Mechatronics Regular Notes

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Let us discuss “Mechatronics” from below example:
Let us consider a human on ground and consider a ball that’s falling down at a
distance of 5m away from human position. Assume that the human tries to catch the
ball, then Its eyes, by which he watches ball first then takes action. Once eyes
recognize the position of the ball then the information of the position of the ball is
sent to brain.
The brain processes this information and send the information to the legs to run.
After reaching the required position on ground then the brain sends information to
hands to catch the ball.In this simple example, eyes detect the position of the ball
and sends information to brain and brain then gives commands to different parts of
the body to take quick action.
Here eyes are acting like sensors.
Brain does the work of processors.
Hands and legs are doing work of actuators.
So sensor is the first device that senses the information, processor does the work on
input and commands the actuators to conduct the desired mechanical outcomes.
Eyes – Brain – Hands/legs.
In mechatronics, all the equipment has sensors, processors and actuators (SPA).
The coordination between all these three ,sensors and processors and actuators will
give a desired outcome.
Working of Robot:
Let us consider a robot on ground, and ball is falling on the ground vertically at a
distance of 5m away from the robot. The positions scanners of the robot detect the
coordinate of the ball in the space, the information of ball position is sent to
processor.
The various motors present on the body of robot will get the signal from processor.
The robot starts moving towards the ball with the help of motor1. As the robot comes
below the ball, then the motor2 gets activated and catches the ball. It is all the job of
processor to command different parts of the robot.
The motor1 and motor2 are well equipped to give both linear and angular
displacement. So these motors are called actuators. All parts taken together called
the Robot.
The combined integration of the Sensor, processor and actuator is called as

Mechatronic system.
The combined integration of Eyes, brain and hand/legs is called Human body.
The combined integration of position scanner, microprocessor and motor is called

robot and it is classical example of Mechatronics system.
Definition:
Mechatronics is the integrated product of ‘Electrical and Electronics sensors’ and
‘Electrical and Mechanical actuators’ and ‘Electronic processor like microprocessor
and microcomputer etc..’.
Sensor is used for detection of physical variable and provides information to

processor.
Processor performs computations on the data received from sensor with the help of
predefined program and delivers command signal to actuator.
Actuator:It provides mechanical output mechatronics system ultimately works like

an automatic control system.
Practical application of mechatronics:

Example1. Washing machine:
When we dump cloths in drum, the weight sensor senses the weight and sends
information to processor and processor sends electrical signals to motor to rotate the
drum according to the weight. The speed of the drum depends on the shaft, and shaft
speed depends the current that is supplied. The processor actually controls the
amount of current supplied for example let us consider
Weight(Sensor) Current(Processor) Speed(Motor)
4N 10 mA 10 RPM
10N 30 mA 30 RPM
This is how the different parameters are controlled.
Sometimes processor has to check the level of dirt in the cloths. To do that another
system called feedback loop is present in the device.
Example2.Car window:
When we press the switch, the actuator gets the information from the processor for
the linear motion of the windows. That is how we control the motion of the window.
Example3 : Washing machine works as feedback control

system.
When the load in the drum keeps increasing, the sensor stars sensing it and sends
information to the micro processor. Then the sensor will send electric signals and
the high amount of current is passed to actuator(motor) and the motor rotates faster.
The total assembly works like Feedback control system.
Advantages of Mechatronics:
1. Cost efficient and good quality products.
2. High grade of flexibility to modify or redesign.
3. Very good performance characteristics.
4. Wide area of application.
5. Great productivity in case of manufacturing organization.
6. Great extent of machine utilization.
Disadvantage of Mechatronics:
1. High initial cost.
2. Multi-disciplinary engineering background required to design and

implementation.
3. Need of highly trained workers.
4. Complexity in identification and correction of problems in the system.
Application/Scope of mechatronics:
There are diverse area of application like transportation, material processing, green
technology, medical etc. In simple words, where ever there is automation there is
requirement of Mechatronics system. The applications of mechatronics engineering
in various fields are mentioned in the figure below.
What is sensor ?
Sensor is a element or device , which reacts to physical quantity and generates some
output signal which can be quantified. Electrical and Electronic sensors generates
electrical output.
We do have mechanical sensors which results in variation of mechanical parameters,

when it senses physical quantity.
Ex: Thermocouple, Piezo electric material, metal rod.
Let us discuss the function of thermocouple
Thermocouple:
It is a device that has a junction formed by two different metals junction and It
actually measures change in temperature and gives the output in the form of voltage.
Sensor is used to measure the physical quantity such as position, force, distance,
strain, vibration, temperature, acceleration etc..
Most of the sensors using these days work by converting some physical parameters
into electrical signals like voltage or current.
Piezo electric effect:
When we apply force on the piezo electric material, then charge / voltage will be
developed. So mechanical input is converted into electric signal.
In case of the metallic rod which has length L , as the temperature increases the
length of the rod also increases according to its thermal expansion.
Using the change is the length we can predicts the change is the temperature of the
rod, so the metallic rod is also called as mechanical sensor.
Electrical or Electronic sensors convert physical quantity into electric output.
Mechanical sensor converts physical quantity into mechanical output.
Electronic sensors are widely used due to its portability and many other
advantages.
What is Actuator ?:
Actuator is a device, which generates mechanical output like linear displacement or
angular displacement. Electrical actuators convert electrical input to mechanical
output.
The cylindrical chamber is equipped with port and a piston as shown in the figure.
As the liquid flows in the port with high pressure, the piston moves with some speed
and if the end of rod is connected to the screw, then it provides the angular motion
of the screw. So If the liquid pressure is converted into linear/ angular displacement,
then we call it as hydraulic actuator.
If the air pressure is converted into linear/ angular displacement, then we call it
pneumatic actuator.
Motor :
In case of the motor, as the electrical input is given to motor and it is converted
into the rotational motion of the shaft, thus it is called as electrical actuators.
What is transducer:
It is a device that converts one form of energy into another form of usable energy.
1. Transducer is much wider term compared to sensors and actuator.
2. Transducer can work like sensor but sensor cannot work like transducer.
In fact, transducer can work both like sensor and actuator.
Lets see and example,
Temp Voltage
Thermocouple
100oc 2 V
Thermocouple Amplifier(106)
100°c 2 V 2V
Here the whole setup of thermocouple and amplifier together is called as transducer
because if we use only thermocouple then it generates very low voltage (which can’t
do any work). So transducer is much bigger term than the sensor.
What is processor?:
Processor is an electronic device that performs digital computations on the digital

data received with the predefined program written in its memory.
Processor is also called as digital device because it receives only digital data from
sensors and also generates only digital data after computations.
The temperature is sensed by the thermocouple and voltage signal is produced. The
voltage of the sensor has to sent to processor but the analog information of the
thermocouple has to be transferred to micro processor in the form of digital data. So
we require to use Analog-to-Digital convertor.
Characteristics of Sensor or Transducer
These are basically the parameters that help us to understand the sensors in a better
way.
Characteristics are as follows:

1. Sensitivity
2. Resolution
3. Linearity and Non-linear error
4. Accuracy
5. Precision
6. Hysteresis
Sensitivity
It is the ratio between the changes in output to the change in input.
For a linear sensors, it is the ratio between change in output span to change in the
input span.
For a Non-linear sensors, it is the ratio between infinitesimal change in output
signal to infinitesimal change in the input signal.
Ideally the sensitivity of sensor should be as high as possible.
Thermocouple is a temperature sensor that converts Temperature difference into

Voltage. It works on Seeback effect.
Note: Reverse seeback effect is Peltier Effect.
If Input Range ( 0°C - 100°C)

And Output Range ( 0mV – 100mV)
Then let us find the input span,

T(min)=0°C T(max)=100°C
Input Span (T) = T(max) – T(min)
= 100°C - 0°C = 100°C
Lets find the out span,
V(min) = 0mV V(max) = 100mV
Output span (V) = V(max) – V(min)
= 100mV – 0mV
= 100mV
𝐎𝐮𝐭𝐩𝐮𝐭 𝐒𝐩𝐚𝐧 𝟏𝟎𝟎𝒎𝑽 𝒎𝑽
𝐒𝐞𝐧𝐬𝐢𝐭𝐢𝐯𝐭𝐲 = = =1
𝐈𝐧𝐩𝐮𝐭 𝐒𝐩𝐚𝐧 𝟏𝟎𝟎°𝑪 °𝑪
Resolution
The smallest change in the input signal that is required to generate detectable change
in the output.
Ideally the magnitude of Resolution should be as low as possible.
Temperature Expected Output Voltage

Output
(Resolution=25°C)
Voltage
0°C 0 mV 0 mV
10°C 10 mV 0 mV
25°C 25 mV 25 mV
35°C 35 mV 25 mV
50°C 50 mV 50 mV
65°C 65 mV 50 mV
75°C 75 mV 75 mV
90°C 90 mV 75 mV
100°C 100 mV 100 mV
Linearity and Non-linear error
Ideally the output of the sensor should be linearly associated with the output of the
sensor irrespective of operating conditions.
If Temperature (T) and Voltage (V) have a relation of,

V = K.T (where K is constant)
It is similar to, Y= mX
And V T
If we plot a graph with Voltage on X-axis and Temperature on Y-axis,

We get a straight line since the relation is linear.
Voltage/Tempatute Graph
120
100 Practical Graph
80
60
Ideal Graph
40
20
0
-20 0 20 40 60 80 100 120
The arrow on the graph, that gives us True value (60) and Measured value (80).
M. V − T. V
%Non Linear error in True value = × 100
T. V
20
= × 100
60
=66.66%
Accuracy
The degree of closeness between true value and measured value of a sensor is
referred as accuracy.
Ideally sensor should be highly accurate.
𝑚𝑉
Sensitivity = 1
°𝐶
At 50°C
𝑚𝑉
Expected Value = 1 ×50 = 50 mV
°𝐶
If measured value = 55mV
𝑀.𝑉−𝑇.𝑉
%Error = × 100
𝑇.𝑉
55𝑚𝑉−50𝑚𝑉
= × 100
50𝑚𝑉
=10%
Then the Accuracy = 100%-10%
= 90%
Precision
It is the ability of an instrument to produce the same output, when sensor is used to
measure same input repeatedly under same operating conditions.
(or)
It is the closeness of “Many output reading taken for same quantity” without
changing operating conditions.
Ideally sensor should be highly precise.
𝑚𝑉
Sensitivity = 1
°𝐶
Let us discuss three different cases or situations , which explains the relation between
accuracy and precision and also the difference between them.
Case 1:
Temperature = 50°C and True Value = 50mV
1st Reading = 25mV
2nd Reading = 25.1mV
3rd Reading = 24.9mV
If we observe the reading, we can say the system is precise because all the readings
are too close but no reading is near to True Value.
So we call the sensor is Precise but not Accurate.
Case 2:
Temperatue = 50°C and True Vaue = 50mV
1st Reading = 50mV
2nd Reading = 50.1mV
3rd Reading = 49.9mV
If we observe the reading, we can say the system is Precise and Accurate.
Case 3:
Temperature = 50°C and True Value = 50mV
1st Reading = 27mV
2nd Reading = 50mV
3rd Reading = 35mV
If we observe the reading, we can say that system is neither Precise nor Accurate.
Hysteresis
It is defined as the difference in the output for a given input when this is approached
from the opposite direction.
Ideally “Hysteresis/Hysteresis error” should be as low as possible.
Y-Values
120
100 Backward
80
60
40
20
Forward
0
0 20 40 60 80 100 120
At 50°C
Hysteresis error = 82 mv(Value on forward curve) – 18 mV (Value on reverse
curve)= 64mV
Qn:For a copper-consttantan thermocouple, the junction potential (E in V) at
100°C is given by E=38.74 +3.3×𝟏𝟎−𝟐 𝟐 +2.07×𝟏𝟎−𝟒 𝟑 - 2.2×𝟏𝟎−𝟔 𝟒 . The
sensitivity of thermocouple at 100°C is approximately.
(a) 45.35 V/°C (b) 42.75 V/°C (c) 38.74 V/°C (d) 0.06 V/°C
Since, Temperature   and EOutput

E 𝑑𝐸
We know that, Sensitivity= =
 𝑑
Now by differentiating the Output with respect to Temperature, we get
𝑑𝐸 3
= 38.74 + 3.3 × 10−2 1 + 2.07 × 10−4 2 − 2.2 × 10−6 
𝑑
𝑑𝐸 ( = 100)
𝑑
= 38.74 + 3.3 × 10−2 1001 + 2.07 × 10−4 1002 − 2.2
3
× 10−6 100
= 38.74+6.6+6.21-8.8
=42.75 V/°C
Qn :A thermocouple is used to measure the temperature in the range of (0°C to

300°C). Ideally the voltage generated by the thermocouple at 0°C and 300°C is
given by 0V and 15V.If the voltage measured at 100°C is 6V, then find the %
non-linear error at 100°C in true value.
(a) 0 (b) 40 (c) 20 (d) 30
Lets consider a thermocouple
Thermocoupl
(0°C to e
300°C) (0mV to 15mV)
V 15mV − 0mV
Sensitivty = =
T 300°C − 0°C
15𝑚𝑉 𝑚𝑉
= = 0.05
300°𝐶 °𝐶
At 100°C,
True value = 0.05 mV/°C ×100°C = 5mV
Measured value = 6mV
𝑀.𝑉−𝑇.𝑉
% Non linearity = ×100
𝑇.𝑉
6𝑚𝑉−5𝑚𝑉
= × 100
5𝑚𝑉
=+20%
Thus the option (c)20 is correct.
Qn :Which of the following statements are true.

Statement 1: Transducer ia s device, which converts one form of energy into
another form of energy.
Statement 2: Sensor is device, which is used in feedback of mechatronics device
and it accepts the current from controller.
Statement 3: Actuator is generally cascaded with controller and receives the
control signal (control) from controller.
(a) Only 1 (b) Only 2 (c) Only 1&3 (d) none of these.
Sensor Processor or Actuator

Controller
From the above flow chart, we understand that the statement 2 is wrong and the
statement 3 is correct and statement 1 is a basic definition.
So Option (c) Only 1 and 3 is correct.
Qn : A force is developed by connecting two sensors as shown in the figure. The

sensitivities and resolution of both the sensors are mentioned below,
Find the output voltage V, when the applied force is 4.75N

(a) 4.74V (b) 4.5V (c) 9V (d) 9.5V
The sensor is such that the mechanical quantity (force) is converted into another
mechanical quantity (dist.) and at last it is converted into electrical
quantity(Voltage).
Force(N) Displacement(cm)
0 0
0.5 1
4 8
4.5 9
4.75 9
5 10
Please note the displacement does not change for 4.75N since the resolution is 0.5N.
Displacement(cm) Voltage(V)
8 8
8.25 8.25
8.5 8.5
8.75 8.75
9 9
So for a displacement of 9cm the output voltage will be 9V.

Thus the option(c)9V correct.
Qn : Match the following
The question can be solved by proper understanding of definition.Answer is (b).
Qn : Which of the following is true about the sensors?
𝑚𝑉
Sensitivity = 1
°𝐶
Temp/Volt Graph
150
100
50
0
0 50 100 150
Since the slope of graph is sensitivity, the slope of a straight line always remains
constant. Thus the first sentence is justified.
If the resolution= 25°C
As the value of resolution is higher, the graph starts appearing like a step showing
the non linear behavior.
Ideally, resolution should be low and sensitivity should be high.
So, Option (a) is Correct.
Qn : A pressure sensor is capable to convert the input pressure of range of (0
MPa- 100 MPa) has a sensitivity of 2 mV/1 MPa and generate 200mV in 200
steps, then the input resolution in MPa is
(a) 1 (b) 0.5 (c) 0.75 (d) 2
(0MPa-100MPa) Pressure Sensor (0mV-200mV)
Sensitivity= 2mV/1 MPa

For, 200 Steps it delivers 200mV
1 Step - 1mV
Pressure (MPa) Voltage(V)
0 0
0.5 1
1 2
1.5 3
From the table, we can conclude Input resolution = 0.5 MPa.
How we can measure physical parameter practically?

