Courses

# Operational Amplifiers Electronics and Communication Engineering (ECE) Notes | EduRev

## Electronics and Communication Engineering (ECE) : Operational Amplifiers Electronics and Communication Engineering (ECE) Notes | EduRev

``` Page 1

ELEC 353  1
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
OPERATIONAL AMPLIFIERS

Operational amplifiers are among the most widely used integrated circuits for analogue electronics.
They are very popular for amplification of small signals, both AC and DC.  They are also used in circuits
performing a variety of other 'operations' such as analogue addition, differentiation and integration.
Elementary Properties of Operational Amplifiers
They are usually purchased as an integrated circuit (IC) with eight pins, a minimum of five of these pins
being connected to the 'chip' (of silicon containing tiny interconnected transistors etc) inside.  The circuit
symbol is shown opposite.  Note that there are two inputs.  The chip (or IC) is a differential amplifier
which has been wired internally to sense the voltage difference (V
+
- V
-
) between the input terminals,
amplify this by a gain, A, and present this at the output;
ie we have     ) (
- +
- = V V A V
o
.

+
-
-E (-15 V)
inputs
output
V
+
A
V
-
V
o
+E (+15 V)

In order to perform this amplification, the IC must, of course, be connected to a DC power supply (or
battery).  Normally this DC is supplied as +E and -E volts as shown.  Supply voltages in the range ±9 V
to ±15 V are common but lower values down to about ±3 V are possible with some sacrifice in
performance.  Because the DC power supply (or battery) is absolutely essential to the operation of the
device, it is assumed to be there and is very seldom shown on circuit diagrams.

The gain of the chip, A, is usually about 2 x 10
5
(at least for DC) though variations of a factor of two or
more must be expected between individual chips of the same type number.  The manufacturer's data sheet
supplies this information.

The small signal resistance between the input terminals is typically 1 - 2 M? , though it may be much higher
in certain devices.  In most of our circuits, we will not need to allow for this input resistance; it will simply
be 'high'.  However, at times we will need to know its approximate size to check what approximations will
be valid.

Note that the two input terminals are not quite identical.  They differ only in terms of sign.  When the '+'
input terminal (ie the voltage V
+
) is made (slightly) more positive, the output, V
o
, will become more
positive but when the '-' input terminal (ie V
-
) is made (slightly) more positive the output voltage will
become slightly more negative, since V
o
= A(V
+
- V
-
).  For this reason the '-' input is known as the
inverting input and the '+' input as the non-inverting input.

The output resistance is low, typically 100? or less, and can thus often be neglected.  (Note, however,
that this low output resistance doesn't mean a typical op amp can drive large currents.  Typically the
manufacturer has added extra circuitry which limits the output current to 25 mA.  This is to prevent
possible destruction of the device by shorting the output.  If you do not try to draw more than 25 mA from
the output this 'current limiting' circuitry has no effect.)
Page 2

ELEC 353  1
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
OPERATIONAL AMPLIFIERS

Operational amplifiers are among the most widely used integrated circuits for analogue electronics.
They are very popular for amplification of small signals, both AC and DC.  They are also used in circuits
performing a variety of other 'operations' such as analogue addition, differentiation and integration.
Elementary Properties of Operational Amplifiers
They are usually purchased as an integrated circuit (IC) with eight pins, a minimum of five of these pins
being connected to the 'chip' (of silicon containing tiny interconnected transistors etc) inside.  The circuit
symbol is shown opposite.  Note that there are two inputs.  The chip (or IC) is a differential amplifier
which has been wired internally to sense the voltage difference (V
+
- V
-
) between the input terminals,
amplify this by a gain, A, and present this at the output;
ie we have     ) (
- +
- = V V A V
o
.

+
-
-E (-15 V)
inputs
output
V
+
A
V
-
V
o
+E (+15 V)

In order to perform this amplification, the IC must, of course, be connected to a DC power supply (or
battery).  Normally this DC is supplied as +E and -E volts as shown.  Supply voltages in the range ±9 V
to ±15 V are common but lower values down to about ±3 V are possible with some sacrifice in
performance.  Because the DC power supply (or battery) is absolutely essential to the operation of the
device, it is assumed to be there and is very seldom shown on circuit diagrams.

