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UNIT IV

PERMANENT MAGNET

BRUSHLESS DC MOTOR

he conventional DC motors are highly efficient and their characteristics make them

suitable for use as servomotors. But the only drawback is that they need a commutator

and brushes which are subjected to wear and requires maintenance. In a conventional

DC motor, commutation is undertaken by brushes and commutator but in brushlesses DC

motor, it is done by using semiconductor devices such as transistors. The commutation refers

to the process which coverts the input direct current to an alternating current and properly

distributes it to each winding in the armature.

4.1 CONSTRUCTION

• The construction of modern brushless DC motor is very similar to the AC motor. It

consists of two important main parts (ie) stator and rotor.

• The stator which is the stationary parts has got stator frame which encloses the internal

parts of the motor and protects it and it is made up of cast iron (or) steel.

• Beneath the stator frame, stator poles are fixed which are of projective type which may

be laminated or solid piece and it is mostly made of silicon steel material.

• The stator windings are placed on the stator poles .They may be copper windings and

may be single (or) Double layer winding.

• The stator windings are excited by the DC supply through the controllable switches.

• The rotor construction is very simple (ie) it has permanent magnets of one (or) two

number with their poles, which may be projecting pole type.

• The rotor doesn’t carry any winding, brushes (or) commutator segments. The

maintenance is less and inertia is low.

T

4.2 Permanent Magent Brushless DC motor

• The rotor positions are sensed by position sensors such as Hall elements (or) optical

encoders which are fixed on the shaft of the motor.

• The constructional details are shown in the figure 4.1

4.1.1 Basic Principle of operation of PMBLDC Motor

• The basic principle of operation of motor can be easily understand by considering simple

three phase unipolar motor as shown in the figure 4.2

• It uses optical sensors (photo transistors) as position detectors. Three photo transistors

PT1, PT2, and PT3 are placed on the end plates at 120° interval and they are exposed to

light in sequence through a revolving shutter coupled to the motor shaft.

• As shown in the figure 4.2, the south pole of the rotor now faces the salient pole P2 of

the stator, and the phototransistor PT1 detects the light and turns transistor Tr1 ON.

• In this state, the south pole which is created by the salient pole P1 by the electrical

current flowing through the winding w1 is attracting the north pole of the rotor to move it

in the direction of the arrow (CCW).

Fig 4.1

P1 P3

P2

N

S

Stator Frame

Stator Pole

Stator Winding

Permanent Magnet Rotor

Optical Disc with Shutter

Special Electrical Machines 4.3

Fig 4.2

• When the south pole comes to the position to face the pole P1 the shutter which is

coupled to rotor shaft will shade PT1 and PT2 will be exposed to the light and current

will flow through the transistor Tr2.

• When a current flows through the winding W2 and creates a south pole on pole P2, then

the north pole of the rotor will revolve in the direction of the arrow and face the pole P2.

At this moment, the shutter shades PT2 and PT3 is exposed to light.

• Thus the P2 is de-energised and P3 is energized. By repeating such a switching action in

the sequence the permanent magnet rotor revolves continuously.

• In order to reverse the direction of rotation of the motor, we can’t change the supply

terminal because most of the semiconductor devices are unidirectional switches.

• Therefore some circuit is necessary when the motor is to be driven in either direction. So

we go for circuit connections change between the phototransistors and the transistors as

given below.

P1 P3

P2

N

S

Tr1 Tr2 Tr3

PT2

PT3

PT1

Shutter

IB

EB


For CCW rotation

PT1 → TR1

PT2 → TR2

PT3 → TR3

For CW rotation

PT1 → TR3

PT2 → TR2

PT3 → TR1

The switching sequence for CCW & CW is given below.

CCW CW

PT1 1 0 0 1 1 0 0 1

PT2 0 1 0 0 0 1 0 0

PT3 0 0 1 0 0 0 1 0

TR1 1 0 0 1 0 0 1 0

TR2 0 1 0 0 0 1 0 0

TR3 0 0 1 0 1 0 0 1

The above switching sequences between the phototransistors and the transistors can be

implemented by using integrated logic gate circuits. The rotation of stator magnetic field with

respect to the excitation of the winding and the respective photo transistor waveforms are

shown in the figure 4.3 for CCW rotational direction.


Fig 4.3

4.1.2 Three Phases Bipolar driven Motor

• When a three phase brushless motor is driven by a three phase bridge circuit, the

efficiency which is the ratio of the mechanical output power to the electrical input power

is the highest, since in this drive, an alternating current flows through each winding as in

an AC motor.

• This drive is referred to as “Bipolar Drive”. Here bipolar means that a winding is

alternatively energised to become south and north poles.

• Here also we use the optical method for detecting the rotor position, using six

phototransistors placed on the end plate at equal intervals.

• Since a shutter is coupled to the shaft, these photo elements are exposed to light in

proper sequence. The problem is the relation between the ON/OFF state of the

transistors and the phototransistors.

• The simplest relation is setting the logic sequencer in such a way that when a

phototransistor is exposed to light then the corresponding transistor is switch ON.

