Phase Controlled Drives

7/28/2019 Phase Controlled Drives

1/26

ELEC 482 Module # 6 Outline

TOPIC: Phase Controlled Converter Drives

Key educational goals:

Evaluate and identify the different types of phase controlled converter (ac-dc converter)

configurations to drive a separately excited dc motor.

Reading/Preparatory activities for class

i) Textbook:

Chapter 6: Controlled bridge rectifiers with DC motor load.

6.1 The principles of rectification.

6.2 Separately excited d.c. motor with rectified single-phase supply (up to and

including equation 6.9 only).

6.3 Separately excited d.c. motor with rectified three-phase supply (up to and

including equation 6.46 only in 6.3.2.1 and exclude 6.3.2.2 and 6.3.2.3).6.1. The

principles of rectification.

Chapter 7: Three-phase naturally commutated bridge circuit as a rectifier or inverter.

7.1 Three-phase controlled bridge rectifier with passive load impedance (only up to

and including equation 7.12 in sub-section 7.1.1, and only up to and including

equation (7.54) in sub-section 7.1.2 ).

7.2 Three-phase controlled bridge rectifier-converter.

ii) Power-point file: Phase_controlled_drives.


2/26

Questions to guide your reading and think about ahead of time.

1. Why is the conduction interval different for an inductive and resistive load in a singlephase half wave controlled rectifier using a single SCR?

2. How does firing angle affect power factor?

3. Why is the power factor in a single phase semi-converter better than a full-converter?

4. Which converter can provide 4 quadrant operation of the drive?

5. A 60 Hz converter produces 360 Hz ripple in the motor current. Is the converter

single- phase or three-phase?

The main concepts for today

1. Analyze why an inductor causes a SCR to conduct even in the negative half of the

voltage cycle in a single SCR, diode less, single phase, half controlled rectifier circuit.

2. Identify the effect of a replacing strategic SCRs with diodes in phase controlled

converters.3. Assess the effect of changing of firing angle on input power factor.

4. Analyze why the input line current looks different for a single-phase semi-converter

and full-converter when driving a DC motor and how it affects power factor.

5. Compare the different phase controlled rectifier configurations and their properties.

6. Recognize the advantages of three phase converters over single phase converters.




3/26

Summary

The knowledge gained from this module helps in analyzing and designing phase

controlled converter circuits for separately excited dc machines.

For next time

We will discuss the basic working principles of an induction motor. This will help in

formulating speed control strategies for these motors.

Sample test/exam questions/problems to help you study

1) A separately excited dc motor (10 kW, 240 V) is supplied from a fully controlledsingle phase bridge. The power supply supplying the bridge is sinusoidal (240 V,

60 Hz). Ra = 0.42 ohm. 2 V-s/rad. Assume Ia to be continuous dc.

Calculate Ia, speed, p.f, efficiency with firing delay angle = 0 degree and 20

degrees with constant load torque equal to rated torque. Assume constant flux

operation. The machine draws rated power at = 0.

2) A 3 phase double converter bridge supplies power to a 560 V, 50 A, DC motorwith Ra = 1.2 ohms. The voltage drop on the bridge thyristor is 20 V at rated

armature current. Supply to the three phase source is 415 V line to line. Find

for

(i) Motoring operation with(ii) Regeneration with(iii) Plugging with and with a current

limiting resistor = 10 ohms.




4/26

Presentation Summary Single phase half-wave controlled rectifier with resistive load

Single phase half-wave controlled rectifier with inductive load

Single phase half-wave rectifier with dc motor load

Different single phase controlled rectifier topologies such as halfwave ,

Semi converter, full converter, double converter

Definition of powerfactor in presence of harmonics

Change of power factor with firing angle

Numerical example on single phase controlled rectifier

Different three phase controlled rectifier topologies such as halfwave ,

Semi converter, full converter, double converter

Numerical example on three phase controlled rectifier to illustrate differentquadrant of operation


5/26

Single-phase half-wave controlled

rectification: resistive load

T1

R

vTh

+ -

vR+

-

vs

is

The SCR has been triggered at an angle

vs

t

is

vTh

t

t

vR

t


6/26


rectification: inductive load

vs

t

is

vTh

t

t

vL

t

2

=2-

=2-

2

The SCR has been triggered at an angle


7/26


rectification: inductive load(2)

= 2 ; extinction angle = = 2 2; conduction angle

= =

= 1 =

+

At = ; = 0

Therefore, =

, =

The current extinct at an angle = 2 . In an inductive circuit, SCR will conduct beyond thevoltage zero crossing.


