of Houston
© University of Houston
Solving Op-Amp Circuits
We will use two assumptions for analysis and
design of op amp circuits where the op amp can be
considered to be ideal.
+V
DC
The Two Assumptions
1) i- = i+ = 0. These currents
are small due to the high input
resistances.
2) If there is negative
feedback, then v- = v+. If there
is no negative feedback, the
op amp output will saturate. If
vi is positive, it saturates at
+VDC, and if vi is negative, it
saturates at –VDC.
iv-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
Solving Op-Amp Circuits
We use these two assumptions for the analysis and design
of op amp circuits where the op amp can be considered to be
ideal. While the equivalent circuit may seem to be easier, we
will see that these assumptions make solving op amp circuits
much easier.
+V
DC
The Two
Assumptions
1) i- = i+ = 0.
2) If there is
negative feedback,
then v- = v+. If not,
the output saturates.
iv-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
First Assumption
The first assumption results from having large resistances
at the input, larger than the resistance values typically
connected to them. This assumption is not conditional; it
happens whether negative feedback is present or not.
The Two
Assumptions
1) i- = i+ = 0.
2) If there is
negative
feedback, then
v- = v+. If not, the
output saturates.
+VDC
iv-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
Second Assumption
The second assumption results from negative feedback
and the very large gain of the op amp. This is called the
virtual short, or the summing-point constraint. The two input
voltages are constrained to be equal by the presence of
negative feedback. Without negative feedback, even a small
input will saturate the output.
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+.
If not, the output
saturates.
+VDC
iv-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
Is This Reasonable?
Many students who are seeing this for the first time have
little trouble accepting the first assumption. It seems
reasonable to be able to have large input resistances.
However, the notion that the input voltage vi will be forced to
zero by something called negative feedback is harder to
accept. Some of these students are troubled by the notion
that the input would be zero, which is then multiplied by a very
large number to get a finite,
+V
nonzero output.
DC
i-
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+. If
not, the output saturates.
v-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
Is This Reasonable? Yes!
The notion that the input voltage vi is forced to zero by
something called negative feedback is an approximation. It is
actually forced to be very small, because the gain is so large.
Thus, a very small input, which is almost zero, is then
multiplied by a very large number, to get a finite, nonzero
output.
To understand this better, we need to understand negative
feedback better.
+V
DC
i-
The Two Assumptions
1) i- = i+ = 0.
2) If there is negative
feedback, then v- = v+. If
not, the output saturates.
v-
vi
vo
+ i+
v+
+
-VDC
Dave Shattuck
University of Houston
© University of Houston
Negative Feedback – Signal
Flow Diagrams
Engineers have developed a way of looking at signals
called the signal flow diagram. This is not a schematic, and
does not represent wire and specific components. A line
represents a path that a signal might follow. The signals can
be voltages or currents. Therefore, we will label the signals
with the symbol x.
In the signal flow diagram shown below, there is an input
signal, xi. This signal flows into an amplifier with gain A, which
is shown with a triangle. This produces an output signal xo.
The input is multiplied by the gain, to give the output.
xi
A
xo=Axi
Signal Flow Diagram
Dave Shattuck
University of Houston
© University of Houston
Negative Feedback – Signal
Flow Diagrams
Now, let’s add negative feedback to our signal flow
diagram. In the signal flow diagram shown below, we add
another amplifier. This amplifier has a gain which is
conventionally called b. This amplifier amplifies the output
signal, to produce a feedback signal, xf. Finally, this feedback
signal is subtracted from the input signal. The symbol for this
action is called a summing point or a summing junction. The
signs at the junction indicate the signs for the summation.
xs
+
xi=xs-xf
S
xo=Axi
A
xf=bxo
b
Dave Shattuck
University of Houston
© University of Houston
Negative Feedback – Definition
At this point, we can define negative feedback. Negative
feedback is when a portion of the output is taken, returned to
the input, and subtracted from this input.
If we were to add it to the input, we would call it positive
feedback.
xs
+
xi=xs-xf
S
xo=Axi
A
xf=bxo
b
Dave Shattuck
University of Houston
© University of Houston
Negative Feedback – Notes
The feedback amplifier, with a gain of b, is typically not an
amplifier per se, but rather is a resistive network. In any case,
the key is that the feedback signal xf is proportional to the
output signal, with a multiplier equal to b.