I1

I2

I3

120 240 360

120 240

240 360

PT1

PT2

PT3

120 240 360

120 240

240 360

P1

P2

P3

Time


Figure 4.4

V

W

U

TR1

TR2

TR3

TR4

TR5

TR6

E

PT1

PT2

PT3 PT6

PT4

PT5 Revolving

Shutter

1

2

3

4

5

6

LO

GIC

SE

QU

EN

CE

R

PT


• Consider that initially PT1, PT4 & PT5 are exposed to the light and electrical current

flows through TR1, TR4 & TR5.

• The terminals U and W have the battery potential and terminal V has zero potential. In

this condition, current will flow from terminal U to V and another current from W to V

as shown in the figure 4.5.

Fig 4.5

• The resultant magnetic field will be at the centre. The rotor is placed in such a position

that the field flux will have a 90° angle with respect to the stator’s magnetic field as

shown in the figure 4.6.

Fig 4.6

• In this condition a clockwise torque will be produced on the rotor and it revolves by 30°,

then PT5 is tuned off and PT6 is ON which makes the stator’s magnetic pole revolve 60°

clockwise.

• Thus when the rotor’s south pole gets nearer to the stator’s south pole, it repeals (or)

goes away further to create a continuous clockwise rotation. The ON/OFF sequence and

the rotation of the transistor are shown in the figure 4.7.

S

N

V

W

U


ON-OFF

Sequence

1 2 3 4 5 6

TR1 1 1 1 0 0 0

TR2 0 0 0 1 1 1

TR3 0 0 1 1 1 0

TR4 1 1 0 0 0 1

TR5 1 0 0 0 1 1

TR6 0 1 1 1 0 0

Fig 4.7

• The rotational direction may be reversed by arranging the logic sequencer in such a way

that when a photo-detector marked with a certain number is exposed to light, the

transistor of the same number is turned OFF and the photo-detector not exposed to light,

the transistor of the same number is turned ON.

• Therefore, in this position, TR2, TR3 and TR6 are ON and the battery voltage appears at

V while U and W has zero electric potential.

• Then the magnetic field in the stator is reversed and rotor’s torque is counter clockwise.

Then the motor rotates about 30 degree and then TR2 turn OFF and TR1 in ON.

• At this point the field revolves by 60 degree as shown in the figure 4.8. So that the rotor

starts rotating continuously in CCW direction

E

D E

O

O E

O

E E

O

E O


ON-OFF

Sequence

1 2 3 4 5 6

TR1 0 1 1 1 0 0

TR2 1 0 0 0 1 1

TR3 1 1 0 0 0 1

TR4 0 0 1 1 1 0

TR5 0 0 0 1 1 1

TR6 1 1 1 0 0 0

Fig4.8

4.1.3 Comparison of Conventional and Brushless DC Motor:

Description Conventional Motor Brushless Motor

Mechanical structure Field magnets on the stator Field magnets on the rotor

Distinctive features Quick response and

excellent controllability

Long lasting, easy

maintenance

E

O E

O

O E E

O

E

O

E O


Winding connection Ring (or) ∆ connection ∆ (or) Υ connected three

phase winding (or) Two

phase winding

Commutation methods Mechanical contact

between brushes and

commutator

Electronic switching using

transistors

Detecting method of rotor’s

position

Automatically detected by

brushes

Hall elements optical

encoders etc.,

Reversing method By reversal of terminal

voltage

Rearranging the logic

sequences

Advantages

1. Brushes maintenance is no longer required and many problems associated with

brushes are eliminated

2. Radio-frequency interference and the sparking associated with the brushes are

eliminated.

3. The conduction of heat through the frame is improved.

4. Increase in the electric loading is possible by providing greater specific torque.

5. The efficiency is likely to be higher than that of DC motor of equal size.

6. Motor length is reduced.

7. The maximum speed of brushless motor is limited by the retention of the magnets

against centrifugal force.

8. Better efficiency, power factor and greater output power.

Disadvantages

1. The need for shaft position sensing

2. Increased complexity in the electronic controller.

3. It is difficult to operate the motor in Field weakening mode providing a constant

power capability at high speed.

4. In very large motors, PM excitation does not make sense because the magnet weight

becomes excessive.

5. Design of logical control circuit is somewhat complex and costlier.


Characteristic features of PMBLDC Motor

1. Rectangular distribution of Magnetic flux in the air gap.

2. Rectangular current waveforms.

3. Concentrated stator windings.

4.2 MAGNETIC CIRCUIT ANALYSIS ON OPEN CIRCUIT

The figure 4.9 shows the cross section of a two pole brushless DC motor having high

energy rare - earth magnets on the rotor. The demagnetization curve of the magnet is shown in

figure 4.10 The axial length of stator and rotor is same. First, we will consider the open circuit

case, ie with no stator current.

