Op-amp.pptx

OPERATIONAL AMPLIFIER
PREPARED BY
DEBASISH MOHANTA
ASSISTANT PROFESSOR
DEPARTMENT OF ELECTRICAL ENGINEERING
GOVERNMENT COLLEGE OF ENGINEERING, KEONJHAR
1

INTRODUCTION
The term “operational amplifier” denotes a special type of amplifier that, by proper selection
of its external components, could be configured for a variety of operations.
First developed by John R. Ragazzine in 1947 with vacuum tube.
In 1960 at FAIRCHILD SEMICONDUCTOR CORPORATION, Robert J. Widlar fabricated
op amp with the help of IC fabrication technology.
In 1968 FAIRCHILD introduces the op-amp that was to become the industry standard.
2
OPERATIONAL AMPLIFIER ANALOG ELECTRONICS CIRCUIT REC3C001

WHAT IS OP-AMP?
An operational amplifier (op-amp) is a DC-coupled high-gain electronic voltage amplifier.
Low cost integrating circuit consisting of: • Transistors • Resistors • Capacitors
Direct- coupled high gain amplifier usually consisting of one or more differential amplifiers.
Output stage is generally a push-pull or push-pull complementary-symmetry pair.
Op amps are differential amplifiers, and their output voltage is proportional to the difference of the
two input voltages.
The two input terminals, called the inverting and non-inverting, are labeled with - and +,
respectively.
3

OP-AMP PIN DIAGRAM
There are 8 pins in a common Op-Amp, like the 741 which is used in many instructional
courses.
Pin 1: Offset null
Pin 2: Inverting input terminal
Pin 3: Non-inverting input terminal
Pin 4: –VCC (negative voltage supply)
Pin 5: Offset null
Pin 6: Output voltage
Pin 7: +VCC (positive voltage supply)
Pin 8: No Connection
4

FUNCTION AND CHARACTERISTICS OF THE
IDEAL OPAMP
The op amp is designed to sense the difference between the voltage signals applied at its two
input terminals (i.e., the quantityv2 − v1), multiply this by a number A, and cause the resulting
voltage A(v2 − v1)appear at output terminal 3.
The voltage 𝑣1means the voltage applied between terminal 1 and ground.
The ideal op amp is not supposed to draw any input current; that is, the signal current into
terminal 1 and the signal current into terminal 2 are both zero. In other words, the input
impedance of an ideal op amp is supposed to be infinite.
The voltage between terminal 3 and ground will always be equal toA(v2 − v1), independent of
the current that may be drawn from terminal 3 into load impedance.
In other words, the output impedance of an ideal op amp is supposed to be zero.
Putting together all of the above, we arrive at the equivalent circuit model shown in Fig.
5

FUNCTION AND CHARACTERISTICS OF THE
IDEAL OPAMP
Note that the output is in phase with (has the same sign as) 𝑣1 and is out of phase with (has the
opposite sign of)𝑣2.
For this reason, input terminal 1 is called the inverting input terminal and is distinguished by
a " - " sign, while input terminal 2 is called the non inverting input terminal and is
distinguished by a "+" sign.
6

FUNCTION AND CHARACTERISTICS OF
THE IDEAL OPAMP
The op amp responds only to the difference signal(v2−v1), and hence ignores any signal
common to both inputs. That is, ifv1 = v2 = 1V, then the output will ideally be zero.
We call this property common-mode rejection, and we conclude that an ideal op amp has
zero common-mode gain or, equivalently, infinite common-mode rejection.
An important characteristic of op amps is that they are direct-coupled or dc amplifiers.
The ideal op amp has a gain A that remains constant down to zero frequency and up to infinite
frequency.
That is, ideal op amps will amplify signals of any frequency with equal gain, and are thus said
to have infinite bandwidth.
The ideal op amp has a gain A whose value is very large and ideally infinite.
7

