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School of Electrical and Computer Engineering

1

Applied Electronics and Electrical

Machines(ELEC 365)

Fall 2015

DC Machines

Key educational goals:

Develop the basic principle of operation of a dc machine. Classify the different

types of dc machines and evaluate their characteristics.

Reading/Preparatory activities for class

i) Textbook:

Chapter 16: 16.1, 16.2, 16.3, 16.4, 16.5, 16.6, 16.7

DC Machines

2

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

1. What is operating principle of a dc machine?

3. What are the different hardware components of dc machine?

4. Name one application of a dc generator.

5. What are the different types of dc motors?

6. How do you distinguish a dc series motor from a separately excited dc motor?

7. What is the effect of armature reaction?

DC Machines

3

Classification of motors

4

Introduction

5

In a transformer energy is transferred from primary to secondary, form of

energy (electrical) is unchanged.

In a ROTATING ELECTRIC MACHINE (M/C or m/c), electrical energy is

converted into mechanical form and vice versa:

Electric Motor Electric energy to Mechanical Energy.

Electric Generator Mechanical Energy to Electrical Energy.

∴ Electric machines are referred to as “ELECTRO-MECHANICAL ENERGY

CONVERTERS”.

7

Introduction

Electromagnetic conversion

Electrical system

V, i

Mechanical system

T, n

Coupling Magnetic Field

Conversion of energy from electrical to mechanical form or vice versa results from

the following two electromagnetic phenomena:

1. When a conductor moves in a magnetic field, voltage is induced in the

conductor.

2. When a current-carrying conductor is placed in a magnetic field, the conductor

experiences a mechanical force.

Note: These two effects occur simultaneously.

Operating Principle of a DC Machine

8

The linear m/c – a simple example dc

machine consisting of a conducting

bar sliding on conducting rails.

There are four basic equations to the m/c. These equations are necessary to

understand the above m/c.

1) The equation for the FORCE on a wire in the presence of a MAGNETIC FIELD:

𝐹 = 𝑖 𝑙 × 𝐵

The direction of the force is given by Fleming’s left hand rule .

Operating Principle of a DC Machine

9

2) The equation for the voltage induced in a wire moving in a magnetic field:

eind = (v × B).l

where v = velocity of wire, l = length of conductor in magnetic field with the vector

pointing along the direction of the wire toward the end assumed to be positive.

3) KIRCHOFF’S VOLTAGE LAW:

VB – iR – eind = 0.

VB = iR + eind

4. NEWTON’S LAW for the bar across the tracks:

Fnet = ma

Equivalent circuit for the linear m/c.

FORE FINGER = MAGNETIC FIELD

900

900

900

MIDDLE FINGER=CURRENT

FORCE = (ilB).sin(θ)

Fleming’s Left Hand Rule Or Motor Rule

10

To find the direction of force:

Where θ is the angle between the wire and the flux density vector. This principle

is basis for MOTOR ACTION. (called Lorentz Force).

FORE FINGER = MAGNETIC FIELD

900

900

900

MIDDLE FINGER = INDUCED VOLTAGE

eind (induced emf) = vBlcos(θ)

Fleming’s Right Hand Rule Or Generator Rule

11

To find the direction of induced emf: right hand rule is applicable.

θ = smallest angle between direction of wire l and (v × B) .

STARTING THE DC MACHINE

13

To start the m/c, close the switch.

(i) A current flows through the bar. Since the bar is initially at rest, eind = 0;

𝑖 =𝑉𝐵 − 𝑒𝑖𝑛𝑑

𝑅=

𝑉𝐵

𝑅(𝐾𝑉𝐿)

STARTING THE DC MACHINE

14

(ii) ∴ A current flows through a wire in the presence of a MAGNETIC FIELD and

hence a FORCE is induced on the wire.

Find = ilB TO THE RIGHT.