Measurement system:
It has group of devices, which collectively works to measure any physical variable
accurately and displays the numerical value of physical variable accurately to end
user.
It consists of devices like Sensor, Amplifier, ADC, Microprocessor, Display device.
Temperature Measurement System:
Suppose the fluid is flowing in the pipe and temperature range is (0°C-100°C).
Sensitivity = 1mV/°C
If the temperature is 50°C, then voltage generated is 50mV.
As the temperature difference is developed at junction, the developed voltage has to
be filtered to avoid NOISE.
The filtered voltage is very small to be detected by ADC, so we use the amplifier is
used to amplify the small amount of voltage.
VfilterVamplified.
After amplifier the analog voltage is converted into digital voltage with the help of
ADC ( Analog to Digital Converter).
Further Microprocessor records the data from ADC and stores it in the memory.
Whenever anybody requires to know temperature at any particular time, a program
has to be written for Microprocessor and the processor will give the data of voltage
on Display device with will be readable by human beings.
This is how atypical industrial temperature measurement system works.
Now let us see the different component in detail.
1. Sensor:
It converts the mechanical quantity(temperature) into electrical quantity (Voltage)
Sensor
2. Filter:
It is used to reduce the noise that is developed during the transfer of signal from
sensor.
Filter
3. Amplifier:
It is used to amplify the small quantity of input into higher valued output
Amplifier
The quantity of amplification can be determined by the ratio called Amplifier

Gain(G).
𝑉(𝑂𝑢𝑡𝑝𝑢𝑡)
Amplifier Gain(G) =
𝑉(𝐼𝑛𝑝𝑢𝑡)
If G  1, VOutput> VInput then we call it as Amplifier.
G < 1, VOutput< VInput then we call it as Attenuator.
G = 1, VOutput = VInput then we call it – Buffer.
4. Analog to Digital Convertor:
It is an electronic device which converts analog input into digital code.
DC Voltage
DC
Analog to b1
From Amplifier Digital b2

Convertor
The above ADC is 2 bit with 4 combinations.
The 3 bit ADC will have 8 combinations.
A N bit ADC will have 2𝑁 combinatios.

𝐹𝑢𝑙𝑙 𝑠𝑐𝑎𝑙𝑒 𝑣𝑎𝑙𝑢𝑒
Resolution =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠
A brief conclusion :
A thermocouple metal is attached to fluid carrying hot fluid the temperature of
which has to be found. The temperature is converted into voltage and the voltage
with noise is filtered at filter. The small amount of voltage is amplified so that
voltage is detectable. The amplified voltage which is analog in nature is converted
into digital with the help of Analog to Digital Convertor. The digital code is then
stored in memory and the required data will be made visible in display on demand
by user.
Block Diagram of Measurement System
Signal conditioning unit
This unit converts the output of sensor into much suitable form such that it can be
processed further easily.
Signal processing unit
This unit converts the output of conditioning unit into much suitable form such that
it can be accurately displayed.
Measuring device
This device should indicate the numerical value of physical variable.
Smart Sensor : If the sensors, signal conditioning system, signal processing

unit and display device all of them a packed into one unit called smart sensors.
1. It has built in processor along with sensor and conditioning unit.

2. As it has built processor, it has te capacity to take decisions.
3. Smart sensor is also called as intelligent sensor.
4. Smart sensor cab be connected to actuator directly.
Ans: Since statement 1 and 2 are independently right and 2 is not right explanation
for 1. So answer is option (a).
Qn : Which of the following electronic device does not allow or resist specified
input to output side.
a) Amplifier b) Attenuator c) Filter d)none of the above.
Ans: Filter is the only device that restrict the flow of signal.So answer is option (c).
Ans: Both the statement are correct and statement 2 is not correct explination for
statement 1. So answer is option (b) is right.
Qn : If the spring sensor deflects 0.075m when a force of 15KN is applied, find
the input force for a displacement of 0.1m
Ans: We know that, F= k.x
Where k= spring constant X= displacement.

𝐹 15 𝑘𝑁
K= = = 200
𝑥 0.075𝑚
So, for x= 0.1,F= 200×.1= 20 KN
Qn: A pressure sensor is connected to 2- bit analog to digital convertor as
shown below. The resolution of the input pressure is______.
a) 6 psi b)5 psi c)7 psi d) 8 psi.

Pressure 2-bit
(0 psi – 20 psi) (2 bit digital output)
sensor
ADC
𝐹𝑢𝑙𝑙 𝑠𝑐𝑎𝑙𝑒 𝑣𝑎𝑙𝑢𝑒
Ans: Resolution = = 20 psi/ 4 = 5 psi.
22
Wheatstone
Sensor Amplifier Pen recorder
Bridge
T R VOB VOA X
The input of the system is temperature and out put is displacement.

𝑋
a. Sensitivity of overall measurement system =
𝑇
𝑅 𝑉 𝑉(𝑎) 𝑋
= ∗ ∗ ∗
𝑇 𝑅 𝑉 𝑉(𝑎)
= 0.3*0.01*80*1.2
= 0.288 mm/°C
b. Temperature corresponding to 30 mm is = 30/.288 = 104.166 °C.
Basics of Electricity:
The flow of electrical charge through a conductor per unit time, is called as
current. It is denoted by letter “I” and units is ampere.
Lets consider a metallic wire with resistance R, length l, and Area A, then the
resistance is given by,
𝑙
R=
𝐴
As the cell is connected and switch is closed, the electrons in the conductor starts
moving which produces electricity in the circuit.
If switch is open = 0A
If switch is closed 0A.
Ohm’s Law:
V=R.I (at constant temperature)
𝑉
I=
𝑅
If R keeps increasing the current in the loop decreases and vice versa.
Lets understand the significance with example.
Consider a metallic wire shown below
Let us consider length li and diameter 𝑑𝑖
Initially, there is on force on the bar,
a) F=0
𝑙
𝑅𝑖 = 𝑅𝑖 = Initial resistance Force F
𝐴
b) when F0, then the length of the bat increases and diameter decreases.
𝑙𝑓𝑖𝑛𝑎𝑙 = li+l
𝑑𝑓𝑖𝑛𝑎𝑙 = 𝑑𝑖 - d
𝐴𝑓𝑖𝑛𝑎𝑙 = 𝐴𝑖 - A
Since there is change in area, the resistance will also change and they are inversely
proportional.
So, 𝑅𝑓𝑖𝑛𝑎𝑙 = 𝑅𝑖 +R.
If we connect this bar is the electric loop, due to change in the resistance the flow
of current also changes.
FRI
So in the industry, the mechanical input should be converted into electrical output.
We have special name for this arrangement called Force sensor or Strain gauge.
Strain gauge
It is a strain sensor, which is made up of elastic material.
It can be made up of metal as well as semiconductor material.
It is used to measure force, acceleration, torque etc.
When the elastic material is acted upon by a load, strain is developed in the
material. Then the change in length is converted into change in resistance.
Thus, R  l
𝑅 𝑙

𝑅 𝑙
𝑅 𝑙
= k. where k is a constant.
𝑅 𝑙
𝑅 𝑙
( )/ ( ) = k = Gf
𝑅 𝑙
The proportionality constant Gf is known as Gauge Factor.

𝑙
Further, = 𝑆𝑡𝑟𝑎𝑖𝑛
𝑙
𝐹
We know that strain =
𝐸𝐴
On rearranging the equation

𝑅∗𝐺
R= ∗𝐹
𝐸𝐴
R, G, E are constants. Since change in area is very less, then A can also be taken as
constant.
Thus R  F.
Physical aspects behind the mathematical formula.
𝑅∗𝐺
R=( )∗𝐹
𝐸𝐴
If R =10
G= 2.0 (metallic gauge)
A= 0.01 𝑚2
E= 108 N/𝑚2
10∗2
R=( )∗𝐹
.01∗108
R/F = 20*10−6 /N
So by this we can understand that the sensitivity of metallic gauge in very low,
since the Young’s modulus is very high and gauge factor is very low.
To increase the sensitivity of conductor we need to decrease the young’s modulus

of the material which is done by doping the material.
Thus semiconductor material has high sensitivity.
L= 20cm , R=10 , R=0.35 , l=0.2mm

𝑅
We know that, Gauge factor, G= 𝑅𝑙
𝑙
= (0.35/100) / (0.0002/.02)
= 3.5
Thus option (a) is right option.
Ans: R=1000
G= +200 at 25°C
Strain= 10−4 m/m

𝑅
G= 𝑅𝑙
𝑙
By rearranging the equation,

𝑙
R= R*G* = 1000*200*10−4 = 20 .
𝑙
1.You can observe that the Gauge factor of the semiconductor material is
quite high, which is desirable.
2. Here Gauge factor is dependent on temperature so here R is not

proportional to force.
3. Advantage with semiconductor is its high sensitivity and disadvantage is

temperature sensitive.
𝑙
Ans: R= R*G* =50*2.1*0.001= 0.105 .
𝑙
Ans:
l=150mm
A=5 sq.cm
𝑁
E=200*109
𝑠𝑞.𝑚
R=220
G=2
R=0.015.
𝑅
G= 𝑅𝑙
𝑙
𝑅
𝑙
= 𝑅
𝑙 𝐺
𝑅
l = l* 𝑅
𝐺
= (150*10−3 )*.015/(2*220)
Change in length = 0.00511 mm.

𝐹
𝐴
To find force E = 𝑙
𝑙
𝑙
F= E*A* = 200*109 *5*10−4 ∗0.00511/150
𝑙
=3406.66 N
A deep discussion into Strain Gauge
Wheatstone
Bridge.
Consider a strain gauge with,
Strain = 10−6
Nominal resistance = 100
Gauge factor = 2.0, due to stain the change in resistance developed is R=2*10−6 .
The mechanical output of the strain gauge has to be sent to amplifier to amplify the
quantity of strain gauge output. But amplifier only accepts the voltage as input. So
we need a special device that converts the change in resistance into change in
voltage, that special device is known as Wheatstone Bridge.
What is Wheatstone bridge: Wheatstone bridge is electrical circuit that
is used to convert the change in resistance into change in voltage. Consider two
resistances 𝑅1 and 𝑅2 connected in series and voltage 𝑉𝑆 is developed between C
and D. The current flowing through 𝑅1 and 𝑅2 are same since they are in series.
So the voltage across C and D is,
𝑉𝑐 – 𝑉𝑑 = 𝑉𝑠 ,
Voltage across A and D,
𝑉𝑎 – 𝑉𝑑 = 𝑉𝑎𝑑 = 𝐼1 ×𝑅2
𝐼1 = 𝑉𝑠 /(𝑅1 +𝑅2 )
𝑉𝑆
𝑉𝑎𝑑 = × 𝑅2 .
𝑅1 +𝑅2
Further consider two more resistances 𝑅3 and 𝑅4 in the circuit that are connected in
the circuit.
Now, The 𝑅1 , 𝑅2 and 𝑅3 , 𝑅4 are connected parallel in the circuit.
Current 𝐼2 flows through 𝑅3 and 𝑅4 ,
Then 𝑉𝑏𝑑 = 𝑉𝑏 – 𝑉𝑑 = 𝐼2 ×𝑅4
𝐼2 = 𝑉𝑆 /(𝑅3 + 𝑅4 )
𝑉𝑆
𝑉𝑏𝑑 = × 𝑅4
𝑅3 +𝑅4
Lets us connect a voltmeter between the point A and B which is used to find output
voltage of the bridge, 𝑉𝑜𝑏
𝑉𝑜𝑏 = 𝑉𝑎 – 𝑉𝑏 = (𝑉𝑎 – 𝑉𝑑 ) – (𝑉𝑏 – 𝑉𝑑 )
𝑉𝑜𝑏 = 𝑉𝑎𝑑 – 𝑉𝑏𝑑

𝑉𝑆 𝑉𝑆
𝑉𝑜𝑏 = × 𝑅2 - × 𝑅4
𝑅1 +𝑅2 𝑅3 +𝑅4
This is the output voltage of wheat stone bridge.
Ans:
a) Applied strain = 0m/m (Unstrained condition)
𝑅2 = 600.
𝑉𝑆 𝑉𝑆
𝑉𝑜𝑏 = × 𝑅2 - × 𝑅4 = 0V (Balanced condition).
𝑅1 +𝑅2 𝑅3 +𝑅4
b) Applied strain = 100/m
R= Nominal resistance * guage factor * strain
=600*2*100*10−6
=0.012
𝑅2 = R+R = 600+0.12 = 600.12.

𝑉𝑆 𝑉𝑆
Now, 𝑉𝑜𝑏 = × 𝑅2 - × 𝑅4
𝑅1 +𝑅2 𝑅3 +𝑅4
600.12 600
𝑉𝑜𝑏 = 4( − ) = 1.63 V.
600+600.12 600+60
Application of strain gauge
Consider a cantilever bear. An elastic

material is pasted on the beam. As the force is applied on the edge of the beam, the
structure deflects, and if the elastic material is pasted on above the neutral axis, it
elongates(tension) and if the strip is pasted below the neutral axis, the strip
contacts(compression), accordingly the resistance of the strip changes.In the actual
case, the strip is actually a strain gauge and whose resistance is governed by the
deflection of the beam. The fourth resistance is actually gauge resistance, 𝑅4 =𝑅𝑔 .
a) If force = 0N,
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅4 – × 𝑅3
𝑅4 +𝑅2 𝑅3 +𝑅1
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 , thus,
𝑉𝑜𝑏 = 0V.
b) If F 0 N.
𝑅4 = R + R
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅4 – × 𝑅3
𝑅4 +𝑅2 𝑅3 +𝑅1
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4
𝑉𝑆 𝑉𝑆
𝑉𝑜𝑏 = × 𝑅 + 𝑅 – *R
2𝑅+𝑅 2𝑅
𝑉𝑜𝑏  0V.
Load cell (Weight measurement system)
It is a device which is used to measure weight of things. There are two gauges that
are kept in the machine, one in lateral and another longitudinal, whenever a weight
is kept on the machine, the compression is developed in on strain gauge and the
tension on another gauge.Let us assume, 𝑅1 is on side face ad 𝑅2 is on front face.
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅4 – × 𝑅3
𝑅4 +𝑅2 𝑅3 +𝑅1
a) If force = 0N,
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅4 – × 𝑅3
𝑅4 +𝑅2 𝑅3 +𝑅1
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 , thus,
𝑉𝑜𝑏 = 0V.
b) If F 0 N.
𝑅4 = R + R
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅4 – × 𝑅3
𝑅4 +𝑅2 𝑅3 +𝑅1
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4
𝑉𝑠 𝑉𝑠
𝑉𝑜𝑏 = × 𝑅 + 𝑅 – *R
2𝑅+𝑅 2𝑅
𝑉𝑜𝑏  0V.
So the basic principle is, the force is recorded in terms of change in resistance
further it is converted into voltage.
1. Piezoelectric Accelerometer:
Piezometric Material: If piezoelectric material is given some input then it will generate
output as voltage.
Note: In this case piezoelectric accelerometer is converting acceleration (input) to

voltage (output)
As acceleration is mechanical quantity & voltage is electrical quantity. So piezoelectric

changes mechanical quantity (Acceleration) into electric quantity (voltage) it is satisfy the
definition of sensors. So piezoelectric accelerometer is a sensor which converts
acceleration into voltage.
Piezoelectric Accelerometer:
1. It is an accelerometer, used to measure the amplitude of acceleration, amplitude of
vibration & shock.
2. It converts ‘acceleration’ into voltage. Ideally the amplitude of output voltage
should be proportional to amplitude of input acceleration.
V0  Ai
V0 = Ks Ai
 
y  mx (linear relationship)
V  V 
Ks  0  
Ai  m/s2 
Where Ks = Sensitivity of sensor