Figure 4.9

g

r1


Figure 4.10

Whenever magnetic circuits are used to analysis a magnetic field, the first task is to

identify the main flux paths and assign reluctance (or) permeances to them. Figure (3) shown

the equivalent magnetic circuit (ie) only half of the equivalent circuit is shown because the

lower half is the mirror image of the upper half about the horizontal axis which is an equi-

potential. In the analysis the steel core of the stator and the rotor shaft are assumed to be

infinitely permeable. Each magnet is represented by a Norton equivalent circuit consisting of a

flux generator in parallel with an internal leakage permeance as shown in Figure 4.11

Fig 4.11

Load line Br

Bm Magnetic Flux

Density B (T)

Operating

Point

Demagnetization

curve

- µoHc - µoHm 0

-µoH (T)

Rotor Shaft

Pr1 Pmo φr

Magnet

φg

Rg

φg/2 φg/2

Stator Yoke Stator Yoke

Fm

φm


The generated flux φr is given as

φr = Br.Am

The internal leakage permeance is given as

µ µ= o rec m

mo

m

AP

l

Where Am → Pole area of the magnet

lm → Magnet length in the direction of Magnetisation

Br → Permanent flux density

µrec → Relative recoil permeability (ie) slope /µo

Let us assume outer pole area and inter-pole area as an average value then

1

2

3 2

= − − ×

mm

lA r g lπ

Most of the Magnet flux crosses the airgap via the airgap reluctance Rg given as

. 0

′=

g

g

gR

Aµ

Where ′g → equivalent air gap length including slotting

′g = Kc x g

Assume Kc=1.05

The air gap Ag is the area through which the flux passes as it crosses the gap. and it is given as

( )1

22 2

3 2

= − + +

g

gA r g l gπ

The remaining permeance in the magnetic circuit is the rotor leakage Permeance Pr1 which


represents the paths of magnet flux components that fail to cross the airgap. The rotor leakage

permeance is difficult to estimate because the flux paths are not obvious. The rotor leakage

permeance is quite small, typically 5-20% of the magnet internal permeance and it can be

added to it.

Then

Pm = Pmo+Pr1 = Pmo(1 + þr1)

where þr1 is the normalised rotor leakage permeance = Pr1/Pmo

Equating the MMF across the magnet to the MMF across the airgap.

−= =r g

m g g

m

F RP

φ φφ

gg

m

g

m

r RPP

Φ=Φ

−Φ

gg

m

g

m

r RPP

Φ+Φ

=Φ

gg

mm

r RPP

Φ+=Φ

)1

(

g

m

mg

m

r

P

PR

PΦ

+=

Φ)

1(

)1

(gm

rg

RP+Φ

=Φ

Divide the above expression by Ag

ggm

r

g

g

ARPA

1.)

1(

+Φ

=Φ

(1)

The flux concentration factor Cφ is ratio of magnet pole area to air gap area given as


= m

g

AC

Aφ

φC

AA m

g =

The Airgap flux density can be written as

g

g

gA

BΦ

=

Put (1) in the above equation and we get

ggm

rg

ARPB

1.)

1(

+Φ

=

Put the expression for Ag in the above equation and we get

mgm

rg

A

C

RPB

φ.)

1(

+Φ

=

( )1= ×

+ ×g r

m g

CB B

P R

φ

Where m

rr

AB

Φ=

Similarly we can get for the magnet flux density Bm which can be written as

r1

1 P

1

+=

+

g

m

m g

RB

P RBr


The magnetizing force Hm in the magnet using demagnetization characteristics is given as

−− = r m

m

o rec

B B AHmµ µ

The negative sign signifies a demagnetizing force and indicates that the magnet operates in the

second quadrant of the B-H curve.

The line drawn from the origin through the operating point is called the “Load line” and the

absolute value of its slope, normalized to µo is called the “Permeance Coefficient” (PC) and

written as

r11 P +

=

g

rec

mo g

RPC

P Rµ

If figure (2) is drawn with µoHm along the x-axis instead of Hm then the units of both axes

are same (Tesla) and the PC is the actual slope of the load line. The PC is ratio of magnet

length to effective airgap length and if Cφ is approximately unity and Pr1 is approximately

zero.

In motors with weak flux concentration factor the magnet should operate on open circuit at a

high permeance coefficient to maximize the airgap flux density and the torque per ampere and

to provide adequate margin against demagnetization by armature reaction.

The airgap flux density on open circuit is plotted is Figure 4.12

Fig 4.12

θ 0 30 90 180

270

360

B(T)

0.543 T

120


In practice ,because of fringing, the distribution is not perfectly rectangular and there are

circumferential as well as radial components of B at the edges of the magnets. There will be

appreciable ripple superimposed on the wave form because of slotting of the stator bore.

4.3 TORQUE-SPEED CHARACTERISTICS OF BLDC MOTOR

The torque speed characteristics of the ideal brushless motor can be derived from the

following equations. If the commutation is perfect and the current waveforms are exactly

rectangular and if the converter is supplied from an ideal direct voltage source V, then at any

instant the following equation can be written for the DC terminal voltage.

V = E + RI

Where R is the sum of two phase resistance in series and E is the sum of two phase Emf’s in

series. This equation is exactly the same as that of the commutator motor.

The voltage drops across two converter switches in series are omitted, but they correspond

exactly to the two brush voltage drops in series in the commutator motor.