The gain A is called the open loop gain, because this
would be the gain if the loop were to be opened, that is, if the
feedback were removed.
xs
+
xi=xs-xf
S
xo=Axi
A
xf=bxo
b
Dave Shattuck
University of Houston
Gain with Negative
Feedback
© University of Houston
Now, let’s solve for the gain with negative feedback, which is xo/xs. We
start by writing an equation for the summing junction, taking into account
the signs, to get
where the second equation comes by using
xi  xs  x f  xs  bxo , the definition of the feedback gain b. Next,
we use a similar definition for the feedforward gain, A, to write
xo  Axi .
We then substitute the first equation into the
second to get
xo  A xs  bx0 .
We can combine terms,
then we can divide
through by xs, and then
by (1+Ab), to get
xs
xo
A

.
xs 1  Ab
+
xi=xs-xf
S
xo=Axi
A
xf=bxo
b
Dave Shattuck
University of Houston
© University of Houston
Gain with Negative Feedback
This is the gain with negative feedback
xo
A

.
xs 1  Ab
If we take the case where A is very large, and it usually is,
we can get a special situation. Specifically, take the case
where Ab >> 1. Then,
1  Ab  Ab ,
and we can use this approximation to simplify the gain
with feedback, which we call Af, to
xo
A
1
 Af 
 .
xs
Ab b
Dave Shattuck
University of Houston
© University of Houston
Gain with Negative Feedback
Thus, the gain with negative feedback, Af, is
Af 
1
b
.
The only requirement is that Ab >> 1. Thus, the gain
is not a function of A at all!?! This is a seemingly
bizarre, but wondrous result, which is fundamental to
the power of negative feedback. The gain of the op
amp, which changes from time to time, and from op
amp to op amp, does not affect the overall gain with
feedback.
Dave Shattuck
University of Houston
© University of Houston
Gain with Negative Feedback
Thus, the gain with negative feedback, Af, is
Af 
1
b
.
Thus, the gain is not a function of A at all!?! The gain of the op
amp does not affect the overall gain with feedback. The overall
gain, Af, is determined by the way feedback is applied.
Feedback is used to allow gain to be traded off for a
variety of desirable results. When we use op amps, we have a
relatively simple way to determine the presence of negative
feedback: If there is a signal path between the output of the op
amp, and the inverting input, there will be negative feedback.
Dave Shattuck
University of Houston
© University of Houston
Gain with Negative Feedback
Thus, the gain with negative feedback, Af, is
Af 
1
b
.
With this result, we can look again at the signal flow diagram.
The input to the op amp, vi, is the output divided by the gain,
vo/A. If A is large, then vi will be much less than vo, and can
usually be neglected.
xs
+
xi=xs-xf
S
xo=Axi
A
xf=bxo
b
This is
what we
call the
virtual
short.
Dave Shattuck
University of Houston
© University of Houston
How do we use this?
• This is a good question.
• We will use the two assumptions to solve op
amp circuits more quickly. We will show how
to do this in the next part.
• The point to recognize here is that negative
feedback can be very useful, and makes op
amps circuits much easier to analyze, and
therefore much easier to design with. Many
circuits that we look at will have negative
feedback. However, we will always
check for negative feedback when we
solve op amp problems.
Dave Shattuck
University of Houston
© University of Houston
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
© University of Houston
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.
Negative Feedback
Identification
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
© University of Houston
Most of the time,
this feedback path
is provided by
using a resistor.
Negative Feedback
Identification
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
Negative Feedback
Identification
In general the rule is
© University of Houston
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
Inverting Input
Output
Noninverting Input
+
Dave Shattuck
University of Houston
© University of Houston
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
next, and then more examples after that.
Dave Shattuck
University of Houston
© University of Houston
Inverting Configuration of
the Op Amp
One of the simplest op amp amplifiers is
called the inverting configuration of the op
amp.
Rf
Ri
+
vi
-
+
+
vo
-
Dave Shattuck
University of Houston
Inverting Configuration of
the Op Amp
© University of Houston
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. The noninverting
input is grounded.