The gain of the chip, A, is usually about 2 x 10
5
(at least for DC) though variations of a factor of two or
more must be expected between individual chips of the same type number.  The manufacturer's data sheet
supplies this information.

The small signal resistance between the input terminals is typically 1 - 2 M? , though it may be much higher
in certain devices.  In most of our circuits, we will not need to allow for this input resistance; it will simply
be 'high'.  However, at times we will need to know its approximate size to check what approximations will
be valid.

Note that the two input terminals are not quite identical.  They differ only in terms of sign.  When the '+'
input terminal (ie the voltage V
+
) is made (slightly) more positive, the output, V
o
, will become more
positive but when the '-' input terminal (ie V
-
) is made (slightly) more positive the output voltage will
become slightly more negative, since V
o
= A(V
+
- V
-
).  For this reason the '-' input is known as the
inverting input and the '+' input as the non-inverting input.

The output resistance is low, typically 100? or less, and can thus often be neglected.  (Note, however,
that this low output resistance doesn't mean a typical op amp can drive large currents.  Typically the
manufacturer has added extra circuitry which limits the output current to 25 mA.  This is to prevent
possible destruction of the device by shorting the output.  If you do not try to draw more than 25 mA from
the output this 'current limiting' circuitry has no effect.)
ELEC 353  2
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr

These operational amplifier IC's are seldom used without feedback because their gains are too large and
too variable (both with temperature and from one chip to the next) to be useful in practice.  With
feedback, the op. amp. chip is the basic building block of many high performance circuits.
Simple Inverting Amplifier
The circuit below, known as the simple inverting amplifier, is a very convenient, stable, and widely used
amplifier for both DC and AC signals.  We will show that its voltage gain is given by
1
R
R
V
V
A
f
in
o
f
- = = to
a (usually) very good approximation.  The stability and convenience come from the fact that the gain is just
the ratio of two resistors.  The minus sign just means that the output is inverted with respect to the input.
Note that the feedback provided by R
f
is negative. (It goes to the '-' input.)

V
o
V
in

R
1
R
f
R
i
I
1
I
f
I
i -
+
V
+
V
-
A

We could, of course, solve the circuit completely, using Kirchhoff's Laws and the manufacturers values of
the gain, A, and internal impedance, R
i
.  (Indeed, you might like to try for yourself; it's not particularly
difficult.)

However, we will prefer to use approximations because this will be simpler, give us more insight, and allow
us to solve more readily a variety of different circuits which use the operational amplifier chip.

We will take A = 2 x 10
5
and R
i
= 1 M? since these are very typical values.  We will think of R
f
= 100
k? , R
1
= 10 k? , and V
in
= 1 volt (DC or AC).  (The above formula for the gain of the amplifier thus
predicts
in
1
0
V
R
R
V
f
- = = -10 volts.)
We will now prove the above formula for the gain, A
f
.

Since V
o
= A (V
+
- V
-
) and V
+
= 0 we have V
-
= -V
o
/A.  But A = 2 x 10
5
, so that V
-
will normally be
very small compared with either V
o
or V
i
.  In particular, in our example, we would have V
-
= -V
o
/A =
10/(2 x 10
5
) = 50 µV which certainly is small compared to V
in
= 1 V and V
o
= -10 V.
Thus 0 ˜
-
V .

Hence the voltage across R
f
= V
-
- V
o
˜ -V
o

and the voltage across   R
1
= V
in
- V
-
˜ V
in
.

Thus the current through R
f
=
f
o
f
R
V
I - =
and the current through R
1

1
1
R
V
I
in
= = .

Page 3

ELEC 353  1
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
OPERATIONAL AMPLIFIERS

Operational amplifiers are among the most widely used integrated circuits for analogue electronics.
They are very popular for amplification of small signals, both AC and DC.  They are also used in circuits
performing a variety of other 'operations' such as analogue addition, differentiation and integration.
