Chapter 7: Operational Amplifier (Op-Amp)

OPERATIONALAMPLIFIER (OP-AMP)
CHAPTER 7
BEKG1123
Principles of Electrical and Electronic

INTRODUCTION
 An op amp is an active circuit element designed to perform
mathematical operations of addition, subtraction, multiplication,
division, differentiation and integration.
 Typical uses: provide voltage amplitude changes (amplitude and
polarity), oscillators, filter circuit and instrumentation circuit.
 Operational amplifier (op amp) are linear integrated circuit and
widely used in analogue electronic system.
 The internal constructions includes a number of transistors, diodes,
resistors and capacitors in a single tiny chip of semiconductor
material.
 It is packaged in a single case to form a functional circuit.

 A typical op amp has eight-pin package where pin/terminal 8 is
unused and pin 1 & 5 are of little concern to us.
 The other 5 are:
 Pin 2 – inverting input
 Pin 3 – noninverting input
 Pin 6 – output
 Pin 7 – +ve power supply
 Pin 4 - -ve power supply

 The circuit symbol is the triangle and it has two inputs and one
output.
 The inputs are marked with minus (-) and plus (+) to specify inverting
and noninverting inputs, respectively.
 An input applied to the noninverting pin will appear with the same
polarity at the output.
 An input applied to the inverting pin will appear inverted at the
output.

Single-Ended Input
TYPE OF INPUT MODE OPERATION
- Input signal is connected to one op-amp input and another op-amp
input is connected to ground.
Input signal that applied to –ve input
 Output is opposite to the input
signal
Input signal that applied to +ve input
 Output has same polarity as input
signal

TYPE OF INPUT MODE OPERATION
Double-Ended (Differential) Input
Both input applied with Vd  amplified
output in phase with input applied
between +ve and –ve inputs.
When input is applied with separate
signals  the difference signal Vi1-Vi2
- Apply signals to each inputs

OP-AMP: TYPE OF INPUT MODE
OPERATION
Common Mode Operation
Same input signals are applied to both inputs of
op-amp
 0V output because the two signals equally
amplified but with opposite polarity signals.

 Here, the basic operation will be divided into two section:
 Non ideal operation
 Ideal operation
 However, the ideal operation will be used throughout the
analysis.
BASIC OPERATION

NON IDEAL OPERATION
 The non ideal op amp has the following characteristics:
 High open-loop gain,A 106
 High input resistance, Ri 2M
 Low output resistance, R0 75
Vd is the differential input voltage
given by:
vd = v2 – v1
where v1 is the voltage between the
inverting terminal and ground and v2
is the voltage between the
noninverting terminal and ground.

IDEAL OPERATION
 The characteristics of an ideal op amp is defined as follows:
 Infinite open-loop gain,A  
 Infinite input resistance, Ri  
 Zero output resistance, R0  0

 For the ideal op amp, the following is considered:
 The current into both input pin are zero:
i1 = 0, i2 = 0
 The voltage across the input terminals are zero:
vd = v2 – v1 = 0
thus
v1 = v2

EXAMPLE:
Calculate the closed-loop gain vo/vs and find io when vs = 1V.
Solution:
Remember that, for ideal op amp
i1 = 0, i2 = 0, v1 = v2
Since i1 = 0, the 40k and 5k resistors are in series; the same current
flow through them.And notice that
v2 = vs

Hence, using the voltage division principle,
Since v1 = v2 , thus
At node O,
When vs = 1V, vo = 9V. Substituting for vo = 9V in the above equation
gives
io = 0.2 + 0.45 = 0.65mA
1
5
5 40 9
o
o
v
k
v v
k k
 

9
9
o o
s
s
v v
v
v
  
0
5 40 20
o o
v v
i
k k k
 


 Figure below shows the connection of inverting amplifier.
 The noninverting input is grounded, vi is connected to the inverting
input through R1 and the feedback resistor Rf is connected between
the inverting input and output.
INVERTING AMPLIFIER