CHARACTERISTICS OF THE IDEAL
OPAMP
In summary, characteristics of ideal op amp are
1. Infinite input impedance
2. Zero output impedance
3. Zero common mode gain or infinite common mode rejection
4. Infinite open-loop gain A
5. Infinite band width
8

DIFFERENTIALAND COMMON-MODE SIGNALS
The difference between two input signals gives the differential input,𝑣𝐼𝑑 and the average of the two input
signals gives the common mode input,𝑣𝐼𝑐𝑚 as given below:
vId = v2 − v1
vIcm =
1
2
(v1 + v2)
Expressing inputs in terms of differential and common mode gains,
we get
v1 = vIcm −
vId
2
v2 = vIcm +
vId
2
These equations can in turn lead to the pictorial representation in Fig.
9

BASIC OPAMP CONFIGURATIONS
Op amps are not used alone; rather, the op amp is connected to passive components in a feedback circuit.
There are two such basic circuit configurations employing an op amp and two resistors: The inverting
configuration, and the non-inverting configuration.
The Inverting Configuration
Inverting Configuration consists of one op amp and two resistors R1 andR2.
Resistor R2 is connected from the output terminal of the op amp, terminal 3, back to the inverting or
negative input terminal, terminal 1, as applying negative feed-back; if R2 were connected between
terminals 3 and 2 we would have called this positive feed-back.
In addition to addingR2, we have grounded terminal 2 and connected a resistor between terminal 1 and
an input signal source with a voltagevI.
10

THE INVERTING CONFIGURATION
The output of the overall circuit is taken at terminal 3 (i.e., between terminal 3 and ground).
The Closed-Loop Gain
The closed-loop gain G, defined as
G =
v0
vI
Assuming the op amp to be ideal. The gain A is very large (ideally infinite).
The voltage between the op amp input terminals should be negligibly small and ideally zero.
v2 − v1 =
v0
A
= 0
11

It follows that the voltage at the inverting input terminal is given by v1 = v2
That is, because the gain A approaches infinity, the voltage v1 approaches and ideally equalsv2.
A virtual short circuit means that whatever voltage is at 2 will automatically appear at 1 because of
the infinite gain A, But terminal 2 happens to be connected to ground; thus, v2 = 0 and v1 = 0.
We speak of terminal 1 as being a virtual ground—that is, having zero voltage but not physically
connected to ground.
12

Applying Ohm's law to find the current𝑖1 throughR1, (see Fig.) as follows, we get
i1 =
vI − v1
R1
==
vI − 0
R1
=
vI
R1
We can then apply Ohm's law to R2and determine𝑣0; that is,
v0 = v1 − i1R2
v0 = 0 −
vI
R1
R2
𝐯𝟎
𝐯𝐈
= −
𝐑𝟐
𝐑𝟏
The minus sign means that the closed-loop amplifier provides signal inversion.
13

THE INVERTING WEIGHTED SUMMER
A very important application of the inverting configuration is the weighted-summer circuit.
Here we have a resistance Rf in the negative-feedback path , but we have a number of input signals
v1, v2 … … , vn applied to a corresponding resistor R1, R2 … … , Rn which are connected to the inverting
terminal of the op amp.
The currents i1, i2 … … , in are given by
i1 =
v1
R1
, i2 =
v2
R2
, … … , in =
vn
Rn
14

THE INVERTING WEIGHTED SUMMER
All these currents sum together to produce the current i
i = i1 + i2 + ⋯ … . . +in
The output voltage 𝑣0may now be determined by another application of Ohm's law,
v0 = 0 − iRf = −iRf
𝐯𝟎 = −(
𝐑𝐟
𝐑𝟏
𝐯𝟏 +
𝐑𝐟
𝐑𝟐
𝐯𝟐 + ⋯ … . +
𝐑𝐟
𝐑𝐧
𝐯𝐧)
That is, the output voltage is a weighted sum of the input signals𝑣1, 𝑣2 … … , 𝑣𝑛. This circuit is therefore
called a weighted summer.
15

