position cs

8/4/2019 Position Cs

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Students :Rabee odeh

Ahmad Anwar

supervisor:

Dr. Raed jaber


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Angular position control of a dc motor


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There are several industrial ,automatic ,and otherprocesses that require the control of angularposition of a motor (or translation position) the

stepper motor achieve this purpose but thestepper motor lack the advantage of controllingthe angular position at wide range of speeds; thisaim is achieved by using dc motor which allow us

to control the position at a wider range of speedsin both reverse and forward direction.


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Introduction

Position control systems are an important

component of many industrial products .

Examples are found in disc drives ,automotive

products ,robotics ,process control and many

others.


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Aim of a position control system


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In all servomechanisms one of the most

important components is the position sensor

it measures the position of the servomotor

and convert it to an electrical signal that the

control system can interpret and use.


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Digital optical position sensors are particularly

important because they are non contact and

thus not subject to the wear and noise of

analog position controllers .

A further advantage of optical sensors is that

they can be used in severe environments

where strong magnetic field exists.


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General information and introduction

We used a dc motor of parameters that will be

studied and analyzed next,an optical shaft

encoder that consist of a transmitter and a

receiver will be studied and analyzed next,pic

16f877 4 MHZ it's cct and code will be

analyzed next.


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Generic block diagram of the project


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Load information and parameters

(DC-MOTOR)

Dc electric motor is used as a manipulativedevice in control systems ; dc motors areextremely versatile drives capable of reversibleoperation over a wide band of speeds ,withaccurate control of the speed at all times ;theycan be controlled from zero speed to full speedin both directions in this project our goal is tocontrol the motor position at a specific speed in

both directions but the advantage of range ofspeeds give the motor a great importance inposition control systems.


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Measurement of motor constants

About used DC motor.

- It is a DC motor (constant flux) ,12 volt rated ,

-We measure its parameters (Ra, La, Kp, J, B) as follows:

: armature resistance (Ra)st1

By an ohm meter we found that Ra =10.5

2nd :induced or back emf constant (Kb)

We applied a 12 v DC voltage to the terminals of themotor and measured Va and the speed (rpm) andrepeated for less voltages as in the following table :


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W=2*pi*N/60

(rad/s)

N(rpm)Ia (mA)Va

314.16300017012

230.384220014510

178.02417001238

141.37213501156

83.776800864

36.652350682


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Now from the relation

Va = e+RaIa +La(dIa/dt)

In steady state Ia is constant so Va =e + RaIa

But e = Kb* *w (: is constant)

e= Kb*w(Va/Ia) = Kb (w/Ia) + Ra so we construct the

following table:


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Wm/IaVa/Ia

184870.58824

1588.85568.96552

1447.3565.04065

1229.32252.17391

974.139546.51163

53929.41176


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We plot (Va/Ia ) vs. (Wm/Ia ) as follows :


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Kb = slope = 0.033 v/rad/

Ra = intersection =12.28Ra = Ra avg =(10.5+12.28)/2 =11.2


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rd : Armature reactance (La )3

By applying a variable low AC voltage to the armatureterminals and increasing the voltage until the motorstart to move we measured the current (Ia) and the

rms voltage (Va) , we found that :The min AC voltage Va = 20 v

Then Ia = 1.7A

Ra = 1.1*11.66 =12.83

we calculated La from the relation:

= 26.8 mH


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4th : J and B :

Rotor mass = 50 gm

Rotor diameter = (shaft diameter+ Rotor

diameter)/2= (0.002+0.02)/2=0.011 m

Jrotor = 0.125* mass*diameter^2

=0.125*0.05*(0.011)^2 =8*10^-7

Now for B :From the relation : (Kb*Ia) =B* (W)We construct the following table and plot (Kb*Ia) vs(W)


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WKb*Ia

314.160.00561

230.3840.004785

178.0240.004059

141.3720.003795

83.7760.002838

36.6520.002244

y = 1E-05x + 0.001

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

050100150200250300350

Series1

Linear (Series1)


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B = slope =1.1*10^-5

Tind rated =Kb* Ia rated =0.0308*0.085 =

0.0056OR:

Tind rated =er*Iar/Wr=(Var-Iar*Ra)*Iar/Wr

= (12-0.085*11.6)*0.085/335=0.00542


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This was the first section which describes the

used dc motor


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driving cct(H-BRIDGE)

An H-bridge is an electronic circuit whichenables a voltage to be applied across a loadin either direction. These circuits are often

used to allow DC motors to run forwards andbackwards. H-bridges are available asintegrated circuits, or can be built fromdiscrete components.