(iii) ∴ Bar will accelerate to the right (by Newton’s Law). As velocity of bar

increases,

eind = vBl (POSITIVE UPWARD)

(iv) This voltage reduces the current flowing through the wire (bar):

𝑖 =𝑉𝐵 − 𝑒𝑖𝑛𝑑 ↑

𝑅𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 (↓)

STARTING THE DC MACHINE

15

𝑣𝑠𝑠 =𝑉𝐵

𝐵𝑙[𝑠𝑖𝑛𝑐𝑒 𝑉𝐵 = 𝑒𝑖𝑛𝑑 = 𝑣𝑠𝑠𝐵𝑙]

(v) ∴ Force induced Decreases. [since F = (i ↓)lB ]. The result is finally a steady state

speed (vss) is reached where the net force on the bar is zero.

i.e. F = (i ↓)lB until F = 0.

At that point eind = VB , i = 0 and the bar moves at a constant speed (NO-LOAD):

THE DC LINEAR MACHINE AS A MOTOR

16

DC Linear m/c as a motor: force applied in the opposite direction of motion.

Assume that the linear m/c is initially running at the no-load steady-state

conditions. Now if an external load is applied to it in a direction opposite to the

direction of motion, what happens?

THE DC LINEAR MACHINE AS A MOTOR

17

(i) Due to application of Fload, net force on the bar will be in the direction opposite

to the direction of motion,

Fnet = Fload – Find

(ii) The resulting acceleration

𝑎 =𝐹𝑛𝑒𝑡

𝑚is negative, so the bar slows down, i.e. v ↓ (v decreases).

(iii) ∴ eind = (v ↓)(Bl) falls, ∴

(iv) Induced force Find = (I ↑)lB increases until Find = Fload and bar runs at a

lower speed (v).

(v) Electrical power, Pconv = eind.i is converted to mechanical power Find.v, and the

m/c is acting as a motor. [Behavior is similar to a DC shunt motor].

𝑖 =𝑉𝐵 − 𝑒𝑖𝑛𝑑↓

𝑅↑

THE DC LINEAR MACHINE AS A GENERATOR

18

Assume again that the linear dc m/c is initially running at no-load steady-state

conditions. Now apply a force in the direction of motion.

DC Linear m/c as a generator: force applied in the direction of motion.

THE DC LINEAR MACHINE AS A GENERATOR

19

(i) Due to application of 𝐹𝑎𝑝𝑝 ; 𝐹𝑛𝑒𝑡 is in the direction of motion.

(ii) acceleration 𝑎 =𝐹𝑛𝑒𝑡

𝑚is positive, so the bar speeds up (𝑣 ↑).

(iii) 𝑒𝑖𝑛𝑑 = 𝑣 ↑ (𝐵𝑙) increase ( 𝑒𝑖𝑛𝑑 > 𝑉𝐵),

𝑖 =𝑒𝑖𝑛𝑑 ↑ −𝑉𝐵

𝑅↑.

(iv) 𝐹𝑖𝑛𝑑 = 𝑖 ↑ 𝑙𝐵 increases until 𝐹𝑖𝑛𝑑 = 𝐹𝑙𝑜𝑎𝑑 and bar runs at a higher

speed v.

(v) Amount of mechanical power = 𝐹𝑖𝑛𝑑 . 𝑣, is converted to electrical power,

𝑃𝑐𝑜𝑛𝑣 = 𝑒𝑖𝑛𝑑 . 𝑖 and the m/c is acting as a generator.

[Behavior is similar to a DC shunt generator].

SUMMARY

20

(1) Same m/c acts as both generator and motor. Only difference is whether the

externally applied force is in the direction of motion (generator) or opposite the

direction of motion (motor).

(2) eind > VB (GENERATOR).

eind < VB (MOTOR).

(3) Whether motor or generator, Find (motor action) and eind (generator action) are

present at all times.

(4) Whether m/c was motor (moved slowly) or generator (moved rapidly), it

always moved in the same direction.

Basics Structure of DC Machines

21

22

Basics Structure of DC Machines

Armature of a DC Machine

23

Basics Structure of DC Machines

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26

Operating Principle of Rotating DC Machine

Operating Principle of Rotating DC Machine

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Action of a Commutator

28

30

Generally, the commutator segments are copper bars insulated from one another

and from the shaft. The brushes contain graphite that lubricates the sliding contact.