The torque speed equation is written as

1

= −

o

o

T

Tω ω

Where the no load speed is φk

Vw o = rad/sec

and the stall torque is given by oo IKT φ=

This is the torque with the motor stalled (ie) at zero speed.

The stalling current is given by =o

VI

R

This characteristic is shown in Figure 4.13


Fig 4.13

If the phase resistance is small, as it should be an efficient design, then characteristic is similar

to that of a DC shunt Motor.

The speed is essentially controlled by the voltage V, and may be varied by varying the supply

voltage. The motor then draws just enough current to drive the torque at this speed. As the

load torque is increased, the speed drops and the drop is directly proportional to the phase

resistance and the torque.

The voltage is usually controlled by chopping or PWM. This gives rise to a family of the

torque-speed curve as shown. Note the boundaries of continuous and intermittent operation.

The continuous limit is usually determined by heat transfer and temperature rise. The

intermittent limit may be determined by the maximum rating of semiconductor Devices in the

controller (or) by temperature rise.

In practice the torque speed curve deviates from the ideal form because of the effects of

inductance and other parasitic influences. The curve also shows the possibility of extending

the speed range by advancing the phase (or) the dwell of the conduction period relative to the

rotor position (ie) Angle control.

4.4 TORQUE AND EMF EQUATION OF BLDC MOTOR

Angle Control

Continuous Rated Torque

Intermittent

T

T0

Continuous

Stall Torque

ω ω0 0

Chopping

Control


The basic Torque & Emf equations of the BLDC motor are quite simple and resemble those of

the DC commutator Motor.

A simple concept machine is shown in figure 4.14 Note that the two pole magnet has a pole

arc of 1800 instead of 120

0 analyzed in the previous section.

Fig 4.14

The airgap flux density waveform is ideally a square wave as shown 4.15a. In practice,

fringing causes to be somewhat rounded. The co-ordinate axes have been chosen so that the

centre of a north pole of the magnet is aligned with x-axis (ie) at θ = 0. The stator has 12 slots

and three phase winding. Thus there are two slots per pole per phase. Each phase winding

consists of two adjacent full pitch coils of N1 turns each, whose axes are displaced from one

another by 30 degree.

The winding is a single layer winding. This winding is equivalent with only one coil per pole

per phase having a fractional pitch of 5/6. This is more practical winding than the one

analyzed because it has less bulky end windings and is generally easier to assemble and its

copper losses are lower.

Consider the flux linkage ψ1 of coil a1A1 as the rotor rotates. This is shown in the figure 4.14

Note that θ now represents the movements of the rotor from the reference position in the

figure 4.14 The flux linkage varies linearly with rotor position because the airgap flux density

is constant over each pole pitch of the rotor. Maximum positive flux linkage occurs at 00 and

maximum negative flux linkage at 1800. By integrating the flux density around the airgap, the

maximum flux linkage of the coil can be found as

S N

A1

A2

a2

a1

300

θ0


∫=Ψπ

θθ0

11max1 )( dlrBN

1 1=g

N B r lπ

The variation of flux linkage with θ as the rotor rotates from 00 to 180

0 is given by

( )1 1max1

2

= −

θψ θ ψ

π

The emf induced in coil a1A1 is given by

1 1 11 = − = − × = −

d d dde

dt d d t d

ψ ψ ψθω

θ θ Volts

which gives

1 1 12 ( )= ωge N B l r volts

Where 2N1 = Number of turns per Coil-1 = Nph

Therefore ω11 rlBNe gph= volts

Similarly ω12 rlBNe gph= Volts

The sum of the EMFs ω121 2 rlBNeee gphph =+= volts


Fig 4.15

This represents the magnitude of the square wave Emf ea1 shown in the figure4.15c The

Emf induced in the second coil of phase A is identical, but rotated in phase by 300. If the two

coils are connected in series, the total phase voltage is sum of the two separate coil voltages

and this is shown in figure4.15e The basic effect of distributing the winding into two coils is

to produce a stepped Emf waveform. In practice, fringing causes its corners to be rounded.

The wave form then has the “trapezoidal” shape, ie., characteristics of the brushless DC

Motor. With 1800 magnet arcs and two slots per pole per phase, the flat top of this waveform

is ideally 1500 wide but in practice the fringing field reduces this to smaller value (ie) 120

0.

The magnitude of the flat topped phase emf is given by

θ

θ

θ

θ

θ

ea = ea1 + ea2

150

ea2 30

ea1

180

360 0

B(θ)

ψmax ψ(θ)

30

ib ic ia

-ib -ic -ia -ic

120

a)

b)

c)

d)

e)

f)


ω121 2 rlBNeee gphph =+= volts

where Nph = 2N1 because two coils are assumed to be in series.

In a machine with P pole pairs, the equation remains valid provided Nph is the number of

turns is series per phase and ω is in mechanical radians per second.

In the figure 4.15f shows an ideal rectangular waveform of phase current, in which the current

pulses are 1200 electrical degree wide and of magnitude I. The positive direction of current is

motoring current. The conduction periods of the three phases are symmetrically phased so as

to produce a three phase set of balanced 1200 square waves.