In this project the first type is used ; a dual fullbridge driver l298.


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L298 is high current high voltage full bridge

driver of a total dc current up to 4 ampere and

of an operating supply voltage up to 46 volts .

The block diagram of the circuit is shown in

the following figure:

From which we use two inputs connected to

the microcontroller and two outputs

connected to the motor(load).


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The alternative was to to build a full bridge

using transistors which is cheaper , but for

purposes of simplifying the work we use this

chip (L298) .


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An H-bridge is built with four switches (solid-

state or mechanical). When the switches S1

and S4 (according to the first figure) are closed

(and S2 and S3 are open) a positive voltage

will be applied across the motor. By opening

S1 and S4 switches and closing S2 and S3

switches, this voltage is reversed, allowingreverse operation of the motor as shown

below .


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Forward direction


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And in this case the applied voltage(Va) at the

motor terminals equal to input voltage(Vi)-

voltage drop on switch 1(VT1)-voltage drop

on switch 2(Vt2)


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shaft encoder and it's cct

Sources and detectors

Source:

The light source for the optical encoder is eitherlight emitting diode (led ) or laser diode , the led

is considerably less expensive than the laser and

used for relatively slow speed or lower frequency

communication application ,the light output fromthe led has a broad spectral bandwidth and emits

from the led as a large cone!


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Detectors:

Optical receivers are critical components in such

systems ,their performance play a dominant role

in determining the spacings and in the flexibility

of the system in terms of its sensitivity.

The received signal must be converted into an

electrical signal , amplified and processed ; togive an estimate of the transmitted signal ;to use

it in directing the motor in the desired direction .


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Basic Light Detector

The following cct features basic, visible lightphoto-detector circuit that can be used todetect trains or other light blocking objects.

The sensor used for this circuit is siliconphototransistor this sensor allow less currentto flow when they are dark. (Phototransistorschange their 'conductance' ).

The phototransistor would normally beplaced between the rails in the circuit .


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The Photo-detector use LM339 voltagecomparator, integrated circuits to detect the change involtage across the sensor. the circuit is configured tohave the LED's turn on when the sensor element isdark (covered by a train.) The LED's can also be madeto turn off when a train is detected.

The supply voltage for the circuit is specified asregulated 12 volts DC but this can be changed if

needed. In some cases the values of some resistorsmay have to be adjusted to compensate(calibrate).


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In this circuit, when the light falling on the

phototransistor (Q1) is blocked, its

conductance will decrease and the voltage

across Q1 will rise. When the voltage rises

above 1/2 of the supply voltage the output of

the comparator will turn ON and the LED will

be lit.


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The only critical part of this circuit is the valueof resistor R1 which in most cases can be 470Kohms but may have to be increase if the room

is dark or decreased if the room is well lit. Increasing the value of R1 will cause the

sensitivity of the sensor to decrease. This maybe necessary when the light falling on the cellis not very strong or shadows can affect thephototransistor.


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`

Pin Diagram For An LM339(Internal Circuitry

For LM339)


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pic16f877

While talking about the pic and its cct ,excuse me Dr.samer and please accept my maximum respect and allof you my classimates , dr.s.

Let me tell a joke about why using this pic(pic 16f877),When I decided to buy the microcontroller (pic) , we look

for the memory of the pic; what will the type whichwill fit the size of the project program , anyhow wedecided to use pic18 you know it has a larger memory

maybe 5 more size we need to control the position atdifferent velocities , we need to do a lot of thingsanyhow students said ... .


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By a childish behavior I switched back to pic 16

despite of my unconveniance .

Yes in the last month we learnned about the pic and

exactly pic 18 but we haven't absorbed it yet in

the matter which enables us to write a program .

However the connection of the basic circuit of the

pic which we used is shown in the next slide theconnection of the input output devices is shown

next:


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The position control mode is the simplest and

least expensive mode of control in which the

controller output has only two possible

values depending on the sign of of the error

(in which a neutral zone is considered around

the zero such that no action take place).


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Such that

SP: step point

en: nth sample of error.

In: integral mode action.

Pn: proportional mode action .

Dn: derivative mode action.

Delta: sample time used to compute thederivative mode action.


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position cs

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