Elementary Properties of Operational Amplifiers
They are usually purchased as an integrated circuit (IC) with eight pins, a minimum of five of these pins
being connected to the 'chip' (of silicon containing tiny interconnected transistors etc) inside.  The circuit
symbol is shown opposite.  Note that there are two inputs.  The chip (or IC) is a differential amplifier
which has been wired internally to sense the voltage difference (V
+
- V
-
) between the input terminals,
amplify this by a gain, A, and present this at the output;
ie we have     ) (
- +
- = V V A V
o
.

+
-
-E (-15 V)
inputs
output
V
+
A
V
-
V
o
+E (+15 V)

In order to perform this amplification, the IC must, of course, be connected to a DC power supply (or
battery).  Normally this DC is supplied as +E and -E volts as shown.  Supply voltages in the range ±9 V
to ±15 V are common but lower values down to about ±3 V are possible with some sacrifice in
performance.  Because the DC power supply (or battery) is absolutely essential to the operation of the
device, it is assumed to be there and is very seldom shown on circuit diagrams.

The gain of the chip, A, is usually about 2 x 10
5
(at least for DC) though variations of a factor of two or
more must be expected between individual chips of the same type number.  The manufacturer's data sheet
supplies this information.

The small signal resistance between the input terminals is typically 1 - 2 M? , though it may be much higher
in certain devices.  In most of our circuits, we will not need to allow for this input resistance; it will simply
be 'high'.  However, at times we will need to know its approximate size to check what approximations will
be valid.

Note that the two input terminals are not quite identical.  They differ only in terms of sign.  When the '+'
input terminal (ie the voltage V
+
) is made (slightly) more positive, the output, V
o
, will become more
positive but when the '-' input terminal (ie V
-
) is made (slightly) more positive the output voltage will
become slightly more negative, since V
o
= A(V
+
- V
-
).  For this reason the '-' input is known as the
inverting input and the '+' input as the non-inverting input.

The output resistance is low, typically 100? or less, and can thus often be neglected.  (Note, however,
that this low output resistance doesn't mean a typical op amp can drive large currents.  Typically the
manufacturer has added extra circuitry which limits the output current to 25 mA.  This is to prevent
possible destruction of the device by shorting the output.  If you do not try to draw more than 25 mA from
the output this 'current limiting' circuitry has no effect.)
ELEC 353  2
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr

These operational amplifier IC's are seldom used without feedback because their gains are too large and
too variable (both with temperature and from one chip to the next) to be useful in practice.  With
feedback, the op. amp. chip is the basic building block of many high performance circuits.
Simple Inverting Amplifier
The circuit below, known as the simple inverting amplifier, is a very convenient, stable, and widely used
amplifier for both DC and AC signals.  We will show that its voltage gain is given by
1
R
R
V
V
A
f
in
o
f
- = = to
a (usually) very good approximation.  The stability and convenience come from the fact that the gain is just
the ratio of two resistors.  The minus sign just means that the output is inverted with respect to the input.
Note that the feedback provided by R
f
is negative. (It goes to the '-' input.)

V
o
V
in

R
1
R
f
R
i
I
1
I
f
I
i -
+
V
+
V
-
A

We could, of course, solve the circuit completely, using Kirchhoff's Laws and the manufacturers values of
the gain, A, and internal impedance, R
i
.  (Indeed, you might like to try for yourself; it's not particularly
difficult.)

However, we will prefer to use approximations because this will be simpler, give us more insight, and allow
us to solve more readily a variety of different circuits which use the operational amplifier chip.

We will take A = 2 x 10
5
and R
i
= 1 M? since these are very typical values.  We will think of R
f
= 100
k? , R
1
= 10 k? , and V
in
= 1 volt (DC or AC).  (The above formula for the gain of the amplifier thus
predicts
in
1
0
V
R
R
V
f
- = = -10 volts.)
We will now prove the above formula for the gain, A
f
.

Since V
o
= A (V
+
- V
-
) and V
+
= 0 we have V
-
= -V
o
/A.  But A = 2 x 10
5
, so that V
-
will normally be
very small compared with either V
o
or V
i
.  In particular, in our example, we would have V
-
= -V
o
/A =
10/(2 x 10
5
) = 50 µV which certainly is small compared to V
in
= 1 V and V
o
= -10 V.
Thus 0 ˜
-
V .

