Pulse Width Modulationand

Motor ControlMark Barnhill

Roy DongAndrew KleevesMicajah Worden

Dave SeatonFacilitator: Professor Strangas

Agenda• Pulse Width Modulation• Brushed DC Motor• How to Code PWM• DACs and PWM Amplification• Back EMF• Ramp Control• PID Controller• Motor Characterization• PID Simulation

• Speed Control• Duty Cycle• Advantages• Disadvantages

Pulse Width Modulation

• Field Magnets• Stator• DC Power Supply• Armature or Rotor• Axle• Commutator• Brushes

Brushed DC Motor

• Example here will cover MSP430– Concepts can be easily extended

How to Code PWM

• One pin has multiple functions– Set PxSEL accordingly– P2DIR |= BIT2; // set P2.2 as output– P2SEL |= BIT2; // use pin as TA1.1

– Why |= operator?

Reading the Datasheet

• Counter counts up each clock cycle• What do the different modes mean?

– CCR0 = 1000-1;

– Why minus 1?

Setting Timer Values

• 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?

Looking into ‘MSP430G2231.h‘

• 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

PWM Output Modes

• We are using Timer A1.1– Recall:– 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?

Setting the Duty Cycle

• 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

DACs and PWM Amplification

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 ofPWM DC Motor Control

• Is an integrator

• Adjusts the set point up to the desired value.

Ramp Control

• e(t)= Setpoint - measured• Kp, Ki and Kd must be tuned according to

desired output characteristics

PID Control

• Basic DC motor systems can be represented by this electromechanical schematic. (bottom-left)

• The motor speed () as a function of input voltage () 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.

𝜃𝑉

=𝐾

( 𝐽 ∙ 𝑠+𝑏 ) (𝐿 ∙ 𝑠+𝑅 )+𝐾 2

DC Motor Model

• 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: to find average resistance

Rwdg = 1.2932 Ω

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

Motor Characterization

• Rotor speed and input voltage are directly related by the motor constant (K) in the equation:

• 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

Motor Characterization Cont.

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: 0.4871SettlingTime: 0.8853 SteadyState: 0.6120 Overshoot: 1.1044

Open 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)

RiseTime: 0.1788SettlingTime: 0.2168 SteadyState: 1.0000 Overshoot: 0

Kp: 20 Ki: 30 Kd: 29

PID/Closed Loop Simulation