If the phase windings are star connected as shown, then at any time there are just two phases

and two transistors conducting.

During any 1200 interval of phase current, the instantaneous power being converted from

electrical to Mechanical is

P = ω Te = 2 e I

The number 2 in this equation arises from the fact that two phases are conducting.

Te = 4Nph Bg l r1 I (Nm)

The above equation is valid for any number of pole pairs. The similarity between the brushless

motor and the commutator motor can now be seen. Writing E = 2 e to represent the

combined Emf of two phases in series, the Emf and torque equations can be written in the

form

E = K φ ω and T = K φ I

Where K = 4 Nph and φ = Bg r1 π l

Where K is the armature constant and φ is the flux. In practice, of course, none of the ideal

conditions can be perfectly realized. The main result of this is to introduce ripple torque but


the basic relationships of Emf proportional to speed and torque proportional to current remain

unchanged.

4.5 Controller for BLDC Motor

• The general structure of a controller for a brushless PM DC Motor is shown in Figure

4.16 This schematic diagram shows the functional blocks required to control the drive in

chopping control range (ie) commutation angle is fixed.

• The rotor shaft position is sensed by a Hall effect sensor a slotted optical disk (or) some

other transducers. These signals are decoded by combinational logic to provide the firing

signal for 120 degree conduction on each of the three phases.

• The commutation logic has six outputs which control the upper and lower phase legs of

the transistors. The basic forward control loop is a voltage control, implemented by a

monostable clocked at a fixed reference frequency, which is typically a few kHz.

• The duty-cycle (or) off time is controlled by an analogue voltage reference that

represents the desired speed.

• The PWM is applied only to the lower phase legs of the transistors and this reduces the

current ripple and also avoid the need for wide bandwidth in the level shifting circuit

that feeds the upper leg transistors.

A B C

1

4

3

6

5

2

V


• The use of AND Gates as a simple way of combining the commutation and chopping

signals to the lower transistors.

• From the control point of view the brushless motor is similar to the dc commutators

motor, as the simple torque & voltage equations are similar.

• It is possible to implement current & speed feed back in the same way as in the dc

motor. It may be necessary in either (or) both loops to improve stability and transient

response. Therefore wide range of speed control and torque control using relatively

simple techniques that are familiar with commutator motors.

• The instantaneous current in the brushless PM motor is regulated in each phase by a

hysteresis type regular which maintain the current with in the adjustable limits.

• This is called “Current mode” control and several algorithms are possible in each phase

and their bandwidth must obviously be wider than that of the sensing resistor as shown.

• The speed feedback from a Tacho-generator TG, can also be derived from the shaft

position sensor by a frequency to voltage converter. This technique only works at high

speed.

• Many of the functions of the circuit can be performed digitally and it is increasingly

common to have a serial communication interface that permits the system to be

computer controlled.

• In high performance systems, the special purpose decoding circuit are used.

• So it is possible to fine tune the firing angles and the PWM control as a function of

speed and load to improve various aspects of performance such as efficiency, dynamic

performance and speed range.


Figure4.16 Structure of controller for PM BLDC Motor drive


4.6 COMMUTATION PROCESS

The function of a brush in a DC motor is replaced with two switches (or) transistors. This is

shown in the below figure4.17

Fig4.17

Consider the commutator segment A which is in any of the three states.

i. Touching the positive brush

ii. Touching the negative brush

iii. Touch neither

Again we have to give importance to the action of the coil inductance in the connected

DC motor. In the figure 4.18 shown below, the switch S1 is closed and the current is supplied

to the coil. In the figure4.19 the current is apparently cutoff when S1 is opened, but because of

inductance ,high voltage across the airgap is produced in the switch and therefore spark occurs

and the current continues to flow through the airgap for a short time. In the figure 4.20 shown

below indicated that the spark can occur when the brushes separate from the commutator

segment A and move to the next segment. If the sparks are weak then it will not produce

serious damage to the commutator segment (or) brush else this spark will damage them in the

repeated operation.

A A A


Fig 4.18 Fig 4.19 Fig 4.20

A smaller inductance will reduce sparking. That is, the number of turns of the coil

connected to one commutator segment should be small for sparkless commutation. In

conventional motors, the windings are separated into many coils, and in some motors, there

are more coils than the commutator segments.

The figure shown above is a commutator circuit which uses two transistors. When

Tr1=S1 is ON and a current is being supplied from the power supply through this transistor to

the coil1. When Tr1 turn OFF the current will flow through diode D2 for a transient period

until it falls to zero. If this diode is not provided, the current will decrease all at once. Hence a

high voltage which is di

Ldt

will be produced in the coil and will be applied across the

collector and emitter of Tr1 and will damage the transistor. Thus, in a brushless DC motor the

electromagnetic energy stored in the coil can return to the power supply through this diode.

Therefore, there is no spark. We see that diode D2 is needed to protect Tr1 from being

damaged and D1 protects Tr2=S2.

