dc drives.pdf

7/25/2019 DC Drives.pdf

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1.1 IntroductionDirect current (DC) motors have variable characteristics and are used

extensively in variable-speed drives, Dc motors can provide a high startingtorque and it is also possible to obtain speed control over a wide range. Themethods of speed control are normally simpler and less expensive than thatof ac drives. Dc motors play a significant role in modern industrial drives.Both series and separately excited dc motors are normally used in variable-speed drives; however, series motors are traditionally employed for tractionapplications. Due to the commutators, dc motors are not suitable for veryhigh-speed applications and require more maintenance than those of ac motors.

Controlled rectifiers provide a variable dc output voltage from a fixedac voltage, whereas choppers can provide a variable dc voltage from a fixeddc voltage. Due to their ability to supply a continuously variable dc voltage,controlled rectifiers and dc choppers made a revolution in modern industrialcontrol equipment and variable-speed drives, with power levels ranging fromfractional horsepower to several megawatts.

DC drives can be classified, in general, into three types

1. Single-phase drives2. Three-phase drives3. Chopper drives

C H A P T E R

1Introduction and Single Phase Converterfed DC Motor Drive


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1.2 Basic Characteristics of DC MotorsThe equivalent circuit for a separately excited dc motor is shown in

Fig. 1-1. When a separately excited motor is excited by a field current of ifand an armature current of iaflows in the armature circuit, the motor developsa back emf and a torque to balance the load torque at a particular speed. The

field current, fi of a separately excited motor is independent of the armature

current, ia, and any change in the armature current has no effect in the fieldcurrent. The field current is normally much less than the armature current.

The equations describing the characteristics of a separately excitedmotor can be determined from Fig. 1.1. The instantaneous field current, if, isdescribed as

dt

fdi

f

L

f

i

f

R

f

V +=

.... 1.1

The instantaneous armature current can be found from

bedt

adiaLaiaRaV ++= .... 1.2

The torque developed by the motor is

affd iiKT = .... 1.3

The developed torque must be equal to the load torque

LTdtdJd BT ++= .... 1.4

Where,= motor speed, rad/s

B = viscous friction constant, N-m/rad/s

vK =voltage constant, V/A-rad/s

aK = Kv= torque constant

aL = armature circuit inductance, H

fL = field circuit inductance, H

Ra= armature circuit resistance,Rf= field circuit resistance,T

L= load torque, N-m


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Figure 1.1: Equivalent circuit of separately excited dc motor

Under steady-state conditions, the time derivatives in these equationsarc zero and the steady-state average quantities are

fff IRV = .... 1.5fvb IKE = .... 1.6

a a a b a a V f V R I E R I K l= + = + .... 1.7

aftd IIKT =

.... 1.8

Ld TBT += .... 1.9

The developed power isdd TP = .... 1.10

The relationship between the field current, If, and the back emf, Eg, isnonlinear due to magnetic saturation. The relationship, which is shown inFig. 1-2, is known as magnetization characteristic of the motor.


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Figure 1.2: Magnetization characteristics

From Eq. (1-3), the speed of a separately excited motor can be foundfrom

f

fv

aaa

fv

aaa

R

VK

IRV

IK

IRV =

=

.... 1.11We can notice from Eq.1.11 that the motor speed can be varied by

(1) Controlling the armature voltage, Va, known as voltage control;

(2) Controlling the field current, If, known as field control; or

(3) Torque demand, which corresponds to an armature current, la, for a fixedfield current l

fthe speed, which corresponds to the rated armature voltage,

rated field current and rated armature current, is known as the base speed.

In practice, for a motor speed less than the base speed, the armaturecurrent and field currents are maintained constant to meet the torque demand,and the armature voltage, V

a, is varied to control the speed. For motor speed

higher than the base speed, the armature voltage is maintained at the rated

value and the field current is varied to control the speed. The power developedby the motor (= torque x speed) remains constant. Figure 1.3 shows thecharacteristics of torque, power, armature current, and field current againstthe speed.


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Figure 1.3: Characteristics of separately excited dc motors

The field of a dc motor may be connected in series with the armature

circuit as shown in Fig. 1.4, and this type of motor is called a series motor.The field circuit is designed to carry the armature current. The steady-state average quantities are

fvg IKE = .... 1.12

gaaa EIRV += .... 1.13

avaaa IKIRV += .... 1.14

aftd IIKT = .... 1.15

Ld

TBT +=

The speed of a series motor can be determined from equation 1-13.


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Figure 1.4:Equivalent circuit dc series motor

The speed can be varied by controlling the (1) armature voltage, Va,

or (2) armature current, which is a measure of the torque demand. Equation

(1-10) indicates that a series motor can provide a high torque, especiallyat starting; and for this reason, series motors are commonly used intraction applications.

For a speed up to the base speed, the armature voltage is variedand the torque is maintained constant. Once the rated armature voltageis applied, the speed-torque relationship follows the natural characteristicof the motor and the power (= torque x speed) remains constant. As thetorque demand is reduced, the speed increases. At a very light load, the

speed could be very high and it is not advisable to run a dc series motoron no-load. Figure 1.5 shows the characteristics of dc series motors.


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Figure 1.5:Characteristics of dc series motors

1.3 Single Phase DrivesIf the armature circuit of a dc motor is connected to the output of a single-

phase controlled rectifier, the armature voltage can be varied by varying the delayangle of the converter,. The forced-commutated ac-dc converters can also be

used to improve the power factor and to reduce the harmonics. The basic circuitagreement for a single-phase converter-fed separately excited motor is shown inFig. 1.6. At a low delay angle, the armature current may be discontinuous, andthis would increase the losses in the motor. A smoothing inductor, Lm, is normallyconnected in series with the armature circuit to reduce the ripple current to anacceptable magnitude. A converter is also applied in the field circuit to control thefield current by varying the delay angle . Depending on the type of single-phaseconverters, single-phase drives may be subdivided into;

1. Single-phase half-wave-converter drives2. Single-phase scmiconverter drives3. Single-phase full-converter drives4. Single-phase dual-converter drives


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Figure 1.6: Basic circuit arrangement of a single-phase dc drive

