| Tutorial:
Closed Loop Speed and Position
Control of DC motors
By Ibrahim Kamal
Last update:
15/4/08
 Overview
Without getting too close to the mathematical nature of
this subject, this tutorial aims to explain what is the
meaning of closed loop control, and how to apply it in your
projects.
As you shall learn in this article, closed
loop control offers new possibilities to a project designer,
it increases accuracy, shorten response time and dramatically
decreases error. |
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1-Closed
loop vs Open loop control
In general open
loop control means that you send electrical signals to an actuator
to perform a certain action, like connecting a motor to a battery
for example. In this scheme of control, there is no any mean
for your controller to make sure the task was performed correctly,
and it often need human intervention to obtain accurate results.
A very simple example of open loop control, is the remote controller
of an RC toy car; you - the human - have to constantly check
the position and the velocity of the car to adapt to the situation
and move the car to the desired place.
But what if you could let the electronics handle a part, if
not all of the tasks performed by a human in an open loop controller,
while obtaining more accurate results with extremely short response
time? This what is called closed loop control. In order to be
able to build a closed loop controller, you need some mean of
gaining information about the rotation of the shaft like the
number
of revolutions executed per second, or even the precise
angle of the shaft. This source of information about
the shaft of the motor is called "feed-back"
because it sends back information from the controlled
actuator to the controller.
Figure 1.A shows clearly the difference between the
two control schemes. Both types have a controller that
gives orders to a driver, which is a power circuit (usually
an H bridge) that drives
the motor in the required direction. It is clear that
the closed loop system is more complicated because it
needs a
|
Figure 1.A: Closed
loop VS Open loop |
'shaft encder' which is a devise that will translate
the rotation of the shaft into electrical signals that can be
communicated to the controller.
In other words, a closed loop controller will regulate the the
power delivered to the motor to reach the required velocity.
If the motor is to turn faster than the required velocity, the
controller will deliver less power to the motor. Controlling
the electrical power delivered to the motor, is usually done
by Pulse Width Modulation.
2-Shaft
encoders
shaped
photo-couple made of an Infra-Red sender and a matching
receiver is positioned in a certain way so that the
beam of infrared light passes through one of the small
openings in the encoder disk.
In reality, photo-couple come in many shapes and sizes,
but most of them are more or less similar to the one
shown in figure 2D. Any photo couple has 4 leads, two
for the sender, which is usually an Infra-Red LED and,
and the two others are for the receiver, which is usually
a photo transistor. You can see the schematic representation
of that photo couple at the lower corner of figure 2.D,
where is is clear that the photo-couple is composed
of a LED and a photo transistor.
The encoder disk is firmly connected to the back-shaft
of the motor, so that both the shaft and the encoder
disk rotates at the same r.p.m. (the back-shaft is an
extension of the output shaft of the |
Figure 2.D
|
motor at its back, usually present
for the sole purpose of adding a shaft encoder). When this encoder
disk is inserted in the configuration shown in figure 2.C, the
rotation of the motor causes the beam of light to be periodically
intercepted by the solid parts of the encoder disk creating
a sequence of pulses of light, that will be translated by the
photo couple's receiver into pulses of electricity.
Those pulses of electricity contain all the information we need
to implement a closed loop control. The frequency of those pulses
is directly proportional the the speed of rotation of the shaft
(RPM) and the number of those pulses correspond to the angular
displacement of the shaft.
The more the number of holes in an encoder disk, the higher
will be the resolution (the slightest angular displacement that
can be detected).
One important factor that affect the performance of shaft encoder
and thus the overall performance of a closed loop control system,
is the position of the encoder disk. Most of the motors are
used with a gearbox designed to reduce the r.p.m. while increasing
the output torque (figure 2.B shows a motor+gearbox assembly).
