DC Motor Drives [Compatibility Mode]

8/7/2019 DC Motor Drives [Compatibility Mode]

1/63

EET 421

POWER ELECTRONIC DRIVES

Indra Nisja


2/63

witched-mode DC Drives


3/63

Advantages of DC motor :

Ease of control

Deliver high starting torque Near-linear performance

Disadvantages:

Hi h maintenance

Large and expensive (compared to induction

motor)

Not suitable for high-speed operation due tocommutator and brushes


4/63

The DC drive is relatively simple and cheap (compared to

induction motor drives). But DC motor itself is more expensive.

Due to the numerous disadvantages of DC motor (especiallymaintenance), it is getting less popular, particularly in high

.

For low power applications the cost of DC motor plus drives.

For servo application, DC drives is still popular because of

good dynamic response and ease of control.

Future Trend? Not so bright prospect for DC, especially in

high power drives.


5/63

The field windings is used to excite the field flux.

Armature current is supplied to the rotor via brush

and commutator for the mechanical work.

Interaction of field flux and armature current in the

ro or pro uces orque.


6/63

When a separately excited motor is excited by a

f athe circuit, the motor develops a back emf and atorque to balance the load torque at a particular speed.

The if

is independent of the ia

.Each windings are

su lied se aratel . An chan e in the armature

current has no effect on the field current.

e f s norma y muc ess an e a.


7/63

Field and armature equations

where Rfand Lfare the field resistor and inductor, respectivelyInstantaneous armature current :

where Ra and La are the armature resistor and inductor,

respectively

, ,is expressed :

e = K i

Kv is the motor voltage constant (in V/A-rad/s)

and is the motor speed (in rad/sec)


8/63

Basic Torque Equation

Td = Kt if ia

t v

Sometimes it is written as :

Td = Kt ia

For normal operation the developed torque must be equal to the load

torque plus the friction and inertia, i.e, :

where

B : viscous friction constant (N-m/rad/s)

TL : load torque (N-m)

J : inertia of the motor (kg.m2)


9/63

Under steady-state operation, time

derivatives is zero. Assuming the motor

is saturated

,

Vf= IfRf

The back emf is given by:

Eg = Kv if

The armature circuitVa = Ia Ra + Eg

=a a a v


10/63

STEADY-STATE TORQUE AND SPEEDThe motor s eed can be easil derived :

If Ra is a small value (which is usual), or when the motor is slightly

loaded, i.e, Ia is small

That is if the field current is kept constant, the motor speed depends

only on the supply voltage.

The developed torque is :

Td = Kt IfIa = B + TL

The required power is :P

d= T

d


11/63

From the derivation, several important facts can be

deduced for steady-state operation of DC motor.

For a fixed field current, or flux (If) , the torque demand

a .

The motor speed can be varied by: a

controlling Vf(field control)

variable DC voltage to control the speed and torque

of DC motor.


12/63

Consider a 500V 10kW 20A rated- DC motor with armature resistanceof 1 ohm. When supplied at 500V, the unloaded motor runs at 1040

rev/min, drawing a current of 0.8A

Estimate the full load s eed at rated values

Estimate the no-load speed at 250V.

At full load and rated value

Va = Ia Ra + Kv If

aaa

fvIK

=

48.0)1(8.0500 ==fv IKno oa an vo age a

(Note : in reality, this equation strictly rad/sec)


13/63

Family of steady-state torque speed curves for a range.

The speed of DC motor can simply be set by applying

.

Note that speed variation from no-load to full load (rated) can


14/63

a

e

a

e

m IKK

=Or

TKK e

a

e

m 2)( =

Shunt and Separately Excited Motor

With a constant field current, the flux can be assumed to be constant. Let

KKe = (Constant)

a

a

m IK

R

K

V

=T

K

R

K

V a

m 2

=

Series Motor


15/63

Base Speed and Fie ld-w eakening

Base speed: base

the speed which correspond to the rated Va, rated Ia andf.

Constant Torque region ( w > wbase)

Iaand Ifare maintained constant to met torque demand.

Va is varied to control the speed. Power increases with speed.

Constant Powerregion ( w > wbase)

V is maintained at the rated value and i is reduced to increase s eed.However, the power developed by the motor (= torque x speed) remains

constant. Known as field weakening.


16/63

DC Shunt Motor


17/63


18/63

mechPMotor efficiency

controlelec

R

PP += ane ec

fincontrol xIVP =


19/63

Example

A 500-V, 60-hp, 600-rev/min d.c. shunt motor has a full-load efficiency of 90%. The

resistance of the field itself is 200 ohm and rated field current is 2 A. R = 0.2 ohm.