1.2.1 Single-Phase Half-Wave-Converter Drives

A single-phase half-wave converter feds a dc motor as shown in Fig.1.7a. The armature current is always discontinuous unless a very large inductoris connected in the armature circuit. A freewheeling diode is always requiredfor a dc motor load and it is a one-quadrant drive, as shown in Fig. 1-7b. Theapplications of this drive are limited to the 1-kW power level. Figure 1-7cshows the waveforms for a highly inductive load. The converter in the fieldcircuit should be a semi-converter. A half-wave converter in the field circuitwould increase the magnetic losses of the motor due to high ripple content on

the field excitation.With a single-phase half-wave converter in the armature circuit, the

average armature voltage is given by

.... 1.16

Where Vmis the peak voltage of the ac supply. With a semiconverter in the

field circuit. the average value of field voltage is given by

.... 1.17


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Figure 1.7: Single-phase half-wave converter drive.1.2.2 Single-Phase Semi converter Drives

A single-phase semi converter feeds the armature circuit as shown inFig. l-8a. It is a one-quadrant drive as shown in Fig. 1-8b and is limited toapplications up to 15 kW. The converter in the field circuit should also be asemi converter. The current waveforms for a highly inductive load are shownin Fig. 1.8c.

With a single-phase semi converter in the armature circuit, the averagearmature voltage is expressed as

.... 1.18


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Where Vmis the peak voltage of the ac supply. With a semiconverter in the

field circuit. the average value of field voltage is given by

....1.19

Figure 1.8: Single-phase semi converter drive

1.2.3 Single-Phase Full-Converter Drives

The armature voltage is varied by a single-phase full-wave converter

as shown in Fig. 1.9a. 1t is a two-quadrant drive as shown in Fig. 1.9b and islimited to applications up to 15 kW. During regeneration for reversing thedirection of power flow, the back emf of the motor is reversed by reversingthe field excitation.


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Figure 1.9: Single-phase full-converter drive

The converter in the field circuit should be a full-wave converter toreverse the polarity of field current. The current waveforms for a highlyinductive load are shown in Fig. l.9c for powering action. With a single-phase full-wave converter in the armature circuit, the average armature voltage

is given by2

cosa a

vV a

= for 0

a .... 1.20


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With a single-phase full-converter in the field circuit, the average valueof field voltage is equal to

2

cos

m

f f

V

V =

for 10 .... 1.21

1.2.4 Single-Phase Dual-Converter DrivesTwo single-phase full-wave converters are connected as shown in Fig. 1.10.Either converter 1 operates to supply a positive armature voltage, Va, orconverter 2 operates to supply a negative armature voltage, - Va. It is a four-quadrant drive and permits four modes of operation: forward motoring, forwardbraking (regeneration), reverse motoring, and reverse braking (regeneration).It is limited to applications up to 15 kW. The polarity of field current is reversedduring the forward and reverse regenerations. The field converter should be a

full-wave one to allow reversing the direction of field current.If converter 1 operating with a delay angle of gives armature voltage as

.... 1.22

If converter 2 operates with a delay angle of gives armature voltage as

.... 1.23

where With a full converter in the field circuit, gives the field voltage as

.... 1.24

Figure 1.10: Single-phase dual-converter drive


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SOLVED EXAMPLES1.1. The full converter is connected to a 120-V, 60-Hz supply. The load

current

aI

is continuous and its ripple content is negligible. The turnsratio of the transformer is unity. (a) Express the input current in a Fou-rier series; determine the HF of the input current, DF, and input PF. (b)

If the delay angle is

/ 3 =

, calculate

, , , ,dc n rmsV V V HF DF

and

.PF

Solution:a. The waveform for input current and the instantaneous input current can beexpressed in a Fourier series as

( ) ( )0 1,2,.... cos sins n nni t a a n t b n t

=

= + +

where

( ) ( ) ( ) ( )2 / 2

01 1 0

2 2

a a a

s a aa a a

a i t d t I d wt I d wt

+ +

+

= = =

( ) ( )2 /1

cosa

n sa

a i t n td t

=

( ) ( )/ 21

cos cosa a

n a aa a

a I n td t I n td t

+

+

=

4

sinaI

nn

= for n=1,3,5,.....

= 0 for n=2,4,..

( ) ( )21

sina

na

b i t n td t

+

=

( ) ( )21 sin sin

aa

a a aa

I n td t I I n td t

++

+

=

4

cosaI

n

n

= for n=1,3,5,.....

0= for n=2,4,.....


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Because

0 0,a =

the input current can be written as

( ) ( )1,3,5,... 2 sins n nni t I n t

=

= +

where1

tan nnn

anb

= =

and

n

is the displacement angle of the nth harmonic current. The rms value

of the nth harmonic input current is

( )

1/22 21 4 2 2

2 2a a

sn n n

I I

I a b nn = + = =

and the rms value of the fundamental current is

1 2 2 a

sII

=

The rms value of the input current can be calculated as

1/2

2

1,3,5,....s sn

n

I I

=

=

sI can also be determined directly from

( )

1/2

22

2

a

s a aaI I d t I

+ = =

the HF is found as

1/22

1

1 0.483s

s

IHF

I

= =

or 48.3%

The DF is

( )1cos cosDF = =

The PF is found as

( )1 2 2

cos coss

s

I

PF I = =


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b. / 3 =

2

cos 54.02

m

dc

V

V V= =

and 0.5n

V pu=

1202

m

rms s

V

V V V= = =

1 2 2 0.90032a

s a

II I

= =

and

s aI I=

1/22

11 0.4834s

s

IHFI

= =

and 48.34%

1 =

and

( )cos cos 0.53DF

= = =

( ) ( )1 cos 0.45ss

IPF lagging

I= =

1.2. A 15-hp 220-V 2000-rpm separately excited dc motor controls a loadrequiring a torque ofT

L= 45 N-m at a speed of 1200 rpm. The field circuit

resistance isRf= 147, the armature circuit resistance isR

a= 0.25, and

the voltage constant of the motor isKv= 0.7032 V/A-rad/s. The field voltage

is Vf= 220 V. The viscous friction and no-load losses are negligible. Thearmature current may he assumed continuous and ripple free. Determinethe (a) back emf, Eg; (b) required armature voltage. Va and (c) ratedarmature current of the motorSolution:

147f

R =

0.25aR = 0.7032 / / c tK K V A rad s= =

Vf= 220 V,

Td= T

L= 45 N-m,


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(a) From Eq. (1-4),

Substitute respective values in above equation

From Eq. (1-2),


(b) From Eq. (1-3),


(c) 1Hp is equal to 746 W



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2.1 Three-Phase DrivesThe armature circuit is connected to the output of a three-phase

controlled rectifier or a forced -commutated three-phase ac-dc converter.Three-phase drives are used for high-power applications up to megawattspower level. The ripple frequency of the armature voltage is higher than thatof single-phase drives and it requires less inductance in the armature circuit

to reduce the armature ripple current. The armature current is mostlycontinuous, and therefore the motor performance is better compared to thatof single-phase drives. Similar to the single-phase drives, three-phase drivesmay also be subdivided into:

1. Three-phase half-wave-converter drives2. Three-phase semi converter drives3. Three-phase full-converter drives4. Three-phase dual-converter drives

2.2 Three-Phase Half-Wave-Converter DrivesThree-phase converters provide higher average output voltage, and in

addition the frequency of the ripples on the output voltage is higher comparedto that of single-phase converters. As a result, the filtering requirement forsmoothing out the load current is simple. For these reasons, three-phaseconverters are used extensively in high-power variable-speed drives.

Where Vm is the peak voltage of a Y-connected three-phase ac supply.

.... 2.1

C H A P T E R2Three Phase AC-DC Converter fed DCMotor Drive


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With a three-phase full converter in the field circuit the average fieldvoltage is given by

..... 2.6

2.5 Three-Phase Dual-Converter DrivesTwo three-phase full-wave converters are connected as in Fig. 1.10a.

Either convener 1 operates to supply a positive armature voltage, Va, orconverter 2 operates to supply a negative armature voltage, -Va. It is a four-quadrant drive and is limited to applications up to 1500 kW. The polarity offield current is reversed during forward and reverse regenerations. Similar tosingle-phase drives, the field converter should be full-wave to allow reversing

the direction of field current. If converter 1 operates with a delay angle of

1a

the average armature voltage is given by

.... 2.7

If converter 2 operates with a delay angle of 2a gives armature voltage isgiven by

.... 2.8

With a three-phase full converter in the field circuit, the average field voltageis given by

.... 2.9

Principle of three-phase half-wave converters:Three-phase converters provide higher average output voltage, and inadition the frequency of the ripples on the output voltage is higher comparedwith that of single phase converters. As a result, the filtering requirements


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for smoothing out the load current and load voltage sre simpler. For thesereasons, three-phase converters are used extensively in high-power variable-spped drives. Three single phase half-wave converters in figure can be con-nected to form a three-phase half-wave converter, as shown in figure.

Figure 2.8: Three-phase half-wave converter


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The rms output voltage is found from

( )25 /6

2 2

/6

3sin

2rms mV V td t

1

+

+

=

1

233 cos2

6mV

1= +

For a resistive load and

/ 6

( )/63 3

sin 1 cos2 2 6m

dc m

V

V V td t

+

= = + +

11 cos

63

dcn

dm

VV

V

= = + +

( )

1

2

2 2/63 sin2rms m

V V td t

+ =

rmsV

1

2

5 13 sin 224 8 3mV

= + +

4

.... 2.12

2.5.1 Gating sequence:The gating sequence is as follows:

1. Generate a pulse signal at the positive zero crossing of the phase voltage

anV

. Delay the pulse by the desired angle

/ 6 +

and apply to the gate and

cathrode terminals of

1T

through a gate-isolating circuit.

2. Generate two more pulses of delay angle

5 / 6 +

and

9 / 6 +

for

gating

2T

and

3T

respectively through gate isolating circuits.

2.6 Three phase full converters:Three-phase converters are extensively used in industrial applicatios

upto the 120-kW level, where a two-quadrant operation is required. Figure


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The maximum average output voltage for delay angle 0 = , is

3 3 m

dm

V

V =

.... 2.20

and the normalized average output voltage is

cosdcn

dm

VVV

= =

The rms value of the output voltage is found from

( )

1

2/22 2

/6

33 sin

6rms mV V t d t

+

+

= +

rmsV

1

2

1 3 33 cos22 4m

V

= +

.... 2.21

Figure shows the waveforms for

/ 3, =

for

/ 3, >

the instanta-

neous output voltage

0

has a negative part. Because the current through

thyristors cannot be negative, the load voltage cannot be negative, and thefull converter behaves as the semi converter.

Gating sequence. The gating sequence is as follows

1. Generate a pulse signal at the positive zero crossing of the phase voltage

anV

. Delay the pulse by the desired angle

/ 6 +

and apply it to the gate

and cathodde terminals of

1T

through a gate-isolating circuit.

2. Generate five more pulses of delayed by

/ 6

from each other for gating

2T

and

3 4,T T

respectively, through gate-isolating circuits.

2.7 Three-phase dual converters:In many variable-speed drives, the four-quadrant operation is general

required and three-phase dual converters are extensively used in applica-tions upto the 2000-01 level. Figure shows three-phase dual converters wheretwo three-phase converters are connected back to back. We have seen insection 104 that due to the instantaneous voltage differences between the


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If the line-to-neutral voltage are defined as

sinan mV t =

.... 2.22

2

sin 3bn mV t

=

.... 2.23

2

sin 3cn mV t

= +

.... 2.24

The corresponding line-to-line voltage are

V 3 sin 6ab an bn mV V t

= = +

.... 2.25

V 3 sin 2bc bn cn mV V t

= =

.... 2.26

5

V 3 sin 6ca cn an mV V t

= = +

.... 2.27

If

01V

and

02V

are the output voltages of converters 1 and 2 respec-

tively instantaneous voltage across the inductor during interval

( ) ( )1 1/ 6 / 2t + +is

01 02rms ab bc = + =

3 sin sin6 2mV t t

= +

rmsV

3 cos 6mV t

=

.... 2.28

The circulating current can be found from

( ) ( )1 1/6 /61 1

3 cos

t t

r r mr ri t d t V tL L

+ +

= =

( )i t

1

3

sin sin6m

r

V

tL

=

.... 2.29


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the switching devices. More current flows from the ac to dc side, and the

capacitor voltage is recovered. In the inverter mode of operation

0I

becomes

negative and the capacitor CD is overcharged. The error signal demands thecontrol to discharge the capacitor and return power to the ac mains.