Rf
Ri
+
vi
-
+
+
vo
-
Dave Shattuck
University of Houston
Inverting Configuration of
the Op Amp
© University of Houston
Note that the feedback resistor, Rf, between
the output and the inverting input, means that
we have negative feedback.
Rf
Ri
+
vi
-
+
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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,
v  v  0.
Rf
Ri
+
vi
-
-
+
v-
+
v+
-
+
+
vo
-
Dave Shattuck
University of Houston
Gain for the Inverting
Configuration
© University of Houston
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
+
vi
-
-
+
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
Gain for the Inverting
Configuration
Next, since we know that the voltage vis zero, we can write that the current ii is
vi  v
ii 

Ri
vi  0 vi
ii 
 .
Ri
Ri
Rf
Ri
+
+
vi
-
-
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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
+
+
vi
-
-
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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
ii  i f , or
vi
vo
 .
Ri
Rf
Rf
Ri
+
+
vi
-
ii
v-
i+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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
Rf
vo
  .
Ri vi
Ri
+
+
vi
-
ii
v-
i+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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 resistor values,
Rf
vo
 .
vi
Ri
Rf
Ri
+
+
vi
-
ii
v-
i+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
Input Resistance for the Inverting
Configuration
Let’s find the input resistance of this amplifier,
which is defined as the Thevenin resistance seen by
the source. The source is not shown here, but is
assumed to be at the input. Here, we will take the
source as the terminals connected to vi.
Rf
Ri
+
vi
-
-
+
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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
vi
ii  . Thus,
Ri
vi
Ri  .
ii
Rf
Ri
+
vi
-
-
+
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
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 not shown here, but is assumed
to be at the output. Here, we will take the load as the
terminals connected to vo.
Rf
Ri
+
vi
-
-
+
ii
v-
+
v+
-
+
if
+
vo
-
Dave Shattuck
University of Houston
© University of Houston
Output Resistance for the Inverting
Configuration
The Thevenin
resistance seen by
the load can be
found by setting all
independent
sources equal to
zero, and then
applying a test
source at the
output. We do this
here, applying a
current source.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© University of Houston
Output Resistance for the Inverting
Configuration
Now, we solve
for vo/it, which is
the output
resistance, Rout.
We know that
v- = 0, due to the
presence of
negative
feedback.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© University of Houston
Output Resistance for the Inverting
Configuration
Now, we solve
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
© University of Houston
Output Resistance for the Inverting
Configuration
Now, we solve
for vo/it, which is
the output
resistance, Rout.
We know that
i- = 0, due our
first assumption.
Thus,
i f  ii  i  0.
Rf
Ri
+
ii
v-
i+
v+
-
if
+
+
vo
it
-
Dave Shattuck
University of Houston
© University of Houston
Output Resistance for the
Inverting Configuration
Now, we solve for vo/it, which is the
output resistance, Rout.
R
Next, we write
KVL around the
R
+
loop marked with
i
i
+
v
+
a dashed green line.
v
Thus,
i
f
if
+
-
i
+
v  i f R f  vo  0, or
0  0 R f  vo  0, or
vo  0.
vo
-
it
-
Dave Shattuck
University of Houston
© University of Houston
Output Resistance for the Inverting
Now, we solve Configuration
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
© University of Houston
Testing the Virtual Short
Assumption
Let’s test the results we have obtained, so test the
virtual short assumption. We found the gain,
input resistance,
R
and output resistance
for this configuration.
R
i
Let’s check our
+
+
i
+
approach, by going
i
v
v
+
v
v
back to the original
+
equivalent circuit for
A(v -v )
the op amp. That is,
we replace the op amp
with a dependent source.
f
i
f
-
i
i
-
o
+
+
-
Dave Shattuck
University of Houston
© University of Houston
Testing the Virtual Short
Assumption
Solving this circuit for the gain, vo/vi, we get
Rf
Ri
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.
+
+
vi
-
ii
v-
if
i-
+
+
v+
-
vo
+
A(v+-v-) -
-
Dave Shattuck
University of Houston
© University of Houston
Is This 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 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
this assumption is.
Overview
slide.