A A A

Spark

S2

S1

B

Direction of commutator

movement

spark


4.7 Detection of rotor position and the use of Hall elements:

A brushless DC motor uses some means of detecting the pole / position on its rotor. The

position sensors used nowadays are:

1. Hall elements

2. Light emitting diodes and phototransistors.

3. Inductor sensitive to inductance variation.

4.7.1 Hall elements:

When an electrical current Ic flows downwards in a semiconductor material which is

placed in a magnetic field perpendicular to the material surface, an electromotive force VH is

created in the material in the direction perpendicular to both current Ic and magnetic Induction

B. Since the electromagnetic force acts on charged particles (electrons (or) holes) according to

Fleming’s Left Hand rule, the charged particles are biased to the left side of the semiconductor

material and the polarity of VH depends upon whether it is p–type (or) n–type. The magnitude

of the electromotive force VH which is called “Hall Voltage” is given by the equation

H C H

1V B I R

d=

Where

RH is the Hall constant (m3c

-1)

IC is the electrical current(A)

B is the flux density (T)

d is the thickness of the material (m)


Fig 4.21

The Hall effect is strong in some specific metal compounds and semiconductos.

Semiconductor devices which are made for use in detecting magnetic fields are called Hall

elements. In modern brushless motors, n-type InSb (Indium – Antimony) is extensively used,

as well as GaAs (Gallium – Arsenide).

4.7.2 Principle of position detection using Hall elements:

An equivalent circuit for a Hall element expressed as a four terminal network as shown

figure 4.22. When a current Ic which is called the control current or bias current, flows from

terminal 3 to 4 in the Hall element exposed to a magnetic field which is perpendicular to the

element plane, a voltage VH is generated across the terminal 1 and 2 as explained before.

When terminal 4 is taken as the reference point, the potentials at terminals 1 and 2 are

VH/2 and -VH/2 respectively where R1 = R2 and R3 = R4 assumed. Moreover the polarity

reverses as the flux direction reverses.

+ _

VH

Ic

d

IH

B


Fig 4.22

Thus when a Hall element is placed near a permanent magnet rotor, the Hall element

can accurately detect the pole position and the flux density providing output voltage VH1 and

VH2.

4.73 Practical methods of position detection:

The simplest brushless DC motor using one Hall element placed is shown in figure

4.23 & 4.24 The output signal from the Hall element operate the two transition Tr1 and Tr2 to

control the electrical currents in stator windings W1 and W2.

Fig 4.23

+ _

W1 W2

Tr1 Tr2

ic

iB

+

_

+ _

VH

Ic

d

B

R1 R4

R3

R2

VH1 VH2

1 2

3

4


Figure 4.24

i. The Hall element detects the north pole of the rotor magnet and W2 is energized to

produce the south pole which drives the rotor in the CCW direction

ii. Since no magnetic field is applied to the Hall element in this positional relation, both

transistors are in the OFF state, and no currents flow in W1 and W2. The rotor

continues to revolve due to inertia.

iii. The Hall element detects the south pole of the rotor and winding W1 is energized to

create the south pole which attracts the north pole of the rotor to continue the CCW

motion.

The state of the rotor as it revolves is shown below.

(a) (b) (c)

4.7.4 Elimination of Dead point in two phase motors

As we have seen before, the two phase motor uses one Hall element to detect the rotor

position. This type of motor has the following two draw backs.

i. There are two dead points at which the hall elements does not experience the

magnetic flux, and as a result no current flows in the windings to produce a torque.

Hence, when the motor carries a frictional load, there is a possibility that it will

N

S N S

ω1 ω2

N

S

S

N

W1

W2

B


stop at a dead point and be unable to start again. When the frictional load is small

the rotor may be able to pass through the dead points due to inertia.

The figure shown below gives the relationship between torque and rotational angle.

ii. Since the back Emf is small at the low torque positions, a large current will flow

and increase the conduction loss. Hence the motor efficiency which is the ratio of

the output mechanical power to the input electrical power is not high.

As far as the elimination of the dead points, there are two major methods.

i. First one uses a polyphase structure (ie) three phase bipolar driven scheme.

ii. Second one , space harmonic magnetic field.

The principle of operation of Three Phase Bipolar driven BLDC motor is already seen using

photo-transistor. These photo-transistors are replaced by Hall element.

4.7.5 Space Harmonic Type:

Consider the motor shown in the figure 4.25 which is an improved type of the plain

two phase two pole motor. An additional four pole magnet is coupled to the rotor and stator

also has a four pole magnet in a positional relationship as shown in the figure4.25. Figure 4.26

shows the relationship between the torque and revolving angle of this motor. The curve A in

the ordinary two pole two phase motor has two dead points. Curve B is the space harmonic

torque curve created by the newly installed permanent magnets.

The two pole torque is always counter clockwise because of the commutation by the

Hall elements. The four pole torque varies alternatively in both directions (ie) CW and CCW.

Since the overall torque, which is shown by curve C, is the sum of both torques, it does not

have any dead points.