Thus, the motor itself can be turning at 4400 r.p.m. for example,
driving a 40:1 gear box, dividing the rpm by 40, giving a final
output speed of 110 r.p.m. You can can take a great advantage
of this to reach very high degrees of accuracy, by connecting
the encoder disk at the back shaft of the motor (which is turning
at 4400 rpm in our example). This way, each turn of the final
output shaft from the gearbox will correspond to 40 turns of
the shaft encoder, and if the encoder disk has 30 holes on its
circumference, a single turn on the final output shaft will
correspond to 1200 pulses, reaching a theoretical precision
of 0.3 degrees (i.e. each pulse correspond to 0.3 degrees of
rotation of the final output shaft). (Depending on the type
of motor and gearbox, it may be difficult to reach exactly that
theoretical precision).
3-The
controller
A closed loop controller can be an analog circuit, a digital
circuit made of logic gates, or a microcontroller. Generally,
a microcontroller is the option that will provide more design
flexibility. Recent microcontrollers running at very high clock
rates can completely replace similar analog controllers, and
can even be cheaper.
In a closed loop system, a microcontroller
will have two main tasks:
1- constantly adjust the average
power delivered to the motor to reach the required velocity.
2- Precisely calculate the position/angle of the motor's
output shaft.
As you can see in figure 3.A, the shaft encoder will
provide the microcontroller's internal counter with
a sequence of pulses that correspond to the rotation
of the motor. A timer is set to execute two software
routine every 1/10 th of a second (which is just an
arbitrary value). One of those software routines is
to recalculate the actual angle of the shaft or the
total number of revolutions.
Then, another software routine is executed to control
the speed of the motor by comparing the number of counted
pulses with a fixed number which is referred to as the
"required pulses". The "required pulses"
corresponds to the desired speed, and |
Figure 3.A
|
the "counted pulses" corresponds to
the actual speed of the motor.
Finally, as you can notice in the schematic, it's all a matter
of comparing those two values and constantly adjusting the power
delivered to the motor. Note that choosing the right timing
between each execution of this routine can dramatically improve
overall system stability and performance, especially on low
quality motors.
Controlling the power delivered to the motor to control
its speed.
Recalling the figure 1.A, a closed loop system contains a controller
and a driver. the driver on its own - which is usually an H
bridge - cannot control the velocity of the motor. The most
common technique to do so is to let the controller turn the
driver ON and OFF at very high rates, changing the ratio between
the ON and OFF time to control the speed of the motor. This
is what is called PWM or Pulse Width Modulation. For more information
about PWM, and to know how to implement it in a microcontroller
program, you may read the "Electronics
and Algorithms" part of this related article.
Example
C source code for a 89S52 microcontroller.
This example
shows how to implement the flow chart shown in figure 3.A into
a microcontroller to control the speed of two motors simultaneously,
which is usually the case in differential drive robots. The
code is taken from a robot's project, where two motors were
controlled by the microcontroller through H-Bridge modules
identical to this one, which are
controlled with only two wires to determine one of the 4 main
orders than can be given to an H-Bridge module: turn clockwise,
turn anticlockwise, break or provide high impendence output
(free or not connected).
In this code, The internals timers of the 89S52 Timer0 and Timer1
are used as counters, to count the pulses coming from the right
and left shaft encoders, while a software timer periodically
executes the functions required to update the calculated position
of the shaft, and rectify the duty cycle of the PWM sent to
each motor to reach the required speed.
Note that this code was simplified from an originally more complicated
code that included complex differential drive line following
routines, so you may encounter some errors due to undeclared
variables or other minor mistakes, but the logic and the main
flow of the program are correct and proved to be working perfectly.
