DC Motors

• Construction very similar to a DC generator

• The dc machine can operate bath as a generator and a motor.

• When the dc machine operates as a motor, the input to the machine is electrical power and the output is mechanical power.

• In fact, the dc machine is used more as a motor.

• DC motors can provide a wide range of accurate speed and torque control.

• Principle of operation – when a current-carrying conductor is placed in magnetic field, it experiences a mechanical force., F = Bli

Eg. Gas turbine,

diesel engine,

electrical motor

DC Motors

• Separately Excited Motors

Field and armature windings are either connected

separately.

•Shunt Motors

Field and armature windings are connected in parallel.

• Series Motors

Field and armature windings are connected in series.

• Compound Motors

Has both shunt and series field so it combines features

of series and shunt motors.

Comparisons of DC Motors

Shunt Motors: “Constant speed” motor (speed regulation is very

good). Adjustable speed, medium starting torque.

Applications: centrifugal pump, machine tools, blowers fans,

reciprocating pumps, etc.

Series Motors: Variable speed motor which changes speed drastically

from one load condition to another. It has a high starting torque.

Applications: hoists, electric trains, conveyors, elevators, electric cars.

Compound motors: Variable speed motors. It has a high starting

torque and the no-load speed is controllable unlike in series motors.

Applications: Rolling mills, sudden temporary loads, heavy machine

tools, punches, etc

Shunt Motor

• The armature circuit and the shunt field circuit are connected across

a dc source of fixed voltage Vt Rfc in the field circuit is used to

control the motor speed by varying if.

= If(Rfc + Rfw)

Power Flow and Efficiency SHORT SHUNT COMPOUND MOTOR

• depend on machine

size

• Range shown for

machine 1 to 100 kW

Power Flow and Losses in

DC Motors

P

out

Protational

Pinput

VtIt

VaIt

VaIa E

aIa

It

2Rsr I

t 2R

tIa

2Ra

– T characteristics

Speed- torque characteristics of DC motors

Example 11 Q. A DC machine (12 kW, 100 V, 1000 rpm) is connected to a 100 V

DC supply and is operated as a DC shunt motor. At no-load condition, the motor runs at 1000 rpm and the armature current takes 6 A. Given armature resistance Ra= 0.1 , shunt field winding resistance Rfw= 80 , and Nf= 1200 turns per pole. The magnetization characteristic at 1000 rpm is shown in the next figure.

a. Find the value of shunt field control Rfc

b. Find the rotational losses at 1000 rpm

c. Find the speed, torque, and efficiency when the rated current flows.

i) Consider the air gap flux remains the same ( no armature reaction) at that at no load

ii) Consider the air gap flux reduces by 5 % when the rated current flow in the armature due to the armature reaction

d. Find the starting torque is the starting current is limited to 150 % of its rated current i) Neglect armature reaction ii) Consider armature reaction, IfAR = 0.16A

Sen pg. 170

Cont. Example

0.99

99.4

Separately Excited DC Motor – Torque

speed characteristic

Vt and flux constant - Drop

in speed as torque increase

is small – good speed

regulation

AR- Improve speed regulation

DC Speed Control

• Can be achieved by

• Armature Voltage Control, Vt

• Field resistance control,

• Armature resistance Control, Ra

• Speed increases as Vt increases, Ra

increases and field flux decreases

Torque

Speed

Torque

Speed

TL = Cm2

Load Torque profile

TL = K

Fans, blowers, centrifugal pumps

Low speed hoist, elevator

T Vt

m m

T = 0

T = 1

T = 2

Vt increasing

Vt1

Vt2

Vt3

Vt4

Armature Voltage control-

Constant load torque – speed

varies linearly as Vt changes

Armature Voltage control-

Terminal voltage (Vt) varies-

speed adjusted by varying Vt

Armature Voltage Control No load

speed

Full load

speed

The speed of DC motor can simply be set by applying the

correct voltage ( fixed flux and Ra). Good speed regulation.

Maintain maximum torque capability. Expensive control.