3. Piezoelectric accelerometer works on “piezoelectric effect”.
Note: Before understanding the concept of piezoelectric effect one should know
working of capacitor.
2. Basics of Capacitor:
Capacitor is a device which stores the electrical charge. Capacitor consists of a voltage
source, a resistor, two parallel metal plates separated by some distance where air is filled.
Q  CV
Where Q = Charge
C = Capacitance
V = Voltage
Capacitance (c) can be written as
A
C
d
Where, C = capacitance
 = Permittivity
A = Area of cross - section of plates
d = distance between plates
So as the plates are charged.
This is similar to voltage source, which is
So a charged capacitor functions as voltage source.
Piezoelectric effect:
1. When piezoelectric material is subjected to force / pressure (mechanical input) then

electrical charge will be accumulated on the surface of material.
2. The developed charged will convert to voltage like in case of capacitor. So

piezoelectric material is called as charge or voltage generator.
Originally the piezoelectric transducer of thickness (t) was kept between two electrodes.
A force is applied on one of electrode. As the piezoelectric material is elastic it will get
compressed by (t) thickness and a polarity of charge will get developed on each
electrode as same of a voltage source and a voltage is developed.
After compression
So here a mechanical force is converting into the voltage by means of piezoelectric

material.
q  F
q = K’ F Here K’ = Charge sensitivity.
qC 
K' 
F  N 
Note:-
If F  t  q  c 
 Piezoelectric material is sensitive to static as well as dynamic loads.
 If it is subjected to static force then the voltage can’t be hold for long time.
 However it provides “Sinusoidal voltage” when it is subjected to dynamic force.
3. Mechanism of developing charge after applying F
For piezoelectric material (SiO2) Quartz.
Initial condition (Before Applying F)
Initially the bonds between silicon and oxygen are very strong. There is no deformation.
Final condition (After applying F)
After applying force ‘F’ the bond b/w silicon and oxygen becomes weakes and positive &
negative charge develops on upper and lower electrodes which is taken out.
Note: As long as force is applied, we will get voltage but when force is removed, SiO 2
will get its original structure and the voltage output will get stopped. When the electrode
is pulled upwards instead of pushing, we will get a negative value and similarly here after
some time we will get no voltage output. Hence to get a constant voltage output, we
need a continuous push & pull force to the electrode.
The output voltage will be AC (Since +ve for the pushing force & -ve for the pulling
force)
This constant pull and push force is known as dynamic load.
4. Seismic Accelerometer:
Seismic: A mass supported by spring.
 It has mass damper spring system, which is enclosed in a casing.
 If the accelerometer is attached to vibrating structure, then the mass present in mass
damper spring system also vibrates about its mean position.
 The amplitude of relative displacement of mass is directly proportional to the
amplitude of applied vibration.
Input displacement  xi  t    xi sin t
Where, xi = Amplitude of vibration

d2 xi  t 
Input Acceleration  ai  t   
dt2
ai  t   2 xi sin t
Here, Ai = 2xi = Amplitude of acceleration.

Relative displacement of mass = x 0  t   x 0 sin  t   
Here x0 = Amplitude of relative displacement of mass

So, xo  Ai
5. Mathematical Analysis:
Using FBD of mass & spring

Md2 x 0  t  dx 0  t 
c  Kx 0  t   Mai  t 
dt2 dt
d2 x 0  t  C dx 0  t  K
  x0  t   ai  t 
dt2 M dt M
By solving above differential equation
x 0  t   x 0 sin  t   
 i 2
Where, x0  ; tan  
2 2 1  2
1   2    2 
  Frequency of vibrating base


n Natural Frequency
K  rad 
n 
M  sec 
  Damping ratio
If   n    1
2  i Ai
x0  2  i 
n 2

n 2  2  i  Ai 
1
xb  Ai  x 0  Ai
n2
Note: To get accurate (linear) acceleration measurement, high natural frequency should
be selected (i.e. K should be high & M should be low)
If  n and ‘’ is low → we can measure accurate vibration

>>1
2
X0  i  x 0  xi
2
6. Piezoelectric Accelerometer:-
1. The sensors consist of a piezoelectric crystal sandwiched between two electrodes
and have a mass placed on it.
2. The mass & piezoelectric material (PEM) setup is enclosed in a frame.
3. The whole unit is attached to the base, whose acceleration is to be measured
4. The frame which is attached/threaded to the base acts as spring & squeezes the
mass against the crystal.
5. When the mass exerts force on the PEM certain output voltage is generated.
6. When the base is now acceleration downward, the internal reaction force on the
base acts upwards against top of the frame & it will be reverse if the base is
accelerated in the upwards.
From basic piezoelectric principle
F  t  q  V0
V0  t V0  t
V0   x 0  Amplitude of Relative displacement
1 
V0   Ai V0  Ai
n2 n2
Vo  A i
Where, V0 = Amplitude of output voltage

Ai = Amplitude of input acceleration.
Advantages of ‘Piezoelectric Accelerometer’
1. The instrument is quite small in size
2. The natural frequency is very high
3. These are useful for high input frequencies and their response is poor at low
frequencies.
Applications of Piezoelectric Accelerometer
1. used to measure vibration of aeroplanes & rockets
2. used to detect the vibration of engine in cars
3. can be used to study the high speed waves which are generated in blasting &
explosions.
Note:
1. Piezoelectric effect is reversible means piezoelectric material converts mechanical
input into electrical output as well as electrical input into mechanical output so it is
called as inverse sensors.
2. Piezoelectric material can fn like sensor as well as actuator.
Q.1 Piezoelectric accelerometer requires the following

(a) Low mass & stiff spring (b) Heavy mass & stiff spring
(c) Low mass & low stiff spring (d) Heavy mass & low stiff spring
Ans:
K
 n 
M
Q.2 An accelerometer has a seismic mass 080.10 kg & a spring constant of 1.5 × 103 N/m. if
the maximum mass displacement is ±0.025 m, then the maximum measurable
acceleration is _________ m/s2.
(a) 375 (b) 350
(c) 475 (d) none
Ans:
x 0  0.025 m K  1.5  10 3 N / m
M  0.10 Kg A i  ____ m / s 2
WKT
1
x0  Ai
n2
1 K 
x0  Ai  Ai    x0
K  M
 
M
1.5  103
Ai   0.025  375 m/ s2
0.10
Q.2 A piezoelectric material which has young’s modules, E = 9 × 1010 N/m2 has a diameter 10
mm under thickness of 2 mm. If the voltage sensitivity is 4500 V / m & the output
voltage generated across the crystal is 127.3 V then the applied load is ___ N.
(a) 100 (b) 200
(c) 110 (d) 210
Ans:
E = 9 × 1010 Nm2
D = 10 × 103 m
t = 2 × 10–3 m
Voltage Sensitivity = 4500 V/m
If t = 1 m then V0 = 4500 V
?  V0  127.3 V
127.3
t   1  10 6  2.82  10 8
4500
E
F / A 
 t / t 
t E  D2 t
F  E.A.  
t 4 t
10 22  10 10  10 3
F  9  10    10 3  2.82  10 8
74 3
2  10
5583.6
  99.70  100
56
Lecture – 05 (Optical Encoder Potentiometer)
1. Optical Encoder Potentiometer:
1. Basics
2. Construction and working
3. Applications
4. Practice questions

e.g. if temp = 70°C
Digital code = 11
2. Optical Encoder:
1. It is a digital sensor, which converts input position information into digital output.
2. It is a position sensor which can be used to measure linear displacement (x) and
angular displacement ().
Optical Encoder are classified in true ways:
(i) Incremental Type Encoder
(ii) Absolute Type Encoder
3. Incremental Type Encoder:
1. It consists of a rotating disc, which has rectangular holes of equal length and
separated at equidistance as shown below.
2. It has LED which pass the light perpendicular to the plane of disc and also a LIGHT
DETECTOR placed back-side of disc.
3. If the light passes through the hole then it will reach the light detector. If the light
falls on the light detector it generates high voltage for some duration (Pulse).
4. If generates pulse for increment in input position.
4. How can we Measure Angular Velocity:
Note: Incremental Encoders are used to detect the angular displacement of stepper
motor pulse counter display.
e.g. 9:00:00 AM  0
9:00:01 AM  40
1 sec  40 Pulses
1 sec  10 rev
1 min (60 sec)  60 rev
 Shaft speed = 600 RPM
5. Disadvantages of Incremental Encoder:
360
Re solution 
N
If N = 4, Resolution = 90°
If N = 100 Resolution = 3.6°
i.e. If shaft rotates 3.6°, then pulse will be generated.
Note:
1. To find very small angular displacement or to improve resolution, we should
increased number of holes on disc.
2. We can measure angular velocity and angular displacement but we can’t identify
the direction of rotation.
3. To identify the direction of rotation, we generally prefer ‘Absolute Encoder’.
(i) Absolute Encoder:
1. Absolute encoder measures the shaft position accurately compared to incremental
encoders.
2. Absolute encoders generates digital code as output.
3. As these encoders generated digital code as output, it is easy to interface to
microprocessor.
(ii) Single Track Absolute Encoder:
360
Re solution   180 (Not a good one)
2'
(iii) Two Track Absolute Encoder:
360
Re solution   90
22
(iv) ‘N’ Track Absolute Encoder:
1. Consists of ‘N’ LED’s; ‘N’ light detectors.
2. Generates ‘N’ bit digital code for each position.
360
3. Re solution 
2N
360
Re solution   1.4
28
6. Application of Optical Encoder:

1 sec  20 m
 60 sec  1.2 km
1 hr  (60 min)  72 km
Speed = 72 km/p/hr
  RPM
x  Linear Velocity
Q.1 Statement 1: The term encoder is used for a device that provides an analogue output as
a result of angular or linear displacement.
Statement 2: An increment encoder defects changes in angular or linear displacement
from some datum position where as an absolute encoder gives the actual angular or
linear position.
Ans: (d)
Statement 12 is false and 2 is true. Consider standard options.
Q.2 A shaft encoder which is attached to a wheel has a sensitivity of 500 pulses/rev. A digit
pulse counter connected to the encoder indicates 5500 pulses in 1 sec then speed of
shaft is rpm is
(a) 660 (b) 650
(c) 680 (d) 640
Ans: (a)
1 Rev.  500 pulses
1 sec  5500 pulses
5500
1 sec   11 rev
500
1 min (60 sec)  660 Rev
Speed of wheel = 660 RPM
Q.3 In a rotary optical encoder, diameter of encoded disc is 60 mm and the encoder
generates 5000 pulses per resolution on then the resolution of the encoder in mm is
Ans: Given: d = 60 mm, 2 πr = πd = π60 mm

1 sec  20 m
5000 pulse  1 rev
5000 pulse  π × 60 mm
 60
1 pules   0.037
5000
Q.4 The resolution of an encoder with 10 tracks will be nearly equals to
(a) 0.15° (b) 0.25°
(c) 0.35° (d) 0.45°
360 360
Ans: Resolution =   0.35
N
2 210
N = Number of tracks
7. Potentiometer:
1. It is a resistive based position sensor.
2. It is capable of measuring linear displacement (x), angular displacement ().
The measured resistance varies linearly with displacement
Construction: It has thin metal film which has distributed resistance through its length. It also has
wiper/jockey which provides metal contact with metal film.
Coefficient of distributed resistance = 1 /1 cm
General Illustration
 Total length of metal film = 10 cm

 Coefficient of distributed resistance = 1 /1 cm
10 V
V0 = I  5   5   5 V
5  5 
Input displacement, x = 5 cm
Output voltage, V0 = 5 V
Working:
1. Wiper is connected to the test object whose displacement is to be measured.
2. For any particular input displacement wiper breaks the metal film into 2 segments and the
voltage is usually collected across the lower segment of the metal film.
 Total length = lT cm

 Coefficient of distributed resistance = 
cm
 Total Resistance, (RT) =  lT ()
 Input displacement = lp cm
 Position length = lp cm
 Position resistance = RP 
Vs
 V0 = IRP   RP
R T  RP   RP 
R l l
V0 = Vs P  Vs P  Vs P
RT lT lT
Vs V
V0 = Vs  lP  V0  s  Xi Vo  xi
lT lT
V Vs
Sensitivity  o   V / cm
xi lT
8. Angular Displacement Measurement
 Total angular position = T (radian)

  
 Coefficient of distributed resistance =   
 radian 
 Total Resistance, (RT) =  T ()
 Input position = p (radian)
 Position resistance = Rp = P 
 Calculations are same as in case of linear
 V
 V0 = Vs P  s  P
T P
V  V
Sensitivity  0   S  V /radian 
 P  T
Q.5 A potentiometer of sensitivity 2 mV/m is cascaded with on amplifier of gain 1000. Find output
voltage of amplifier. When the potentiometer is used to measured displacement of 5 m.
(a) 4 V (b) 1V
(c) 20 V (d) 10 V
Ans: (d)
If xi = 5 mm, then V0 = 10 mV
VaA
Gain (G) =
V0
V0A = G × V0 = 1000 × 10 V
V0A  10 V
Q.6 A potentiometer, resistive based sensor has a total resistance of 8 kW and a maximum
displacement of 5 cm. Determine the output voltage for 2 cm displacement. Given that the
supply voltage is 20 V. Find sensitivity of sensor also.
(a) 4 V, 4 V/cm (b) 8 V, 4 V/cm

(c) 12 V, 8 V/cm (d) 10 V, 4 V/cm
VS 20 V
Ans: Sensitivity =   4 V / cm
lT 5 cm
1 cm  4 V
Input displacement  2 cm  8 V
Q.7 Consider the following statements regarding electromechanical devices:
1. A potentiometer has an input of rotation and output of a potential difference.
2. An electric motor has on input of a potential difference and on output of a rotation of

shaft.
3. A generator has on input of rotation of a shaft and on output of a potential difference.
Which of above are correct?
Ans: All 1, 2, and 3 are correct.
Proximity Sensor
1. Proximity Sensors
1. Promixity sensors are used to detect the presence of nearby objects without any
physical contact
2. These sensors are capable of measuring linear displacement (X), angular
displacement (θ)
3. These sensors find the application in automatic positioning system
Types of proximity sensors:

1. Capacitive type sensors
2. Inductive type sensors
3. photoelectric sensors
2. Capacitive proximity sensor:
EA
Capacitance (c) =
d
E : permitivity (Farad/meter)
A : Area of overlapping of both the plates (meter2)
d : Gap between the plates (meter)
  c = f (, A, d)
If ‘d’ varies, then ‘c’ varies inversely i.e.,
1. If x = 0 mm, then
EA
c  co 
d

nominal capaci tance
A
2. If x  0 mm and moving upwards, then c = c =
d  x1
1
 A /d co  x 
c =   co  1  1 
21  x   d 
1  1 1 
d  d 
x  x  x
If 1 ≪ 1; c = co  1  1   co  co 1  co  AC,
d  d d
co x
Where, AC =
d
i.e., the capacitance is increased when the distance between the plates is decreased
Note:Similarly, the converse is also possible i.e., if the distance b/w the plates is
increased then the capacitance is decreased.
Summary:
1. If the object is not moving, then capacitance = co (F)

2. If object comes close to the fixed plate then capacitance increases
3. If object moves away from the fixed plate then capacitance decreases
3. Real time capacitive type proximity sensors:
4. Inductive type proximity sensor:
Inductance - Amount of magnetic flux linked to the coil per 1 ampere current. If more
amount of magnetic flux is linked per 1 ampere current then we can say that the
inductance of coil is high.
total Total time linked