The PWM can control both the active power and reactive power. Thusthis type of rectifier can be used for PF correction. The ac current wave formscan also be maintained almost sinusoidal, reducing harmonic contaminationto the mains supply. The PWM turns on and off the switches in a preestab-lished form, usually a sinusoidal waveform of voltage or current. An ex-

ample the modulation of one phase is shown in figure with amplitude of

modV

for the modulating signal.

Figure 2.11: Basic topologies for force-commuted PWM rectifiers (a)current source rectifier; (b) voltage source rectifier.


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Twelve-pulse converters:A three-phase bridge gives a six-pulse output voltage for high-power

applications such as high-voltage dc transmission and dc motor drives, a 12-pulse output is generally required to reduce the output ripples and to increasethe ripple frequencies. Two & pulse bridges can be combined either in series

or in parallel to produce an effective 12-pulse output. Two configurations areshown in figure. A 300phase shift between secondary windings can be ac-

complished by connecting one secondary in y adn the other in delta

( )

.

Configurations for 12-pulse output


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3. Modeling of AC-DC convertor fed DC drive components & design ofcontroller3.1 DC Motor and Load

The dc machine contains an inner loop due to the induced emf it is not

physically seen; it is magnetically coupled. The inner current loop will crossthis back-emf loop, creating a complexity in the development of the model.

Figure 3.1: DC motor and current-control loop

It is shown in Fig. 3.1; the interactions of these loops can be decoupled

by suitably redrawing the block diagram. The development of such a block

diagram for the dc machine is shown in Fig.3.2, step by step. The load is

assumed to be proportional to speed and is given as

I l mT B =

3.1

C H A P T E R3Modeling of AC-DC Converter fed DCdrive components & design of controller


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Figure 3.2: Step by step reduction of Block diagram

2 2

11 1 1 1, 2 2 4

t a t a b

a a

B R B R KT T J L J L JL

+ = + +

1 2

t

b a t

B

K K R B=

+


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3.6 Design of ControllersThe overall closed-loop system is shown in Figure 3.3, It is seen that

the current loop does not contain the inner induced-emf loop. The design of

control loops starts from the innermost (fastest) loop and proceeds to the

slowest loop, which in this case is the outer speed loop. The reason to proceedfrom the inner to the outer loop in the design process is that the gain and time

constants of only one controller at a lime are solved, instead of solving for the

gain and lime constants of all the controllers simultaneously. Not only is that

logical; it also has a practical implication. Note that every motor drive need

not be speed-controlled but may be torque-controlled, such as for a traction

application. In that case, the current loop is essential and exists regardless of

whether the speed loop is going to be closed. Additionally, the performance

of the outer loop is dependent on the inner loop; therefore, the tuning of theinner loop has to precede the design and tuning of the outer loop. That way,

the dynamics of the inner loop can be simplified and the impact of the outer

loop on its performance could be minimized. The design of the current and

speed controllers is considered in this section.

Figure 3.3: Block diagram of the motor drive

3.6.1 Current Controller

( ) ( )( )

( )( )( )1

1 2

1 1

1 1 1c mc r c

i

c r

sT sT K K K HGH s

T s sT sT sT

+ + =

+ + + 3.10


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Tells us that K is approximated as

2

2 2I I

I r t

T T

K T T T

3.24

By equaling (3.13) to (3.24), the current-controller gain is evaluated as

1 1

2I c

c

r I r c m

T TK

T K K H T

=

3.25

3.6.2 First-Order Approximation of Inner Current LoopTo design the speed loop, the second-order model of the current loop is

replaced with an approximate first-order model. This helps to reduce the order

of the overall speed-loop gain function.

Figure 3.5: Simplified Current control loop

The current loop is approximated by adding the time delay in the

converter block to T; of the motor: because of the cancellation of one motor

pole by a zero of the current controller, the resulting current limp can he

shown in Figure 3.5. The transfer function of the current and its commanded

value is


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The step response is given by

With a rise time of 3.1T4a maximum overshoot of 43.4%, and a

settling time 16.5T4.

iii. Since the overshoot is high, it can be reduced by compensatingfor its cause, i.e., the zero of a pole in the speed command path,

as shown in Figure 3.7. The resulting transfer Function of thespeed to its command is

( )

( )* 2 2 3 34 4 4

1 1

1 4 8 8m

r

s

s H T s T s T s

= + + +

3.46

( )4 42 4

4

1 2 31 sin 43

t t

T T

r

t

t e eH T

= +

3.47

Whose step response is

Figure 3.7: Smoothing of the overshoot via a compensator

With a rise time of 7.6T4, a maximum overshoot of 8.1% and a settling

time of 13.3T4. Even though the rise time has increased, the overshoot has

been reduced to approximately 20% of its previous value, and the settling

time come down by 19%.iv. The poles of the closed-loop transfer function are

4 4 4

1 1 3

;2 4 4s jT T T=

3.48

The real parts of the poles are negative, and there are no repeated polesat the origin, so the system is asymptotically stable. Hence, in thesymmetric optimum design, the system stability is guaranteed, and there


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is no need to check for it in the design process. Whether this is true forthe original system without approximation will be explored in thefollowing example.

v. Symmetric optimum eliminates the effects due to the disturbance veryrapidly compared to other optimum techniques employed in practical

systems, such as linear or modulus optimum. This approach indicatesone of the possible methods to synthesize the speed controller. Thai the

judicious choice of approximation is based on the physical constantsof the motor, on the converter and transducer gains, and on time delaysis to he emphasized hurt.