THE NON INVERTING CONFIGURATION
Here the input signal 𝑣𝐼is applied directly to the positive input terminal of the op amp while one
terminal of 𝑅1is connected to ground.
The Closed-Loop Gain
Assuming that the op amp is ideal with infinite gain, a virtual short circuit exists between its two input
terminals. Hence the difference input signal is
vId =
v0
A
= 0 for A = ∞
Thus the voltage at the inverting input terminal will be equal to that at the non-inverting input terminal,
which is the applied voltage vI .
16

THE NON INVERTING CONFIGURATION
The current through R1can then be determined as
vI
R1
.
Because of the infinite input impedance of the op amp, this current will flow through 𝑅2.
Now the output voltage can be determined from
v0 = vI + (
vI
R1
)R2
𝐯𝟎
𝐯𝐈
= 𝟏 +
𝐑𝟐
𝐑𝟏
17

UNITY FOLLOWER
The unity-follower circuit, as shown provides a gain of unity (1) with no polarity or phase
reversal.
The output is the same polarity and magnitude as the input.
𝐕𝐎 = 𝐕𝐢
The circuit operates like an emitter- or source-follower circuit except that the gain is exactly
unity.
18

DIFFERENTIALAND COMMON-MODE OPERATION
Differential Inputs
When separate inputs are applied to the op-amp, the resulting difference signal is the difference between
the two inputs.
Common Inputs
When both input signals are the same, a common signal element due to the two inputs can be
defined as the average of the sum of the two signals.
Output Voltage
Since any signals applied to an op-amp in general have both in-phase and out-of-phase
components, the resulting output can be expressed as
where Ad=differential gain of the amplifier 𝑉𝑑=difference voltage
Ac=common mode gain of the amplifier 𝑉
𝑐=common voltage
OPERATIONALAMPLIFIER ANALOG ELECTRONICS CIRCUIT REC3C001
19
𝐕𝐝 = 𝐕𝐢𝟏
− 𝐕𝐢𝟐
𝐕𝐜 =
𝟏
𝟐
𝐕𝐢𝟏
+ 𝐕𝐢𝟐
𝐕𝐨 = 𝐀𝐝𝐕𝐝 + 𝐀𝐜𝐕𝐜

COMMON MODE REJECTION RATIO
we can now calculate a value for the common-mode rejection ratio (CMRR), which is defined
by the following equation:
The value of CMRR can also be expressed in logarithmic terms as
dB
We can express the output voltage in terms of the value of CMRR as follows:
Vo = AdVd + AcVc
= AdVd 1 +
AcVc
AdVd
20
𝐂𝐌𝐑𝐑 =
𝐀𝐝
𝐀𝐜
𝐕𝐨 = 𝐀𝐝𝐕𝐝(𝟏 +
𝟏
𝐂𝐌𝐑𝐑
𝐕𝐜
𝐕𝐝
)
𝐂𝐌𝐑𝐑 𝐥𝐨𝐠 = 𝟐𝟎 𝐥𝐨𝐠𝟏𝟎
𝐀𝐝
𝐀𝐜

EXAMPLE
Calculate the CMRR for the circuit measurements shown in Fig.
21

EXAMPLE
Determine the output voltage of an op-amp for input voltages of Vi1
= 150μV and Vi2
= 140μV.
The amplifier has a differential gain of Ad = 4000 and the value of CMRR is
a. 100
b. 105
22

DIFFERENCE AMPLIFIER
A difference amplifier is one that responds to the difference between the two signals applied at its
input and ideally rejects signals that are common to the two inputs.
The gain of the non inverting amplifier configuration is positive,(1 + 𝑅2
𝑅1
), while that of the inverting
configuration is negative,(− 𝑅2
𝑅1
).
Combining the two configurations together we are getting the difference between two input signals.
The two gain magnitudes should be made equal in order to reject common-mode signals.
23