#include <REGX52.h>
#include <math.h>
unsigned char req_right_pulses,req_left_pulses,speed_chk_counter;
unsigned char right_pwm,left_pwm,max_pwm,pwm_counter,right_dir,left_dir;
unsigned char pwm_adjust_delay,req_delay,max_speed;
unsigned int position_counter;
setup_timers(){
EA = 1;
TMOD = 0X55; // counters 1 and 2 in mode 1 (16 bit counter)
ET0 = 1; //Enable the Timer/counter 0 interrupt
TR0 = 1; //Enable Timer/counter 0 to count
ET1 = 1; //Enable the Timer/counter 1 interrupt
TR1 = 1; //Enable Timer/counter 1 to count
}
timer0_overflow() interrupt 1{}
timer1_overflow() interrupt 3{}
pwm_check_and_adjust(){
speed_chk_counter++; //This is what makes the between
if (speed_chk_counter > pwm_adjust_delay){ //each two executions of this
speed_chk_counter = 0; //function
if (req_right_pulses > TL1){ // Compare TL1 (which contains the
if (right_pwm < max_pwm){ // value of counter 1) and either
right_pwm++; // increase of decrease the pwm
} // of the right motor, while making
}else{ // sure the value of the pwm stays
if (right_pwm > 0){ // between 0 and max_pwm.
right_pwm--;
}
}
if (req_left_pulses > TL0){ // The same that applies
if (left_pwm < max_pwm){ // to TL1 and right_pwm,
left_pwm++; // applies TL0 and
} // left_pwm.
}else{
if (left_pwm > 0){
left_pwm--;
}
}
position_counter += TL1; // Update the position of the shaft
TL1 = 0; // Reset the counters to 0.
TL0 = 0;
}
}
pwm_generator(){
pwm_counter++; // This is Just a counter
if (pwm_counter > max_pwm){ pwm_counter = 0; } // From 0 to max_pwm
if (right_pwm > pwm_counter){ // Right Pwm, ON period
if (right_dir == 1){ // Depending on the value of
P2_0 = 0; // the variable right_dir
P2_1 = 1; // a corresponding order will be
}else if(right_dir == 2){ // given on the pins P2.0 and P2.1
P2_0 = 1; // that are connected to the H-Bridge
P2_1 = 0; // that drives the motor
}else{
P2_0 = 1;
P2_1 = 1;
}
}else{
P2_0 = 0;
P2_1 = 0;
}
if (left_pwm > pwm_counter){ // Same applies for the left motor.
//ON period
if (left_dir == 1){
P2_2 = 0;
P2_3 = 1;
}else if(left_dir == 2){
P2_2 = 1;
P2_3 = 0;
}else{
P2_2 = 1;
P2_3 = 1;
}
}else{
P2_2 = 0;
P2_3 = 0;
}
}
void main(){ // This is the main part of the program,
pwm_adjust_delay = 12; // where main variables are initialized
max_pwm = 25;
setup_timers(); // timers are also initialized here
right_dir = 1;
left_dir = 1;
req_right_pulses = 15;
req_left_pulses = 15;
while(1){ // This is the main loop
pwm_check_and_adjust(); // where those twu finctions are constantly called
pwm_generator();
}
} |
Preview of the last 15
messages discussing this page. Messages are sorted from the newest to
the oldest. |
Posted
by:
yonas tesfaye
on:
15 Jan 2010 |
Closed Loop Speed and Position Control of DC motors |
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please please give me a detail of the mathematical modeling and the controller design of closed loop speed control of dc motor using digital control as soon as possible untile tomorrow since this mini project is submitted on monday.
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Posted
by:
belaid
on:
23 Feb 2009 |
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Posted
by:
egehaner
on:
02 Sep 2008 |
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Closed Loop Speed and Position Control of DC motors |
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Hi,
I have been studying on a project named as ''dc motor speed and position control with encoder''. I want to use a PIC 16f84A or a 16f877A as a controller and i'm using Basic programming language (Pic Basic Pro program). I have a dc motor and 64 pulse optic encoder. I don't exactly know how can i use encoders' A and B phases. In my project i have been planning to use a pic and inputs are Porta.0 and Porta.1 as encoder inputs. Motor inputs are pics' Portb.0 and Portb.1 outputs. I want to control motor position and speed according to encoders' inputs. In some projects people EXOR the A and B phases but i couldn't understand how do they make it.