 Our goal is to obtain the relationship between the input voltage vi and
the output voltage v0.
 Applying KCL at node 1,
 But for an ideal op amp, v1 = v2 = 0, since the noninverting pin is
grounded. Hence,
1 1
1 2
1
i o
f
v v v v
i i
R R
 
  
1 1
f
i o
o i
f
R
v v
v v
R R R

   

 The voltage gain is
 The designation of the circuit as an inverter arises from the negative
sign.
1
f
o
v
i
R
v
A
v R
  

EXAMPLE:
If vi = 0.5V, calculate: a) the output voltage vo and b) the current in the
10k resistor.

Solution:
a)
b)The current through the 10k resistor is
 
1
25
2.5
10
2.5 2.5 0.5 1.25
f
o
i
o i
R
v
v R
v v V
     
     
1
0 0.5 0
50
10
i
v
i A
R k

 
  

 Figure below shows the connection of noninverting amplifier.
 The input voltage vi is applied directly at the noninverting input pin
and resistor R1 is connected between the ground and the inverting pin.
NONINVERTING AMPLIFIER

 Applying KCL at the inverting pin gives
but v1 = v2 = vi, the above equation becomes
or
1
1 2
1
0 i o
f
v v
v
i i
R R


  
1
i i o
f
v v v
R R
 

1
1
f
o i
R
v v
R
 
 
 
 

 The voltage gain is
 The output has the same polarity as the input, thus it is named as
noninverting amplifier.
1
1
f
o
v
i
R
v
A
v R
  

VOLTAGE FOLLOWER/BUFFER
 If we let Rf = 0 and R1 = 
the circuit will become as shown figure which is called a voltage
follower (or unity gain amplifier) because the output follows the input.
Thus,
1
o
o i
i
v
v v
v
  

EXAMPLE:
Calculate the output voltage vo
Solution:
Applying KCL at node a,
6
4 10
a a o
v v v
k k
 


But va = vb = 4, and also
4
6 4
1V
4 10
o
o
v
v
k k


   

 Figure below shows the connection of summing amplifier.Also called a
summer.
 A summing amplifier is an op amp circuit that combines several inputs
and produces an output that is the weighted sum of the inputs.
SUMMING AMPLIFIER

 It is an inverting amplifier with multiple inputs. It can have more than
three inputs.
 Applying KCL at node a gives
i = i1 + i2 + i3
but
note that va = 0, thus
SUMMING AMPLIFIER
1 2 3
1 2 3
1 2 3
, , ,
a a a a o
f
v v v v v v v v
i i i i
R R R R
   
   
1 2 3
1 2 3
f f f
o
R R R
v v v v
R R R
 
   
 
 

EXAMPLE:
Calculate vo and io of the following op amp circuit
Solution:
There are two inputs.
SUMMING AMPLIFIER
1 2
1 2
f f
o
R R
v v v
R R
 
  
 
 

The current io is the sum of the currents through the 10k and 2k
resistors. Both of these resistors have voltage vo = -8V across them, since
va = vb = 0. Hence,
SUMMING AMPLIFIER
     
10 10
2 1 4 4 8V
5 2.5
o
v
k k
 
       
 
 
0 0
0.8 4 4.8
10 2
o o
o
v v
i m m mA
k k
 
      

 This type of amp is used to amplify the difference between two input
signals.
DIFFERENCE AMPLIFIER

 Applying KCL at node a gives
or
 Applying KCL at node b gives
or
1
1 2
a a o
v v v v
R R
 

2 2
1
1 1
1
o a
R R
v v v
R R
 
  
 
 
2
3 4
0
b b
v v v
R R
 

4
2
3 4
b
R
v v
R R
 
  

 

since va = vb
or
2 4 2
2 1
1 3 4 1
1
o
R R R
v v v
R R R R
 
 
  
 
 

  
 