Inductance, L = 
i Unit current
L  Amount of flux  total 
 Reluctance - It is the opposition offered by medium to magnetic flux.
Note: Air has more reluctance compared to magnetic material (ex. Soft iron).
 From the above figure, the magnetic flux has to travel through different mediums i.e.,
M-1, M-2, M-3. The strength of magnetic flux varies. Initially, the distance between
object & core is ‘d’.
 If ‘d’ is increased, then reluctance faced by magnetic flux increases and then amount
of flux decreases i.e., inductance decreases.
i.e., d ↑↑, R↑↑, total ↓↓, L↓↓
 If x  0 mm (→) Air gap becomes d → d + x, then L ↓↓
 If x  0 mm (←) Air gap becomes d → d - x, then L ↑↑
Real time inductive type proximity sensor:
Note: In case of inductive proximity sensor, the object should be of material which
allows magnetic flux to pass through it.
5. Photoelectric proximity sensor:

 Photoelectric proximity sensors use a beam of light to detect the presence of
objects that block (or) reflect the light beam.
 A beam of light passes from the light source and a photodetector detects the
presence (or) absence of light from source.
 Incandescent lamps (or) infrared LEDs may be used as light sources
 Photodiode, photo transistors are used as light detectors.
Problems:
Q.1 A capacitive transducer consist of two parallel plates of diameter 1.5 cm each separated
by an air gap of 0.5 mm and has nominal capacitance of 300 pf. The displacement
sensitivity of the transducer is _ PF/mm
(a) – 200 (b) – 150

(c) – 300 (d) – 600
Ans. (d)
EA
We know that, c =
x
Where x is distance between the plates,

dc EA EA 1
Sensitivity    .
dx x2 x x

nomin al capaci tance
1 1 300PF 600PF
= co  = co  = 
x x 0.5mm mm
Q.2 Statement 1: capacitive proximity sensor can only be used for the detection of metal
objects and it is best with ferrous metals.
Statement 2: One form of capacitance proximity sensors consists of a single capacitor
plate probe with the other plate being formed by the object, which has to be metallic
and earthed.
Ans. ()
Both the statements are individually correct.
6. Basic of magnetism:
Magnetic flux lines always originates at north pole and travels to south pole.
(a) Magnetic flux (; unit – water

(b) 
Magnetic field B . unit – water/meter
From above figure,  = 3.x

 If Area 1 = Area 2 = A
1 2x  x
Then, B1 =  and B2 = 2 
A1 A A A
B1 > B2
7. Electromagnet :
“Current carrying conductor generates magnetic flux”
8. Solenoid:
“Consider current direction has four fingers direction & the thumb indicates north pole
(Right hand)
Lecture:7 – Hall Sensor
Topics:
1. Definition of Hall effect 2. Basics required for Hall effect
3. Derivation of Hall voltage 4. Application of Hall sensor
5. Practice questions
1. Hall Sensor
(a) It is proximity sensor, and it is used to measure at the strength of the magnetic
field.
(b) Hall sensor can be made up of conductor as well as semiconductor.
(b) Hall sensor works on the principle of Hall effects.
2. Hall Effect:
When current carrying conductor or semi-conductor is placed in perpendicular magnetic
field then transverse voltage will be developed. This generated voltage, is called as Hall
voltage.
3. Current Density: Represented by (J)

I I
J = 
A  t
 QT
 I =
time
I I T
 J =  
A  t  t  time
I Ne l
 = (Vd = Drift velocity)
 t  t  l  time
I N
  =  e  vd  n.e.v d (n = Carrier density)
 t v
1 I
So, Vd =  … (1) (Vd = Drift velocity)
n.e  t
VH = ?
dq
I =
dt
QT
I = ; (Q T  Total charge)
time
 So, when 1 e– travel from 2 to 1 then
QT = 1.e
  When 2e– travel from 2 to 1 then
QT = 2.e
  and when 3 electron travel from face 2 to face 1 then
QT = 3.e.
 So, if N-electron flow from face 2 to face 1 then total charge (QT)
QT  N e
N  Number of free electrons those are travel from face 2 to face 1.
4. Lorentz Force:
If a charge ‘q’ is moving with velocity ‘v’ in magnetic field ‘B’ then it will experience
‘Lorentz force’
  

FB  q v *B 
q : Charge, V : Velocity of charge, B : Magnetic field
So, Magnitude of force, FB = q vd B sin ( = 90°)
 FB = q vd B
FB  qvdB … (2)
5. Electric Field developed in conductor:
Voltage gradient available between 2 opposite faces inside the conductor is called as
electric field.
v
Electric field  E 
d
Force because of electric field
FE  q  E … (3)
 v
E =
d
v
So, FE  q  … (iii)
d
6. Role of electric field in Hall Effect:
VH
Electric field, E 

7. Lets Summarize all the forces acting inside the conductor:
At steady state, FB  FE
Wb
 At starting when B = 0 ; then no electron occupies lower face of conductor.
m2
Wb
 If B  0 ; then after some time many electrons are present on lower face i.e.
m2
electric is developed.
At steady state, FB = FE [FB = q.VdB, from equation (2)]
V
 qVdB = qE [FE = qE = q , from equation (3)]
d
 V 
E = Electric field,  E  H 
  
 VH 
 VdB = E E   
 
VH
 VdB =

 VH = .VdB
1 I   1 I 
 VH =     B  Vd  ne   t  From equation (1)
 ne  t   
1 IB IB 1
VH =   kH  ; kH 
ne t t ne
VH: Hall voltage, I : Current applied, B : Magnetic field applied
kH : Hall coefficient
8. How we should understand the Mathematical Formula:

IB  1
VH  kH  ; kH    Basic Physics
t  ne 
VH  f kH ,I,B, t 
 EC- Students:
× × ×
Do not change I, B and t.  V H = f (k H, I, B, t)
So graph made between
VH Vs kH
kH = ?
‘n’ = ?
 1
kH =
n q  e
For P-type semi-conductor:
q > 0; kH > 0
For N-type semi-conductor:
q < 0; kH < 0
EE- Students:
× × ×
Do not change the value of kH, B and t.  V H = f (k H, I, B, t)
So graph plotted between
 VH Vs I
I = VH
kH, B, t  Constant
ME & IN- Students:
× × ×
They do not change the value of I, kH & t.  VH = f (k H, I, B, t)
So graph plotted between 

 VH Vs B
i.e. I, kH, t are constant
So, VH = B
Two blades are fixed on the shaft and that arrangement are in-front of 3D-conductor.
Shaft is rotated. Shaft is situated between conductor and magnetic field.
When Blade-1 and Blade-2 are comes in front of conductor then magnetic field does not
reached to conductor faces – 3 & 4. Because at that time Blade – 1 & 2 stopped the
magnetic field and voltmeter shows 0 Volts.
VH = 0 V; 4 times in 1 second.
 1 sec  2 Rev.
 60 sec  120 Rev.
That means
So, Speed of shaft =120 RPM
x  Linear velocity,   Angular position,   RPM
 90°  2 sec
 360°(1 Rev.)  4 sec
So, 15 Rev 60 sec (1 min)
 Hall Sensor is also known as/called as position sensor.
9. Application of Hall Sensor:

1. Carrier density (n)
2. Type of semiconductor (i.e. P-type, N-type)
3. Amplitude of current
4. Strength of magnetic field
5. Position [ – RPM, x – Linear Velocity]
Q.1 If B is the magnetic flux density at right angles to a plate, I is the current flow through
the plate, t is the thickness of the plate and kH is the Hall coefficient, the resultant
transverse potential difference VH of Hall Sensor is given by:
Bt IB
(a) kH  (b) kH 
I t
t I
(c) kH  (d) kH 
BI Bt
[IES-2018 ME(P)]
Ans: (c)
 IB
VH = kH 
t
Q.2 A strip of copper 150 m thickness is placed in a magnetic field B = 0.65T perpendicular
to the plane of the strip and a current I = 23 A is setup in the strip. What is the
magnitude of Hall voltage developed, if the carrier density is 8.49 × 1025 electrons/m3,
charge of carrier = –1.60 × 10–19 C
(a) 7.3 V (b) 9.3 V
(c) 5.3 V (d) None of these
[ESE-2011 EC(P)]
Ans: (a)
Given: t = 150 × 10–6 m, B = 0.65 Wb/m2 or T, I = 23 A, n = 8.49 × 1028 electrons/m3, e
= –1.60 × 10–19 C, VH = ?
 IB
VH = kH 
t
1 IB
 VH =   7.3  10 6 V Magnitude of hall voltage
ne t
So, VH = –7.3 × 10–6 V
Q.3 For the liquid level measurement, hall sensor setup is used as shown below. The sensor
carrier a constant current of 2 A. Perpendicular to the magnetic field. The magnetic field
associated with the sensor charge with the liquid level as (0.2h + 0.1) Wb/m2. If the
output voltage of the hall sensor is perpendicular to both applied magnetic field and
current, then find the change in the output voltage when the water level in the tank
increased from 1 m to 3 m. Given that thickness of the sensor = 100 m, Hall coefficient
= 2 × 10–10.
(a) 1.6 V (b) 2.8 V
(c) 1.2 V (d) 4 V
[GATE-2005 IN]
Ans: (a)
Given: I = 2 A, B(h) = (0.2h + 0.1) Wb/m2, t = 100 m, kH = 2 × 10–10
 If, h = 0 meter.
 Then ‘B’ (Magnetic field) received by sensor is very low.
 The VH is low.
If h; then magnet close to the sensor.
Then B, then automatically VH.
IB h 
VH(h) =
t
2  kH  10 10 * 2I  0.2hB  h   0.1
VH(h) =
100  106
= 4 × 10–6 (0.2h + 0.1)
@h = 1 m; VH(h) = 1.2 × 10–6 V
@h = 3 m; VH(h) = 2.8 × 10–6 V
So, VH = VH(3) – VH(1)
= 1.6 × 10–6 V
= 1.6 V
 If piston move x-distances upward then VH. Because at that time magnet close to Hall
Sensor and magnetic field.
So, VH es.
 If piston move x-distances downward from initial position then VHes. Because distance
between Hall Sensor and magnetic field B strength .es. So VH will be also es.
Application that means measures:

Temperature, pressure, linear velocity discharge, force (torque/acceleration), position etc.
Angular displacement linear displacement etc.
1. Basics of Magnetism
1. Magnetic flux lines always travels from North Pole to South Pole.
2. Number of flux lines per unit area, is called as magnetic field and it denoted by (B) where
  Wb 
(B) =  2 
A m 
3. Magnetic field (B) is a vector quantity and it follows vector calculus.
4. Magnetic flux density (field) is very high near to magnet and decreases if we go away
from the magnet.
5. Magnetic flux always enters at South Pole and exits from North pole
6. Effective magnetic field vector is always drawn from the North Pole.
2. Electromagnet
1. When the current flows through the conductor for some time then the conductor
behaves like "Electro-Magnet". Current flows from positive terminal of the battery to
negative terminal of the battery through conductor.
2. If current flows in the down ward direction then circular magnetic flux develops around
the conductor.
3. From right hand rule if place our right hand in the direction of current such that thumb
faces down ward and curl remaining 4 fingers then fingers turns from left side of
conductor to right side of the conductor.
In the figure shown
(×) Magnetic flux line enters into the page

() Magnetic flux line comes out of the page
Around the conductor magnetic flux rotates in clock wise direction.
3. Solenoid
o A conductor which is twisted in the form of number of turns then it is called as solenoid.
o If the currents flows through each turn of coil then magnetic flux gets developed in each
turn and altogether very high amount of magnetic flux develops around the coil.
o If the current direction is as shown in the figure then magnetic flux develops as shown
.Therefore solenoid behaves like electro magnet.
o If the current flows in anti clockwise direction in the loop as shown then North Pole gets
developed left side and South Pole gets developed right side.
Basics of electro magnetic induction:
When ever there is a change in the magnetic flux linked with the coil, an E.M.F is induced.
When the magnet displaces near the coil then magnetic flux likened to the coil changes. This
leads to the induction of voltage across the coil. The magnitude of induced voltage V0 =
4. Basics of transformer
Primary Voltage Vin (t) = Vi Sint
Vi : Amplitude of input voltage
 : Frequency of input voltage
Secondary Voltage V0 (t) = Vo sin(t – ) … (i)
Vo N2
  k Transformation ratio
Vi N1
If k > 1 then it is step-up transformers
If k < 1 then it is step down transformers
So, Vo  kVi … (ii)
If we keep (ii) in (i) then V0(t) = kVi sin (t - )
When the time varying current flows through the primary coil then time varying magnetic flux
generates in the primary coil. This time varying flux generated in the primary coil links to the
secondary coil. So because of time varying flux in secondary coil voltage will be induced across
the coil. This voltage is called as "Secondary voltage".
5. Linear variable differential transformer (LVDT)

Construction:
1. LVDT is a displacement sensor, which is used to measure linear displacement (X), angular
displacement (θ).
2. LVDT works on the principle of "Electromagnetic induction"(mutual inductance variation).
3. LVDT consists of single primary coil and two secondary coils, which are connected in
"series opposition".
4. LVDT has "Movable magnetic core" which couples the magnetic flux generated by
primary coil to secondary coil.
6. Working of LVDT:
Input voltage Vi(t) = Vi sin t
1st secondary,
Output Voltage, Vs1  Vs1 sin(t  )
2nd secondary,
Output Voltage, Vs2  Vs2 sin(t  )
Vo (t)  Vs1 (t)  Vs2 (t)
Vo (t )  Vs1 sin(t  )  Vs2 sin(t  )
Vo (t )  (Vs1  Vs2 )sin(t  )
Amplitude of output voltage, Vo  Vs  Vs

1 2
7. Working of LVDT - Case 1.
If the core is at the centre then amount of flux linked to both the secondary coils is same and
the amplitude of voltages developed across is both the secondary coils is same.
Vs1 = Vs2
Amplitude of output voltage, V0  Vs  Vs  0

1 2
8. Working of LVDT-Case 2
If the core moves in upward direction then the amount of flux to the 1 st secondary coil is more
than 2nd secondary coil. So the amplitude of voltage developed across 1st secondary coil is more
than that of 2nd secondary coil.
Vs1  Vs2
Amplitude of output voltage, V0  Vs1  Vs2  0 .
9. Working of LVDT-Case 3.
If the core moves in downward direction then the amount of flux linked to the 1st secondary coil
is less than 2nd secondary coil. So the amplitude of voltage developed across 1st secondary coil
is less than that of 2nd secondary coil.
Vs1  Vs2
Amplitude of output voltage, V0-= Vs1 — Vs2 < 0.
Q.1 LVDT is used as
(a) Impedance Transformer (b) Displacement Transducer

(c) Temperature Sensor (d) None of the above
Ans. (b)
Q.2 LVDT consists of
(a) Two primary coils and single secondary coil
(b) Single primary coil and two secondary windings connected in series opposition
(c) Single primary winding and two secondary winding which is connected in series support
(d) None of the above
Ans. (b)
Q.3 An LVDT has an output range of -50 mV to 50 mV when the core movement is -5mm to +5mm.
The output of the LVDT is connected to an amplifier of gain 100. The output voltage of amplifier
for core displacement of 2.5mm is
(a) 25V (b) 0.25V
(c) 2.5V (d) 2.5 mV
Ans. (c)
Input displacement (Xi) : (–5 mm to +5 mm)
Output voltage of LVDT (V01) : (–50 mV to +50 mV)
Output voltage of amplifies, V0 = Gain × V01
If Input displacement (Xi) = 2.5 mm
Then output voltage of LVDT (V0L) = 25 mx.
Output voltage of amplifies, V0 = 100 × 25 × 10–3 = 2.5 V
10. Construction Resolver
1. Resolver is a position sensor, which is able to to measure linear displacement (x), angular
displacement (θ).
2. Resolver works on electromagnetic induction (Mutual inductance variation).
3. Resolver has single primary coil (which acts like rotor) and two secondary coils (acts like
stator).
Let's analyze the below diagram:
o When supply voltage is connected to primary coil then magnetic field (B) vector develops
as shown in the figure.
o Cosine component of magnetic field vector (B) passes through 1'st secondary coil and
Sine component of magnetic field vector (B) passes through 2'nd secondary coil.
o The field (flux) passes through the secondary coils generates or induces voltage across
secondary coils Vs1, Vs2 respectively.
o At each position of the primary winding (rotor/ shaft), the component of magnetic field
is coupled with secondary windings. As a result we will get voltage at both the
secondary windings.
o The ratio between the amplitude of output voltage across secondary windings provides
the information shaft position.
Primary or Input voltage, Vi(t) = Vi sin t
Vs1(t), Vs2(t) are secondary/output voltages
Vs1(t) = Vs1 sin(t – )
Vs1(t) = (kVi cos) sin(t – )