Torque

(CCW)

Torque

(CW)

Dead point Dead point θ


This type of motor in which two additional magnets are set up has not been used in practical

application. The methods of exploiting this effect are as follows.

i. The method which has auxiliary salient poles in its stator and an auxiliary magnet

in the rotor, the magnet having twice as many magnetic poles as the main magnet.

ii. The method which uses a second harmonic magnetization in the rotor, the winding

pitch being moved from 1800 on the stator.

iii. The method which uses a non uniform airgap to create a second harmonic torque.

Figure 4.25

Fig 4.26

Dead point Dead point

Curve A

Curve B

Curve C

θ

(CCW)

↑ Torque

↓ (CW)

N

N

S S

N

N

S S

N

S


PROBLEM

1. A brushless DC. motor has the cross section r1 = 30 mm, lm = 80mm, g = 0.8 mm

and axial length l = 35mm. slotting is rejected and a single full – pitch stator coil

with 30 turns. The cross magnet has Br = 0.35T, Prec = 1 and H0 = 278 KA/m. The

current in the stator cell is zero.

i. Calculate Bg in the average at the centre of the magnet arc ie on the direct

axis.

ii. Determine the m.m.f across the magnet and the internal magnetizing force

Hm.

iii. Calculate the flux linkage of the coil.

iv. Determine the peak value of e.m.f induced in stator coil if the rotor rotates at

4000 r.p.m.

April / May 2003

Given data r1 = 30 mm, lm = 8 mm,

g = 0.8 mm, l = 35 mm,

N = 30, Br = 0.35T,

µrec = 1, Hc = 278KA/m.

To find i. Bg in the airgap

ii. MMF across the magnet and internal magnet force

iii. Magnetizing force Hm

iv. Flux linkage in the coil (stator)

v. Peak value of e.m.f induced.

Solution

r

g

m g

C BB

1 P R

φ=+

cg

0 g

K gR

A=

µ

( ) ( )g 1 g32A r Z 2g

3 2 = π − + +

l

( ) ( )2 30 0.4 1.6 35 1.63

= π − + +

Rg = 23275 mm2

3

og 7 6

o g

K g 1.05 0.8 10R

A 4 10 0.2327 10

−

− −

× ×= =

µ π× × ×


5

gR 2.87 10= ×

o rec mmo

m

AP

µ µ=

l

mm 1

2A r g

23

= π − −

ll

[ ]230 0.8 4 35

3π − − ×=

2

mA 1847mm=

o rec mmo

m

AP

µ µ=

l

7 6

3

4 10 1 1847 10

8 10

− −

−

π× × × ×=

×

7

mA 2.9 10−= ×

[ ]m mo r1P P 1 P= +

m

g

A 1847C 0.7935

A 23275φ = = =

[ ]72.9 10 0.1−= × +

7

mP 3.19 10−= ×

r

m g

C BB

1 P R

φφ =

+

( )7 7

0.7935 0.35

1 3.19 10 2.89 10− −

×=

+ × × ×

Bg = 0.248 T

Flux crossing the airgap φr = 0.35 x 1847 x 102 = 0.646 x 10

3.

b. MMF across the magnet

r g

m

m

FP

φ − φ=

rg 7 5

m g

0.646

1 P R 1 3.19 10 2.85 10−

φφ = =

+ + × × ×

g 0.5918φ =

m 7

0.646 0.5918F

3.19 10−

−=

×

mF 165.9 AT=


C. r mm

0 rec

B BH

−− =

µ µ

r1 m moP P P= −

7 73.19 10 2.9 10− −= × − × 7

r1P 0.29 10−= ×

r1 g

m r

m g

1 P RB B

1 P R

+= ×

+

( )

( )

7 5

7 5

1 0.29 10 2.8780.35

1 3.19 10 2.87 10

−

−

+ × ×= ×

+ × × ×

Bm = 0.323

m 7

0.35 0.323H

4 10 1−

−− =

π× ×

Hm = 20.7 kA/m

d. Induced emf E = 2Bg l r Nph ω

640002 0.248 35 3030 10 2

60

−= × × × × × × π

E = 6.5445 volts

2. A PM brushless DC motor has a torque constant of 0.12 Nm/A referred to the DC

supply.

a. Estimate its no load speed in rpm when connected to a 48 V DC supply.

b. If the armature resistance is 0.15ΩΩΩΩ/phase and the total voltage drop in the

controller transistor is 12 V. Determine the stall current and stall torque.

c. The DC current is 8.2 amps when the motor is star connected and has two

phases on at any instant with a total of 2V dropped across the two conducting

transistor in series; this voltage drop can be assumed constant. The friction

torque has been separately measured as 0.046 Nm at this speed. If the supply

voltage is 48 V dc. Calculate the efficiency of the complete drive and the

separate power loss components due to

i. Voltage drop in the transistor

ii. Winding resistance

iii. Friction and

iv. Iron loss


If the iron loss is modeled by means of a resistor connected in series parallel with

each phase of the motor determine the value of this resistance.