Would you please answer my questions and sent me a program about this project.
Regards,
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Posted
by:
ikalogic
on:
07 Apr 2008 |
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Re: Closed Loop Speed and Position Control of DC motors |
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| Quoting AndreAhmed: So what's really the purpose of that article ? please clarify using a problem ... |
Well, it's purpose is to explain the concept of shaft encoders in a closed loop control system.
About the concept of "reducing the fastest motor's speed just until it is equal to the slowest motor speed" well, I assumed you usually use motors of the same speed for a certain application, and if they are not of the same speed, you won't usually ask for the motors to run at "impossible" speeds.
But as i said in my previous post, this could be an update/enhancement to your controller, letting detect this kind of situation.
My idea was to give an introduction for the use of the shaft encoders, this will also give me the ability to write more complicated articles in the near future, that would be some kind of a "part 2" to that article.
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Posted
by:
ikalogic
on:
06 Apr 2008 |
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Re: Closed Loop Speed and Position Control of DC motors |
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| Quoting AndreAhmed: yes , that's what i want do !! , so i didnt understand the aricl prolly , cuz i thought the code in the article does this !! |
well.. no it does not...  maybe you'll do it, and send it to me, and i'll add it on the page a user contribution 
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Posted
by:
andreahmed
on:
05 Apr 2008 |
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Re: Closed Loop Speed and Position Control of DC motors |
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unless you implement some ingenious algorithm that detects this situation and reduce the fastest motor's speed just until it is equal to the slowest motor speed..
yes , that's what i want do !! , so i didnt understand the aricl prolly , cuz i thought the code in the article does this !!
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Posted
by:
ikalogic
on:
05 Apr 2008 |
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Re: Closed Loop Speed and Position Control of DC motors |
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Quoting AndreAhmed: Hi , i liked the tutorial but i've a question, what if the lets say motor 2 has a 6V battery , and the other motor works at 12V battery , is the close loop which you discussed should reduce the faster motor to equal it to the lower speed motor ? and if not , how do i do that ?
Thanks |
It all depends on how you deal with the data you gather from the shaft encoders..
Here is how I would imagine a simple example system: I want the tow motors to turn at 50 RPM. IF one motor come to slow down, I increase average voltage on that motor to compensate the speed difference and reach the 50 RPM again.
Now, if the two motors don't have the same speed (or the same voltage), and the slow motor has reached its maximum speed, but still cannot satisfy the 50 RPM speed requested by the controller, there would be a serious problem.. unless you implement some ingenious algorithm that detects this situation and reduce the fastest motor's speed just until it is equal to the slowest motor speed..
hope you see what i mean
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Posted
by:
andreahmed
on:
05 Apr 2008 |
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Closed Loop Speed and Position Control of DC motors |
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Hi , i liked the tutorial but i've a question, what if the lets say motor 2 has a 6V battery , and the other motor works at 12V battery , is the close loop which you discussed should reduce the faster motor to equal it to the lower speed motor ? and if not , how do i do that ?
Thanks
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Posted
by:
ikalogic
on:
28 Feb 2008 |
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Closed Loop Speed and Position Control of DC motors |
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Quote: Dear Sir,
I want to work on Position and speed control releated applications. My aim is to move a target of man having 25 Kg weight with speed of 10 m/Sec on a straight track of 500 metre.I want to stop at 100m,200m 300m 400m and 500m and then I want to move back. Please suggest type of motor, specification of motor, manufacturer/supplier and if possible circuit and software for it.
Regards |
Any regular geared DC motor, with a power of more or less 450W, is capable of moving a 25Kg charge horisontally. The more the mechanical parts are frictionless, the less you have power losses, the less you will need a powerfull motor, and the more you have seamless control on your motor. In a more scientific fashion, reducing friction and losses, helps to reduce the 'equivalent inertia' that appears at the end of your motor's shaft
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really a nice project to carry on......thanks a lot 