Significant disadvantage of dc machines?

Action of a Commutator

EMF EQUATION OF A GENERATOR

33

Let φ = flux/pole Weber (Wb)

Z = Total number of armature conductors.

= (No. of slots) × (No. of conductors/slot)

P = Number of Generator poles (Even number)

a = Number of parallel paths in the armature winding.

𝐸𝑎= Average voltage induced (V)

N = Armature speed Revolutions per second (rps)

Ω = Angular velocity, = 2πN (Rads/s.)

Speed regulation :Ω𝑛𝑜−𝑙𝑜𝑎𝑑−Ω𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑

Ω𝑓𝑢𝑙𝑙−𝑙𝑜𝑎𝑑∗ 100%

EMF EQUATION OF A GENERATOR

34

The voltage induced in armature:

𝐸𝑎 = 𝐾𝜙Ω 𝑉

Where K is a machine constant which depends on the design parameters of the

machines.

Where 𝐾 =𝑍𝑃

2𝜋𝑎

Torque equation

𝑇𝑑 = 𝐾𝜙𝐼𝑎 Nm.

𝑬𝒂𝑰𝒂 is the air-gap power of the machine.

𝐸𝑎𝐼𝑎 = (KɸΩ) 𝐼𝑎 Watts

= (𝐾𝜙Ω)𝑇𝑑

𝐾𝜙= (𝑇𝑑) Ω = 𝑃𝑑 Watts This is the gross mechanical power in Watts.

Where Ω is in Rads/s.

CLASSIFICATION OF DC MACHINES

Based on excitation.

I. Permanent magnet:

• No field windings.

• Usually, fractional horsepower rating m/c.

• With the development of permanent magnet materials, higher power

available.

35

36

separately excited DC motor.

II.

(1) Separately excited: No electrical interconnection between the field and

armature winding.

Separately Excited DC Machine

Vt = Ea + IaRa V

= KφΩ + IaRa

Ω = Vt − Ia RaKφ

rads/sec.

(2) Shunt excited (Self): Parallel connection between the field and armature

winding.

38

Shunt excited

Shunt excited DC m/c.

(3) Series excited: Series field winding is connected in series with the armature

and therefore, carries the same current as the armature.

39

Series excited

Series excited DC m/c.

(4) Compound excitation:

40

Compound excitation

Short shunt: Cumulative Compound Long Shunt: Differential Compound.

41

MAGNETIZATION CURVES

Circuit diagram to vary the field current (in turn flux) to obtain the magnetization curve.

Air-gap line approximates the magnetization curve.

constant speed

Relation between Speed and backemf:

Armature Reaction (AR)

42

In a DC machine, two kinds of magnetic

fluxes are present; 'armature flux' and 'main

field flux'. The effect of armature flux on the

main field flux is called as armature

reaction.

• AR is the magnetic field produced by the

armature current

• AR aids the main flux in one half of the

pole and opposes the main flux in the

other half of the pole.

• However due to saturation of the pole

faces the net effect of AR is demagnetizing.

COMPENSATING WINDINGS

43

One way to reduce the effects of armature reaction is to connect winding called

“compensating winding" in series with armature windings so that it’s mmf is

proportional to the armature mmf.

These pole face windings are so arranged that

the mmf produced by currents flowing in these

windings opposes the armature mmf. These pole

face windings are expensive. Therefore they are

only used in large machines or in machines that

are subjected to abrupt change of armature

current

Interpoles

44

Another way to reduce the effects of armature reaction is to place small auxiliary

poles called "interpoles" between the main field poles. The interpoles have a few

turns of large wire and are connected in series with the armature. Interpoles are

wound and placed so that each interpole has the same magnetic polarity as the

main pole ahead of it, in the direction of rotation.

45

EQUIVALENT CIRCUIT OF A DC GENERATOR

Equivalent circuit of a separately excited generator.