DIFFERENCE AMPLIFIER
To apply superposition, we first reduce 𝑣𝐼2 to zero—that is, ground the terminal to which 𝑣𝐼2 is applied and
then find the corresponding output voltage, which will be due entirely to 𝑣𝐼1.
We denote this output voltage 𝑣01 and its value may be found as
𝑣01 = −
𝑅2
𝑅1
𝑣𝐼1
Next, we reduce 𝑣𝐼1 to zero and evaluate the corresponding output voltage𝑣02.
The output voltage 𝑣02 is therefore given by
𝒗𝟎𝟐 = 𝒗𝑰𝟐
𝑅4
𝑅3 + 𝑅4
1 +
𝑅2
𝑅1
=
𝑅2
𝑅1
𝑣𝐼2
The superposition principle tells us that the output voltage 𝑣0 is equal to the sum of 𝑣01𝑎𝑛𝑑 𝑣02Thus we have
𝑣0 = 𝑣01 + 𝑣02
𝒗𝟎 =
𝑹𝟐
𝑹𝟏
(𝒗𝑰𝟐 − 𝒗𝑰𝟏)
24

INSTRUMENTATION AMPLIFIER
The low-input-resistance problem of the difference amplifier of can be solved by buffering the two input
terminals using voltage followers
A voltage follower is connected between each input terminal and the corresponding input terminal of the
difference amplifier.
It consists of two stages.
The first stage is formed by op amps A1 and A2 , and their associated resistors, and the second stage is the
difference amplifier formed by op amp A3 and its four associated resistors.
25

The output of amplifier A1 and A2 are labelled as V01and V02respectively.
Since the amplifier 𝐴3 is a difference one, the output voltage is given by,
V0 =
R2
R1
(V02 − V01)
From amplifier A1 V1 = Va
From amplifier A2 V2= Vb
The current
i =
Va − Vb
R4
=
V1 − V2
R4
The differences between output voltages of amplifiers A1 and A2 is
V02 − V01 = i(2R3 + R4)
26

Now, the output voltage is,
V0 =
R2
R1
(V02 − V01)
𝐕𝟎 =
𝐑𝟐
𝐑𝟏
(
𝐕𝟏 − 𝐕𝟐
𝐑𝟒
)(𝟐𝐑𝟑 + 𝐑𝟒)
Advantages
i. Very high (ideally infinite) input resistance
ii. High differential gain
iii. High CMRR
iv. Low Output Resistance
27

OP-AMP SPECIFICATIONS—DC
OFFSET PARAMETERS
Offset Voltage
Because op amps are direct-coupled devices with large gains at dc, they are prone to dc problems.
The first such problem is the dc offset voltage.
If the two input terminals of the op amp are tied together and connected to ground, it will be found that a
finite dc voltage exists at the output.
The op-amp output can be brought back to its ideal value of 0 V by connecting a dc voltage source of
appropriate polarity and magnitude between the two input terminals of the op amp.
This external source balances out the input offset voltage of the op amp.
The output offset voltage can be shown to be affected by two separate circuit conditions: (1) an input
offset voltage VIO and (2) an offset current due to the difference in currents resulting at the plus (+) and
minus (-) inputs.
28

INPUT OFFSET VOLTAGE VIO
The manufacturer’s specification sheet provides a value of VIO for the op-amp.
We can write Vo = AVi = A(VIO − V0
R1
R1+Rf
)
V0 = VIO
A
1 + A[
R1
(R1+Rf)]
≈ VIO
A
A[
R1
(R1+Rf)]
𝐕𝟎 𝐨𝐟𝐟𝐬𝐞𝐭 = 𝐕𝐈𝐎
𝐑𝟏 + 𝐑𝐟
𝐑𝟏
29

EXAMPLE
Calculate the output offset voltage of the circuit in Fig.. The op-amp specification lists VIO = 1.2mV.
Solution
V0 offset = VIO
R1 + Rf
R1
=(1.2mV)(
2𝑘Ω+150𝑘Ω
2𝑘Ω
)=91.2mV
30

