Transcript technical lecture

Pulse Width Modulation
and
Motor Control
Mark Barnhill
Roy Dong
Andrew Kleeves
Micajah Worden
Dave Seaton
Facilitator: Professor Strangas
Agenda
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Pulse Width Modulation
Brushed DC Motor
How to Code PWM
DACs and PWM Amplification
Back EMF
Ramp Control
PID Controller
Motor Characterization
PID Simulation
Pulse Width Modulation
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Speed Control
Duty Cycle
Advantages
Disadvantages
Brushed DC Motor
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Field Magnets
Stator
DC Power Supply
Armature or Rotor
Axle
Commutator
Brushes
How to Code PWM
• Example here will cover MSP430
– Concepts can be easily extended
Reading the Datasheet
• One pin has multiple functions
– Set PxSEL accordingly
– P2DIR |= BIT2;
– P2SEL |= BIT2;
– Why |= operator?
// set P2.2 as output
// use pin as TA1.1
Setting Timer Values
• Counter counts up each clock cycle
• What do the different modes mean?
– CCR0 = 1000-1;
– Why minus 1?
Looking into ‘MSP430G2231.h‘
• We are using Timer A
• We must set TACTL
– TACTL = TASSEL_2 + MC_1; // SMCLK, up to CCR0
– Which clock do you want to use?
PWM Output Modes
• We are using Timer A1.1
• CCTL1 = OUTMOD_7;
• ;
// reset at CCR1
// set at CCR0
• OUTMOD_1 sets at CCRx
• OUTMOD_2 toggles at CCRx, resets at CCR0
Setting the Duty Cycle
• We are using Timer A1.1
– Recall:
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TACTL = TASSEL_2 + MC_1; // SMCLK, up to CCR0
CCR0 = 1000-1;
CCTL1 = OUTMOD_7; // reset at CCR1
;
// set at CCR0
– Now:
– CCR1 = 200-1;// 20% duty cycle
– What will this do?
DACs and PWM Amplification
• DACs are used to convert a digital signal to
analog
– Why does a PWM signal become a steady DC
value?
• Microprocessors can’t provide enough current
to drive a motor
Back Electromotive Force (EMF)
• A motor converts electrical energy to
mechanical energy
• This conversion can go both ways
• If a motor is spinning it will generate electrical
energy
– Called back emf
Example of Back EMF
Example of BEMF with a Load
Functional Block Diagram of
PWM DC Motor Control
Ramp Control
• Is an integrator
• Adjusts the set
point up to the
desired value.
PID Control
• e(t)= Setpoint - measured
• Kp, Ki and Kd must be tuned according
to desired output characteristics
DC Motor Model
• Basic DC motor systems can be represented by this
electromechanical schematic. (bottom-left)
• The motor speed (𝜃) as a function of input voltage (𝑉)
is governed by an open loop transfer function.
(bottom-right)
• It is helpful to characterize the motor to obtain
simulations/projected results along with PID
estimates for the system.
𝜃
𝑉
𝐾
=
𝐽 ∙ 𝑠 + 𝑏 𝐿 ∙ 𝑠 + 𝑅 + 𝐾2
Motor Characterization
• In order to obtain the motor parameters, basic
DC machine tests must be used.
• To get an estimate for Rwdg :
– The rotor must be locked.
– 5 different voltages are supplied to the windings.
– The current is measured.
– Ohm’s Law:
𝑉
𝐼
= 𝑅 to find average resistance
Voltage (Volts)
Current (Amps)
Resistance (Ohms)
0.30 V
0.23 A
1.304 Ω
0.50 V
0.39 A
1.282 Ω
0.70 V
0.56 A
1.250 Ω
1.00 V
0.79 A
1.266 Ω
1.20 V
0.88 A
1.364 Ω
Rwdg = 1.2932 Ω
Motor Characterization Cont.
• Rotor speed and input voltage are directly related by the
motor constant (K) in the equation:
–𝐾=
𝑉𝑆𝑈𝑃𝑃𝐿𝑌 − 𝑉𝐵𝐸𝑀𝐹
𝜔𝑅𝑂𝑇𝑂𝑅
=
𝑉𝑆𝑈𝑃𝑃𝐿𝑌 − 𝐼𝐴𝑅𝑀 ∙𝑅𝑊𝐷𝐺
𝜔𝑅𝑂𝑇𝑂𝑅
• A no-load test supplying 12.0 Volts to the motor
results in 830 mA drawn at a speed of ~14,200 rpm
(1,487.0205 rad/s).
• Using the winding resistance from before, the Back
EMF is subtracted from the supply which results in:
K = 0.007348 V/rad
Open Loop Simulation
J=0.002;
b=0.00924;
K=0.007348;
R=1.2932;
L=0.05;
step(K,[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)]);
RiseTime:
SettlingTime:
SteadyState:
Overshoot:
0.4871
0.8853
0.6120
1.1044
PID/Closed Loop Simulation
J=0.002;
b=0.00924;
K=0.007348;
R=1.2932;
L=0.05;
Kp=20;
Ki=30;
Kd=29;
num_PID=[Kd, Kp, Ki];
den_LOOP=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)];
num_B=conv(K,num_PID);
den_B=conv(den_LOOP,[1 0]);
[num_SYS,den_SYS]=cloop(num_B,den_B);
step(num_SYS,den_SYS)
Kp: 20
Ki: 30
Kd: 29
RiseTime:
SettlingTime:
SteadyState:
Overshoot:
0.1788
0.2168
1.0000
0