Industrial Electronics DEK 3113

DC motor speed control

 

    

    

    

   

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APPLICATIONAPPLICATION

Multi-Axis Conveyor System

Vertical Operation

Feeding Materials

Rotational Operation

Variable Speed Conveyor

Applications for Speed Control Systems Applications for Speed Control Systems

Normal method of controlling speed of DC motor

There are three method used:

1) Adjustment of field current control

2) Adjustment of resistance

3) Adjustment of armature terminal voltage

But all this method is improved by using

PWM(pulse width modulation).

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

-if DC level is greater than triangle waveform voltage , 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.

In waveform 1b, the signal has a mark-space ratio of 3:1, with the signal 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.

In waveform 1c, the signal has a mark-space ratio is 1:3, with the signal of 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= tON/(tON+tOFF) x 100%

Example: 1) tON+tOFF=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

tON=3us x 10/100

=0.3us

tOFF=2.7us

Speed control circuit

-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.

-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.

-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.

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)

Vave arm=Vsupply x (t1/(t1+t2) = 500x0.2 =100V

Note: Vave arm is an average voltage that cause motor to move.In PWM usage, the voltage value to move the motor is lower than Vsupply. This is power saving and security.

Voltage time area apply to inductor=(Vsupply-Vaverage)xTON=(500-100)t1

=400t1

ChopperGiven t1/(t1+t2)=0.2 ,Imax=10A

The rise current:

Ia=400t1/0.06=10A V=Ldi/dt

t1=tON=1.5ms

t1/(t1+t2)=0.2

1.5m/(1.5m+t2)=0.2

t2=6ms

t1+t2=1.5ms+6ms=7.5ms

pulse=1/T=1/7.5m=133pulse/sec

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