4

3

4

1

1

i

fi

l r

fi

T T T

TT

K

T TT T

K

= +

= ++

+= +

+

3.49

That the speed-loop transfer function is expressed in terms of T4 is

significant in That it clearly links the dynamic performance to the speed-feedback and current-loop time constants, that a faster current loop with asmaller speed-filler time constant accelerates the speed response is evidentfrom this. Expressing T

4in terms of the motor, the converter and transducer

gains, and the time delays by using expressions (3.28) and (3.34) yields

Since Kfi>> 1, T

4is found approximately after substituting for K

fifrom

equation (3.29) in terms of gains and time delays as

( ) 24

1l r

m l c r c

T T TT T

T K K K H

+ + 3.50

This clearly shows the influence of the subsystem parameters on the

system dynamics. A clear understanding of this would help the proper selectionof the subsystems to obtain the required dynamic performance of the speed-

controlled motor-drive system. Further, this derivation demonstrates that the

system behavior to a large degree depends on the subsystem parameters rather


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loop. The outer induced-emf feedback loop enforces a constant induced emffor speeds higher than the base speed. This amounts to

If the induced emf en is kept at rated value, say 1 p.u. then the field fluxis inversely proportional to the rotor speed. This condition also enables

constant-air gap-power operation.

3.8 Four-Quadrant DC Motor DriveA four-quadrant dc motor drive has a set of dual three-phase converters

for the power stage. Its control is very similar to that of the two-quadrant dc

motor drive. Converters have to be energized depending on the quadrant of

operation. Converters 1 and 2 are for forward and reverse directions of rotation

of the motor, respectively. The changeover from one converter to another is

safely handled by monitoring speed, current-command, and zero-crossingcurrent signals. These signals form the inputs to the selector block, which

assigns the pulse-control of signals to the appropriate converter. The converters

share the same current and speed loops. A control schematic of the four-

quadrant dc motor drive is shown in Figure 3.9.

Figure 3.9: A four-quadrant DC motor drive

The selector block will switch the converters over only when the currentin the outgoing convener has come to zero. Apart from that, some otherconditions have lobe satisfied to transfer the control of converters. Assuming


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that a positive speed command is required for forward running and a negativespeed command is required for reverse running, the crossover control forconverters is discussed. The machine is running at rated speed and the speedcommand is changed lo reverse rated speed. The current command becomesnegative, indicating that the machine has to become regenerative in order to

decelerate the motor in the forward direction. Forward regeneration is possibleonly in quadrant IV, and the converter to provide this quadrant of Operationis converter 2. Before the current is reversed, it has to go through zero. Thai isachieved by increasing the triggering angle of converter 1. When the currentis zero, a certain dead time is given to enable the thyristors in converter 1 torecover reverse blocking capability. After this interval, converter 2 is enabled.At this time, the armature, current has been forced to zero, but the speed isstill positive. The triggering angle of converter 2 is set such that its outputvoltage equals and opposes the induced emf. Then, slowly decreasing the

triggering angle increases the armature current in the opposite direction. By avarying of the triggering angle, the armature current is fully reversed andthen is maintained at the reference level. When the motor reaches zero speed,the operation of converter 2 is continued by bringing it into rectification mode.i.e. with less than900, that will accelerate the motor in the reverse directionuntil it matches the speed reference.

The function of the selector block is to determine which converter hasto be operating. In the previous discussion, as soon as the current command

goes negative, the selector block will transfer the control from converter 1 toconverter 2 with proper initial triggering angle. If the triggering angle of

converter 1 is

1

then the initial triggering angle 2 of converter 2 is ( )2180 to match the output voltage of converter 1. If circulating current is not allowedbetween the converters, zero crossing of the armature current is required totransfer the control from one converter to further. The actual rotor speed isrequired to determine the quadrant of operation. Based on the rotor speed, thearmature current, and its command, the selector block identifies the converter

for control and operation. The method dis-cussed here with Figure 3.9 doesnot allow circulating current between converters 1 and 2. This type of controlhas a drawback: slower current transfer from one to the other bridge, due tothe dead time: but it had the advantage that no additional passive componentin the form of an interphase reactor is needed to limit the circulating current.


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The other type of crossover control uses a circulating current between thethyristor bridges. It has the advantage of faster current crossover and highdynamic response, but it has the disadvantages of requiring additionalinterphase reactors and losses associated with (hem during the operation.

3.9 Simulation of the One-Quadrant DC Motor DriveThe equations for various subsystems are derived and then assembled

for computer simulation in this section. Key results are discussed. The

simulation for either a two-or four-quadrant dc motor drive is very similar to

the present development. In the present one-quadrant speed-controlled motor-

drive simulation, it is assumed that the field current is constant in the constant

torque mode and is varied through a three-phase controlled rectifier for the

field-weakening mode of operation to provide constant power over a wide

speed range.

3.9.1 The Motor Equationsa

a a a f m

diR i L Mia dt

V = + + 3.52

The motor equations, including that of a simple load modeling, are givenbelow

f

f f f

diR i L

f dtV = + 3.53

mf a l m

dMi i T J B

dt

= + 3.54

These equations can be rearranged in the following form lo facilitatetheir solution by numerical integration

a a aa f m

a a a

di R M V i i

dt L L L= + .... 3.55

f f f

ff

di R V

idt L Lf = +

.... 3.56


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m lf a m

d M B T i i

dt J J J

= .... 3.57

1 ax i=

.... 3.58

2 fx i=

.... 3.59

3 mx =

.... 3.60

Choose ia, i

r. and

m

as state variables and denote them as

From the motor equations and the above set of definitions, the equations

of the motor are written as

.

1 1 2 3a a

a a a

R M Vx x x xL L L= +

3.61

.

2 2f f

f

R Vx x

L Lf= + 3.62

.

3 1 2 3l

M B Tx x x x

J J J= 3.63

3.9.2 Fitter in the Speed-Feedback LoopFigure 3.10 shows the speed-feedback filler. The transfer function of

the filter and tacho generator can be represented as

Figure 3.10: Speed-feedback filter


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Figure 3.12: State diagram of Speed-controller

( )* * 4 5c ps r isT K x K x= + 3.67

The torque command signal is derived as*

max0 eT T + 3.68

In order to maintain the drive system in the safe operating region, thetorque reference is limited to allowable maximum limits determined by theconverter and motor peak capabilities. In this case, let that be +Tmax. Thistorque reference limit is integrated into the simulation as

3.9.4 Current-Reference GeneratorThe current reference is derived from the torque reference by using the

relationship

**

*4 5

2 2 2

. . .

ea

f

ps psr is

Ti

Mi

K K x K x

M x M x M x

=

= + 3.69


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Figure 3.12:Flowchart for the simulation of a single-quadrant phase-controlled dc motor drive


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3.10.1 Power controlTransistor Q

1and diode D

1operate. When Q

1is turned on, the supply

voltage Vsis connected to the motor terminals. When Q

1is turned off, the

armature current which flows through the freewheeling diode D1, decays.