Field Control

m

Flux decreasing

if1

if2

if3

if4

Rfc max

Rfc=0

T

Field Control

The speed of DC motor can simply be set by applying the

correct field resistance (Rfext) ( fixed Va and Ra). Slow/sluggish

transient respond. Unable to maintain maximum torque

capability. Simple and cheap control.

f

2

a

aea

a

t

i where

TΦ)(K

)R(R

ΦK

Vω

Armature Resistance Control

m

Ra increasing

Raemax

Rae=0

T Resistance control

The speed of DC motor can simply be set by applying the

correct armature resistance (Raext) ( fixed Va and Rf). Poor speed

regulation. High Losses. Unable to maintain maximum torque

capability ( TL rated). Simple and cheap control.

No load

speed TΦ)(K

)R(R

ΦK

Vω

2

a

aea

a

t

TL

rated

Example 12

• A variable speed drive system uses a dc motor which is supplied from a variable-voltage source. The drive speed is varied from 0 to 1500 rpm (base speed) by varying the terminal voltage from 0 to 500 V with the field current maintained constant.

(a) Determine the motor armature current if the torque is held constant at 300 N-m up to the base speed.

(b) Determine the torque available a speed of 3000 rpm if the armature current is held constant at the value obtained in part (a).

Neglect all losses.

Sen pg 180

Series Motor

T – characteristics

Speed- torque characteristics of DC motors

Example 13

• A 220 V, 7 hp series motor is mechanically coupled to a fan and draws 25 amps and runs at 300 rpm when connected to a 220 V supply with no external resistance connected to the armature circuit (Rae= 0 ). The torque required by the fan is proportional to the square of the speed. Ra= 0.6 and Rsr= 0.4 . Neglect armature reaction and rotational loss.

(a) Determine the power delivered to the fan and the torque developed by the machine.

(b) The speed is to be reduced to 200 rpm by inserting a resistance Rae in the armature circuit. Determine the value of this resistance and the power delivered to the fan.

PC Sen pg 182

Motor Starter

• If a DC motor directly connected to a DC supply, the

starting current will be dangerously high

a

ta

a

ata

R

V I start;at 0

R

EVI

aa KE

•Ra small, Ia large. Ia can be limited to a safe value by:

•Insert an external resistance, Rae

•Use a low dc voltage (Vt) at starts, which require a

variable-voltage supply

•With external resistance,

aea

ata

RR

EVI

Motor Starter

Development of a DC motor starter

Ea speed (). As

speed increases Rae

can be gradually

taken out without the

current exceed a

limit ( starter box).

Initially at position 1,

as the speed

increases, the

starter move to

position 2,3,4and 5,

Example 13.1

A 10 kW , 100 V , 1000 rpm dc machine has Ra=0.1

ohm and is connected to a 100 V dc supply.

a) Determine the starting current if no starting

resistance is used in the armature circuit

b) Determine the value of the starting resistance if

the starting current is limited to twice the rated

current

c) This dc machine is to run as a motor, using starter

box. Determine the values of resistance required

in the starter box such that the armature current Ia

is constraint within 100% to 200% of its rated

value during start-up.

Permanent Magnet DC motor

• Widely used in low power application

• Field winding is replaced by a permanent magnet (simple construction and less space)

• No requirement on external excitation

• Limitation imposed by the permanent magnet themselves such as demagnetization and overheating)

Equation Ea = Kadm becomes Ea = Kmm

Example 14

• A permanent magnet DC motor has Ra = 1.03 .

When operated at no-load from a DC source of 50

V, its operates at 2100 rpm and draw a current of

1.25 A. Find:

i. The torque constant, Km

ii. The no-load rotational losses

iii. The armature current and the motor power

output when it is operating at 1700 rpm from a 48 V

source

Fgrt; pg 389: 0.22 V/(rad/sec), 61 W, 8.54A, 274 W

Speed Control

• Numerous applications require control of speed, as in

rolling mills, cranes, hoists, elevators, machine tools,

and locomotive drives.