Under STEADY-STATE:

𝑣𝑓 = (𝑅𝑓)(𝐼𝑓) V

𝑣𝑡 = 𝐾𝜙Ω − 𝑅𝑎𝐼𝑎 V

𝑇𝑠ℎ𝑎𝑓𝑡 = 𝐾𝜙𝐼𝑎 + 𝑇𝑙𝑜𝑠𝑠 N.m.

J = Polar moment of inertia (Kg.m²)

𝑣𝑓 = 𝑅𝑓𝑖𝑓 + 𝐿𝑓𝑑𝑖𝑓

𝑑𝑡V

𝑣𝑡 = 𝐾𝜙Ω − 𝐿𝑎𝑑𝑖𝑎

𝑑𝑡− 𝑅𝑎𝑖𝑎 V

𝑇𝑠ℎ𝑎𝑓𝑡 = 𝐾𝜙𝑖𝑎 + 𝐽𝑑Ω

𝑑𝑡+ 𝑇𝑙𝑜𝑠𝑠 N.m.

These equations are used under

transient conditions.

46

PERFORMANCE OF GENERATORS

(a) External characteristics [potential difference (pd) Vt versus load current, iL].

(b) Efficiency.

(c) Voltage Regulation = [(No Load Voltage) – (Full Load Voltage)]/[Full load Voltage].

1) Separately Excited Generator

Separately excited DC generator. Re is the external resistance added in the field circuit.

47

PERFORMANCE OF GENERATORS (Separately Excited)

External characteristic of a separately excited generator.

Steady-state: 𝐼𝑓 =𝑉𝑓𝑠

(𝑅𝑒+𝑅𝑓)A

𝐸𝑎 = 𝐾𝜙Ω V

𝑅𝐿 =𝑉𝑡

𝐼𝐿Ω

𝐼𝐿 = 𝐼𝑎 A

𝐸𝑎 = 𝑉𝑡 + 𝐼𝑎𝑅𝑎 V

𝑉𝑡 = 𝐸𝑎 − 𝐼𝑎𝑅𝑎 = 𝐸𝑎 − 𝐼𝐿𝑅𝑎 V

𝐼𝑎 = 𝐼𝐿=𝐸𝑎

(𝑅𝑎+𝑅𝐿)=

𝐾𝜙Ω

(𝑅𝑎+𝑅𝐿)A

48

PERFORMANCE OF GENERATORS (Shunt)

2) Shunt Generator (self-excited)

When m/c is driven under NO LOAD:

A small Ear due to residual magnetism appears at the terminals. This voltagecirculates a current through field winding which in turn increases generated voltage… (cumulative build up) … until a voltage Ea is reached, where: Ea = If.Rf = Vt

(Assume Ra is neglected)

Schematic of shunt dc machine. Voltage buildup in a shunt dc generator.

49

PERFORMANCE OF GENERATORS (Shunt)

Shunt generator under no-load.

Shunt generator with a load.

When the m/c is LOADED:

Under NO LOAD:Ea = (Ra + Rfc + Rfw)If

Vt = (Rfc + Rfw)If

Vt ≅ Ea = F(If)

Vt = (RL)IL

Vt = Ea - (Ia Ra)

Ia = IL + If

50

PERFORMANCE OF GENERATORS (Shunt)

Effect of field resistance.

51

PERFORMANCE OF GENERATORS (Shunt)

A shunt generator can not build up if:

(a) There is no residual magnetism,

(b) If the field is connected in a wrong-way around, opposing the permanent

magnetism, &

(c) The value of the field resistance is greater than a value called critical

resistance.

Example 1

54

A shunt generator delivers 450 A at 230 V and the resistances of the shunt field

and armature are 50 Ω and 0.03 Ω, respectively. Calculate the generated emf.

Field current, 𝐼𝐹 =230

50= 4.6 𝐴

Load current, IL = 450 A with

load voltage, Vt = 230 V.

∴ Armature current,

Ia = IL + IF = 450 + 4.6 = 454.6 A.

∴Generated emf, Ea = Vt + IaRa = 230 + (454.6)(0.03) = 243.6 V.