3.10.2 Regenerative controlTransistor Q

2and diode D

2operate. When Q

2is turned on, the series

motor acts as a generator and the armature current rises. When Q2is turned

off, the motor, acting as a generator, returns energy to the supply through the

regenerative diode D2.

In industrial applications, four-quadrant operation, as shown in Figure

10.14a is required. A transistorized four-quadrant drive is shown in Figure10.14b.

Figure 3.14: Four-quadrant transistorized chopper drive

3.10.3 Forward power controlTransistors Q

1

and Q2

operate. Transistors Q3

and Q4

are off. When Qland Q

2are turned on together, the supply voltage appears across the motor

terminals and the armature current rises. When Qlis turned off and Q

2is still

turned on, the armature current decays through Q2and D

4.


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4.1 IntroductionWhenever the source is a constant-voltage dc, such as a battery or diode-

bridge rectified ac supply, a different type of converter is required to convertthe fixed voltage into a variable-voltage/variable-current source for the speedcontrol of the dc motor drive. The variable dc voltage is controlled by choppingthe input voltage by varying the on- and off-times of a converter, and the type

of converter capable of such a function is known as a chopper.

4.2 Principle of Operation Of The ChopperA schematic diagram of the chopper is shown in Figure 4.1. The

control voltage to its gate is vc. The chopper is on for a lime t

on, and its off

time is toff

, Its frequency of operation is

.......4.1

And its duty cycle is defined as.......4.2

The output voltage across the load during the on-time of the switch isequal to the difference between the source voltage Vs and the voltage dropacross the power switch. Assuming that the switch is ideal with zero voltagedrop. the average output voltage V

dcis given as

........4.3

C H A P T E R4DC-DC Converter Drive fed DC Motor Drive


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Where Vs is the source voltage.

Figure 4.1: chopper schematic and its waveforms

Varying the duty cycle changes the output voltage. Note that the outputvoltage follows the control voltage as shown in Fig 4.1 signifying that the

chopper is a voltage amplifier.The duty cycle d can be changed in two ways:(i) By keeping the switching/chopping frequency constant and

varying the on- time, to get a changing duty cycle.(ii) Keeping the on-time constant and varying the chopping frequency

to obtain various values of the duty cycle.

A constant switching frequency has the advantages of predeterminedswitching losses of the chopper, enabling optimal design of the cooling forthe power circuit, and predetermined harmonic contents, leading to an optimalinput filter. Both of these advantages are lost by varying the switchingfrequency of the chopper hence, this technique for chopper control is notprevalent in practice.


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4.3 Four-Quadrant Chopper CircuitA four-quadrant chopper with transistor switches is shown in Figure

4.2. Each transistor has a freewheeling diode across it and a snubber circuit tolimit the rate of rise of the voltage. The snubber circuit is not shown in thefigure. The load consists of a resistance, an inductance and an induced emf.

The source is dc, and a capacitor is connected across it to maintain a constantvoltage. The base drive circuits of the transistors are isolated and theyreproduce and amplify the control signals at the output. For the sake ofsimplicity it is assumed that the switches are ideal and hence, the base drivesignals can be used to draw the load voltage.

4.3.1 First-Quadrant OperationFirst-quadrant operation corresponds to a positive output voltage and

current. This is obtained by triggering T1 and T2 together, as is shown inFigure 6.3: then the load voltage is equal to the source voltage. To obtain zeroload voltage, eitherT

1orT

2can be turned off.

Assume that T1is turned off; then the current will decrease in the power

switch and inductance. As the current tries to decrease in the inductance itwill have a voltage induced across it in proportion to the rate of fall of currentwith a polarity opposite to the load-induced emf, thus forward-biasing diodeD

4. D

4provides the path for armature current continuity during this time.

Because of this, the circuit configuration changes as shown in Fig 4.4. Theload is short-circuited reducing its voltage to zero.

The current and voltage waveforms for continuous and discontinuouscurrent conduction are shown in Figure 4.5. Note that in the discontinuouscurrent conduction mode the induced emf of the load appears across the loadwhen the current is zero.

The load voltage, therefore, is a stepped waveform. The operationdiscussed here corresponds to motoring in the clockwise direction, or forwardmotoring. It can be observed that the average output voltage will vary from oto Vs the duty cycle can be varied only from 0 to 1.


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might not be considerable or cost-effective .When the current in the load isdecreasing, T

2is turned on. This allows the short-circuiting of the load through

T2and T

4, resulting in an

increase in the load current. Turning off T

2results in

a pulse of current flowing into thesource via D

3and D

4.

Figure 4.6:Second-quadrant operation, with negative load voltage andpositive current

Figure 4.7:Second-quadrant operation of the chopper

This operation allows the priming up of the current and a building upof the energy in the inductor from the loads emf. thus enabling the transfer ofenergy from the load to the source. Note that it is possible to transfer energy


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Figure 4.11:Front-end of the chopper circuit

Figure 4.12: Chopper with regeneration capability

It is usual to set V to be 15 or 20% of V. In such a case, there is a need

for a step-up transformer in the path of the phase-controlled converter, tomatch the dc link voltage Vs. The phase converter is disabled when V

s is

slightly greater than 1.35 V, where V is the line-to-line rms voltage ,to preventenergy flow from source to dc link and from dc link to source via the phase-controlled converter.


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Figure 4.14: Applied voltage and armature current in choppercontrolled of DC motor drive

4.8.3 Continuous and discontinuous Current ConductionThe relevant electrical equations of the motor for ON and OFF times are asfollows:

..... 4.13

..... 4.14

The solution of equation (4.13) is

...... 4.15


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Where

....... 4.16

Similarly, the solution of equation (4.14) is

...... 4.17

And

....... 4.18

From Figure 4.14, it is seen that

....... 4.19

By using this boundary condition,

0aI

and 1aI are evaluated as

......... 4.20

........ 4.21

Having evaluated Ia0

and la1

.you can use equations (4.15) and (4.17) toevaluate the instantaneous armature current in steady state.