• DC motors are extensively used in many of these

applications.

• Control of dc motors speed below and above the base

(rated) speed can easily be achieved.

• The methods of control are simpler and less expensive

than ac motors.

• Classis way used Ward-Leonard System, latest used

solid-state converters.

Ward-Leonard System

• In the classical method, a Ward-Leonard system(1890s) with rotating machines is used for speed control of dc motors. The system uses the motor-generator set ( M-G set) control the speed of a DC motor. Normally AC motor runs at constant speed is used as prime mover.

The system is operated in two control methods:

• Vt Control; In the armature voltage control mode, the motor current Ifm is kept constant at its rated value. The generator field current Ifg is changed such that Vt changes from zero to its rated.

• If Control; The field current control mode is used to obtain speed above the base speed. In this mode, the armature voltage Vt remains constant and the motor field current Ifm is decreased to obtain higher speeds .

Constant

Torque Region

Constant

Power Region

Ward-Leonard System

Prime mover

Solid-State Control

• In recent years, solid state converters have been used (replace motor- generator set) to control the speed of dc motors.

The converter used are controlled rectifiers or choppers:

Controlled Rectifiers

• If the supply is ac, controlled rectifiers can be used to convert a fixed ac supply voltage into variable-voltage dc supply (using SCR). High ripple, slow response

Choppers

• A solid state chopper converts a fixed-voltage dc supply into a variable-voltage dc supply(Using controllable swithes such as Power Mosfet, Power BJT, IGBT, GTO etc).Low ripple, fast response

Single phase rectifier

Controlled Rectifier

cos3 llV

V mt

Three phase rectifier

cos2 m

t

VV

1-phase or 3-phase

Eg: Va and Vf Control using solid state

devices – single phase supply

+

vs

_

Ia

Ta1

Ta2

Ta3

+

+

V a

Ra

Ta4

La

E

g

If

+

vs

_

Tf1

Tf2

Tf3 +

Lf

Tf4

Lf

Vf

ARMATURE FIELD

Chopper Control

inon

int VT

TDVV

Closed-loop Operation

• Open loop operation: If load torque changes, the speed will change too – not satisfactory.

• May not be satisfactory in many applications where a constant speed is required

• Close loop operation: the speed can be maintained constant by adjusting the motor terminal voltage as the load torque changes.

(eg. Load torque increases, speed decreases, speed error eN increases, results in control signal Vc increases – decrease in the converter firing angle (controlled rectifeir), or increases in duty cycle (chopper) to restore back the speed)

Closed loop speed control system (basic system)

Chopper or Control

rectifier

Speed

demand

Voltage

control

Speed

demand

Current/torque

demand

Closed-loop speed control with inner current loop

Example 15

• The speed of a 10 hp, 220 V, 1200 rpm

separately excited DC motor is controlled by a

single-phase full-controlled converter. The rated

current is 40 A. Ra= 0.25 ohm, and La= 10mH.

The AC supply voltage is 265 V. Motor constant

is Ka=0.18 V/rpm. Assume the motor current is

constant and ripple free. For firing angle = 30

degree, determine:

(a) Speed of the motor

(b) Motor Torque

(c) Power to the motor.

Sen pg 191

Example: 2009/10 Question 3

(a) Describe briefly classification of self-excited DC motor based on connections of field

circuit and armature circuit. Sketch the torque speed profile for this type of motor.

(b) A 500 V shunt motor takes a current of 21 A and runs at 400 rpm on full load. The

armature resistance and field resistance are 0.3 and 500 respectively. In order to

control the speed, an additional resistance is added in series in the armature circuit.

The flux remains constant in the machine.

(i) Sketch the new schematic diagram with armature resistance motor speed control.

(ii) Find the motor speed at full load when an additional resistance of 2 is added in

armature circuit.

(iii) Find the motor speed at double full-load with added resistance of Qb(ii).

(iv) Find the motor starting current with an additional resistance of 2 .

(v) Find the required value of the additional armature resistance to reduce the speed to half

of its rated speed at full load.