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Pulse Width Modulation and Motor Control

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Presentation on theme: "Pulse Width Modulation and Motor Control"— Presentation transcript:

1 Pulse Width Modulation and Motor Control
Mark Barnhill Roy Dong Andrew Kleeves Micajah Worden Dave Seaton Facilitator: Professor Strangas

2 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

3 Pulse Width Modulation
Speed Control Duty Cycle Advantages Disadvantages

4 Brushed DC Motor Field Magnets Stator DC Power Supply
Armature or Rotor Axle Commutator Brushes

5 How to Code PWM Example here will cover MSP430
Concepts can be easily extended

6 Reading the Datasheet One pin has multiple functions
Set PxSEL accordingly P2DIR |= BIT2; // set P2.2 as output P2SEL |= BIT2; // use pin as TA1.1 Why |= operator?

7 Setting Timer Values Counter counts up each clock cycle
What do the different modes mean? CCR0 = ; Why minus 1?

8 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?

9 PWM Output Modes We are using Timer A1.1 OUTMOD_1 sets at CCRx
CCTL1 = OUTMOD_7; // reset at CCR1 ; // set at CCR0 OUTMOD_1 sets at CCRx OUTMOD_2 toggles at CCRx, resets at CCR0

10 Setting the Duty Cycle We are using Timer A1.1 Recall: Now:
TACTL = TASSEL_2 + MC_1; // SMCLK, up to CCR0 CCR0 = ; CCTL1 = OUTMOD_7; // reset at CCR1 ; // set at CCR0 Now: CCR1 = 200-1; // 20% duty cycle What will this do?

11 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

12 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

13 Example of Back EMF

14 Example of BEMF with a Load

15 Functional Block Diagram of
PWM DC Motor Control

16 Ramp Control Is an integrator
Adjusts the set point up to the desired value.

17 PID Control e(t)= Setpoint - measured
Kp, Ki and Kd must be tuned according to desired output characteristics

18 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 + 𝐾 2

19 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 = Ω

20 Motor Characterization Cont.
Rotor speed and input voltage are directly related by the motor constant (K) in the equation: 𝐾= 𝑉 𝑆𝑈𝑃𝑃𝐿𝑌 − 𝑉 𝐵𝐸𝑀𝐹 𝜔 𝑅𝑂𝑇𝑂𝑅 = 𝑉 𝑆𝑈𝑃𝑃𝐿𝑌 − 𝐼 𝐴𝑅𝑀 ∙ 𝑅 𝑊𝐷𝐺 𝜔 𝑅𝑂𝑇𝑂𝑅 A no-load test supplying 12.0 Volts to the motor results in 830 mA drawn at a speed of ~14,200 rpm (1, rad/s). Using the winding resistance from before, the Back EMF is subtracted from the supply which results in: K = V/rad

21 Open Loop Simulation RiseTime: 0.4871 SettlingTime: 0.8853
SteadyState: Overshoot: J=0.002; b= ; K= ; R=1.2932; L=0.05; step(K,[(J*L) ((J*R)+(L*b)) ((b*R)+K^2)]);

22 PID/Closed Loop Simulation
RiseTime: SettlingTime: SteadyState: Overshoot: 0 J=0.002; b= ; K= ; 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

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