Given data

Torque constant, Kφ = 0.12 N-m

Supply voltage V = 48 volts

Armature resistance/ phase, Ra = 0.15 ohm

Voltage drop in the controller transistor Vdrop = 2 volts

Mechanical power to the load = 3000 rpm

DC supply current = 8.2 amps

Friction torque = 0.046 N-m

To find a. No load speed , no in rpm

b. Stall torque To and stall current Io.

c. Efficiency of the3complete drive

i. Power loss due to voltage drop in transistor

ii. Power loss due to winding resistance

iii. Power loss due to friction

iv. Total iron loss

Solution

a. No Load speed, o

V

Kω =

φ

48 rad400sec0.12

= =

o

400 60n 3820 rpm

2

×= =

π

no = 3820 rpm

b. During this operation two transistor will be connected in series

R = 0.15 + 0.15 = 0.3

Stall current, o

V 48 2I 153.3 amps

R 0.3

−= = =

Io = 153.3 amps

[48-2 = subtract the voltage drop from supply voltage]

Stall torque , To = KφIo

= 0.12 x 153.3

To = 18.396 Nm

c. Efficiency of the complete drive Mechanical output

100Electrical input

= ×


( )330

10048 8.2

η = ××

[Q V = 48V, I 8.2 A]

83.84 %η =

i. Power loss due to voltage drop in the transistor = Voltage drop x current

= 2 x 8.2

= 16.4 watts

ii. Due to winding resistance = I2R

= 8.22 x 0.15

= 20.2 watts

iii. Power loss due to friction 2 NT

P60

π=

2 3400 0.046

60

π× ×= [Friction Torque T = 0.046 Nm given ]

iv. Iron loss = Electrical input – [power loss due to [transistor + winding resistance

+ friction + mean output]

= 4.8 x 1.2 – [16.4 + 20.2 + 330]

= 393.6 – 383.6

Iron loss Pi = 10.6 watts

v. The value of resistance Pi = I2R = 10.6

2

10.6R 0.157 omhs

8.2= =

3. A permanent magnet DC commutator motor has a no load speed of 6000 rpm when

connected to a 120 V supply. The armature resistance is 2.5 ΩΩΩΩ and rotational and

iron losses may be neglected. Determine the speed when the supply voltage is 60V

and the torque is 0.5 Nm.

Given data

No load supply no = 6000 rpm

Supply voltage = 120 volts

Armature Resistance = 2.5 ohm

To Find

i. Speed when the supply voltage = 60 Volts and torque 0.5 Nm

Solution

Stall current oa

VIR

=


12048 amps

2.5= =

Stall torque, To = Kφ Io

But, o

VKφ =

ω

1200.1909

60002

60

= =π×

oo

2 N

60

π ω =

Torque constant Kφ = 0.1909

To = 0.1909 x 48

To = 9.167 Nm

The speed at V = 120 volts

T = 0.5 Nm

o

o

T1

T

ω = ω −

o

o

TN n 1

T

= −

0.5N 6000 1

9.167

= − [no = 6000 and T = 0.5 Nm given]

N = 5672 rpm

At 60 V and 0.5 Nm

1 1V Kφ= ω

120 K 5672φ= × _____ (1)

2 2V Kφ= ω _____ (2)

From equation 1 and 2 we write

2

60N 5672

120= ×


N2 = 2863 rpm

4. A permanent magnet DC commutator motor has a stall torque of 1 NM with a stall

current of 5 A. Estimate it’s no load speed in rpm when fed from a 28 V DC voltage

supply.

Given data Stall torque To = 1 Nm

Stall current Io = 5 amps

DC supply voltage V = 28 V

To find

No load speed (no) in rpm

Solution

V = Kφωo To = KφIo

28 = Kφωo 1 = Kφ x 5

o

28

0.2ω = Kφ=0.2 Nm/A

ωo = 140 rad/sec

o2 n140

60

π=

o

140 60n

2

×=

π

no = 1336.9 rpm

5. In problem 4 during an overload the motor temperature becomes excessive and the

magnet lose 12 percent of their permanent flux density. If the armature resistance

is 0.8 ΩΩΩΩ. Determine the speed at which the motor will run when the load torque is

0.3 Nm. Assume a total brush voltage drop of 2 V and a supply voltage of 28 V DC.

Ignore friction and other losses

Nov/Dec 2005

Given Data

Reduction of permanent flux density = 12 % = 0.12

Torque constant Kφ` = 0.12 x 0.88

New torque constant Kφ` = 0.176

Load Torque T = 0.3 Nm


Total brush voltage drop Vdrop = 2 volts

To find : Determine the motor speed in rpm

Solution:

drop'

o '

V V 28 2147.7 rad/sec

K 0.176

− −ω = = =

φ

'

o '

o

T1

T

ω = ω −

_____ (1)

( )' ' '

o oT K I= φ

28 2

0.1760.8

− = ×

dop'

o

V VI

I

− =

'

oT 5.726 Nm=

(1) ⇒ 0.3

147.7 15.726

ω = −

ω = 140 rad/sec

2 n

14060

π=

140 60

n2

×=

π

N = 1337 rpm

Brushless DC motor cross section

r1

lm

microsoft word - unit iv blpmdc

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