Hence the voltage across R
f
= V
-
- V
o
˜ -V
o

and the voltage across   R
1
= V
in
- V
-
˜ V
in
.

Thus the current through R
f
=
f
o
f
R
V
I - =
and the current through R
1

1
1
R
V
I
in
= = .

ELEC 353  3
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
Now I
i
, the current into the op amp chip, is very small indeed because

pA 50
MO 1
µV 50 ) (
= = =
-
=
- + -
i i
i
R
V
R
V V
I .     In comparison,

A 100
kO 100
V 10
m =
-
= - =
f
o
f
R
V
I and

A 100
kO 10
V 1
1
1
m = = =
R
V
I
in
.
Thus, because I
i
is negligible, we have
1
I I
f
= to a very good approximation.  Hence (substituting the
expressions for I
f
and I
1
from above):

1
0
R
V
R
V
in
f
= - which, when rearranged, gives

1
R
R
V
V
A
f
in
o
f
- = = for the voltage gain of the simple inverting op amp amplifier.
Two Important  Principles:
The above analysis brought to light two very important principles in such circuits as the one above
with high gain op amps:

(1) 0 ˜
-
V This is often expressed by saying the '-' (inverting) input terminal is a 'virtual ground'.
When V
+
? 0 this excellent approximation becomes    0 ˜ -
- +
V V
because V
+
- V
-
= V
o
/A  and A  is normally very large.

(2)
1
I I
f
˜ Essentially all the current going through R
1
(from the amplifier input) continues on through R
f
.
(Only a tiny fraction of I
1
goes into the chip itself.)
The Approximations:
Generally the approximations will be valid if, for the whole (inverting) amplifier, we have
A
R
R
f
«
1

ie the approximations, V
-
˜ 0, and I
f
= I
1
, will be valid provided our (inverting, feedback) amplifier is set
up to achieve a gain,
1
R
R
f
, which is much less than the gain, A, of the chip alone.  (Obviously we can't
expect more gain from our feedback amplifier than from the chip alone when the gain is negative as above.)

Page 4

ELEC 353  1
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
OPERATIONAL AMPLIFIERS

Operational amplifiers are among the most widely used integrated circuits for analogue electronics.
They are very popular for amplification of small signals, both AC and DC.  They are also used in circuits
performing a variety of other 'operations' such as analogue addition, differentiation and integration.
Elementary Properties of Operational Amplifiers
They are usually purchased as an integrated circuit (IC) with eight pins, a minimum of five of these pins
being connected to the 'chip' (of silicon containing tiny interconnected transistors etc) inside.  The circuit
symbol is shown opposite.  Note that there are two inputs.  The chip (or IC) is a differential amplifier
which has been wired internally to sense the voltage difference (V
+
- V
-
) between the input terminals,
amplify this by a gain, A, and present this at the output;
ie we have     ) (
- +
- = V V A V
o
.

+
-
-E (-15 V)
inputs
output
V
+
A
V
-
V
o
+E (+15 V)

In order to perform this amplification, the IC must, of course, be connected to a DC power supply (or
battery).  Normally this DC is supplied as +E and -E volts as shown.  Supply voltages in the range ±9 V
to ±15 V are common but lower values down to about ±3 V are possible with some sacrifice in
performance.  Because the DC power supply (or battery) is absolutely essential to the operation of the
device, it is assumed to be there and is very seldom shown on circuit diagrams.

The gain of the chip, A, is usually about 2 x 10
5
(at least for DC) though variations of a factor of two or
more must be expected between individual chips of the same type number.  The manufacturer's data sheet
supplies this information.

The small signal resistance between the input terminals is typically 1 - 2 M? , though it may be much higher
in certain devices.  In most of our circuits, we will not need to allow for this input resistance; it will simply
be 'high'.  However, at times we will need to know its approximate size to check what approximations will
be valid.

Note that the two input terminals are not quite identical.  They differ only in terms of sign.  When the '+'
input terminal (ie the voltage V
+
) is made (slightly) more positive, the output, V
o
, will become more
positive but when the '-' input terminal (ie V
-
) is made (slightly) more positive the output voltage will
become slightly more negative, since V
o
= A(V
+
- V
-
).  For this reason the '-' input is known as the
inverting input and the '+' input as the non-inverting input.

