EE3092 -Laboratory Practice V

CHOPPER CONTROL DC DRIVE

Instructor: - Mr. R.D. Wijesinghe

Name: - V.I.P. Dasanayake

Index Number: - 090075M

Group: - 3

Submission date: -27/03/2012

OBSERVATION SHEET

Name : V.I.P. Dasanayake

Index No : 090075M

Practical : Chopper Control DC Motors

Date : 09/03/2012

Field : Electrical

Instructed by: Mr. R.D. Wijesinghe

b). Measurement of chopper performance with non-back emf load (under no load condition)

Speed-input potentiometer

settingVL (V) IL (A) T(on)(μs) δ(Duty ratio)

Calculated VL(V)VL= δVDC

0 8 0.3 6.2*20=124 0.0264 3.4961 8 0.3 6.2*20=124 0.0264 3.4962 10 0.38 3.6*50=180 0.0383 5.0743 17 0.64 8.9*50=445 0.0947 12.554 23 0.88 6.9*100=690 0.147 19.455 30 1.16 4.7*200=940 0.200 26.5006 38 1.44 6.1*200=1220 0.260 34.3947 45 1.72 7.5*200=1500 0.319 42.2878 53 2.04 8.8*200=1760 0.374 49.6179 61 2.32 4*500=2000 0.426 56.38310 62 2.36 4.1*500=2050 0.436 57.793

Time period Ts= 4.7ms

Chopper input DC voltage Vdc= 132.5V

c). Measurement of DC motor speed control characteristics

Speed-input potentiometer setting

Speed N (rpm)

1 1852 1853 3344 5135 6986 9057 11068 13389 156210 1600

Figure 1 shows waveforms of armature voltage [upper graph] and the armature current [lower graph]

Figure 2 shows waveforms of field voltage [upper graph] and the field current [lower graph]

Figure 1

Figure 2

Figure 3 shows waveforms of pulse generator output, gate waveform of TH2, saw tooth waveform and comparator control input respectively

Figure 4 shows waveforms of comparator output [upper most graph] and gate of TH1 [lower graphs]

Discussion

Figure 3

Figure 4

1. There is a voltage at the load even when the speed-input potentiometer is at minimum

Here we are using PWM switching circuit with fixed switching frequency. When the input

voltage is changing, gate input is also changing with that. Theoretically in the beginning speed input

potentiometer should be zero, but it is very unlike to happen, because there may be small steps exits

even in the little time. The reason for that would be the time difference between thyristor 1 and

thyristor 2. So because of such practical limitations, voltage won’t be zero even the potentiometer is

at minimum.

2. Why the speed input potentiometer, in the first few steps didn’t respond to the motor; how the problems can possibly be eliminated?

Practical is done under no load condition at the start, which means (load torque =friction

torque at shaft bearings). At low speed friction torque is also low. So friction torque will not be

increasing, unless speed is increasing.

So the motor generating balancing torque won’t change considerably, at low speed. Then the

converter supplies voltage to get the motor to required speed. Therefore in the beginning motor will

not accelerate significantly. When friction torque increases with the speed, it will go up to value

such that, giving balance torque with significant change in speed.

To eliminate this problem, practical should be done, under loaded condition, where load is

significantly large, so friction torque can be neglected.

3. Why the field current waveform contain less ripple compared with the field voltage waveform

Field current is proportional to the integration of field voltage for a given field inductance.

So even there voltage waveform consists with high number of ripples, in filed current there are less

ripple compared with it. Mathematically it is clear that, when there are lots of ripple in voltage wave

form, according to the relationship of (if ∝∫V f ) there will be less ripples in current waveform.

(Since the ∫V f means the cumulative summation of voltage)

If it is explained in some other way, if we imagine the system is a spring mass system and

mass represents the inductance. Since mass has inertia, even there is a high kinetic energy, it

doesn’t like to change the velocity suddenly. Similarly in here even there are ripples (sudden

changes) in the voltage waveform, current will have lesser ripples.

4. Comment on the shape of the armature voltage and current waveform of part(c)?

Here dotted line represents the current waveform when the load torque is reduced by half of its

original value.

Armature voltage waveform

According to the figure 5 it can be clearly seen that TON and TOFF time of the PWM

waveform can be clearly identified. Normally Waveform is given like a square waveform as

given above.

current waveform

Because of the armature voltage variation, we can see some ripples in the current

waveform. From the figure 5 it can be clearly seen that, there are two impulses, because of

the dropping and rising edges in PWM.

5. Comment on the control circuit waveform in part (d)?

This consists with several waveforms, and those are explained below

1) Pulse generator output-square wave generate by the pulse generator, with a constant frequency

2) Trace gate wave form of TH2- waveform is generated , by detecting the dropping edge of

square waveform (this is inverted before it goes to the TH2)

3) Saw tooth waveform- this is obtained from the integration of the square form, with same

frequency. This is used to generate control signal to TH1. Shape of the waveform is shown in

the figure 6.

4) Comparator control input waveform-when the switching control signal to TH1 is generated this

is considered as the reference waveform. Some variation can be observed in this waveform,

because of the feedback getting from armature voltage and current.

Sawtooth waveform Control voltage

5) Comparator output- A square waveform generated using saw tooth waveform and comparator

control input

6) Gate waveform of TH1- this is generated, by detecting dropping edge of square waveform at

comparator output. (this is inverted before it goes to the TH1)

Upper graph-sawtooth waveform and control voltage (VC)

Lower graph-Drive of the switch Vg

6. State functions of potentiometer VR1, VR2, VR3 VR4& VR5?

VR1:-adjust the armature voltage feedback signal

VR2:-adjust the error between armature voltage feedback & set speed input

VR3:-adjust the armature current feedback signal

VR4:-adjust error between armature current feedback signal and VR2 output

VR5:-adjust signal range set value

7. State few practical applications of chopper drives?

Chopper diver circuit is popular of using for electronic switching devices. The reason for that is, high efficiency (since it involves one stage conversion), smooth control, fast operation of the chopper driver circuit.

Battery operated vehicles

DC motor drives when DC supply is available (use as a speed controller as well)

Switched mode power supplies

Use as a switched frequency devices

Figure 6

For mine haulers

For marine hoist

For class D electronic amplifiers

Battery charges where uncontrolled rectifier give DC to choppers

Lighting and lamp controls also prefer choppers

Traction drives use four quadrant choppers for energy saving

For variable frequency devices

References

Power Devices And Machines By Dr.J.S.Chitode, U.A.Bakshi

http://www.eetimes.com/design/industrial-control/4214636/DC-Chopper-control-vs--Phase-Angle-

control-for-appliance-and-power-Tool-Applications-Who-wins-

http://www.mathworks.in/help/toolbox/physmod/powersys/ref/onequadrantchopperdcdrive.html