 
2 1 2 2
2 1
1 3 4 1
1
1
o
R R R R
v v v
R R R R

 


EXAMPLE:
Obtain io in the instrumentation amplifier circuit below
Answer:2A

Example 1:
Calculate the output voltage of the circuit for resistor
component of value RF = 470 kΩ, R1 = 4.3 kΩ, R2 = 33
kΩ, r3 = 33 kΩ for an input of 80 µV

Example 2:
Determine the output voltage of the circuit in figure below.
Answer:
Vo1 = 0.587 mV
Vo2 = - 18.13mV
Vo3 = 412mV

Exercise:
Based on the multiple stages operational amplifier (op-amp)
shown in Figure 1:
(a) Name the type of the amplifier of circuit 1,2 and 3.
(b) Determine the output voltage of the amplifier,VO1,VO2 andVO3.

Op-Amp Applications:
Control Sources
 Op-Amp can be used to form a various types of controlled
sources:
1. Voltage-ControlledVoltage Source
2. Voltage-Controlled Current Source
3. Current-ControlledVoltage Source
4. Current-Controlled Current Source
 Voltage/Current can be used to control an output voltage or
current
BEKG1123
37

Voltage-Controlled Voltage Source
 Vo is controlled byV1
 The output voltage see to be dependent on the input voltage
(scale factor k)
 Inverting input
 Non-inverting input
BEKG1123
38

Voltage-Controlled Current Source
 Provide output current controlled by input voltage.
BEKG1123
39

Current-Controlled Voltage Source
 The output voltage is dependent on the input current
BEKG1123
40

Current-Controlled Current Source
 Output current is dependent on the input current
BEKG1123
41

Control Sources
 Example:
 Calculate IL.
BEKG1123
42

Control Sources
 Example:
 CalculateVo.
BEKG1123
43

Instrument Amplifier
 Provides an output based on the difference betweenTWO
inputs.
 Potentiometer = provide the scale factor adjustment.
BEKG1123
44

Instrument Amplifier
 Example:
 Calculate the output voltage expression of the circuit:
BEKG1123
45

Active Filters
 To provide voltage amplification and signal isolation or
buffering.
 Low-Pass Filter – provides a constant output from DC up to
cut-off frequency, fOH and then passes no signal above that frequency.
 High-Pass Filter – provides or passes signals above a cut-off frequency, FOL.
 Band-Pass Filter – filter passes signals that above one cut-off frequency and
below another cut-off frequency.
BEKG1123
46

Low-Pass Filter
 First-Order Low-Pass Filter
 Using single R and C
 Has practical slope of -20 dB per decade
BEKG1123
47

Low-Pass Filter
 Second-Order Low-Pass Filter
 2 sections of filter
BEKG1123
48

Low-Pass Filter
 Example:
 Calculate the cut-off frequency of a first-order low pass filter
for R1 = 1.2 kΩ and C1 = 0.02 µF
 Answer:
BEKG1123
49

High-Pass Filter
 First- Order
 Second-Order
BEKG1123
50

High-Pass Filter
 Example:
 Calculate the gain and cut-off frequency of a second-order high-
pass filter for R1 = R2 = 2.1 kΩ, C1 = C2 = 0.05 µF and RG
= 10 k Ω, RF = 50 kΩ
 Answer:
BEKG1123
51

Band-Pass Filter
 Combine high-pass filter and low-pass filter
BEKG1123
52

Band-Pass Filter
 Example:
 Calculate the cut-off frequencies of the band-pass filter with R1
= R2 = 10 kΩ, C1 = 0.1 µF and C2 = 0.002 µF
BEKG1123
53

Chapter 7: Operational Amplifier (Op-Amp)

More Related Content

What's hot (20)

Similar to Chapter 7: Operational Amplifier (Op-Amp) (20)

More from JeremyLauKarHei (6)

Recently uploaded (20)

Chapter 7: Operational Amplifier (Op-Amp)