8/26


rectification: dc motor load

AC

T1

vs

vThM

+

ith

- +

-

vM

iD

D

The motor draws ripple free current due to large

inductance inside.

T1 does not need auxiliary commutation

circuit. As Vs tries to go negative with

T1,D turns on as it gets forward biased. Motor

current starts freewheeling through D.

ThusT1 becomes reverse biased and turns off.

.

vs

t

vM

iTh

t

t

t

2

2

iD

2


9/26

Single-phase naturally commutated phase

controlled converters and their properties

Note: isthe peak of

phase voltage

Average output

voltage

Ripple

frequency

Maximum

Drive rating

Half wave =

2

1+cos Few

hundred

watts

Semi-

converter=

1+cos 2 75 kW

Full

converter=

2

cos 2 75 kW

Double

converter=

cos 2 75 kW


10/26

Full-converter waveforms for

= 600

Note:

The current ripple is due to non-ideal filtering

by inductance in motor circuit.


11/26

Semi-converter waveforms for

= 600

Note:

The current ripple is due to non-ideal filtering

by inductance in motor circuit.

f h ll d


12/26

Waveforms in phase-controlled converter

and calculation of power factor

(a) Full and semi-converter circuits, (b)Full-converter waveforms

(c) Semi-converter waveforms. The current waveforms assume infinite filterinductance.

Power factor =Average input power

RMS input volt ampreres=

V1 cos1

OR

Power factor =

where

= Displacement or Fundamental Power Factor cos1DF= Distortion Factor= 1

1+2=1 ;

= Total Harmonic Distotion = 2 121 = 2=21

S1, D2

S1, D1S2, D1

S2, D2


13/26

Variation of DPF and PF with Firing Angle

DPF Power Factor

Full Converter cos 22 cosSemi Converter

cos2

2(1 + cos)1

E l I


14/26

Example IQuestion :A separately excited dc motor (10 kW, 240 V) is supplied from a fully

controlled single phase bridge. The power supply supplying the bridge is sinusoidal (240

V, 60 Hz). Ra = 0.42 ohm. = 2 V-s/rad. Assume Ia to be continuous dc. Calculate Ia,

speed, p.f, efficiency with firing delay angle = 0 degree and 20 degrees with constant loadtorque equal to rated torque. Assume constant flux operation. The machine draws rated

power at = 0.

Solution:

The output of a fully controlled single phase full bridge rectifier is given by

() =2

For = 0,

() =2

= 22240

cos00 = 216.1 .

() = + ; = ()Multiplying the above equation by Ia,

() = + 2 . The equation holds because for ripple free dc current =

= ( ).or

216.1 = 10000 + 0.422; 0 = = 10000W

or0.42

2 216.1 + 10000 = 0


15/26

Example I (2)

=216.1216.1240.4210000

0.84=

216.1172.9

0.84= 51.4, 463.1 A.

If = 51.4, = () = 216.1 51.4 0.42 = 194.51 V.If = 463.1, = () = 216.1 463.1 0.42 = 21.59 V.

Clearly the second case is unrealistic as is too high and too small, implying eitherfield failure or a very low speed as though the motor is starting up. Thus = 51.4 isthe only acceptable choice.

Speed of the motor =

= 194.51

2= 97.25 rad/s =97.25 60

2= 928.7 rpm.

Input power = = () = + 2 = 10000 + 51.42 0.42 = 11107.54 W.

Efficiency =

= 10000

11107.54= 90%

Power factor =

=

11107.54

240 51.4= 0.9


16/26

Example I (3)

Alternatively,

Using Fourier series (#) DF=1

=22

= 0.9.

DPF= cos 00 = 1.

Power factor = DPF*DF=1*0.9=0.9.

For = 20,

() =2

= 22240

cos 200 = 203.04 V.

= () = 203.04 51.4 0.42 = 181.45 (The current is the same as the

load torque is constant)

Speed =

= 181.45

2= 90.76 rad/s= 90.76 60

2= 866.7 rpm.