Amplitude of Vs1(t) is Vs1
Vs1  kVi cos 
Vs2(t) = Vs2 sin(t – )
Vs2(t) = (kVi sin) sin(t – )
Amplitude of Vs2(t) is Vs2
Vs2  kVi sin 
 Vs 
Absolute position of the shaft ()  tan1  2 
 Vs 
 1 
Q.4 Which of the following statements are correct related to resolver
Statement-1: Resolver generates digital data and operates on "Electromagnetic induction"
principle.
Statement-2: Resolver has secondary coils which acts like rotor and primary coil acts like stator.
Statement-3: A resolver is an electro-mechanical transformer whose analog output voltage is
a function of shaft angle.
Statement-4: Resolver is an absolute position transducer, producing true angular information
of shaft.
(a) 1, 2 and 4 (b) 3 only
(c) 3 and 4 (d) 4 only
Ans: ()
Q.5 In a shaft position (θ) measurement scheme, a resolver is used, the resolver generates two
components of output at each position of the shaft. If the supply voltage to the shaft is Vin(t) =
10sint and transformation ratio (k) = 1, if the CRO display is as shown in the figure, then the
position of the shaft (θ) is
Note: Assume frequency of the supply is more than that of shaft speed.
(a) 60° (b) 1200
(c) 2400 (d) 300°
Ans: ()
11. Construction and Working

1. Inductosyn is a position sensor, which is used to measure linear displacement (x), angular
displacement.
2. Inductosyn works on the principle "Electromagnetic induction (Mutual inductance
variation).
3. Inductosyn consists of two parts Scale (Static) Slider(Movable element).
4. Scale: It has trace of conductive material, which is regular rectangular shape as shown in
figure.
5. Slider box: It has two sliders and each slider is in regular rectangular shape as shown in the
figure.
6. When the slider box displaces laterally on the scale, then voltage will be generated across
the slider because of induction.
7. When the conductor of slider lies exactly on the conductor of scale then voltage generated
across slider will be maximum.
P : Scale pitch
S : Slider pitch
Slider displaces laterally above the scale.
Let’s discuss if p = s the voltage generates across sliders Vs1(t), Vs2(t) are as shown below.
12. Inductosyn
Amplitude of Vs1(t),
 x
Vs1  kVi cos 2  
 s
Amplitude of Vs2(t),
 x
Vs2  kVi sin  2  
 s
Let’s conclude Indutosyn
13. Technometer
Construction
1. It consists of a rotor which rotates with rotating shaft and a coil which is winded on magnet.
2. The rotor has teeth (made up of magnetic material so it is called as toothed rotor.
3. A transformer is used to collect the voltage from the coil.
4. Electronic counter counts number of pulses in 1 sec.
5. Pulse frequency is used to calculate to the speed of shaft.
14. Working
1. The magnetic flux generated by magnet links to the coil.
2. When the rotor rotates inside the magnetic flux then the magnetic flux linked to the coil
changes.
3. When the teeth comes close to the coil then magnetic flux linked to the coil will be
maximum as a result generated voltage is also maximum.
4. Voltage generated by coil is connected to transformer so that it can be processed further.
5. Each rotor teeth, when it comes close to the magnet and coil the voltage generated across
the coil will be maximum and it will be considered as 1 pulse.
6. Electronic counter counts the number of pulses in 1 sec. This information is used to measure
the shaft speed.
Q.6. A measurement system for rotational speed of the shaft is as shown below. Shaft is attached to
disc to which magnetic pieces are attached as shown. If the pulse counter displayed 80 pulses
per 1 sec, then the rpm is ______
(a) 8000 rev/min (b) 300 rev/min

(c) 600 rev/min (d) 200 rev/min
Ans: (c)
o If the teeth (Magnet) is inline with
o Hall sensor then electrical (Voltage) pulse will be generated
o As the wheel has 8 teeth then per 1 revolution 8 pulse will be generated
o The pulse counter is showing 80 pulse/sec this means 10 rev/sec so the shat speed is
600 rev/min.
Stepper Motor
Actuator:
It is a device, which generates mechanical output line linear displacement (x),
angular displacement ().
We have two different types of actuators.
 Electrical Actuator.
 Mechanical Actuator.
Electrical actuator: It converts the voltage or current into linear displacement or

angular displacement.
Ex: DC Motor.
Mechanical actuator : converts the liquid pressure or air pressure into linear
displacement or angular actuator.
Ex: Hydraulic or pneumatic actuators.
Basics of Electromagnetism:
Explanation of above figure:
Consider the voltage source connected to a conductor. When the switch is closed,
the current flows in the conductor due to which magnetic flux is developed around
the conductor and the direction of the flux can be found by “Right hand thumb
rule”.
Near to conductor, we have a cylinder on which there is a rectangular iron piece.

The cylinder houses a shaft . When the electromagnetic filed is strong enough, it
attracts the soft iron towards it and the iron piece rotates by some angle along with
cylinder and the shaft.
When the switch 1 is connected then the current in first conductor starts flowing
and magnetic flux is developed and further the iron piece on cylinder is then
attracted towards the first conductor, and total rotation of the cylinder will be 90o.
When 𝐈𝟏  0 A 𝐁𝟏  0Wb/m2  = 90°.
When the switch 1 is disconnected and the switch 2 is on, then current in the
second conductor starts flowing and magnetic flux is developed and further the
iron piece on cylinder is then attracted towards the second conductor, and total
rotation of the cylinder will be 180°.
When 𝐈𝟐  0 A 𝑩𝟐  0 Wb/m2  = 180°.
So here the electrical input is converted into magnetic flux, further it is converted
into angular displacement.
Here, the cylinder is called as rotor, the soft iron is called as rotor teeth.
Step angle () = 90°
Number of steps = 360/90 = 4.
Practically the step angle is smaller than 90°.
If we use 4 stator coil, the step angle = 90°
If we use 6 stator coil, the step angle = 60°
For smooth run of stepper motor, we need to have lesser step angle.
Step angle may be desired to be 1°, to have this step angle we need to have 360
stator coils which is quite impossible. So, let us see how we can have such a small
step angle.To understand this one must understand the working of stepper motor in
detail.
Stepper Motor:
 A stepper motor is a rotating machine, and functions as electrical actuator.
 It converts DC voltage pulses into a series of discrete rotational steps.
 The total angular rotation is directly proportional to the number of pulses.
 Stepper motor on predefined switching sequence.
360
Step angle =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑒𝑝𝑠 𝑝𝑒𝑟 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛
If 4 steps per revolution,
Step angle = 360/4 = 90°
Construction of stepper motor:
Consider a stepper motor with 4 stator coils, 4 stator pole and 2 rotor teeth. Outer
most cylinder is called as stator core and inner most cylinder is called as rotor.
So it is called as 4/2 stepper motor.
Working of stepper motor: If the switch S𝐴 is closed, the current flows

through the A coil and goes to ground. The stator pole behaves like a solenoid
(electromagnet) and magnetic flux is developed around stator pole.
The rotor teeth is attracted towards the axis of stator pole because of magnetic flux.
If both the switch A and B are on, the stator pole A and B will have magnetic flux
of same quantity. Because of this the rotor teeth is attracted by both A and B and
stops between A and B. It will be 135°(from starting position) between A and B
stator.
Following to it, the switch A is opened and only stator B is having magnetic flux,
thus the rotor teeth stops at stator B. Now the rotor teeth have moves from its
initial position to 180°.
The same is repeated, and rotor teeth keeps moving to that stator which is having
magnetic flux around it. By changing the different switches, a complete 360°
rotation is achieved.
The rotor pole 1 will be initially magnetized so as to solve a confusion between

which pole has to be attracted towards the stator pole.
The clock wise and anti-clock wise rotation of the rotor is governed by
switching the different switches and its sequence.
The reason for having to rotor teeth is to balance the rotation of shaft. This
process of balancing is known as damping which reduces the oscillatory
behavior of the rotor.
Consider, 4phase 4/2 stepper motor. The table below explains how the current
flows in the stator coils and change in rotor position accordingly.
O – Open switch
C – Closed switch
Switch positions Current (A) Position

()
S𝐴 S𝐵 S𝐶 S𝐷 𝐼𝐴 𝐼𝐵 𝐼𝐶 𝐼𝐷
O O O C 0 0 0 0 0°
C O O O 0 0 0 0 90°
C C O O 0 0 0 0 135°
O C O O 0 0 0 0 180°
O O C O 0 0 0 0 270°
O O O C 0 0 0 0 360°
If we follow the sequence of D-A-B-C-D, we get Anticlockwise rotation.
And if we follow the sequence of D-C-B-A-D, we get a clockwise rotation.
Switching sequence:
Lets consider closed switch as – 1
Lets consider open switch as – 0
S𝐴 S𝐵 S𝐶 S𝐷 
0 0 0 1 0
1 0 0 1 45
1 0 0 0 90
1 1 0 0 135
0 1 0 0 180
0 1 1 0 225
0 0 1 0 270
0 0 1 1 315
0 0 0 1 360
If the rotor goes from 0° to 90°, we call it full step angle.
If the rotor goes from 0° to 45°, we call it half step angle.
Stator pitch:
The angular separation between two successive stator poles is called as “stator
pitch”.
360°
Stator pitch(𝑺 ) =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑡𝑎𝑡𝑜𝑟 𝑝𝑜𝑙𝑒𝑠(𝑁𝑠)
In the previous example stator pitch = 360/4 = 90°.
Rotor pitch:
The angular separation between two successive teeth on the rotor is called as
“Rotor pitch”.
360°
Rotor pitch (𝑹 ) =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑜𝑡𝑜𝑟 𝑡𝑒𝑒𝑡ℎ (𝑁𝑟)
In the previous example rotor pitch = 360/2 = 180°.
Full step angle:
The angular rotation of the rotor/shaft, when only one phase is activated at a time
is called as “full step angle”.
Full step angle (𝑭𝑺 ) = r - s
In previous example, 𝐹𝑆 = 180° - 90° = 90°
Half step angle:
The angular rotation of the rotor/shaft, when two successive phases are activated at
a time is called “Half step angle”.
𝑟− 𝑠 𝑓𝑠
Half step angle (𝑯𝑺 ) = = .
2 2
In previous example, 𝐻𝑆 = 90°/2 = 45°.
Digital pulse logic:
The energy supplied for the coil for some small amount of time is known as
pulse.
The flow of pulses is controlled by micro processor.
If we apply 1 pulse = 1 step angle.
The time required to each pulse is 1 second.
Case 1:
4 sec – 4 pulses – 4 steps – 1 revolution.
1 sec – 1 pulse – 1 step – ¼ revolution
So, for 1 minute – 60* ¼ revolution
1 min – 15 revolution,
Thus Shaft speed = 15 RPM (revolution per minute).
Now we have arrived to the required mechanical rotation of shaft which is used in
many mechanical machines.
Case 2:
4  sec – 4 pulses – 4 steps – 1 revolution.
1  sec – 1 pulse – 1 step – ¼ revolution
1 sec - 1×106 pulses - 1×106 steps – ¼ ×106 rev
So, for 1 minute – 60× ¼ ×106 revolution
1 min – 15 ×106 revolution,
Shaft speed = 15×106 RPM.
Here we get great change in the speed of rotation, we don’t change the magnitude
of the current but we are going to reduce the time of application of the pulses.
If we increase the number of pulses per sec then speed is automatically increased.
Pulse rate:
The number of pulses applied to the stepper motor in 1 sec is called as input pulse
rate or input pulse frequency.
Note:
1. Shaft speed is directly proportional to input pulse frequency.
2. Total angular rotation depends on number of pulses.
1 pulse – 1 step - 90°.
2 pulses – 2steps - 180°.
Number of stator poles 𝑁𝑆 = 8
Number of rotor teeth 𝑁𝑅 = 6
Stator pitch 𝑆 = 360/8 = 45°
Rotor pitch 𝑅 = 360/6 = 60°
Full step angle 𝐹𝑆 = 𝑅 − 𝑆 = 60° - 45° = 15°.
Half step angle 𝐻𝑆 = 𝐹𝑆 /2= 15/2 = 7.5°.
And 7.5° is also the resolution of the stepper motor.

360°
Number of full steps per revolution = = 24.
15°
360°
Number of half steps per revolution = = 48
7.5°
130 steps – 1 revolution 10.5 revolution – 1 Sec
1 pulse -1 step 10.5×130 pulses – 1 sec
130 pulses – 130 steps – 1 revolution
Input pulse rate = 1365 pulses/sec.
The right answer is option (C).
Half step operation should be in a incremental order of 0° - 45° - 90° - 135°
So the right option is (d).
Right option is option (b).
200 steps – 1 revolution
1 pulse – 1step
200 pulses – 200 steps – 1 revolution.
1 min – 10,000 pulses – 50 revolution.
Due to gear box the speed is reducted to 10 revolution/ min.
1 revolution – 5 mm
1 min – 10 revolution – 50 mm
10,000 pulses – 50 mm
50 𝑚𝑚
1 pulse – = 5 m
104
For each pulse there is movement of 5 m.
So the right option is (a) 5 m.
Lets consider Case 1:
200 steps/ rev
Step angle = 360°/200 = 1.8°.
1 pulse - 5m.
Lets consider Case 2:
100 steps/rev
Step angle = 360°/100 = 3.6°.
1 pulse - 10m.
Note:
1. Stepper motor can provide liner displacement also.
2. To get highly precise motion, the step angle should be as low as possible.
3. Stepper can be operated in open loop control system.
Torque Vs Speed characteristics:
At low speed the torque will be high and can lift heavy load, known as holding
torque.
At very high speed, the torque will be very low.
Types of Stepper Motor:

1. Variable reluctance type stepper motor.
2. Permanent magnet type stepper motor.
3. Hybrid type stepper motor.
Variable reluctance type stepper motor:

Reluctance is resistance by the medium to the magnetic flux. Consider a reluctance
motor. On the core there is wire winding which is connected to voltage.
Due to application of the voltage the core develops the magnetic flux, one end of
core is north pole and other is south pole, flux needs to complete its cycle. Due to
air presence there will be high reluctance to magnetic flux so the rotor with high
permeability will be attracted towards the pole and path is closed. Due to this the
rotation of the rotor is achieved.
The stepper motor that’s uses this theory to run is called Reluctance type
stepper motor.
(𝐴+ − 𝐴− ) This is called one pair,
(𝐵+ − 𝐵− )
(𝐶 + − 𝐶 − )
(𝐷+ − 𝐷− ), there are 4 such phases, so its 4 phase stator.
A voltage source is connected to the coil of 𝐴+ and 𝐴− , like that all the four stators
are connected to the voltage source. By alternatively switching different stator
switches, the rotor will rotate in the desired direction depending on switching
sequence
360°
Step angle =
3∗𝑁𝑟
m - Number of phases.
𝑁𝑟 - Number of rotor teeth.
Here, m=4 𝑁𝑟 = 6
360°
𝑠 = 4∗6
=15°.
Advantages :
1. Rotor has low mass.
2. It is more responsive.
Disadvantage:
1. It cannot provide high torque.
This disadvantage is over come by the following stepper motors.
Permanent magnetic type stepper motor:
In this case the rotor is replaced by magnet. Magnet will rotate rotate since there is
no requirement of polarization. As the phase 1 and 2 are applied by the voltage, the
magnetic flux gets developed in the windings. As the coils develop magnetic flux
the north pole and south pole are achieved on the coil and the magnetic bar is
attracted or repulsed depending on the polarity. The polarity can be changed using
the battery switch.
Advantages:
1. It requires very low current because the rotor is magnet.
2. It requires very low electrical power.
3. It provides higher torque compared to variable reluctance type stepper motor.
Hybrid type stepper motor:
There is a magnetic bar on which two rotor cups are arranged. Stator pole is
arranged on the magnetic bar. One part of the magnetic bar is north pole and other
will be south pole. After the current is passed in the coil, the magnetic flux is
induced in the stator, and north pole of the stator attracts the south pole of the
rotor. By doing this the rotor starts rotating.
Number of rotor teeth on cup 1 = 100
Number of rotor teeth on cup 2 = 100
Due to overlapping of both north and south pole, the number of rotor teeth will be
200.
360°
Thus, step angle = = 1.8°.
200
Hybrid stepper motor provides very less stepper angle.
Application of the stepper motors:

1. Stepper motor is used in precise positioning is required, in combination with the
microprocessor controller.
2. The stepper motor can accelerate its load easily as maximum occurs at low pulse
rate.
3. It is used in computer printers, robotics, machine tools, etc.
Ans: Since only option 1 is right, the correct option is (a).
Ans: Since the hybrid type has two cups and each cup can give the rotation, so the
right option is (c).
Ans:
For, 0.5mm liner resolution,

0.5𝑚𝑚
× 360° = 6°.
30𝑚𝑚
So, 1 pulse - 6° - 0.5mm.
Mechanical actuators:
Mechanical actuators converts liquid/air pressure into mechanical power.
1. Hydraulic actuators.
2. Pneumatic actuators.
Hydraulic actuator:
It is a device which converts liquid pressure into mechanical output i.e, either
linear displacement or angular displacement.
The output mechanical power depends on flow rate applied and pressure on piston.
Hydraulic actuators are classified into 2 types;
1. Hydraulic cylinders (linear actuator).
2. Hydraulic motor (rotator actuator).
Types of Hydraulic cylinders:

1. Single acting type cylinder.
2. Double acting type cylinder.
3. Telescopic type cylinder.
4. Tandem type cylinder.
Single acting cylinder:

1. It can provide only one side stroke with the help of liquid pressure energy.
2. Energy stored in the spring can be used to get return stroke.
As shown in the figure below ,this actuator basically works on the liquid pressure,
as the fluid is made to enter the cylinder chamber in the port, the piston slowly
moves forward and spring is compressed. And after that as the fluid moves out
from the port, the spring releases the energy and return stroke is achieved
.
𝑄𝑖𝑛
Power (P) = 𝐹1 *𝑉1 = (𝑃𝑖𝑛 ∗ 𝐴𝑝 )×
𝐴𝑃
= 𝑃𝑖𝑛 *𝑄𝑖𝑛 (Watts).
Double acting cylinder:

It provides two sides strokes with the help of liquid pressure energy. As shown in
the figure below ,this cylinder is provided with two ports that have facility of both
inlet and outlet motion. Once liquid flows in, the fluid in the other port goes out
and vice verse providing the two way motion of the piston.
Area during extension stroke = 𝐴𝑝
Area during return stroke = 𝐴𝑝 - 𝐴𝑟 .
Telescopic type cylinder:
1. Both extension and retraction will be done stage by stage.
2. When we require large stroke length, we use telescopic type cylinders.
Tandem Cylinder:
1. More than one cylinder connected in cascade configuration forms Tandem type
cylinder.
2. It transmits heavy force without changing the diameter of cylinder.
𝐹𝑇 = 𝑀𝑎 + 𝐹𝑊 (=0.1)
= 50000*a + 0.1*𝐹𝑊 -------(1)
To find acceleration ‘a’
Lets use v2-u2=2as--------(2)
From (2) find the value of ‘a’ and use in (1) and find FT.
AP = FT / P .
We can get the required diameter from the value of AP.
𝐹𝑇 = 𝑀𝑎 + 𝐹𝑊
= (𝐹𝑊 /𝑔)𝑎 + 𝐹𝑊
= 0.15
𝐹𝑊 = 13𝑘𝑁
g= 9.81 m/𝑠 2
to find a,
v= u + at
0.13 = 0 + a*0.5
a = 0.26 m/𝑠 2
𝐹𝑇 = 344.54 + 1950
= 2294.54 N
𝐹𝑇
P = 𝐹𝑇 /𝐴𝑝 = 𝜋
∗(𝐷𝑝 )2
4
P= 11.68 MPa.
Rotary actuators:
Rotary actuators are the hydraulic or pneumatic and they are equivalent of electric
motors which are used when a twisting or turning motion is required.
Rack and pinion type rotary actuator:
There is cylindrical barrel that houses piston, rack and pinion. As the fluid enters
any of the inlet, the piston moves accordingly and the pinion starts rotating.
Pressure is converted into linear displacement further it is converted into angular

displacement.
Gear motor:
A high pressure fluid is passed to the junction to two gear. As the fluid hit the
mating gears, the gears starts rotating and the low pressure fluid exits from the
bottom. That is how the rotation of gears is achieved.
Vane motor:
The high pressure fluid is passed on the vanes and due to the force of the fluid, the
vane rotates and the rotation motion of the shaft is achieved.
Electrical actuators Vs Mechanical actuators:

Electrical actuators:
1. They are portable and low power devices.
2. These actuators can be programmable hence they can be used in “Digital control
applications”.
3. They provide very low output. Hence cannot be used to drive heavy loads.
Mechanical actuators:
1. These are little bit complicated to build.
2. It is comparatively difficult to program.
3. These actuators provide very high mechanical output. Hence they can be used to
drive heavy loads also.
4. Maintenance of these actuators is quite difficult.
𝑃𝑖𝑛 = 100 𝑏𝑎𝑟
𝑑𝑝= 0.05𝑚
𝑚3
Q = 0.3 .
𝑚𝑖𝑛
0.3
𝑄 60
Velocity = = 𝜋 = 2.546 m/s.
𝐴𝑝 (50∗10−3 )2
4
F= 10×103 N
𝑑𝑝= 100×10−3
𝐹
Pressure 𝑃𝑖𝑛 = F/A = 𝜋
∗(𝐷𝑝 )2
4
10∗103
=𝜋 = 1.27 MPa.
∗(100∗10−3 )2
4
50 mm – 10 sec
V = 5 mm/s
𝜋
Q = 𝐴𝑝 ∗ 𝑉 = ( ) ∗ 𝑑𝑝 2 ∗ 𝑉
4
𝜋 𝑚3
= ( ) ∗ 0.12 ∗ 0.005 = 3.93×10−5 .
4 𝑠
Lecture 13 Control System
Topics:
 Basics
 Mathematical modelling of physical system
 Transfer function
 Block diagram reduction
 Controllability and observability
1. Mathematical modelling of physical system:

The process of finding mathematical expression for a given system is called as “Mathematical
modeling. Mathematical model explains the relation between input and output of a component
or system.
Example:
1.
Fk = kx
2.
F = Bdx/dt
3.
F = Bdx/dt
Md2s  t 
F =
dt 2
Ex: Mass Damper spring system:
Input = f1(t) [Force]

Output = x0(t) [Displacement]
Fi(t) = FM(t) + FB(t) + FK(t)
d2x 0 (t) dx 0 (t)

Fi(t) = M.  B. + kx0 (t)
2 dt
dt
d2 x6 (t) dx 0 (t)
M.  B.  kxo  t   Fi  t 
2 dt
dt
2nd order differential equation.

If we solve this differential equation:
x0(t) = _____
F0(t) = Fk(t)
dy(t)
B = k[x(t) - y(t)]
dt
dy(t)
B  ky  t   kx  t 
dt
1st order differential eqn.
FK2 (t) = FM(t) + FB(t) + Fk1 (t)
Md2y(t) Bdy(t)
k2 [x(t) × y(t)] =  + Ky(t)
dt dt
Md2 y(t) Bdy(t)

   k1  k 2  y  t   k 2 x  t 
dt2 dt
2nd order differential equation.
Q.1 Define the term Transfer function and find the transfer function of the mechanical
system shown below
Ans: It is the ratio between the output and input in laplace domain by considering initial
conditions are equal to zero.
Ex 1: Find the transfer function of the differential equation shown below. It is given that Y(0)
= X(0) = 0
Ans: Consider the differential equation

dy(t)
+ 5y (t) = x (t)
dt
Ans: Apply L.T on both sides
sY(s) – Y(0) + 5Y(s) = X(s) (Given)
(s + 5) Y(s) = X(s)
Y(s) 1

X(s) (S  5)
Ex2: Find the transfer function of differential equation given below Y(0) = X(0) = 0 consider
the differential equation
dY(t2 ) dY(t) dx
5  7Y(t)  3 (t)  2x(t)
dt dt dt
x(t) : Input, y(t) : output
Ans: Apply LT on both side
[S2Y(s) – 5Y(0) – Y(0)] + 5 [SY(s) – Y(0)]
7Y(s) = 3[5 × (s) - X(0–)] + 2 × (S)
(s2 + 5s + 7) Y(s) = (3s + 2) × (s)
Y(s) 3s  2

2
X(s) s  5s  7
Ex3: Find the transfer function of mechnanical system shown below:
Given that X(0) = f(0) = 0
Ans: F(t) : force applied or input
X(t) → output
FM(t) + FC(t) + FK(t) = F(t)
Md2x(t) dx(t)
 C. + Kx(t) = F(t)
2 dt
dt
Apply L.T on both sides
M[s2 × (s) - X(0) s – X (0–1)] + X(0-1) + C (s × (s) – X(0–1) + KX(s) = F(s)
(Ms2 + (s + K) × (s) = F(s)
X(s) 1

F(s) Ms2  Cs  K
Find the ratio C/R of block diagram shown below

Ex 1:
Concept:
C = G (R - CH)
C = GR - GCH
(1 + GH) C = GR
C G
 Ans.
R 1  GH
Ideally (at steady state)
C =R
Example:
G = 10 and H = 1
C 10

R 11
10
1
11
Ex: 2
C = G2 [G1 (R - CH1) - CH2]

C [1 + G1G2H1 + G2H2] = G1G2R
C G1G2

R 1  G1G2H1  G2H2
Ex3: Find C/R of the block diagram shown below
C = G2 [G1(R - CH1 H2) - CH1]

C = G1G2 (R - CH1H2) - CG2H1
C[1 + G1G2H1H2 + G2H1] = G1G2R
C G1G2

R 1  G1G2H1H2  G2H1
2. Controllability and Observability:
Let’s consider the example given below
SP(t) = f (Ti(t), SN(t), Sa(t)

“SN(t), Sa(t)  State variables (Here 2 state variables)
Initial values:
SN(0) = 10%
Sa (0) = 5%
 S (40)  100%
Desired values  N
 Sa (40)  100%
3. State Variables:
 The minimum set of variables required to fully describe the system are called
state variables.
 The minimum set of variables, which gives enough information to predict future
behavior.
 Number of state varibles are equals to the “Order of the system”.
Controllability:
If we can change the state variables to any desired values in finite time, by finite input
then we can say the system is controllable.
Observablity:
If we can determine the behaviour of entire system or state variables or initial values of
state variables by observing the output of system for some finite interval of time from
system’s output then we can say the system is observable.
Let us consider mass damper spring system:
What is the value x(5)?
Initial values
x(0) = 2 mm
dx(t) 1mm

dt sec
1mm
x(5) = x(0) + 5x × × sec
sec
x(5) = 2 + 5 = 7 mm
x(t) → state variable 1
dx(t)
→ state variable 2
dt
State vector
 x(t) 
 x(t) 
   21
Input = F(t), Output = x(t)
d2 x(t) dx(t)
M B + Kx(t) = F (t)
2 dt
dt
d2 x(t) B dx(t) K F(t)
  x(t) 
2 M dt M M
dt
d2 x(t) K B dx(t) 1
x(t)   F(t) (2)
2 M M dt M
dt
dx(t) dx(t)
= 0.x (t) + 1 + 0.F (t) (1)
dt dt
 dx(t) 
 dt   0 1   x(t)   0 
    K B   dx(t)    1  F(t)
 d2 x(t)  
  M M   dt   M 
 dt2 
 x  t   = [A]2×2 [X(t)]2×1 + [B]2×1 [U(t)]

21
 x(t) 
 x  t    1 012  dx(t)    011 F  t 
 
 dt  21
dx(t)
For x(t) = 1 x(t) + 0. + 0 F(t)
dt
Input F(t) = U(t), Output x(t) = y(t)
[X(t)] = [C]1×2 [x(t)]2×1 + [0]1×1 [U(t)]
Standard state space model:
X  t  = A X(t) + B u (t)
Y(t) = C X(t) + D V (t)

State space representation
Nth order system:
[A]N×N [X(t)]N×1
How to check controllability mathematically?
If we can change the state variables to any desired values, in finite time, by finite input
then we can say the system is controllable
To check controllability matrix, [C] = [B AB]2×2
A2×2 → system matrix,B2×1 → input matrix C1×2 → output matrix
If rank of [C] = 2 : Then system is controllable.
MOS → 2nd order system.
nth order system:
[C] = [B AB A2B …. AN-1 B]N×N
If rank of [C] = N; Then the system is controllable.
How to check observability mathematically?
If we can determine the behaviour of entire system or state variables by observing the
output of system for some finite interval of time from system’s output then we can say
the system is observable.
To check observability, matrix [0] = [C AC A2C….AN-1]N×N
If rank of [0] = N: Then the system is observable.
 1 2
Q.2 A system is represented in state - space as x = Ax + Bu where A =   and B =
  6  2 2
1
1 . The value of α for which the system is not controllable is:
 21
1 3 
Ans: Controllability matrix [C] = [B AB] =  
1   6  22
To be not controllable rank of C  2
|[C]| = 0
1 ( + 6) – 3 × 1 = 0
–3 = 0
 = 3
If  = 3 then the system is not controllable.
Q.3 A certain linear time in variant system having the state and the output equation given
below:
 x 1  1 1  x1   0 
 x      4
 2  0 1   x 2  1 
x 
Y = [1 1]  1 
 x2 
1. Controllable or not
2. Obsrvable or not
Ans: It is 2nd order system
 1 1 
A =  
 0 1  22
0
B =  
1 21
C = [1 1]1×2
To check controllability
1 1 
[o] = [CT ATCT] =  
 0 0  22
Rank of [o] = –1  0
So, Rank of [o] = 2 (the system is observable)
MICROPROCESSOR -Basics
Basics of digital systems
 The role of digital system is extended to many fields like data processing, signal processing, automation,
process industry
 Microprocessor, Micro controllers are widely used devices in “Mechatronics”, which falls under the
category of digital systems.
 At fundamental level, digital systems operates between two possible states, which can be considered as
either “ON” or “OFF”, high or low, 1 or 0.
I = 0 mA BULB OFF I = 3 mA BULB ON

OFF
SWITCH BATTERY BATTERY
SWITCH
DIGITAL LOGIGIC ‘0’ DIGITAL LOGIGIC ‘1’
BULB OFF BULB ON
What is complement?
 Complement of “ON” is “OFF”
 Complement of “OFF” is “ON”
For single bit
Data A: 1, then complement of A is 0
For multi bit NOT GATE
Data A: 1010, then complement of A is 0101

Not gate is made up of transistors
Basics of numbers systems
 Decimal (485)10 BASE

BASE
 Binary (01011)2
 Octa (467)8 BASE
 Hexa decimal (467)16

BASE
In Decimal, we have 10 symbols : 0,1,2,3,4,5,6,7,8,9

In Binary, we have 2 symbols : 0, 1
In Octa, we have 8 symbols : 0,1,2,3,4,5,6,7
In Hexa, we have 16 symbols : 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,
How to understand number systems?
Any number to decimal number system
1. (467)10 = 4*102 + 6*101 + 7*100 = (467)10
2. (467)8 = 4*82 + 6*81 + 7*80 = (311)10
3. (467)16 = 4*162 + 6*161 + 7*160 = (1127)10
4. (0110)2 = 0*23 + 1*22 + 1*21 + 0*20 = (6)10
Decimal ----------- Binary --------- Hexadecimal