The limiting or minimum value of duty cycle for continuous current isevaluated by equating I

a0to zero. This value is termed the critical duty cycle,

dcand is given by

........ 4.22

Duty cycles lower than dcwill produce discontinuous current in themotor. Note that this critical value is dependent on the ratio between choppingtime period and armature time constant and also on the ratio between inducedemf and source voltage.


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4.8.4 Discontinuous Current ConductionThe relevant equations for discontinuous current-conduction mode are

obvious from Fig 4.14

...... 4.23

...... 4.24

With

....... 4.25

...... 4.26

Hence

...... 4.27

...... 4.28

This equation is equal to zero, by the constraint given in equation(4.25), and, from that, t

xis evaluated as

...... 4.29

The solution for the armature current in three time segments is

...... 4.30

.... 4.31

...... 4.32

The steady-state performance is calculated by using equations (4.30)to (4.32).


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The minimum voltage rating for both the devices is,

...... 4.35

Whenever regeneration occurs in the motor drive, the current in thefreewheeling diodes would change, and, depending on the frequency and

duration of regeneration, the diode currents have to be recalculated. To optimizethe chopper rating in comparison to the motor and load demands, the operatingconditions have to be known beforehand. The extreme operating conditionswould then prevail on the design and hence on the final rating of the chopper.

4.10 Pulsating TorquesThe armature current has ac components. These ac components or

harmonics pro-duce corresponding pulsating torques. The average of the

harmonic torques is zero and they do not contribute to useful torque andpower. Some high-performance applications, such as machine tools androbots, require the pulsating torque to be a minimum so as not to degradethe process and the products. In that case, an estimation of the pulsatingtorques is in order.

Figure 4.17: Fourth-quadrant operation of DC drive


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...... 4.51

And the fundamental-harmonic armature current is given by

..... 4.52

The fundamental armature current can be alternately expressed interms of duty cycle as

...... 4.53

When the duty cycle is 50%, the fundamental armature current, and

hence the pulsating torque, is maximum. For a duty cycle of 100%, thereare no pulsating-torque components. The instantaneous electromagnetictorque is the sum of the dc and harmonic torques, written as

....... 4.54

Where h is the harmonic order. The harmonic torques, Teh

, do notcontribute to the load: their averages are zero,

The fundamental pulsating torque is expressed as a fraction ofaverage torque, to examine the impact of duty cycle on the pulsating torque.It is facilitated by the following development.

...... 4.55


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Where 1 is the fundamental power-factor angle, which can beextracted from

..... 4.56

And

...... 4.57

A set of normalized curves is shown in Fig 4.18 to indicate theinfluence of duty cycle and of the ratio between the induced emf and thesource voltage on the magnitude of the pulsating torque.

By noting that the fundamental is the predominant component amongthe ac components, the rms value of the armature current is approximated

as...... 4.58

Where I1is the fundamental rms current, given by

...... 4.59

Figure 4.18: Normalized fundamental torque pulsation vs.duty cycle as a function ol the ratio between induced emf and source

voltage


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The armature resistive losses are

...... 4.60

This indicates that the thermal capability of the motor is degraded

by the additional losses produced by the harmonic currents.

Minimization of the dominant-harmonic torque is of importance inmany applications, mainly in positioning of machine tool drives the keyto mitigation of the harmonic torque is revealed by the expression forharmonic current given by equation (4.53). Given a fixed voltage source,there are only two variables that could be utilized to reduce harmoniccurrent: the chopping frequency, and the machine inductance. Note that d

cannot be used it is a variable dependent on the speed and load. Thechopping frequency is limited by the selection of power device, by itsswitching losses, and by other factors, such as electromagneticcompatibility. The advantage of varying the carrier frequency is that it ismachine-independent, and hence the solution is contained within thechopper. This may not be feasible, as in the case of large (>100 hp) motordrives, but it is possible to increase the armature inductance of the machineduring the design of the motor or to include an external inductor to increase

the effective inductance in the armature path. The latter solution is theonly practical approach in retrofit applications.


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Closed loop control of chopper fed DC motor DrivesDesign of current controller, design of speed controller, modeling

of current and speed controller, introduction to simulation of speed controlleddc motor drive.

5.1 Closed-loop Operations5.1.1 Speed-Controlled Drive System

The speed-controlled dc-motor chopper drive is very similar to thephase-controlled dc-motor drive in its outer speed-control loop. The innercurrent loop and its control are distinctly different from those of the phase-controlled dc motor drive. This difference is due to the particular characteristicsof the chopper power stage. The current loop and speed loop are examined.and their characteristics are explained. in this section. The closed-loop speed-controlled separately-excited dc motor drive is shown in Figure 4.20 foranalysis. but the drive system control strategy is equally applicable to a series

motor drive.

5.1.2 Current Control LoopWith inner current loop alone, the motor drive system is a torque

amplifier. The commanded value of current is compared to the actual armaturecurrent and its error is processed through a current controller. The output ofthe current controller.in conjunction with other constraints, determines thebase drive signals of the chopper switches. The current controller can be eitherof the following types:

(i) Pulse- Width-Modulation (PWM) controller(ii) Hysteresis controller

C H A P T E R5Closed Loop Control of Chopper fed DC Motor Drives


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Figure: 5.1 speed controlled dc motor chopper drive

The selection of the current controller affects its transient response (andhence the Overall speed loop bandwidth indirectly). These two controllersare described in the following sections.

5.1.3 Pulse-Width-Modulated Current ControllerThe current error is fed into a controller, which could be proportional

(P), proportional plus integral (PI), or proportional, integral, and differential(PID). The most commonly used controller among them is the PI controller.The current error is amplified through this controller and emerges as a controlvoltage, ve. It is required to generate a proportional armature voltage fromthe fixed source through a chopper operation. Therefore, the control voltageis equivalent to the duty cycle of the chopper. Its realization is as follows. Thecontrol voltage is compared with a ramp signal to generate the 00- and off-times, as shown in Figure 5.21. On signal is produced if the control voltageisgreater than the ramp (carrier) signal; olfsignal is generated when thecontrolsignal is less than the ramp signal.