55

A separately excited generator is rated at 10 kW, 200 V, 1000 rpm. Rf = 80 Ω,

Ra= 0.4 Ω and Ia = 50 A. Predict the no-load voltage at 1000 rpm and the full-

load voltage at 800 rpm if the field current is kept constant.

Example 2

56

Example 2

At no-load, Ia = 0 and Vt = Ea = VNL

∴ VNL = Ea = VFL + IaRa = 200 + (50)(0.4) = 220 V.

If the field current is constant, the generated emf Ea is directly proportional to

speed; therefore, at 800 rpm, induced (or generated voltage) E’a is given by

The terminal voltage at 800 rpm under full-load (Ia = 50 A) is

58

Equivalent circuit of a separately excited dc motor.

J=Polar moment of inertia (Kg.m²)

𝑣𝑓 = 𝑅𝑓𝑖𝑓 + 𝐿𝑓𝑑𝑖𝑓

𝑑𝑡V

𝑣𝑡 = 𝐾𝜙Ω + 𝐿𝑎𝑑𝑖𝑎

𝑑𝑡+ 𝑅𝑎𝑖𝑎 V

𝑇𝑙𝑜𝑎𝑑 = 𝐾𝜙𝑖𝑎 − 𝐽𝑑Ω

𝑑𝑡− 𝑇𝑙𝑜𝑠𝑠 N.m.

These equations are used under transient conditions.Under STEADY-STATE:

𝑣𝑓 = (𝑅𝑓)(𝐼𝑓) V

𝑣𝑡 = 𝐾𝜙Ω + 𝑅𝑎𝐼𝑎 V𝑇𝑙𝑜𝑎𝑑 = 𝐾𝜙𝐼𝑎 − 𝑇𝑙𝑜𝑠𝑠 N.m.

𝐸𝑎 = 𝐾𝜙Ω = F 𝑖𝑓

Illustrated below for a separately excited motor.

Ω= ΩR

EQUIVALENT CIRCUIT OF A DC MOTOR

Shunt Connected DC Motor

59

A shunt-connected motor is similar to a separately excited dc motor except that:

Speed versus torque characteristic for a shunt motor.

Series Connected DC Motor

60

Speed versus torque for a series motor.

Since, φ is a function of Ia, for light loads,

speed becomes very high. Therefore,

SERIES MOTORS MUST NEVER BE STARTED

WITHOUT ANY LOAD.

Torque – Speed relationship in Series DC Motor

61

Apply Kirchhoff's Voltage Law to equivalent circuit:

Assumption:

Compound Excited DC Motor

• If the shunt and series field aid each other it is called a cumulatively excited

machine.

• If the shunt and series field oppose each other it is called a differentially excited

machine.

63

64

Torque – speed characteristics of different dc motors.

Speed – Torque characteristics of different machines

65

In a permanent-magnet (PM) dc motor, the field is supplied by magnets mounted on the

stator rather than by field coils. PM motors are common on fractional or subfractional

horsepower sizes.

permanent-magnet DC Motor

66

Advantages:

• No power is required to establish the field leading to better efficiency.

• Smaller

• cheaper

Disadvantages:

• Magnets can become demagnetized by overheating or because of excessive

armature currents.

• The flux density magnitude is smaller than in wound- field machines.

Consequently, the torque produced per ampere of armature current is smaller

in PM motors than in wound- field motors with equal power ratings.

Compound Excited DC Motor

67

Motor Types CharacteristicsPower

RangeApplications

Separately excited

motors

Can be controlled either by varying

the voltage applied to the field

winding or by varying the voltage

applied to the armature; can

produce high torques

Up to

100 hp

Traction applications, to control

the speed and torque of the

motor by changing both

armature

voltage and stator current

DC shunt motor

Constant flux constant speed motor

creates moderate torque at startUp to

200 hp

Machine tools like lathes, milling

machines, grinding machines,

centrifugal and reciprocating

pumps, blowers, and fans

DC series motor

Creates dangerously high torques at

low speeds and should be always

connected to the load; speed can be

varied

Up to

200 hp

Preferred for traction-type

loads; employed in electric

locomotives, conveyors, cranes,

elevators, trolleys

Permanent magnet

DC motor

Higher efficiency, smaller size and

simpler architecture; magnets can

become demagnetized due to

excessive use and overheating;