23 22 21 20
(0)10 0 0 0 0 (0)16
(1)10 0 0 0 1 (1)16
(2)10 0 0 1 0 (2)16
(3)10 0 0 1 1 (3)16
(4)10 0 1 0 0 (4)16
(5)10 0 1 0 1 (5)16
(6)10 0 1 1 0 (6)16
(7)10 0 1 1 1 (7)16
(8)10 1 0 0 0 (8)16
(9)10 1 0 0 1 (9)16
(10)10 1 0 1 0 (A) 16
(11)10 1 0 1 1 (B) 16
(12)10 1 1 0 0 (C) 16
(13)10 1 1 0 1 (D) 16
(14)10 1 1 1 0 (E) 16
(15)10 1 1 1 1 (F) 16
Explanation for above number system conversions
Binary to Hexadecimal
Consider last 4 digits and give its hexadecimal equivalent and then continue for next pair to convert data into
binary to hexadecimal
Same is applicable if we want to convert hexadecimal data to binary data, take LSB first to convert it to binary
and then continue till MSB
4- Bit binary representation (0000 to 1111)
8 – Bit representation (00000000 to 11111111)
16- Bit Representation (0000000000000000 to 1111111111111111 )
Number of total combinations are 65,536
LOGICAL OPERATIONS
These all operations are binary operations
OR Operation
AND Operation
Exclusive – OR operation
Role of logical operations
Let us take a practical example of logical gates
Take solenoid valve which operates door, Sensor 1 is humidity sensor and sensor 2 is temperature sensor,
Condition – If temperature is greater than 600 c then the door should be open
e.g. if temperature is greater than 600 C then sensor generates a digital code (00110101)2 =(35)H and with above
given conditions and binary conditions solenoid will get signal from processor then with mechanical link Door
operates.
Arithmetic operations
Binary addition
Miscellaneous
 Instructions stores in “memory
 Data is transferred into “Internal Memory”
 “ALU” collects data from internal memory and “Instructions” from external memory then computes
the instruction
MICROPROCESSORS
Before going to the actual topic, we need to understand a common term called as “Integrated Circuit(IC)”. So,
what actually is an Integrated Circuit(IC)?
1) An Integrated is a semiconductor device, which consists of billions of transistors packed in a small area.
2) The transistors and electronic components inside the chip are fabricated in such a way that they can
perform logical(ex: AND, OR etc), arthemetic(ex: ADD, SUB etc) operations.
3) Every IC is specified by a “specific number” or “series”(ex: Intel @1976 P8085AHetc). Here in the given
picture “P8085AH” is the specific series number.
4) Every IC requires “voltage” for its operation. For example, a smartphone requires a charger to store the
energy, which will be used by its IC’s for its operation.
5) Every IC has a silver coated pin , which are used to connect to the external world.
So, basically a Micro-processor is an Integrated circuit. Some of the examples of microprocessor(fig-1.1) and
microcontroller(fig-1.2) are
Fig-1.1 fig-1.2
Microprocesssor vs Mircoprocesssor based system

Microprocessor is a digital semiconductor based device ,it takes data from the sensor and computes the
instructions in the main memory, after computation it provides the digital data to the actuators.
These are components of the microprocessor,which is also know as “Central
Processsing Unit(CPU)” or “Microprocessor Unit(MPU)”. Here ALU means
“Arthemetic Logical Unit” and CU means “Control Unit”.
Mircoprocesssor based system is combination of Microprocessor, sensors, actuators, ADC and DAC.
Consider an example of temperature measurement which uses a thermocouple as a sensor, which reads a output
voltage of 3.6V when connected to a amplifier. This analog output voltage is converted to digital output(say
35H(00110101 in binary)) using ADC. This digital data is tranferred to the microprocessor for its computation
using a bus, after computation it is back converted to analog using a DAC ,to feed it to the actuator.
BUS:
1. A bus is an information path made of group of conducting wires.
2. Bus has the communication lines between the function logical systems inside as well as outtside the
processor.
Before the
data
transmission.
After the data transmission. Consider a

data in each 4-bits of device-1(A0 A1 A2
A3) consists of data (0010) is used to
transmit to the 4-bits of the device-2(B0
B1 B2 B3) using 4-individual wires. So,
these group of 4- individual wires
pointing from device-1 to device-2 is
called as bus
Here the device-1 and device-2 are connected through a group of 4-wires, so it is called as 4-bit bus. This will be
used to transfer 4-bit data between those devices at a particular time. For a “N” bit bus ,it has a capacity to transfer
2^n combinations of data at different time instants.
MODES OF TRANSMISSION IN A BUS:

1. Serial mode: In “serial bus” information may pass-on the information bit-by-bit.
Here the D7 bit will

be first reached to
the receiving side
and the D0 bit will
be last reached to
the receiving side.
Here all the bits

(D0-D7) will reach
the receiving side
at the same time
MEMORY:
1. Memory is electronic space ,in which binary data as well as information code stores.
2. In memory at each memory location it consists of register, so memory is a collection o registers.
3. Eaach register has a flipflop. Each flipflop can store either 1 or 0.
4. For example 8-bit register consists of 8-flipflops.
5. In memory each register has a specific address.
Example of 2-bit address for a 4-byte memory unit.
These b0,b1 are the 2-bits These 4-registers ,each

which are helpful to form having 8-bits to store
a unique address by the data will form a
different combinations. 4*8-bits of data, which
is 4bytes of memory(as
1byte= 8bits).
Example of 3-bit address for a 8-byte memory unit.
These 8-registers ,each

These b0,b1,b2 are the 3- having 8-bits to store
bits which are helpful to the data will form a
form a unique address by 8*8-bits of data, which
different combinations. is bytes of memory(as
1byte= 8bits).
Similarly example of 16-bit address for a 64 kilo-byte memory unit.
 So, if a processor has “n” address bit, then the number of memory locations = 2^n.
 Let capacity of each memory location/register = p bytes.
 Total memory capacity = (no. of memory locations)*(capacitty of each register).= (2^n)*p.
MEMORY TYPES
They are several types of memory in the memory systems of microcomputer systems. They are divided into
two major categories:
1. Volatile memory.
2. Non-volatile memory.
The volatile memory loses its information if power is disrupted and can easily be altered , whereas non-volatile
memory retains the information even is the power is disrupted.
Some of the microcmputers are used to hold the system memory and some to hold user memory.
Important memory types as follows:

1. Randomaccess memory(RAM): RAM is used to store the temporary data. It is designed so that
information written into or read from memory . dyanamic RAM’s refreshed typically by external refresh
circuits can hold million second. Whereaas static RAM’S STORE THE DATA IN FLIPFLOP’S.
2. Read-only memory(ROM): Internal use and operation of the microcontroller requires ROM to give the
storage for the operation of the system. This non-volatile memory is designed so that the information
stored in refresshed to as firmware.
3. Programmable read-only memory(PROM): It is the specail type of ROM that allows intial and /or add
information to be written into the chip. It can be written into only once after being received from the
manufacture.
4. Erasable and programmable ROM(EPROM): It is the specially designed PROM and is programmed
using programmer. It can be reprogrammed after gaving been entirely eraseeed with the use of an
ultraviolet light source.
5. Electrically erasable andd programmable ROM(EEPROM): It is a non-volatile memory that can be

electrical pulses rather than ultraviolet light. The EEPROM can be electrically overwritthen with new
data.
Applications of Micro-controller
Consider a process plant,which is controlling the flowrate by having a desired setpoint. In this a
sensor/transmitter is used to measure the process variable , and this difference between the setpoint and
transmitter output is feeded to the block controller(this block is mainly the
micorcontroller/microprocessor/PLC etc) which takes the error input and produce the controlling output. This,
controller output is given as input to the actuation system before feeding to the plant to control the flow rate.
MICROPROCESSOR VS MICROCONTROLLER:
MICROPROCESSOR MICRO-CONTROLLER
Microprocessor is heart of computer system Microcontroller is a heart of embedded

system.
Since memory an i/o components are all Since memory an i/o components are all
external each instruction will need external internal each instruction will need
operation, hence it is relatively slower external operation, hence it is relatively
faster
It is just a processor, Memory and I/O Microcontroller has external processing
components have to be concerned externally. along with internal memory and i/o
components
Since memory and i/o has to be connected Since memory and i/o are present
externally , this circuit becomes large initially, the circuit is small
Cannot be used in compact systems and Can be used in compact systems and
hence infficient hence it is an efficient technique
They have less number of registers,hence More number of registers, hence the
more operatins are emory based programs are easier to write
Costof the system increases Cost of the system is low

Due to external components, the entire Since external components are low, total
power consumption is high power is less.
These are based on von-newmann model These are based on harvard model where
where program and data are stored in same data and memory are separate blocks
memory blocks
Mainly used in personal computers Mainly used in washing machine, mp3
player.
Let us consider an IC, to work with this IC we need to first activate the IC by “EN” pin. So for activating IC,
we need to give voltage to that pin. There are two types of EN pin: Active high pin, Active low pin
For selecting IC with active high pin, we need to give positive voltage to that pin and for selecting an IC with
active low pin, we need to give negative voltage (commonly ground) to that pin
8085 Microprocesor (Manufactured by INTEL)

It is 8 bit processor, all ALU and other operations are carried out in 8 bit, it has 40 pins.
Give high voltage (5V) to pin 40, ground is connected to pin 20

Clock signal is given across pin 1 and pin 2
16 pins for addressing, 65,536 memory locations can be addressed
Out of which 8 are multiplexed and used for data pins along with address pins. Clock signal is generated by
oscillator circuit. Each activity in microprocesor is controlled by clock signal
Pin diagram, microprocessor with external memory

All 16 address pins are connected to external memory address lines
On External pc, when programmer writes program he can make any particular pin high or low
If all 16 lines are ‘0’, i.e (0000)H then 1st location is selected.
WR’ is active low pin, so with external programmer if we make it as ‘0’, then we can write data into the external
memory/internal memory from microprocessor
OUT = [35]H it is instruction which is used to send data from MPU to external I/O
Then AD0 to AD15 are address pins so first they all (0000)H for selecting address of external memory and then
in next cycle, AD0 to AD7 they are now data lines so data (35)H is now on AD0 to AD7 for data transfer
For selecting next location now AD0 to AD15 is (0001)H and then taking data from external memory to
microprocessor, we use RD’ pin which is also active low pin and remaining all same as above.
AD0 to AD15 is 16 bit address bus, they are unidirectional
AD0 to AD7 is 8 bit data bus, they are bidirectional
Instruction – ALE, address latch enable,
If ALE is 1, all AD0 to AD15 are address lines and they carry address bits, if ALE =0, AD0 to AD7 are data
lines which carry data.
If IO/M’ = 1, i.e IO = 1 and M’ = 1 or M = 0

For selecting internal memory, M= 1
For selecting Input or output devices, IO=1
If programmer want memory write based operation
If programmer want memory read based operation
If programmer want I/O read/write based operation

Just change pin – 34 should be 1 and remaining all are same
TYPES OF BUSES IN MICROPROCESSOR SYSTEM
ADDRESS BUS
MPU to memory/sensors/actuators/input or output devices
1. This bus is used to transmit the address of a location in memory or I/O parts.
2. The address is transferred from MPU to memory or I/O devices but not from Memory or I/O devices to
MPU.
3. Address bus is “Unilateral bus”.
4. “16 bit address bus” can access 216 memory locations.
DATA BUS
1. This bus is used to transfer the data from MPU to memory, I/O devices and memory, I/O devices to MPU.
2. Data Bus is “Bi directional bus”.
3. “8 bit data bus” transfers “8 bit data” at a time.
MPU TO MEMORY
MPU TO SENSORS
MPU TO ACTUACTORS
CONTROL BUS
This carries control signal such as read or write from the control unit to the memory and I/O devices and the
other parts of the microprocessors
ARCHITECTURE OF 8085
Total 5 parts
5 parts are:
PART1 – ALU
PART 2 DATA REGISTERS
PART 3 INSTRUCTION REGISTERS/INSTRUCTION DECODERS
PART 4 TIMING AND CONTROL UNIT
PART 5 INTERRUPT CONTROL
Arithmetic logic unit (ALU)
1 It process the data by fetching instructions from memory,

decoding them and executing them
2 MPU (MICROPROCESSOR UNIT) consists of ALU,
data registers, instruction registers, instruction decoders,
interrupt control.
Architecture of 8085
SID (Serial input data line)

1. This pin is used for receiving the data into microprocessor serially
2. The data enters into D7 bit accumulator
SOD (Serial output data line)

1. The pin is used for sending the data from the microprocessor serially.
2. The data is sent from D7 bit of accumulator to peripheral.
SID, SOD are used for “serial communication”.
Example – A digital temperature sensor measures the temperature and gives data in digital format to
microprocessor(ACCUMULATOR) through SID pin serially and microprocessor executes the logic then it send
digital data from accumulator to motor or control valve
through SOD pin serially.
Accumulator
1. Accumulator is 8-bit programmable register
2. 2. After the execution of logical or arithmetic
operation the result will be stored in “Accumulator”.
3. The data from the input device enters into “Accumulator”. After the computations the output will be
transferred from accumulator to output device.
Example: Accumulator data is [05] H
Program: ADD A, [53] H
Accumulator data is now: [05]H +[53]H= [58]H
Temporary register
1. It is 8 bit non-programmable register
2. It holds the data temporarily during the execution of large computations
3. W, Z are temporarily and non – programmable
4. B, C, D, E, H, L are called as general purpose registers of 8 – bit size
Example 1: ADD A, B
Then after execution [88] H stored in accumulator

Example 2: ADD A, B
Flag register
1. Flag register is 8-bit register, which has 5 flags.
2. After the execution of logical or arithmetic instructions the content of flag register gets updated
3. Flag register is also called as “status register”.
4. Bits can be transferred from ALU to flag register and flag register to ALU
Example 3: ADD A, B
Example 4: ADD A,B
Example 5: XOR A, B
Register array
1. It consists of group of registers, which constitutes “internal memory” of the processor.
2. W, Z are temporary registers and non- programmable
3. B,C, D, E, H, L are called as general purpose registers of 8-bit size.
4. B, C, D, E, H, L are programmable registers they can be programmed individually or as registers pairs
“BC”, “DE”, “HL”
5. HL register is called as memory pointer and stores the address of memory location.
Interrupt control
Maskable interrupt: these interrupts can be ignored by processor.
Vectored interrupt: vectored interrupts is a processing technique in which the interrupting
device directs the processor to the appropriate interrupt service routine.
INTR: It is only non – vectored interrupt in 8085 microprocessor and also maskable interrupt
RST 5.5: Vectored address – (002C) H
I.e. 5.5*8 = (44) D = (002C) H
RST 6.5: Vectored address – (0034) H
I.e. 6.5*8 = (52) D = (0034) H
RST 7.5: Vectored address – (003C) H
I.e. 7.5*8 = (60) D = (003C) H
TRAP (RST 4.5)

It is the only non – maskable interrupt in 8085 microprocessor. If this pin is activated then the 8085 processor
gets interrupted respective of its state.
It is the highest priority interrupt pin
I.e. 4.5*8 = (36) D = (0024) H
INTA’: When microprocessor receives interrupt signals, it sends an acknowledgement (INTA) to the peripheral
which is requesting for its service.
Interrupt priority
TRAP> RST 5.5> RST 6.5> RST 7.5 > INTR
PIN DESCRIPTION
RESET IN’
1. This is active low pin
2. If this pin is activated then “Then contents of all registers will be cleared and becomes (00)H
RESET OUT:
1. It indicated that the MPU (Microprocessor unit) is being reset.
2. This signal can be used to rest the other peripheral devices
READY
1. It is used to interface with the slow peripheral devices(memory or I/O device) to the microprocessor
2. When ready signal is low, the microprocessor waits until the READY pin goes HIGH
HOLD and HLDA

1. These signals are used for peripherals such as DMA(Direct memory access)
2. To perform “DMA” operations external device requires address and databus but normally they are under
the control of microprocessor.
3. For this reason, an external devices will generate “HOLD” signal to microprocessor then after receiving
HOLD signal microprocessor firstly completes current operation and releases the address and data bus
devises by using “HLDA” signal (hold acknowledgement).
IO/M’
1. This is the status signal used to differentiate between I/O and memory operations.
2. When this pin is high -> an I/O operation is performed.
3. When this pin is low -> a memory operation is performed.
RD’
1. It is an active low signal. When the signal is low on this pin, the microprocessor performs memory reading
or I.O reading/
WR’
1. It is an active low signal. When the signal is low on this pin, the microprocessor performs memory writing
or I/O writing operation.
STATUS SIGNALS
Programming
Program
Set of instructions written in a specified manner, to execute a given project is called as program
Instruction
Instruction is a binary command written by programmer, to perform a given task on specified data.
Each instruction has two parts
1. OPCODE(operational code)
2. OPERAND
Example 1:
BASIC SEQUENCE OF PROGRAMMING

Programmer will write program of 8085 microprocessor using screen of PC.
Let us consider the instructions like MOV A,B and ADD A, B.
In the front screen programmer written the code as shown above.