OUTPUT OFFSET VOLTAGE DUE TO INPUT
OFFSET CURRENT IIO
An output offset voltage will also result due to any difference in dc bias currents at both inputs.
Since the two input transistors are never exactly matched, each will operate at a slightly
different current.
The output offset voltage can be determined by replacing the bias currents through the input
resistors by the voltage drop that each develops.
𝐕𝐎 𝐨𝐟𝐟𝐬𝐞𝐭 𝐝𝐮𝐞 𝐭𝐨 𝐈𝐈𝐎 = 𝐈𝐈𝐎𝐑𝐟
31

TOTAL OFFSET DUE TO VIO AND IIO
The total output offset voltage can be expressed as
VO(offset) = VO(offset due to VIO) + VO(offsetdue to IIO)
Example
Calculate the total offset voltage for the circuit of Figure for an opamp with specified values of input offset
voltage VIO = 4 mV and input offset current IIO = 150 nA.
32

INPUT BIAS CURRENT, IIB
A parameter related to IIO and the separate input bias currents IIB
+ and IIB
- is the average bias current
defined as 𝐈𝐈𝐁 =
𝐈𝐈𝐁
+
+𝐈𝐈𝐁
−
𝟐
 The separate input bias currents can be determined using the specified values IIO and 𝐼𝐼𝐵.
It can be shown that for 𝐼𝐼𝐵
+
> 𝐼𝐼𝐵
−
𝐈𝐈𝐁
+
= 𝐈𝐈𝐁+
𝐈𝐈𝐎
𝟐
𝐈𝐈𝐁
−
= 𝐈𝐈𝐁 −
𝐈𝐈𝐎
𝟐
Example
Calculate the input bias currents at each input of an op-amp having specified values of IIO = 5
nA and IIB = 30 nA.
33

OP-AMP SPECIFICATIONS—
FREQUENCY PARAMETERS
An op-amp is designed to be a high-gain, wide-bandwidth amplifier.
This operation tends to be unstable (oscillate) due to positive feedback.
To ensure stable operation, op-amps are built with internal compensation circuitry, which also causes
the very high open-loop gain to diminish with increasing frequency.
This gain reduction is referred to as roll-off .
In most op-amps, roll-off occurs at a rate of 20 dB per decade (-20 dB/decade) or 6 dB per octave (-6
dB/octave).
Op-amp specifications list an open-loop voltage gain ( AVD ), the user typically connects the
op-amp using feedback resistors to reduce the circuit voltage gain to a much smaller value
(closed-loop voltage gain, ACL ).
34

OP-AMP SPECIFICATIONS—
FREQUENCY PARAMETERS
A number of circuit improvements result for this gain reduction.
I. The amplifier voltage gain is a more stable, precise value set by the external resistors;
II. The input impedance of the circuit is increased over that of the op-amp alone
III. The circuit output impedance is reduced from that of the op-amp alone
IV. The frequency response of the circuit is increased over that of the op-amp alone.
Gain–Bandwidth
Because of the internal compensation circuitry included in an op-amp, the voltage gain drops
off as frequency increases.
At low frequency down to dc operation the gain is that value listed by the manufacturer’s
specification AVD (voltage differential gain) and is typically a very large value.
35

GAIN–BANDWIDTH
As the frequency of the input signal increases, the open-loop gain drops off until it finally
reaches the value of 1 (unity).
The frequency at this gain value is specified by the manufacturer as the unity-gain bandwidth,
B1.
Although this value is a frequency at which the gain becomes 1, it can be considered a
bandwidth, since the frequency band from 0 Hz to the unity-gain frequency is also a
bandwidth.
36

GAIN–BANDWIDTH
The point at which the gain reduces to 1 as the unity-gain frequency ( f 1 ) or unity-gain
bandwidth ( B1 ).
At which the gain drops by3 dB (or to 0.707 the dc gain, AVD), this being the cutoff frequency of
the op-amp, fC.
In fact, the unity-gain frequency and cutoff frequency are related by
𝒇𝟏 = 𝑨𝑽𝑫𝒇𝑪
The unity-gain frequency may also be called the gain–bandwidth product of the op-amp.
Example
Determine the cutoff frequency of an op-amp having specified values B1 = 1 MHz and AVD = 200
V/mV.
37