Output power= = = 181.45 51.4 = 9326.53W.Input power = = () = +

2 =81.45 51.4 + 51.42 0.42 =10440 W.


17/26

Example I (4)

Efficiency = =9326.53W

10440= 89.37%

Power factor =

=

10440

240 51.4= 0.85

Alternatively,

Using Fourier series DF(#) = 1= 22 = 0.9.

DPF= cos 200 = 0.94.

Power factor = DPF*DF=0.94*0.9=0.85.

# Note: The Fourier series of rectangular shaped current is given by

= 4=1,3,5.. sin. = 1 =

1

1 + 132

+1

52+

=22


18/26

Comparison of phase-controlled and

chopper drives

Variation of peak-to-average motor

current with speed. The separately

excited dc motor operates at a

constant torque ( 10 N-m) of rated

load.

Variation of rms-to-average motor

current with speed. The separately

excited dc motor operates at a

constant torque ( 10 N-m) of rated

load.


19/26

Three-phase naturally commutated phase

controlled converters and their properties

Note: isthe peak of

phasevoltage

Average output

voltage

Ripple

frequency

Maximum

Drive

rating

Half wave =

332 cos

3 7.5-37.5

kW

Semi-

converter=

332 1+cos

3 10- 110 kW

Full

converter=

33 cos

6 75-110 kW

Double

converter=

33 cos

6 150-1500kW


20/26

Characteristics of full converter

in CCM

Th1Th6

Th1Th2

Th3Th2

Th3Th4

Th5Th4

Th5Th6

Th1Th6

Motor vo tage an ine current


21/26

Motor vo tage an ine currentwith a 3 phase double converter working

both as an controlled rectifier and inverter

Note: (i) >1500 is not usually used in practice because an SCR needs

sufficient reverse voltage over sufficient time (tq) to commutate.(ii) The change of power factor due to change in (DPF=cos ) .

Motor Voltage Line Current


22/26

Example II

A 3 phase double converter bridge supplies power to a 560 V, 50 A, DC motor with Ra =

1.2 ohms. The voltage drop on the bridge thyristor is 20 V at rated armature current.

Supply to the three phase source is 415 V line to line. Find and for(i) Motoring operation with ( ) = 500 V, ( ) = 50A(ii) Regeneration with ( ) = 500 V, ( ) = 50A(iii) Plugging with ( ) = 500 V, ( ) = 50A and with a current limiting

resistor = 10 ohms.

E l II (2)


23/26

Example II (2)i) Quadrant 1 operation (forward motoring):

+

- DC DC

DC

vconv

vTh drop

Eb

Ra

va(avg)

+

-

ia(avg)

( ) = 500 V = + ( )

Therefore, = 440 V. =

or 500 V = 20V

or = 520 V

=33

V =

32

cos V =32 415

cos = 520 V

cos =520

560.45= 0.9278.

= cos1 0.9278 = 21.91.

E ample II (3)


24/26

Example II (3)ii) Quadrant 4 operation (Regenerating):

+

-

vconvC DC

DC

vTh drop

Eb

Ra

va(avg)

+

-

ia(avg)

vconv

In regeneration, > 90implying > ( ). Also by regenerating in this mode the motor

terminals do not have to be flipped.

=

= + = 560 V.

= + .

or 500 V = + 20V

or = 480 V

In this quadrant of operation, the converter voltage has to reverse its polarity for inverter

operation. Thus a negative sign has to be used for .

=33

V =

32

cos V =32 415

cos = 480 V

cos = 480

560.45= 0.8565.

= cos1

0.8565 = 148.92.

Example II (4)


25/26

Example II (4)

+

-

iii) Quadrant 4 operation (plugging):

During plugging, the armature voltage and back emf are aiding each other. Thus the equivalentcircuit is modified as shown above.

=

= + = 560 V.

= + + .

or 500 V = + 20V + 50 10 V;or = 20 V

or32

cos = 20 V

cos =20

560.45= 0.03569.

= cos1 0.03569 = 87.95.

vconv DC DC

DC

vTh drop

Eb

Rs Ra

va(avg)

+

-

vconv

ia(avg)


26/26

Additional References Other

than the textbook

P.C.Sen : Thyristor DC Drives, Wiley-Interscience, 1981.

Documents

Phase Controlled Drives