The output resistance is low, typically 100? or less, and can thus often be neglected.  (Note, however,
that this low output resistance doesn't mean a typical op amp can drive large currents.  Typically the
manufacturer has added extra circuitry which limits the output current to 25 mA.  This is to prevent
possible destruction of the device by shorting the output.  If you do not try to draw more than 25 mA from
the output this 'current limiting' circuitry has no effect.)
ELEC 353  2
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr

These operational amplifier IC's are seldom used without feedback because their gains are too large and
too variable (both with temperature and from one chip to the next) to be useful in practice.  With
feedback, the op. amp. chip is the basic building block of many high performance circuits.
Simple Inverting Amplifier
The circuit below, known as the simple inverting amplifier, is a very convenient, stable, and widely used
amplifier for both DC and AC signals.  We will show that its voltage gain is given by
1
R
R
V
V
A
f
in
o
f
- = = to
a (usually) very good approximation.  The stability and convenience come from the fact that the gain is just
the ratio of two resistors.  The minus sign just means that the output is inverted with respect to the input.
Note that the feedback provided by R
f
is negative. (It goes to the '-' input.)

V
o
V
in

R
1
R
f
R
i
I
1
I
f
I
i -
+
V
+
V
-
A

We could, of course, solve the circuit completely, using Kirchhoff's Laws and the manufacturers values of
the gain, A, and internal impedance, R
i
.  (Indeed, you might like to try for yourself; it's not particularly
difficult.)

However, we will prefer to use approximations because this will be simpler, give us more insight, and allow
us to solve more readily a variety of different circuits which use the operational amplifier chip.

We will take A = 2 x 10
5
and R
i
= 1 M? since these are very typical values.  We will think of R
f
= 100
k? , R
1
= 10 k? , and V
in
= 1 volt (DC or AC).  (The above formula for the gain of the amplifier thus
predicts
in
1
0
V
R
R
V
f
- = = -10 volts.)
We will now prove the above formula for the gain, A
f
.

Since V
o
= A (V
+
- V
-
) and V
+
= 0 we have V
-
= -V
o
/A.  But A = 2 x 10
5
, so that V
-
will normally be
very small compared with either V
o
or V
i
.  In particular, in our example, we would have V
-
= -V
o
/A =
10/(2 x 10
5
) = 50 µV which certainly is small compared to V
in
= 1 V and V
o
= -10 V.
Thus 0 ˜
-
V .

Hence the voltage across R
f
= V
-
- V
o
˜ -V
o

and the voltage across   R
1
= V
in
- V
-
˜ V
in
.

Thus the current through R
f
=
f
o
f
R
V
I - =
and the current through R
1

1
1
R
V
I
in
= = .

ELEC 353  3
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
Now I
i
, the current into the op amp chip, is very small indeed because

pA 50
MO 1
µV 50 ) (
= = =
-
=
- + -
i i
i
R
V
R
V V
I .     In comparison,

A 100
kO 100
V 10
m =
-
= - =
f
o
f
R
V
I and

A 100
kO 10
V 1
1
1
m = = =
R
V
I
in
.
Thus, because I
i
is negligible, we have
1
I I
f
= to a very good approximation.  Hence (substituting the
expressions for I
f
and I
1
from above):

1
0
R
V
R
V
in
f
= - which, when rearranged, gives

1
R
R
V
V
A
f
in
o
f
- = = for the voltage gain of the simple inverting op amp amplifier.
Two Important  Principles:
The above analysis brought to light two very important principles in such circuits as the one above
with high gain op amps:

(1) 0 ˜
-
V This is often expressed by saying the '-' (inverting) input terminal is a 'virtual ground'.
When V
+
? 0 this excellent approximation becomes    0 ˜ -
- +
V V
because V
+
- V
-
= V
o
/A  and A  is normally very large.

(2)
1
I I
f
˜ Essentially all the current going through R
1
(from the amplifier input) continues on through R
f
.
(Only a tiny fraction of I
1
goes into the chip itself.)
