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    Chapter 6 edit Chapter 6 edit Presentation Transcript

    • DC motor speed control
    • APPLICATION                                                                                                                                                                                                                 Downsized Conveyor   Material Conveyor   Fluid Injection   Material Elevator                                                                                                                                                                                                                                     Conveyor Belt   Conveyor System   Bi-Directional   Uni -Directional                          
    •                                                                                                                                                                    Dual Axis Control   Conveyor with Positioning Sensor   Belt & Pulley                                                                                                                                                                                                                                     Conveyor Positioned with CCD Camera   Conveyor w/ Frequent START/STOP   Vertical Conveyor   Showering Application
    • Multi-Axis Conveyor System Vertical Operation Feeding Materials Rotational Operation Variable Speed Conveyor Applications for Speed Control Systems
    • Normal method of controlling speed of DC motor
      • There are three method used:
      • Adjustment of field current control
      • Adjustment of resistance
      • Adjustment of armature terminal voltage
      • But all this method is improved by using
      • PWM(pulse width modulation).
    • A Simple Block Diagram of the Speed Controller
    • Speed, Supply Voltage Vs. Time
    • PWM(pulse width modulation)
      • - PWM is pulse width modulation
      • -It is generated by comparing a triangular wave signal with a DC
      • signal.The DC signal can range between the maximum and
      • minimum voltages of the triangular voltage
      When triangle waveform voltage is greater than the DC level, the output of the op-amp swings high, and when it is lower, the output swings low the way it controls the motor by the timing ON and OFF duration that can be control by the user.
    • Analogy to PWM
      • A good analogy is bicycle riding
      • You peddle (exert energy) and then coast (relax) using your
      • momentum to carry you forward
      • As you slow down (due to wind resistance, friction, road shape) you
      • peddle to speed up and then coast again. 
      • -The 'duty cycle' is the ratio of peddling time to the total time (peddle
      • + coast time). 
      • -A 100% duty cycle means you are peddling all the time, and a 50%
      • duty cycle means you are peddling only half the time.
      • PWM for motor speed control works in a very similar way.
      • -Instead of supplying a varying voltage to a motor , it is supplied with a fixed
      • voltage value (such as 12V) which starts it spinning immediately . 
      • -The voltage is then removed and the motor 'coasts '. 
      • By continuing this voltage on/off cycle with a varying duty cycle, the motor
      • speed can be controlled.
    • Duty cycle
      • In waveform 1a
      • the signal has a mark-space ratio of 1:1. With the signal at 12V for 50% of the time
      • the average voltage is 6V,
      • so the motor runs at half its maximum speed.
    • Duty cycle
      • In waveform 1b
      • the signal has a mark-space ratio of 3:1, which means that the output is at 12V for 75% of the time. 
      • This clearly gives an average output voltage of 9V,
      • so the motor runs at 3/4 of its maximum speed.
    • Duty cycle
      • In waveform 1c
      • the signal has a mark-space ratio is 1:3, giving an output signal that is 12V for just 25% of the time. 
      • The average output voltage of this signal is just 3V,
      • so the motor runs at 1/4 of its maximum speed.
    • PWM(Pulse width modulation)
      • for PWM can use MOSFET for switch
      -use concept ON and OFF signal Duty cycle= t ON /(t ON +t OFF ) x 100% Example: 1) t ON +t OFF =100 % duty cycle=10% means total ON=10% 2) duty cycle=90% means total ON 90% Note: To control speed, when %ON increase, the motor faster
      • Example:
      • Given f=500kHz. Find time to generate:
      • 1) 10% duty cycle
      • T=1/500kHz
      • =3us
      • t ON =3us x 10/100
      • =0.3us
      • t OFF =2.7us
    • Speed control circuit 1) When Q1 is ON,Ia will go through La and motor will start to move.At this time La will start to charge and magnetic field exist. 2) When Q1 is OFF after ON, energy in La need to discharge. This is done by discharge through D1. This kind of circuit is for safety so Q1 is not damage by very high voltage or current. -Please refer to Figure 6.9 for full bridge circuit for reverse and forward movement of motor.
    • Chopper the relatives ON-OFF thyristor determine the average of the motor. -when thyristor is ON,Ia will flow through L and motor will start to move. -when motor is move until around 75% speed of motor,commutating switch will closed. This will caused thyristor off. -when thyristor off after time t1,armature current decline through freewheeling diode as energy stored in a circuit inductance is applied to the armature.
    • Chopper -motor will slow down(since thyristor OFF),then commutating switch will be open and thyristor ON again. -diode serve a path to maintain the armature current which no longer flow through external circuit. -as the ratio ON/OFF increased,the average motor Vm rises. -This process will be repeated.
    • Example : Chopper
      • A 100 hp series motor rated 180A is operating in a chopper circuit from a
      • 500V dc source ( Vsupply=500V ).
      • The armature and field inductance is 0.06H(L=0.06H). At the minimum ratio
      • t1/(t1 + t2) of 0.20 . Find the pulse frequency to limit the amplitude of the
      • armature current excursion to 10A.( Ia max=10A )
    • Example : Chopper
      • A 100 hp series motor rated 180A is operating in a chopper circuit from a
      • 500V dc source ( Vsupply=500V ). The armature and field inductance is 0.06H(L=0.06H). At the minimum ratio t1/(t1 + t2) of 0.20 .
      • Find the pulse frequency to limit the amplitude of the armature current excursion to 10A.(Ia max=10A)
      1)Vave arm=Vsupply x (t1/(t1+t2) = 500x0.2 =100V Note: Vave arm is an average voltage that cause motor to move.InPWM usage,the voltage value to move the motor is lower than Vsupply.This is power saving and security. 2) Voltage time area apply to inductor =(Vsupply-Vaverage)xT ON =(500-100)t1 =400t1
    • Chopper
      • 3) Given t1/(t1+t2)=0.2 ,Imax=10A
      • The rise current:
      • Ia=400t1/0.06=10A V=Ldi/dt
      • t1=t ON =1.5ms
      • t1/(t1+t2)=0.2
      • 1.5m/(1.5m+t2)=0.2
      • t2=6ms
      • t1+t2=1.5ms+6ms=7.5ms
      • pulse frequency =1/T=1/7.5m=133pulse/sec
    • Vo Vo Vm t1 t2 vm (on pulse) Ia ia Vm (i) Waveforms of motor voltage and current at low speed vm (on pulse) ia t1 t2 Ia (ii) Waveforms of motor voltage and current at high speed
    • PWM Basics
      • The higher the voltage seen by the motor, the higher the speed .
      • We’ll manipulate the PWM Duty Cycle ( The ratio of the “on” time divided by the period