Calculate die full-load (rated) current IaR

and in subsequent calculations, maintain this

value. Determine the loss torque.

e spee s o e ncrease up o rev m n y e wea en ng. a cu a e e

Extra resistance, over and above the field winding itself to cover the range 600-1000

rev/min. Determine the output torque and power at the top speed, assuming that the

loss torque varies in proportion to speed. For the magnetisation curve use the empirical

expression below, which is an approximation to the curve shape.

where the flux ratio is that between a particular operating flux and rated flux .

The field-current ratio is that of the corresponding field currents.


20/63


21/63

Say the motor running at position A. Suddenly va

is reduced (below e).The current ia will reverse direction. Operating point is shifted to B.

Since ia is negative, torque Te is negative.

Power is also ne ative which im lies ower is enerated back to

the supply.

In other words, during the deceleration phase, kinetic energy from the

.

This is known as regenerative braking-an efficient way to brake a motor.

Widely employ in electric vehicle and electric trains. If we wish the motorto operate continuously at position B, the machine have to be driven by

mechanical source.

. We must force the prime mover it to run faster so that the generated eg

will be reater than va.


22/63

Braking circuits


23/63


24/63


25/63

Example 3.20

, . .resistance of 0.13 ohm and takes an armature current

of 60A when delivering rated torque at rated flux. If flux is

,

braking torque equal in magnitude to the full-load torque isdeveloped when:

(a) regeneratively braking at normal terminal voltage;

(b) plugging, with extra resistance to limit the peak torque on

(c) dynamically braking, with resistance to limit the current to

2 per unit;

.(e) What terminal voltage would be required to run the motor in

reverse rotation at rated torque and half rated speed?


26/63


27/63

SCR hase-an le controlled drive- By changing the firing angle, variable DC output voltage can beobtained.

ng e p ase ow power an ree p ase g an very g power

supply can be used

The line current is unidirectional but the out ut volta e can reverse

polarity. Hence 2- quadrant operation is inherently possible.

4-quadrant is also possible using two sets of controlled rectifiers.

Switched-mode drive

Using switched mode DC-DC converter. Dc voltage is varied by

u y cyc e. Mainly used for low to medium power range.

Single-quadrant converter (buck): 1- quadrant

Half bridge: 2-quadrant

Full bridge: 4-quadrant operation


28/63

Mains operated.

.

Slow response.

orma y e rec er ave muc ower ra ngs an e arma ure

rectifier. It is only used to establish the flux.


29/63

Continuous/Discontinuous current

armature inductance La.

Large Laallows for almost constant armature current (with small ripple)due to current filtering effect of L. (Refer to notes on Rectifier).

Average value of the ripple current is zero. No significant effect on the

tor ue.

IfLa is not large enough, or when the motor is lightly loaded, or if

supply is single phase (halfwave), discontinuous current may occur.

Effect of discontinuous current: Output voltage of rectifier rises; motor

speed goes higher. In open loop operation the speed is poorly regulated.

Worthwhile to add extra inductance in series with the armature

inductance.


30/63

Armature Field

For continuous current, armature voltage is :

Field voltage


31/63

1. Single-Phase Half-Wave Converter Drives

= mV

for 02

2. Single-Phase Semiconverter Drives

)cos1(

+= mV for 0

3. Single-Phase Full-Converter Drives

cos2 mVV = for 0

for 04. Single-Phase Dual-Converter Drives

cos2 mVV =


32/63

Armature volta e :

Armature (DC) current is :

If single phase is used for field is :


33/63

1. Three-Phase Half-Wave Converter Drives

for 033 mV=

2. Three-Phase Semiconverter Drives

2

for 0

3. Three-Phase Full-Converter Drives

)cos1(2

+= mV

for 0

cos33 mVV =

for 04. Three-Phase Dual-Converter Drives

cos33 mVV =


34/63

A separately excited DC motor has a constant torque load of 60 Nm. The

motor is driven by a full-wave converter connected to a 240 V ac supply.

The field constant of the motor KI = 2.5 and the armature resistance is 2

ohm. Calculate the triggering angle for the motor to operate at 200 rpm.

Assume the current is continuous.

,

and aaga RIEV +=a

maV

cos=

Where Eg is the back emf,i.e

and

ffg .==

+

= m KIRTV

cos2

af a

f

a

KI


35/63

A rectifier-DC motor drive is supplied by a three-phase, full controlled SCRbridge 240Vrms/50Hz per-phase. The field is supplied by a single-phase 240V

rms/50Hz, with uncontrolled diode bridge rectifier. The field current is set as

max mum as poss e.