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Figure: 5.2 Generation of base-drive signals from current error forforward motoring when One is using the chopper

This logic amounts to the fact that the duration for which the controlsignal exceeds the ramp signal determines the duty cycle of the chopper. Theon- and off- time signals are combined with other control features, such asinterlock, minimum on- and off-times, and quadrant selection. The interlockfeature prevents the turning on of the transistor (top/bottom) in the same legbefore the other transistor (bottom/top) is turned off completely. This is ensured

by giving a time delay between the turn-off instant of one device and the turn-on instant of the other device in the same phase leg. Simultaneous conductionof the top and bottom devices in the same leg results in a short circuit of thede source it is known as shoot-through failure in the literature.


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Figure 5.2 corresponds to the forward motoring operation in the fourquadrant chopper shown in Figure 5.2. When the motor drive is to operate inthe third and fourth quadrants, the armature current reverses. This calls for achange in the current-control circuitry. A block diagram of the current controllerIs shown in Figure 7.3 ,including all the constraints pertaining to the

operational quadrant .The speed and current polarities, along with that of thecontrol voltage, determine the quadrant and hence the appropriate gatingsignals. The on-time is determined by comparing the ramp signal with theabsolute value of the control voltage. The current-error signal, whichdetermines the control voltage V

c is rectified to find the intersection point

between the carrier ramp and VcA unidirectional carrier-ramp signal can be

used when the control voltage is also unidirectional, but the control voltagewill be negative when the current error becomes negative. It happens forvarious cases, such as reducing the reference during transient operation and

changing the polarity of the reference to go from quadrant one to three orfour. Taking the polarity of the control voltage and

Figure: 5.3 PWM current controller implementation with ramp carriersignal combining it with the polarities of the current and speed gives theoperational quadrant. The chopper on and off pulses generated with theintersection of rectified Vc and carrier ramp will then be combined with thequadrant-selector signals of speed, current, and control-voltage polarities togenerate the base-drive signals to the chopper switching devices. An illustrationis given in the drive-system simulation section. Instead of a ramp signal forcarrier waveform ,a unidirectional sawtooth waveform could be used. It is


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advantageous in that it has symmetry between the rising and falling sides ofthe waveform, unlike the ramp signal. Its principle of operation is explainedin the following. The voltage applied to the load is varied within one cycle ofthe carrier signal. This is illustrated in Figure 7.4 . The switching logic issummarized as follows: i

a* -i

ae+ carrier frequency saw tooth wave form

magnitude.T

p= = 1, V

a= = V

s........ (5.1)

Figure: 5.4 principle of PWM operationi: - ia < carrier frequency saw tooth waveform magnitude, Tp = 0, Va =0.....(5.2)

For a fast response ,the current error is amplified so that a small currenterror would activate the chopper control. The PWM controller has theadvantage of smaller output ripple current for a given switching frequency,

compared to the hysteresis current controller described later. The pulsesgenerated from the PWM controller are substituted for Tp in the block diagramshown in Figure 5.3.Theyare then processed for quadrant selection, interlock.and safety features, and appropriate base-drive signals are generated forapplication to the chopper circuit.


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5.1.4 : Hysteresis-Current ControllerThe PWM current controller acts once a cycle, controlling the duty

cycle of the chopper. The chopper then is a variable voltage source withaverage current control. Instantaneous current control is not exercised in thePWM current controller. In between two consecutive switchings , the current

can exceed the maximum limit if the PWM controller is sampled and heldonce a switching cycle, then the current is controlled on an average but not onan instantaneous basis. The hysteresis controller overcomes such a drawbackby converting a voltage source into a fast-acting current source. The currentis controlled within a narrow band of excursion from its desired value in thehysteresis controller. The hysteresis window determines the allowable or presetdeviation of current, i Commanded current and actual current are shown inFigure 7.5 with the hysteresis windows. The voltage applied to the load isdetermined by the following logic:

Iad+i

a*-i ,set V

a=V

s............. (5.3)

Iae+ i

a*+ i ,set V

a=0 ............. (5.4)

Figure: 5.5 hystersis controller operation

Figure: 5.6 realization of hysteresis controller


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The realization of this logic is shown in Figure 5.6. The windows is,can either be externally set as a constant or be made a fraction of armaturecurrent, by proper programming. The chopping frequency is a varying quantity,unlike the constant frequency in the PWM controller .This has the disadvantageof higher switching losses in the devices with increased switching frequency

.5.1.5: Modeling of Current Controllers

The current-error amplifier is modeled as a gain and is given by

Gc(s)=k

c........... (5.5)

lhe chopper is modeled as a first-order lag, with a gain given by

....... 5.6

The PWM current controller has a delay of half the time period of thecarrier wave-form, and its gain is that of the chopper. Hence, its transferfunction, including that of the chopper, is

..........5.7

where Kcis the gain of the PWM current controller, K

ris the gain of the

chopper, and the time constant T is given by,

......... 5.8

The gain of the PWM current controller is dependent on the gain of thecurrent error amplifier. For all practical purposes, the PWM current controlloop can be modeled as a unity-gain block if the delay due to the carrierfrequency is negligible. The hysteresis controller has instantaneous response:hence, the current loop is approximated as a simple gain of unity.

5.6 Design of Current Controller

The current loop is not easily approximated into a first-order transferfunction. unlike the case of the phase-controlled-rectifier drive system. Thechopping frequency is considered to be high enough that the time constant ofthe converter is very much smaller than the electrical time constants of the dc


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motor. That leads to the converter model given by the product of the converterand current-controller gains. Then the closed-loop current-transfer functionis written as

5.9

Where and Kcand K

r are the current-controller and chopper gains,

respectively. The chopper gain is derived as

...... 5.10

where Vs is the de link voltage and Vcm is the maximum control voltage.

The gain of the current controller is not chosen on the basis of the dampingratio. because the poles are 010stlikely to be real ones. Lower the gain of thecurrent controller; the poles will be far removed from the zeros. The higherthe value of the gain, the closer will one pole move to the zero, leading to theapproximate cancellation of the zero. The other pole will be far away fromthe origin and will contribute to the fast response of the current loop.

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