The Approximations:
Generally the approximations will be valid if, for the whole (inverting) amplifier, we have
A
R
R
f
«
1

ie the approximations, V
-
˜ 0, and I
f
= I
1
, will be valid provided our (inverting, feedback) amplifier is set
up to achieve a gain,
1
R
R
f
, which is much less than the gain, A, of the chip alone.  (Obviously we can't
expect more gain from our feedback amplifier than from the chip alone when the gain is negative as above.)

ELEC 353  4
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
Simple Non-Inverting Amplifier
When a non-inverting gain is required the circuit shown below can be used.

V
0
V
in
R
1 R
f
I
1
-
+
V
+
V
-
A
I
f
R
i
I
i

The first of our op amp principles above tells us that V
+
- V
-
˜ 0  (ie V
+
- V
-
is very much less than V
in
or
V
o
) so that V
-
= V
+
= V
in
.  Our second op amp principle gives I
f
= I
1
(since the current into the chip itself
is very tiny indeed).

Hence, voltage across R
1
= V
-
= V
in
= I
1
R
1

and voltage across R
1
+ R
f
=  V
o
=  I
1
(R
1
+ R
f
).

Dividing the two equations (on the right) gives:

1 1
1
1
R
R
R
R R
V
V
A
f f
in
o
f
+ =
+
= =
for the voltage gain of our non-inverting amplifier.
Input Resistance:
The current flowing into the non-inverting amplifier, I
in
, is just the small current going into the chip itself.
This is given by
i
in f
i
o
i
in
AR
V A
AR
V
R
V V
I = =
-
=
- +

where
1
1
R
R
V
V
A
f
in
o
f
+ = = , the voltage gain of the non-inverting amplifier as above.
Hence, we have
f
i
in
in
in
A
A
R
I
V
R = = for the input resistance of our non-inverting amplifier.  This is normally
rather high since R
i
, the internal resistance of the chip itself, is at least 1 - 2 M? and usually we will have A

»A
f
.
Voltage Follower:
If we make R
f
= 0 and R
1
= 8 (by removing it) then we get the circuit below and, from above, we have

1 = =
in
o
f
V
V
A and  A R R
i in
= .

This is called a voltage follower ; it has a gain of one,
very high input impedance and very low output
impedance.  It is very useful as a buffer between a high
V
0
V
in
-
+
V
+
V
-
A R
i
I
in
Page 5

ELEC 353  1
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
OPERATIONAL AMPLIFIERS

Operational amplifiers are among the most widely used integrated circuits for analogue electronics.
They are very popular for amplification of small signals, both AC and DC.  They are also used in circuits
performing a variety of other 'operations' such as analogue addition, differentiation and integration.
Elementary Properties of Operational Amplifiers
They are usually purchased as an integrated circuit (IC) with eight pins, a minimum of five of these pins
being connected to the 'chip' (of silicon containing tiny interconnected transistors etc) inside.  The circuit
symbol is shown opposite.  Note that there are two inputs.  The chip (or IC) is a differential amplifier
which has been wired internally to sense the voltage difference (V
+
- V
-
) between the input terminals,
amplify this by a gain, A, and present this at the output;
ie we have     ) (
- +
- = V V A V
o
.

+
-
-E (-15 V)
inputs
output
V
+
A
V
-
V
o
+E (+15 V)

In order to perform this amplification, the IC must, of course, be connected to a DC power supply (or
battery).  Normally this DC is supplied as +E and -E volts as shown.  Supply voltages in the range ±9 V
to ±15 V are common but lower values down to about ±3 V are possible with some sacrifice in
performance.  Because the DC power supply (or battery) is absolutely essential to the operation of the
device, it is assumed to be there and is very seldom shown on circuit diagrams.

The gain of the chip, A, is usually about 2 x 10
5
(at least for DC) though variations of a factor of two or
more must be expected between individual chips of the same type number.  The manufacturer's data sheet
supplies this information.

The small signal resistance between the input terminals is typically 1 - 2 M? , though it may be much higher
in certain devices.  In most of our circuits, we will not need to allow for this input resistance; it will simply
be 'high'.  However, at times we will need to know its approximate size to check what approximations will
be valid.

Note that the two input terminals are not quite identical.  They differ only in terms of sign.  When the '+'
input terminal (ie the voltage V
+
) is made (slightly) more positive, the output, V
o
, will become more
positive but when the '-' input terminal (ie V
-
) is made (slightly) more positive the output voltage will
become slightly more negative, since V
o
= A(V
+
- V
-
).  For this reason the '-' input is known as the
inverting input and the '+' input as the non-inverting input.

The output resistance is low, typically 100? or less, and can thus often be neglected.  (Note, however,
that this low output resistance doesn't mean a typical op amp can drive large currents.  Typically the
manufacturer has added extra circuitry which limits the output current to 25 mA.  This is to prevent
possible destruction of the device by shorting the output.  If you do not try to draw more than 25 mA from
the output this 'current limiting' circuitry has no effect.)
ELEC 353  2
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr

These operational amplifier IC's are seldom used without feedback because their gains are too large and
too variable (both with temperature and from one chip to the next) to be useful in practice.  With
feedback, the op. amp. chip is the basic building block of many high performance circuits.
Simple Inverting Amplifier
The circuit below, known as the simple inverting amplifier, is a very convenient, stable, and widely used
amplifier for both DC and AC signals.  We will show that its voltage gain is given by
1
R
R
V
V
A
f
in
o
f
- = = to
a (usually) very good approximation.  The stability and convenience come from the fact that the gain is just
the ratio of two resistors.  The minus sign just means that the output is inverted with respect to the input.
Note that the feedback provided by R
f
is negative. (It goes to the '-' input.)

V
o
V
in

R
1
R
f
R
i
I
1
I
f
I
i -
+
V
+
V
-
A

We could, of course, solve the circuit completely, using Kirchhoff's Laws and the manufacturers values of
the gain, A, and internal impedance, R
i
.  (Indeed, you might like to try for yourself; it's not particularly
difficult.)

However, we will prefer to use approximations because this will be simpler, give us more insight, and allow
us to solve more readily a variety of different circuits which use the operational amplifier chip.

We will take A = 2 x 10
5
and R
i
= 1 M? since these are very typical values.  We will think of R
f
= 100
k? , R
1
= 10 k? , and V
in
= 1 volt (DC or AC).  (The above formula for the gain of the amplifier thus
predicts
in
1
0
V
R
R
V
f
- = = -10 volts.)
We will now prove the above formula for the gain, A
f
.

Since V
o
= A (V
+
- V
-
) and V
+
= 0 we have V
-
= -V
o
/A.  But A = 2 x 10
5
, so that V
-
will normally be
very small compared with either V
o
or V
i
.  In particular, in our example, we would have V
-
= -V
o
/A =
10/(2 x 10
5
) = 50 µV which certainly is small compared to V
in
= 1 V and V
o
= -10 V.
Thus 0 ˜
-
V .

Hence the voltage across R
f
= V
-
- V
o
˜ -V
o

and the voltage across   R
1
= V
in
- V
-
˜ V
in
.

Thus the current through R
f
=
f
o
f
R
V
I - =
and the current through R
1

1
1
R
V
I
in
= = .

ELEC 353  3
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
Now I
i
, the current into the op amp chip, is very small indeed because

pA 50
MO 1
µV 50 ) (
= = =
-
=
- + -
i i
i
R
V
R
V V
I .     In comparison,

A 100
kO 100
V 10
m =
-
= - =
f
o
f
R
V
I and

A 100
kO 10
V 1
1
1
m = = =
R
V
I
in
.
Thus, because I
i
is negligible, we have
1
I I
f
= to a very good approximation.  Hence (substituting the
expressions for I
f
and I
1
from above):

1
0
R
V
R
V
in
f
= - which, when rearranged, gives

1
R
R
V
V
A
f
in
o
f
- = = for the voltage gain of the simple inverting op amp amplifier.
Two Important  Principles:
The above analysis brought to light two very important principles in such circuits as the one above
with high gain op amps:

(1) 0 ˜
-
V This is often expressed by saying the '-' (inverting) input terminal is a 'virtual ground'.
When V
+
? 0 this excellent approximation becomes    0 ˜ -
- +
V V
because V
+
- V
-
= V
o
/A  and A  is normally very large.

(2)
1
I I
f
˜ Essentially all the current going through R
1
(from the amplifier input) continues on through R
f
.
(Only a tiny fraction of I
1
goes into the chip itself.)
The Approximations:
Generally the approximations will be valid if, for the whole (inverting) amplifier, we have
A
R
R
f
«
1

ie the approximations, V
-
˜ 0, and I
f
= I
1
, will be valid provided our (inverting, feedback) amplifier is set
up to achieve a gain,
1
R
R
f
, which is much less than the gain, A, of the chip alone.  (Obviously we can't
expect more gain from our feedback amplifier than from the chip alone when the gain is negative as above.)

ELEC 353  4
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
Simple Non-Inverting Amplifier
When a non-inverting gain is required the circuit shown below can be used.

V
0
V
in
R
1 R
f
I
1
-
+
V
+
V
-
A
I
f
R
i
I
i

The first of our op amp principles above tells us that V
+
- V
-
˜ 0  (ie V
+
- V
-
is very much less than V
in
or
V
o
) so that V
-
= V
+
= V
in
.  Our second op amp principle gives I
f
= I
1
(since the current into the chip itself
is very tiny indeed).

Hence, voltage across R
1
= V
-
= V
in
= I
1
R
1

and voltage across R
1
+ R
f
=  V
o
=  I
1
(R
1
+ R
f
).

Dividing the two equations (on the right) gives:

1 1
1
1
R
R
R
R R
V
V
A
f f
in
o
f
+ =
+
= =
for the voltage gain of our non-inverting amplifier.
Input Resistance:
The current flowing into the non-inverting amplifier, I
in
, is just the small current going into the chip itself.
This is given by
i
in f
i
o
i
in
AR
V A
AR
V
R
V V
I = =
-
=
- +

where
1
1
R
R
V
V
A
f
in
o
f
+ = = , the voltage gain of the non-inverting amplifier as above.
Hence, we have
f
i
in
in
in
A
A
R
I
V
R = = for the input resistance of our non-inverting amplifier.  This is normally
rather high since R
i
, the internal resistance of the chip itself, is at least 1 - 2 M? and usually we will have A

»A
f
.
Voltage Follower:
If we make R
f
= 0 and R
1
= 8 (by removing it) then we get the circuit below and, from above, we have

1 = =
in
o
f
V
V
A and  A R R
i in
= .

This is called a voltage follower ; it has a gain of one,
very high input impedance and very low output
impedance.  It is very useful as a buffer between a high
V
0
V
in
-
+
V
+
V
-
A R
i
I
in
ELEC 353  5
Operational Amplifiers 2006-04-04  © Neil R Thomson & John L Bähr
impedance source and low impedance load.
Equivalent Circuit

V
0
V
in
R
i
R
o
V
+
V
-
A(V
+
- V
-
) I
i

To find the output resistance we connect a test voltage (V
t
) to the output and short the input to earth, so
the equivalent circuit becomes:

V
t
a

R
i
R
o
V
+
V
-
A(V
+
- V
-
)
I
t
+
--

At the node “a” we sum over all the currents:
( )
0 =
- -
+
-
+
- + - +
o
t
i
t
R
V V V A
R
V V
I
Now V
+
= 0 and V
-
= V
t
so
( )
or
1
t
i o
i o
t
V
R R
R A R
I
+ +
=
( )
i o
i o
t
t
out
R A R
R R
I
V
R
1 + +
= =
eg for the OP-177:  A = 12x10
6
, R
i
= 45 MO, and R
o
= 60 O so  R
out
= 0.000005 O = 5 µO
Other 'Operations' with the Op Amp
V
o
R
R
f
I
1
I
f
-
+
V
+
V
-
V
1
V
2
V
3
I
2
I
3
R
R

The principle of negligible current into the op amp chip gives us I
f
= I
1
+ I
2
+ I
3

The virtual ground principle gives V
-
˜ V
+
= 0, so that the voltage across each of the three input resistors
(R) is V
1
, V
2
, and V
3
, and the voltage across the feedback resistor (R
f
) is -V
o
.

```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;