The separately excited DC motor characteristics is given as follows :Armature resistance:Ra= 0.3 ohm

e res s ance: f= o m

Motor constant: KV=1.5 V/A-rad/s

Assume the inductance of the armature and field circuit is lar e enou h to ensure

continuous and ripple-free currents. If the delay angle of the armature converter (a) is

45 degrees and the required armature current is 30A,

a) Calculate the developed torque, Td.

b) Speed of the motor, (rad/s)

c) If the polarity of the field current is reversed, the motor back emf will reverse.

For the same armature current of 30A determine the re uired dela an le of the

armature converter.


36/63

Since field current is maximum

= 0.

The armature is su lied b three- hase with = 45o a


37/63

ow e po ar y o e s reverse , en

and

also


38/63

DC motor in inherently bi-directional. Hence no problem to reverse

the direction. It can be a motor or generator.

But the rectifier is unidirectional, because the SCR are unidirectionaldevices.

However, if the rectifier is fully controlled, it can be operated to become

negative DC voltage, by making firing angle greater than 90 degrees,

eversa can e ac eve y:

armature reversal using contactors (2 quadrant)

field reversal using contactors (2-quadrant)

double converter(full 4-quadrants)

Reversal using armature or field contactors


39/63

Reversal using armature or field contactors

DRIVE REVERSING USING ARMATURE OR FIELD CONTACTORS

CONTACTOR AT THE ARMATURE SIDE

SINGLE PHASE SYSTEM

Reversing using double converters


40/63

Reversing using double converters

Practical circuit


41/63

Supply is DC (maybe from rectified-filtered AC, or some other DC sources).

DC-DC converters (choppers) are used.

su a e or app ca ons requ r ng pos on con ro or as response, or

example in servo applications, robotics, etc. Normally operate at high frequency

the average output voltage response is significantly faster

the armature current ripple is relatively less than the controlled rectifier

In terms of quadrant of operations, 3 possible configurations are possible:

single quadrant,

twoquadrant


42/63

Unidirectional speed. Braking not required.

For 0 < t < T

The armature voltage at steady state :

Armature (DC) current is :

DVVT

tdtV

TV ona

on

=== 0

.1

ga EVI

=

And speed can be approximated as :

a

R

v

a

IK

V=


43/63

FORWARD MOTORING T1 and D2 o erate

T1 on: The supply is connected to motor terminal.

T1 off: The armature current freewheels through D2.

Va(hence speed) is determined by the duty ratio.

REGENERATION (T2 and D1 operate)

T2 on: motor acts as a generator

T2 off: the motor acting as a generator returnsenergy to the supply through D2.


44/63

- - .

Vo,dc

Io dc

12

3 4


45/63

T1 and T2 operate; T3 and T4 off.

T1 and T2 turn on together: the supply voltage appear across the motor

. .

T1 and T2 turn off: the armature current decay through D3 and D4


46/63

T1, T2 and T3 turned off.

When T4 is turned on, the armature current rises through T4 and D2.

When T4 is turned off, the motor, acting as a generator, returns energyto the supply through D1 and D2.


47/63

T3 and T4 operate; T1 and T2 off.

When T3 and T4 are on together, the armature current rises and flows

.

Hence the motor rotates in reverse direction.

, .


48/63

T1, T3 and T4 are off.

When T1 is on, the armature current rises through T2 and D4.

When T2 is turned off, the armature current falls and the motor returns

.


49/63

Quadrant First Second Third Fourth

E 0 >0 0


50/63

When IGBT is turning ON:

gEdt

diLRiV ++= 22

w c n a curren11 t == gives the load current as:

1 t

gt

eEV

eIti

+=R


51/63

This mode valid for10 tt at the end of this mode load current becomes:

211 )( Itti ==

The load current for mode 2 can be found from:

gEdt

LRi ++= 220

with initial current and redefining the time origin (i.e t=0) at the beginning of mode 2, we have:22 )0( Iti ==

)1()( 21

tg

t

e

R

EeIti

=

This mod is valid for21 ttt

At the end of this mode the load current becomes:

122

Example:


52/63

Example:

An electric train is powered by four 750-V d.c. series motors. The motor resistance

an n uc ance are respec ve y . an m . o a ne n uc ance o m anresistance of 0.1 is in series with the supply. The fixed voltage from the traction

supply is regulated by the chopper of Figure 1 below, operating at 200 Hz. When the

machine is runnin at 800 rev/min the enerated e.m.f. er am ere can be taken

as an average value of 0.79 V/A. If the modulation factor (D) is 80%,

a. calculate the maximum and minimum currents, allowing 2V for semiconductor loss,

.

b. The mass of the fully loaded train is 120 tonnes and its resistance to

motion on level track is 500N. Each motor is geared to the wheel of the motor

coach by a 3:1 ratio and the coach wheel tread diameter is 1.0m. Estimate thetrain's rate of acceleration under the above conditions.