SLEW RATE
Another parameter reflecting the op-amp’s ability to handle varying signals is the slew rate,
defined as
Slew rate = maximum rate at which amplifier output can change in volts per microsecond (𝑉 𝜇𝑠)
The slew rate provides a parameter specifying the maximum rate of change of the output
voltage when driven by a large step-input signal.
SR =
∆Vo
∆Vi
Example
For an op-amp having a slew rate of SR = 2 𝑉 𝜇𝑠, what is the maximum closed-loop voltage gain that
can be used when the input signal varies by 0.5 V in 10 𝜇𝑠?
38

MAXIMUM SIGNAL FREQUENCY
The maximum frequency at which an op-amp may operate depends on both the bandwidth
(BW) and slew rate (SR) parameters of the op-amp.
For a sinusoidal signal of general form
𝑣𝑜 = 𝐾 sin(2𝜋𝑓𝑡)
signal maximum voltage rate of change =2𝜋𝑓𝐾 𝑉/𝑠
To prevent distortion at the output, the rate of change must also be less than the slew rate,
that is,
2𝜋𝑓𝐾 ≤ 𝑆𝑅
𝜔𝐾 ≤ 𝑆𝑅
So that 𝑓 ≤
𝑆𝑅
2𝜋𝑓𝐾
Hz
𝜔 ≤
𝑆𝑅
𝐾
rad/s
39

EXAMPLE
For the signal and circuit of Fig., determine the maximum frequency that may be used. Op-amp
slew rate is SR = 0.5 𝑉 𝜇𝑠.
40

THE OPAMP INTEGRATOR
By placing a capacitor in the feedback path and a resistor at the input, we shall now show that this circuit
realizes the mathematical operation of integration.
Let the input be a time varying function𝑣𝐼(𝑡).
The virtual ground at the inverting op-amp input causes 𝑣𝐼(𝑡) to appear in effect across R, and thus the
current will be 𝑣𝐼(𝑡)/𝑅.
This current flows through the capacitor C, causing charge to accumulate on C.
OPERATIONAL AMPLIFIER ANALOG ELECTRONICSCIRCUITREC3C001
41

THE OPAMP INTEGRATOR
The current through the resistor R
i1 = vI(t)−0
R = vI(t)
R
The current through the capacitor C iC = C
dV
dt
= C
d(0−v0)
dt
iC = −C
dv0
dt
Both currents are equal because ideal op amp draws no current to its input terminals.
So, vI(t)
R = −C
dv0
dt
dv0 = −
1
RC
vI t dt
𝐯𝟎 = −
𝟏
𝐑𝐂
𝐯𝐈 𝐭 𝐝𝐭
OPERATIONALAMPLIFIER ANALOGELECTRONICSCIRCUITREC3C001
42

THE OP-AMP DIFFERENTIATOR
Interchanging the location of the capacitor and the resistor of the integrator circuit results in the circuit in
Fig., which performs the mathematical function of differentiation.
Let the input be the time-varying function vI t .
The virtual ground at the inverting input terminal of the op amp causes vI t to appear in effect across
the capacitor C.
Thus the current through C will be C
dvI
dt
and this current flows through the feedback resistor R providing
at the op-amp output a voltagev0 t .
OPERATIONALAMPLIFIER ANALOGELECTRONICSCIRCUITREC3C001
43

THE OP-AMP DIFFERENTIATOR
The current through the capacitor C
iC = C
d(vI − 0)
dt
= C
dvI(t)
dt
The current through the resistor R
iR =
0 − v0
R
= −
v0
R
 Both currents are equal because ideal op amp draws no current to its input terminals.
So,
−
v0
R
= C
dvI t
dt
𝐯𝟎 = −𝐑𝐂
𝐝𝐯𝐈(𝐭)
𝐝𝐭
OPERATIONAL AMPLIFIER ANALOGELECTRONICSCIRCUIT REC3C001
44

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