produces lower torque

Up to

10 hp

Power windows in automobiles,

computer peripherals

Comparison

Example 5

70

A 50-hp shunt-connected dc motor has the magnetization curve shown in

Figure. The dc supply voltage is VT = 240 V, the armature resistance is RA =

0.065Ω , the field resistance is RF = 10 , and the adjustable resistance is Radj =

14Ω . At a speed of 1200 rpm, the rotational loss is Prot = 1450 W. If this motor

drives a hoist that demands a torque of Tout = 250 Nm independent of speed,

determine the motor speed and efficiency.

Example 5

71

The equivalent circuit is shown in Figure. The field current is given by

Next, we use the magnetization curve

to find the machine constant K for this

value of field current.

Assuming constant torque for the rotational loss.

Thus, the developed torque is

Example 5

72

Then, applying Kirchhoff s voltage law to the armature circuit, we have

To find efficiency, we first compute the output power and the input power, given by

Speed Control

73

1 - Varying armature terminal voltage (Vt) while holding the field constant.

2 - Varying field current (If ) and therefore φ.

3 - Varying Ra (Insert resistance in series with the armature circuit.)

OR combination of these.

Speed Control by varying Ra

By inserting additional resistance in series with armature circuit we can control the

speed of DC machine. This method is applicable for various type of DC machine.

Speed Control by varying Ra

74

• Add external resistance Rd in series with the armature.

• INEFFICIENT method due to losses in added resistance.

Speed versus Torque characteristics of a separately excited DC motor with armature resistance control.

Speed Control by Variation of Terminal Voltage Vt

75

P0 No-load operating point.

P1 Operating point for a given load.

φ1 = Maximum (or rated) flux for which the

m/c is designed.

Speed versus Torque characteristics of a separately

excited DC motor with voltage

control.

Speed Control by varying If

76

Speed control above base (rated) value is obtained by reducing VF and therefore

If and called FIELD WEAKENING.

Speed versus Torque characteristics of a separately

excited DC motor with field control.

77

Vt Control: Speed control from zero to base speed (Ω𝑏) with rated field current.

Over this speed range, Ia must normally be limited to its rated value, and hence

developed torque will also be limited,

Since T = KɸIa Nm.

P = ΩT W.

[Vt is limited to rated value because of commutator insulation and because of the

limit on the available supply, further speed increase requires field weakening].

For negligible, 𝑅𝑎 ,Ω ∝

1

𝜙

𝑟𝑎𝑑𝑠

𝑠.

𝑇 = 𝐾𝜙𝐼𝑎 ∝1

Ω[since 𝐼𝑎 is limited to rated value ]

𝑃 = Ω𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑊 .

Speed Control by varying If

Early variable DC system: Ward Leonard

79

1- three-phase induction motor drives a dc generator.

2- Produced dc voltage supplies to the motor to be controlled.

80

Starting of DC Motors

Therefore, at starting, very high current (since Ra is very small) flows through the

armature windings. Therefore, limit starting current by inserting series resistance

or applying reduced voltage. As speed picks up, Ea increases, reduce the

resistance. This large current is approximately seven times greater than actual full-

load current for motor. The high current would, in all probability, cause severe

damage to the brushes, commutator, or windings. Starting resistors

are usually incorporated into the motor design to limit starting current to 125

to 200 percent of full load current .

Ea = KφΩ

Vt = Ea + IaRa

Under standstill (at starting), Ω = 0, i.e., Ea = 0 and ∴ Ia =𝑣𝑡

𝑅𝑎

∴ Ia =(𝑣𝑡− 𝐸𝑎)

𝑅𝑎

Losses and efficiency of DC machines

Power flow in a shunt-connected dc motor

Power flow in a dc generator

81