53/63

Figure 1: Separately excited DC motor

Solution:

2+++= ga

a Edt

diLRiV

2016.035.079.0750 +++=dt

ii aaa

Rearranging:


54/63

di a=dt

a..

For steady state condition:

0=di

L a

t

So, Ampia 14.656==

L= .

R

)1(14.656)( 0457.00457.012

tt

eeiti

+=

At 200Hz chopping frequency with D=0.8

So, tON = 4ms

004.0004.0

. ..12 += eeiti

)9162.01(14.6569162.0)( 12 += iti

. 12 = (1)

Mode 2 (switch OFF)


55/63

Mode 2 (switch OFF)

gEdt

diLRi ++=2

20

2006.025.079.00 +++=dt

diii aaa

2006.004.10 +=dt

i aa

For steady state condition:

0=dt

L a

Ampia

923.1=So,

Time constant ( )=0.024R

L=

)1(923.1)( 024.0024.02

tt

a eeiti

=

At 200Hz chopping frequency with D=0.8So, tOFF = 1ms

00100010


56/63

)1(923.1)( 024.0001.0

024.0

001.0

2

= eeiti a

0788.0959.0)( 21 = iti

Eliminate i1, substitute eq (2) into eq (1):

(2)

55)0788.0959.0(9162.0)( 22 += iti

mp.2 =

Ampi 82.4331 =

For a series motor, torque = Kq=Kv x Ifaa

aq

xiixik

79.0=


57/63

m

n s case, e average orque w e propor ona o e average va ue o a , .e. emean-square current. As the electrical L/Rtime constant is considerably larger than tON or

toff, it is reasonable to assume that the graph of current against time roughly follows straight

lines.

The area under the current squared curve for tON:

3/))(( 22212

1 ixiiitON ++

Ax 59.7853/)45.452)45.45282.433(82.433(004.0

22 =++

The area under the current squared curve for tOFF:

22

2211OFF

Ax 4.1963/)45.452)45.45282.433(82.433(001.0 22 =++

The mean current squared is then,

419659758 +


58/63

51.1963974.19659.758=

+

.

mN

x

xTorque == 853,1

60

2800

51.19639779.0

Tractive force at wheel = Nx

shheelradiu

rratiotorquexgea118,11

5.0

3853,1==

For four motors, total tractive force = 44,472N. With 500 N rolling resistance, the tractive force available for acceleration is 43972N.

.

Acceleration:2

36643.0120

972.43

s

m=

A separately excited DC motor has rating 220 hp, 230Vdc and 2000 rpm.


59/63

A separately excited DC motor has rating 220 hp, 230Vdc and 2000 rpm.

Armature volta e su lied b full brid e control rectifier with in ut volta e

tVoltsvs 314sin5.346=

tVoltsvs 314sin5.346= Field voltage supplied by diode rectifier with input

- cons an vo age v = .

- armature resistance Ra

= 5 Ohm

- field resistanced Rf= 150 Ohm

0=a

b. If armature voltage reduced such that the motor run at a speed of 1200 rpm,

calculate the value of and developed torque. Armature current is 10 Aa

a. a cu ate t e oa requ r ng torque

Solution


60/63

Out ut of control rectifier for=0:

a

m

a

VV

cos

2=

0cos5.3462

xV

a

=

=a

Output of diode rectifier:

mV2

a =

5.3462 xV a =

VV a 220=

Equation of armature separately excitation DC motor:


61/63

aag RIEVa +=

aaag RIVE = (1)

fvg IKE = (2)

Substitute (2) into (1):

aaafv RIVIK =

Equation of field winding of DC motor:

fff RIV =

Vf 220mp

R ff .

150

===

So,

fv

aaa

IK

=

1736.1

a=

2201736.1 + I


62/63

=aI

4423472.0 += aI

mNTL += )4423472.0(1736.1

Torque requiring by the load is:

b.

from equation (2)

60

212001736.1

xE g =

4.147=E

504.147 +=Va

VoltVa 4.197=

Firing angle (a) is:


63/63

2204.1971= Cosa

0.=a

Documents

DC Motor Drives [Compatibility Mode]