In the address location 3000H, code is MOV A, B which is 1 byte instruction and after execution of this code
contents of B will go into A.
In the address location 3001H, code is ADD A, B which is also 1 byte instruction and after execution of this
code contents of B will add to contents of A and then final data will get stored in A.
Generally in old days, programming is done in ‘1’ and ‘0’ by converting instructions into binary form which is
called opcode.
Example 2:
Example 3:
Example 4:
SHORTCUT FOR UNDERSTANDING INSTRUCTIONS
1. Input device sends data to accumulator which needs to have instruction
2. MPU sends data to output device which needs to have instruction.
3. General purpose registers to accumulator and vice versa data transfer instructions.
4. Accumulator to memory (RAM) and vice versa needs instruction.
Tips to remember instruction

1. If the OPCODE consists of MOV or MV then it is “moving/copy” operation.
2. If the OPCODE consists of LD or L then it represents loading operation. Similarly if the OPCODE
consists of ST or S then it represents storing operation.
3. If the OPCODE consists of letter ‘I’ then it is immediate transfer of data.
4. If the OPERAND consists of letter ‘M’ then it is memory based operation.
5. If the OPCODE consists of letter ‘X’ then it is represents operation has to be performed with register
pair
DATA TRANSFER INSTRUCTION SET

1. These instructions are used to transfer the data from
 Register to register
 Memory to register and register to memory
 I/O device to accumulator
 Directly loading 8-bit data into register
2. After the execution of data transfer instructions the content of the source will not be changed but
destination will be changed
3. After the execution of data transfer instructions the content of flag register will not be affected.
Example:
MOV A, B
Where A is destination and B is source
MVI R, [XX]H
[R][XX]H
MOV R1, R2
[R1][R2]H
MOV R, M
[R] [[M]]
Example: How to get memory from memory to accumulator

In 3000H memory location, data is [12] H and we need to transfer the data to accumulator
LDA 3000H
Here contents of 3000 H is now load into accumulator
Finally contents of accumulator is [12] H
Example: LXI H 3000 H
H L is a register pair and if we use collectively it can store 16-bit data.
For understanding LXI instruction we can take use of other instructions like
MVI H, 30H
MVI L, 00H
MOV B, M
HLT
ARITHMETIC INSTRUCTION SET
1. These instructions performs the arithmetic operations like addition, subtraction, increment and
decrement.
2. Addition, subtraction will be performed only with the content of accumulator and the result will be
stored back in the accumulator.
3. After the execution of “addition, subtraction” instructions the content of flag register will be updated as
per the result.Example:
Example:
LOGICAL INSTRUCTIONS SET

1. These instructions performs logical operations like AND, OR and exclusive- OR operations.
2. Logical operations will be performed only with the content of accumulator and after the execution of
logical instruction, the result will be stored in accumulator.
3. After the execution of logical operation, the content of the flag register will be updated as per the result.
Example: find the content of accumulator and the flag register after the execution of 8085 microprocessor with
[A] = [53]H
Example: Find the content of HL register pair after the execution of 8085 microprocessor
Example: A single instruction to clear the lower 4- bits of the accumulator in 8085 microprocessor is
A. XRI [0F]H
B. ANI [0F] H
C. XRI [F0] H
D. ANI [F0] H
INSTRUCTION SET BASED ON LENGTH OF WORD SIZE

1 – Byte or one word size: if the operational code (OPCODE) and the operands are accommodated in same
byte in memory then it is called as 1- byte instruction.
Example: MOV A, B ADD B
2-Byte instruction set: In the instruction if the OPCODE occupies 1st byte and operand occupies 2nd byte in
memory then it is called as 2-byte instruction.
Example: MVI A, 32H ADI 35H
3-Byte instruction set: In the instruction if the OPCODE occupies 1st byte and 16 bit address in the instruction
occupies in subsequent 2 bytes of memory then it is called as 3- byte instruction.
Example: LDA 2500H, LXI H, 2400H
INSTRUCTION SET BASED ON ADDRESSING MODES

DIRECT ADDRESSING MODE: If the address of operand is directly mentioned in the instruction itself then
it is called as direct addressing mode
Example: LDA 2600H, STA 2700H
REGISTER ADDRESSING MODE: If the operands are present in the microprocessor’s register then it is
called as register addressing mode.
Example: MOV A, B ADD B
REGISTER INDIRECT ADDRESSING MODE: If the operands are present in the memory and address of
memory location is present in HL register pair then it is called as register indirect addressing mode.
Example: MOV A, M ADD M
IMMEDIATE ADDRESSING MODE: If the operands and data are present in the instruction itself then it is
called as immediate addressing mode.
Example: MVI A, 25H ADI 36H
IMPLIED/IMPLICIT ADDRESSING MODE: If the operations of microprocessor are performed by
accumulator alone then it is called as implied addressing mode.
Example: CMA
INSTRUCTION SET
In 8085 microprocessor, the instructions are classified on 3 basic categories.
In big program consists of hundreds of lines in program, using JMP instruction we can move the pointer to that
particular memory location.
SPECIAL PURPOSE REGISTER : Program counter
1. Consider that an instruction is being executed by processor. As soon as the ALU finished
executing the instruction, the processor looks for the next instruction to be executed.
2. So, there is a necessity for holding the address of the next instruction to be executed in order
to save time. This is taken care by the program counter.
3. A program counter stores the address of the next instruction to be executed.
4. Microprocessor increments the program whenever an instruction is being executed, so that
the program counter points to the memory address of the next instruction that is going to be
executed. Program counter is a 16- bit register.
STACK
1. Reserved memory space in RAM, which is used for general purpose (to store the data) is called as stack.
2. The highest memory address location of stack is actually gets stored in “stack pointer”.
3. Data storage begins at an address 1 less than the specified beginning of the stack.
4. PUSH RP and POP RP are the instructions used to enter the data into stack and to retrieve the data from
the stack
5. Every time when “PUSH RP” instruction is executed the content of stack pointer is decreased by 2.
STACK FOLLOWS LIFO – LAST IN FIRSTOUT MECHANISM
Example: PUSH INSTRUCTION
Example: POP INSTRUCTION
INCREMENT/DECREMENT REGISTER
1. The 8- bit contents of a register or a memory location can be incremented or decremented by 1.
2. This 16-bit register is used to increment or decrement the content of program counter and stack pointer
register by 1.
3. Increment or decrement can be performed on any register or a memory location.
ADDRESS/DATA BUFFER AND ADDRESS BUFFER
1. The contents of the stack pointer and program counter are loaded into the address buffer and address
data buffer.
2. These buffers are then used to drive the external address bus and address-data bus.
3. As the memory and I/O chips are connected to these buses, the CPU can exchange desired data to the
memory and I/O chips.
4. The address-data buffer is not only connected to the external data but also to the internal data bus which
consists of 8-bits.
5. The address data buffer can both send and receive data from internal data bus.
Input/output Interfacing
A processor has to process data available from any system. To feed in the data and to store the processed data,
the peripheral devices, also called the I/O devices. The process of connecting I/O devices to processor, is
called as interfacing.
Challenge 1
The main reason is that the speed of the I/O devices is very less when compared with that of the processor.
Since the I/O device is very slow, the processor will be idle most of the time waiting for the data to come in.
Challenge 2
Not all I/O devices will produce information or data in binary form. So it becomes necessary to convert each
of these data into a processor understandable form.
Challenge 3
How to handle more sensors and actuators at a time?
Challenge 4
Electrical compatibility is biggest challenge.
Solution
To overcome the above problems, the I/O devices are interfaced to microprocessors using their own special
interfacing. Interfacing between the I/O device and the computer is done by the “interfacing systems”.
Interface is the mid system between I/O and a microprocessor and takes care of code conversion,
synchronisation and other associated problems. In other words, interface is the device which resolves the
functional and constructional difference between I/O device and microprocessor. The differences are
enumerated below.
The I/O interfacing performs the following tasks:
1. It provides ways through which data from each external device will be transferred properly with the
microprocessor without causing interference to other devices connected to the system buses.
2. It resolves any difference that may exist regarding the timing between the microprocessor and the
peripheral device.
3. It converts the format of the data of the peripherals to the format that is acceptable by the
microprocessor and vice-versa.
4. Also produce interrupt signals to force the microprocessor to react immediately in peripherals demand
immediate action.
Let’s say, there are four sensors and four actuators,
1st sensor is temperature sensor.
2nd sensor is pressure sensor.
3rd sensor is flow sensor.
4th sensor is proximity sensor.
These analog sensors generates analog voltage and to interface with processor we need to have an amplifier
(to amplify small signals), ADC (analog to digital converter), multiplexer, timing control circuit and altogether
it is called as input module/interface circuit.
For processor to communicate with actuator, we need amplifier, DAC (digital to analog converter), time
managing unit, de-multiplexer and altogether it is called as output module/interface circuit.
Specific address is assigned to all I/Os that address is called port address.
Intel has designed an interface/controller for each one of the I/O devices separately. Some of the most popular
interface chips used with 8085 microprocessors are:
1. 8279: Keyboard/display interface
2. 8255: Programmable peripheral interface
3. 8253: Programmable interval timer/counter
4. 8257: DMA(Direct memory access) controller
5. 8251: USART(universal serial asynchronous receiver/transmitter)
6. 8259: Programmable interrupt controller
7. 8212: I/O port.
Instructions used to access IO devices in 8085 microprocessor

1. In 8085 microprocessor input and output devices are communicated by 8-bit address, whereas memory
location is communicated by 16- bit address.
2. We can connect maximum 256 input devices and 256 output devices to 8085 processor through input
port.
RELAY
1. It is an electromagnetic device. It consists of switch and coil and armature
2. It is used for remote and automatic control
Normally open (NO), no power to coil if we give power to coil, now loop
closes
If we give power to coil, loop opens Normally closes (NC), if there is no
power to coil, now loop closes
Relay systems
1. Relays are commonly controlled by switches, push buttons, sensors
2. It consists of control circuit and load circuit.
Example:
Challenges facing with relay systems

1. Relays consume lot of electricity and generate a lot of heat in the operation.
2. The operation time, space, wiring is high in case of relay.
3. For remote control it is still a difficult challenge and even maintenance is difficult.
Programmable logic controller (PLC)

1. PLC is a computer based system, used for industrial automation.
2. PLC receives information from connected sensors or input devices, process the data, triggers output
based on pre-programmed parameters.
3. PLCs are flexible and robust controller and adaptable to almost any application.
4. It is a sequential controller.
Architecture of PLC
It requires power supply .Programming device used to dump code into the PLC or get program from PLC
PLC consists of processor which have memory (program data and other data) and CPU, then input modules,
output modules.
Advantages of PLC
1. It has very strong “INPUT/OUTPUT interface”.
2. PLC has very good “HUMAN INTERFACE”.
3. PLC can withstand even moderately hazardous conditions also.
4. Programming and reprogramming both will be easy in PLC.
5. It can be used to control “Electro-mechanical devices”.
PROGRAMMING
LADDER LOGIC
1. The programming of PLC looks like ladder diagram. Hence it is called as ladder logic.
2. The main advantage of ladder logic is it diagrammatic and it can be easily understandable to
anyone.
3. Programming and reprogramming both are very easy with ladder diagram.
Symbols of ladder programming

1. Normally opened switch: if it is triggered then it will be closed.
2. Normally closed switch: if it is triggered then it will be opened.
3. Output devices: if it is triggered then it will be operated.
Let’s consider an example
DEVELOP LADDER LOGIC FOR THE FOLLOWING OPERATIONS

1. If temperature and flow sensor both are on then turn on Motor.
2. Turn on motor, if at least one of “temp/flow” sensors is/are on.
3. Turn off control valve, if pressure sensor is On
Steps involved in PLC programming

Four steps in the PLC operations
1. Input scan
Scan the state of the inputs
2. Program scan
Processes the program code
3. Output scan
Energize/de-energize the outputs
4. Housekeeping
This step includes communications, internal diagnostics etc.

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Here eyes are acting like sensors.

Brain does the work of processors.

Hands and legs are doing work of actuators.

Eyes – Brain – Hands/legs.

The combined integration of the Sensor, processor and actuator is called as

The combined integration of position scanner, microprocessor and motor is called

Sensor is used for detection of physical variable and provides information to

Actuator:It provides mechanical output mechatronics system ultimately works like

Practical application of mechatronics:

Weight(Sensor) Current(Processor) Speed(Motor)

This is how the different parameters are controlled.

Example3 : Washing machine works as feedback control

The total assembly works like Feedback control system.

1. Cost efficient and good quality products.

2. High grade of flexibility to modify or redesign.

3. Very good performance characteristics.

4. Wide area of application.

5. Great productivity in case of manufacturing organization.

6. Great extent of machine utilization.

1. High initial cost.

2. Multi-disciplinary engineering background required to design and

3. Need of highly trained workers.

4. Complexity in identification and correction of problems in the system.

We do have mechanical sensors which results in variation of mechanical parameters,

Ex: Thermocouple, Piezo electric material, metal rod.

Let us discuss the function of thermocouple

Piezo electric effect:

Electrical or Electronic sensors convert physical quantity into electric output.

Mechanical sensor converts physical quantity into mechanical output.

1. Transducer is much wider term compared to sensors and actuator.

In fact, transducer can work both like sensor and actuator.

Lets see and example,

Processor is an electronic device that performs digital computations on the digital

Characteristics are as follows:

Thermocouple is a temperature sensor that converts Temperature difference into

If Input Range ( 0°C - 100°C)

Then let us find the input span,

Temperature Expected Output Voltage

100°C 100 mV 100 mV

If Temperature (T) and Voltage (V) have a relation of,

If we plot a graph with Voltage on X-axis and Temperature on Y-axis,

100 Practical Graph

Ideally sensor should be highly precise.

Since, Temperature   and EOutput

Qn :A thermocouple is used to measure the temperature in the range of (0°C to

Lets consider a thermocouple

Qn :Which of the following statements are true.

Sensor Processor or Actuator

Qn : A force is developed by connecting two sensors as shown in the figure. The

Find the output voltage V, when the applied force is 4.75N

So for a displacement of 9cm the output voltage will be 9V.

The question can be solved by proper understanding of definition.Answer is (b).

Qn : Which of the following is true about the sensors?

(0MPa-100MPa) Pressure Sensor (0mV-200mV)

Sensitivity= 2mV/1 MPa

From the table, we can conclude Input resolution = 0.5 MPa.

How we can measure physical parameter practically?

Temperature Measurement System: