DC Machinery Fundamentals

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Chap. 8: DC Machinery Fundamentals
This chapter will firstly explain the principles of dc machine
operation using simple examples before considering some of thecomplications that occur in real dc machines.
8.1. A simple rotating loop between curved pole facesThe simplest rotating dc machine is shown below.
It consists of:
A rotating single loop of wire in a slot carved in acylindrical ferromagnetic core rotor
A pair of magnetic poles (north and south) statorThe magnetic field is supplied by the stator, i.e. N and S poles.
The air gap between the rotor and stator is of constant width
(i.e.gap is the same everywhere under the pole faces).
magnetic flux density is constant everywhereunder the pole surface.
From Chapter 1, we know thatirongap >> , hence
magnetic flux takes shortest possible path through
the air gap (i.e. perpendicular to the rotor surface).

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The voltage induced in a rotating loop
If the rotor is rotated, a voltage will be induced in the wire
loop. (Note: No voltage source applied to the rotor terminals.)
To determine the total voltage etot on the loop, each segment of
the loop (as shown in the figure above) has to be examined
separately and the resulting voltages summed up (see page 33).
The voltage on each segment is given by ( ) lBveind
rr
r
=
Hence, the total induced voltage in the loop:
=
+++==
edgespolethebeyond0
facespoleunder the2vBl
e
eeeeee
ind
addccbbatotind
When the loop rotates through 180,
segment ab is under the north pole instead of the southpole
direction ofeba and edc but magnituderemains constant
(8.1)
To and front view of the sim le dc machine

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The resulting voltageetotis shown in the figure below.
An alternative expression for eind that relates the behaviour of
the single loop to the behaviour of larger, real dc machines can
be obtained by examining the figure below:
The tangential velocity v of the loop edges can be expressed as v
= r. Therefore,
=
edgespolethebeyond0
facepoleunder the2 Blre
ind
Output voltage of the simple dc machine
Derivation of an alternative form of induced voltage equation.

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For a 2pole machine, if we assume that the gap between the
poles is negligible (the gaps at the top and bottom of the
diagram) then the surface area of the pole can be written as
rlrl
AP
== 2
2
Therefore,
=
edgespolethebeyond0
facepoleunder the2
BAe
P
ind
Since the air gap flux density B is constant everywhere under
the pole faces, the total flux under each pole is:
BAP=
Thus, the final form of the voltage equation is:
=
edgespolethebeyond0
facepoleunder the2inde
In general, the voltage in any real machine will depend on the
same three factors:
the flux in the machine
the speed of rotation a constant representing the machine construction
Getting DC voltage out of the rotating loop
The voltage out of the loop is alternating between a constant
positive value and a constant negative value, i.e. ac voltage.
How can this machine be modified to produce a dc voltage?
(8.2)

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By using a mechanism called commutator and brushes
(see below)
two semicircular conducting segments (commutatorsegments) are added to the end of the loop
two fixed contacts (brushes) are placed at an angle suchthat at the instant the voltage in the loop is zero, the
contacts shortcircuit the two segments
Every time voltage of the loop changes direction, the
contacts also switch connections and the output of
the contacts is always built up in the same way (as
shown below).
This connectionswitching process is known as
.
Output voltage of a dcmachine with
commutator and brushes.

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The induced torque in the rotating loop
If the simple machine is connected to a battery, as shown below,
how much torque will be produced in the stationary loop
when the switch is closed and current is allowed to flow?
As before, the approach is to examine each segment of the loop
and then sum the effects of all segments (see page 34).
The force oneach segment is given by BliFrrr
= .
And the torque on the segment is given by
sinrFFrT ==r
r
, where is the angle between rr
and Fr
.
When the loop is beyond the pole edges, T= 0 (since Br
= 0).
The simple dc
machine
connection for
derivation of
induced torque
equation.
Front view ofthe machine
(the iron core
is omitted for
clarity).

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Hence, the total induced torque in the loop is:
=
+++==
edgespolethebeyond0
facespoleunder the2rilBT
TTTTTT
ind
dacdbcabtotind
By employing the facts that rlAP
and BAP= , the torque
expression can be reduced to:
=
edgespolethebeyond0
facepoleunder the2 iT
ind
In general, the torque in any real machine will depend on thesame three factors:
the flux in the machine the current in the machine a constant representing the machine construction
(8.3)
(8.4)

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commutators

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8.2. Commutation in a simple fourloop DC machine(see page 35 for the detailed explanation)
The resulting terminal voltage as a function of time is shownbelow:
Note: this is a better approximation to a constant dc level thanthat produced by the single rotating loop of Section 8.1.
As the number of loops on the rotor increases, the
approximation to a perfect dc voltage gets better and better.
In summary:
Commutation is the process ofswitching the loop connections
on the rotor of a dc machine just as the voltage in the loop
switches polarity, in order to maintain an essentially constantdc output voltage.
Commutator segments = rotating segments to which the loops
are attached. Typically made of copper bars.
Brushes = stationary pieces that ride on top of the moving
segments. Typically made up of a mixture containing graphite,
so that they cause very little friction.
The resulting output voltage of the fourloop twopole dc machine.

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8.3. Commutation and armature construction in real DCmachines
In real dc machines, there are several ways in which the loops
on the rotor (also called the armature) can be connected to itscommutator segments.
These different connections affect:
Number of parallel current paths within the rotor The output voltage of the rotor Number and position of the brushes riding on he
commutator segments
The rotor coils
Regardless of the way thewindings are connected to the
commutator segments, the rotor
windings consist ofdiamond
shaped preformed coils which
are inserted into the armature
slots as a unit (see figure).
Each coil consists of a number
of turns (loops) of wire, each turn taped and insulated from
the other turns and from the rotor slot.
Each side of a turn is called a conductor and the number of
conductors on a machines armature is given by:
cCNZ 2=
whereZ = number of conductors on rotor
C = number of coils on rotor
Nc = number of turns per coil
Normally, a coil spans 180 electrical degrees. This means that
when one side is under the centre of a given magnetic pole, theother side is under the centre of a pole ofopposite polarity.
(8.8)
The shape of a
typical
preformedrotor coil.

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The physical poles may not be 180 mechanical degrees apart
but the magnetic field has completely reversed its polarity in
travelling from under one pole to the next.
The relationship between the electrical angle and mechanical
angle in a given machine is given by:
me
P
2=
where e = electrical angle, in degreesm = mechanical angle, in degreesP = number of magnetic poles on the machine
If a coil spans 180 electrical degrees, the voltages in theconductors on either side of the coil will be exactly the same in
magnitude but opposite in direction at all times. Such a coil is
called a fullpitch coil.
A fractionalpitch coil spans less than 180 electrical degrees,
and a rotor winding wound with fractionalpitch coils is called achorded winding.
The amount of chording in a winding is described by a pitchfactorp, defined by:
%100
180
coilofangleelectrical
=p
Most rotor windings are twolayer windings, meaning that sidesfrom two different coils are inserted into each slot. One side of
each coil will be at the bottom of its slot, and the other side will
be at the top of its slot.
(8.9)
(8.10)

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Connections to the commutator segments
There are a number of ways in which the rotor windings can be
connected to the commutator segments. The different winding
arrangements have different advantages and disadvantages.
Commutator pitch yc = distance (in number of segments)
between the commutator segments to which the two ends of acoil are connected.
Progressive winding (yc =1) =
the end of a coil (or a set number
of coils) is connected to acommutator segment ahead of
the one its beginning is
connected to.
Retrogressive winding ( yc =1)
= the end of a coil (or a set
number of coils) is connected toa commutator segment behind
the one its beginning is
connected to.
If everything else is identical, the rotation direction of a
progressivewound rotor is opposite to that of a retrogressive
wound rotor.
Rotor (armature) windings are further classified according to the
plex of their windings:
1.A simplex rotor winding is a single, complete, closedwinding wound on a rotor.
2.A duplex rotor winding is a rotor with two complete andindependent sets of rotor windings. Each of the windingswill be associated with every other commutator segment:
one winding will be connected to segments 1, 3, 5, etc.,
A coil in a progressive winding. A coil in a retrogressive winding.

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and the other winding will be connected to segments 2, 4,
6, etc.
3.A triplex winding will have three complete andindependent sets of windings, each winding connected toevery third commutator segment on the rotor.
4.All armature with more than one set of windings are calledmultiplex windings.
Finally, armature windings are classified according to thesequence of their connections to the commutator segments:
Lap winding Wave winding Frogleg winding (combines lap and wave windings in a
single rotor)
Note: for individual characteristics, advantages and
disadvantages of these windings, please refer to Chapman
textbook page 493 502.
8.4. Problems with commutation in real machinesIn practice, the commutation process is not as simple as
described theoretically in Sections 8.2 and 8.3. There are twomajor effects that disturb the commutation process:
I. Armature reactionII.L di/dt voltagesThis section explores the nature of these problems and the
solutions employed to mitigate their effects.

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I. Armature reactionIn real DC machines, the magnetic field is provided by field
windings on the stator, i.e. no magnets.
If the magnetic field windings of a dc machine are connected
to a power supply and the rotor of the machine is turned by
an external source of mechanical power, then
a voltage will be induced in the conductors of the
rotor.
This voltage will be rectified into a dc output by the action ofthe machines commutator.
When a load is connected to the terminals of the machine:
current will flow in its armature windings. this current produces
which will distort the original magnetic field from the
machines poles
Armature reaction = of the flux in the
machine as the load is increased.
It causes two serious problems in real dc machines.
Problem 1: Neutralplane shift
Magnetic neutral plane = the plane within the machine where
the velocity of the rotor is exactly parallel to the magnetic
flux lines, so that in the conductors in the plane.
To understand the problem of neutral plane shift, the twopole
dc machine shown below is employed.

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The flux is distributeduniformly under the polefaces.
The rotor windings havevoltages built up as shown on
the left.
Hence, the neutral plane isexactly vertical.
Now suppose a load is connected such that the machine actsas a generator (i.e. no power supplied to rotor windings).
Current flows out of thepositive terminal of the
generator, as shown on the
left.
This current flow produces a magnetic field from the rotorwindings, as shown in picture (c). (direction: righthand rule)..
This rotor magnetic field affects the original magnetic fieldfrom the poles that produced the generators voltage in the
first place.
In some places under thepole surfaces, it subtractsfrom the pole flux.
In other places, it adds to
the pole flux.

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The overall result is that themagnetic flux in the air gap of
the machine is skewed as
shown on the left.
Notice that the place on the rotor where induced voltage in aconductor would be zero (i.e. the neutral plane) has
.
In a dc generator magnetic neutral plane shifts
the direction of rotation.
In a dc motor magnetic neutral plane shifts
to the direction of
rotation (due to reverse in rotor currentdirection, hence flux add and subtract
at opposite corners from that shown in
picture (d) above).
Note: amount of shift depends on the machine load.
What is the effect of neutralplane shift?
The brushes must short out commutator segments just atthe moment when the voltage across them is zero.
If the brushes are set to short out conductors in the verticalplane, then the voltage between the segments is zero until
the machine is loaded.
When the machine is loaded, the neutral plane shifts andthe brushes short out commutator segments withfinite
voltage across them.

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The result is:
Current flow circulating between the shorted segments Large sparks at the brushes when the current path is
interrupted as the brush leaves the segment
End result: at the
brushes.
This is a very serious problem. It leads to:
Drastically reduced brush life. Pitting of the commutator segments. Greatly increased maintenance costs.
Notice that this problem cannot be fixed by placing the
brushes over the fullload neutral plane, as it will then causesparks at no load.
In extreme cases, the neutralplane shift can lead to flashover
in the commutator segments near the brushes.
(The air near the burshes in a machine is normally ionized as aresult of sparking on the brushes. Flashover occurs when the
voltage of adjacent comutator segments gets large enough to
sustain an arc in the ionized air above them. If flashover occurs,the resulting arc can even melt the commutators surface.)
Problem 2: Flux weakening
To understand flux weakening, refer to the magnetisation curve
shown below:

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Most machines operate near the saturation point (knee ofcurve).
At locations on the pole surface where rotor mmfadds to thepole mmf, only a small increase in flux occurs.
But at locations on the pole surface where rotor mmfsubtracts from the pole mmf, a decrease influx occurs.
The net result:
The total average flux under the entire pole face is
decreased (see figure below).

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Flux weakening causes problems in both generators and motors.
In generators, the effect is reduced voltage supplied by the
generator for any given load.
The fluxweakening effect can be more serious for motors:
when flux is decreased, its speed but increase in motor speed can increase its load, resulting
in more flux weakening (hence, speed increase further)
it is possible for some shunt dc motors to reachcondition, where motorspeed just keeps
on increasing until the machine is disconnected from the
power line or until it destroys itself.

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II.L di/dt voltagesThe L di/dt voltages occur in commutator segments that are
being by the brushes.
Also sometimes known as inductive kick.
To understand this problem, refer to the figures belowrepresenting a series of commutator segments (a, b, c, d ..) and
the conductors (or coils) connected to them.
Assuming that the current in the brush = 400A, the current in
each path is 200A (since current split into two coils).
When a commutator segment is shorted out, the current flow
through the commutator segment must reverse.
How fast must this reversal occur?
Assuming the machine is rotating at 800 rpm and that there are50 commutator segments (a reasonable number for a typical
motor), each commutator segment moves under a brush andclears it again in t= 0.0015s.
Calculations:
800 rpm = 83.78 rad/s
1 segment takes up 0.126 rad ( = 50 / 2 )
Time to clear 1 segment = 0.126 / 83.78 = 0.0015s

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Therefore, the rate of change in current with respect to time in
the shorted loop must average:
With even a tiny inductance L in the loop, a very significant
inductive voltage kickv = L di/dtwill be induced in the shorted
commutator segment.
This high voltage naturally causes sparking in the brushes ofthe machine, resulting in the same arcing problems as caused
by the neutralplane shift.
Solution to the problems with commutation
The following three approaches have been developed to partiallyor completely correct the problems of armature reaction and L
di/dtvoltages:
I. Brush shiftingThis method attempts to stop the sparking at the brushes caused
by the neutralplane shifts andL di/dteffects.
A/s667,266s0015.0
)200(200=
=
dt
di
The current reversal in the
coil undergoing
commutation as a function
of time for both ideal and
real commutation, with thecoil inductance taken into
account.
ideal
commutation
real
commutation
 200 A

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Since the neutral plane shifts, hence shift the brushes (to the
new neutral plane position) to stop the sparking.
But there are several serious problems associated with this
method:
The neutral plane moves with changes in load and the shiftdirection reverses when going from motor to generatoroperation.
Even though brush sparking is stopped, brush shiftingaggravates the fluxweakening effect of armature reaction
in the machine. This is true because of two effects:
i. The rotor mmf now has a vector component that opposesthe mmf from the poles (see figure below).
ii. The change in armature current distribution causes theflux to bunch up even more at the saturated parts of the
pole faces.
However, this method is obsolete.
Before brush shifting (i.e.
brushes over the vertical lane)
After brush shifting (i.e. brushes
over the shifted neutral lane)

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It is only used in very small motors where other better solutions
are not economical.
II. Commutating poles or interpolesBasic idea:
If the voltage in the wires undergoing commutation can be
made zero, then there will be no sparking at the brushes.
Method:
Small poles calledcommutating poles or interpoles are placed
midway between the main poles anddirectly
over the conductors being commutated.
The flux provided by the commutating poles will exactly
the voltage in the coils undergoing
commutation. Therefore,
there will be no sparking at the brushes.
The commutation poles do not change the operation of themachine because they are so small and only have effect on the
few conductors undergoing commutation.
Hence, armature reaction under the main pole faces is
unaffected, i.e.
flux weakening in the machine is unaffected by
the commutating poles (i.e. flux weakening problem isnot resolved).
How is cancellation of voltage in the commutator segments
accomplishes for all values of load?
This is done by connecting the interpole windings in series
with the windings on the rotor.

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As the load increases
The rotor current (IA) increases
Magnitude of neutral plane shift
and
size ofL di/dt effects increases
Increases voltage in conductors undergoing commutation
(which was supposed to be zero)
BUT
Interpole flux increases as well
(due to series connection, i.e. IA, Iinterpole)
Gives larger voltage in the conductor which
opposes voltage due tothe neutralplane shift
The net result is:
Commutating pole cancels the neutralplane
shift andL di/dt effects over a broad range of loads.
Note: Interpoles work for both motor and generator
operation.
A dc machine
with interpoles

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What polarity must the flux in the interpoles be?
Interpoles must induce voltage in the conductors undergoing
commutation that is opposite to the voltage caused by the
neutralplane shift andL di/dteffects.
For a generator,
neutralplane shifts in the direction of rotation. conductors undergoing commutation have the same
voltage polarity as the pole they just left.
to cancel this effect:Interpoles must be of the samepolarity as the
next upcoming main pole in a generator.
For a motor,
reverse in rotor current direction compared to generatormode
neutralplane shifts opposite to the direction of rotation. conductors undergoing commutation have the same
voltage polarity as the pole they are approaching.
to cancel this effect:Interpoles must be of the same polarity as the
previous main pole in a motor.
Determining the required polarity
of an interpole. The flux from the
interpole must produce a voltage
that opposes the existing voltage in
the conductor.

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Use of commutating poles:
very common they correct sparking problems at low cost
BUT fluxweakening problem is till present!
III. Compensating windingsFluxweakening problem can be very serious in very heavy,
severe duty cycle motors.
Compensating windings are:
placed in slots carved in the faces of the poles parallelto the rotor conductors.
Connected in series with the rotor windings such thatload changes will change the currents in the compensating
windings.
Figures below shows the basic concept of compensating
windings.
1. Here, the pole flux is shownby itself.
2. The rotor flux and compensatingwinding flux is shown.
3. The sum of the fluxes is just equal to the original pole flux
by itself.

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A more careful development of the compensating winding effect
is shown below:
Mmf due to compensating windings is equal and oppositeto the mmf due to the rotor at every point under the polesurface.
The resulting net mmf is just the mmf due to the poles.Flux in the machine is unchanged regardless of load.
The flux and magnetomotive forces in a dc machine with compensating
windings.

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The major disadvantage of compensating windings:
expensive must be machined into pole faces does not cancelL di/dt effects also requires interpoles
8.5. Internal generated voltage and induced torqueequations for real DC machines
Internal generated voltage equation
The voltage out of the armature of a real machine is:
( )
( )vBla
ZE
E
A
A
=
= conductoreachonvoltage
pathcurrentper
conductorsofnumber
whereZ= total number of conductor
a = number of current paths
The velocity of each conductor in the rotor can be expressed as v= r, where r is the radius of the rotor.
Therefore,
( )Blra
ZE
A=
Since the area per pole PrlAP
2= , hence the total flux per
pole in a P pole machine is:
BP
rlBA
P
2==
The voltage out of the armature of a real dc machine is:
(8.11)
(8.13)
(8.12)
( )
KE
a
ZP
P
rlB
a
ZPrlB
a
ZE
A
A
=
=
==
2
2
2

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Induced torque equation
The torque on the armature of a real machine is:
( ) ( )
=
=
a
lBrIZT
T
A
ind
indconductoreachontorqueconductorsofnumber
The flux per pole in the machine can be expressed as:
( )
P
rlBBA
P
2==
Therefore, torque on the armature of a real dc machine is:
( )
Aind
AAind
IKT
Ia
ZPIrlB
a
ZT
=
==2
8.6. The construction of DC machinesA simplified diagram of a dc machine is shown below.
(8.14)
(8.15)

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The physical structure of the machine consists of two parts:
o Stator, consists of : the frame,which provides physical support. pole pieces, which project inward and provides path
for the machine magnetic flux. pole shoes, i.e. the ends of the pole pieces that
spread out near the rotor surface for even flux
distribution. pole face, i.e. exposed surface of a pole shoe.
o RotorAir gap = distance between stator and rotor.
There are two principal windings on a dc machine:
o Armature windings windings in which voltage isinduced (i.e. rotor windings in the dc machine)
o Field windings windings that produce the mainmagnetic flux in the machine (i.e. stator windings in thedc machine)
Note: Since the armature windings are on the rotor, a dc
machines rotor is sometimes called an armature.
8.7. Power flow and losses in DC machinesThe efficiency of a dc machine:
%100%100%100
=+
==in
lossin
lossout
out
in
out
P
PP
PP
P
P
P

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32Dr. Ungku Anisa, UNITEN 2006
Losses in DC machines
1.Electrical or copper losses (I2R loss):Armature loss:
AAARIP2
=
Field loss :FFF
RIP2=
2.Brush losses power loss across the contact potential at themachine brushes.
Brush drop loss: ABDBD IVP =
3.Core losses hysteresis and eddy current losses occurring inthe metal of the motor.
4.Mechanical losses losses associated with mechanicaleffects, i.e. friction and windage losses.
5.Stray losses (or miscellaneous losses) losses that cannotbe placed in any of the above categories.
The power flow diagram
For a dc motor:
Pout = Tappm
T

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33Dr. Ungku Anisa, UNITEN 2006
8.1 A simple rotating loop between curved pole facesDetailed segment analysis to obtain the voltage induced in a
simple rotating loop
Segment ab:
Velocity vab tangential to rotation path
=edgespolethebeyond,0
poleunder the,surfacetoBr
=edgespolethebeyond,0
facepoleunder the,pageintopositivevBle
ba
Segment bc:
direction of( )Bv rr either into or out of page ( ) lBv rrr 0=cbe Segment cd:
Velocity vcdtangential to rotation path
=edgespolethebeyond,0
poleunder the,surfacetoBr
=edgespolethebeyond,0
facepoleunder the,pageofoutpositivevBledc
Segment da:
( ) lBv
rr
r
0=ade

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34Dr. Ungku Anisa, UNITEN 2006
Detailed segment analysis to obtain the torque induced in a
simple rotating loop
Segment ab:Current directed out of the page
=edgespolethebeyond,0
poleunder the,surfacetoBr
=edgespolethebeyond,0
facepoleunder the,directionmotionotangent tilBF
ab
r
=edgespolethebeyond,0
facepoleunder the,ckwisecounterclorilBTab
Segment bc:
Current flowing from upper left to lower right ( ) BlBliFbc rrrrr toparallelissince0== 0=bcT Segment cd:
Current directed into the page
=edgespolethebeyond,0
poleunder the,surfacetoBr
=edgespolethebeyond,0
facepoleunder the,directionmotionotangent tilBFcdr
=edgespolethebeyond,0
facepoleunder the,ckwisecounterclorilBTcd
Segment da:Current flowing from upper left to lower right ) BlBliFda rrrrr toparallelissince0== 0=daT

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35Dr. Ungku Anisa, UNITEN 2006
8.2 Commutation in a simple fourloop DC machineCommutation = process ofconverting the ac voltages
and currents in the rotor of a dc machine to dc voltages and
currents in its terminals.
A simple fourloop, twopole dc machine (see figure below) has:
Four complete loops buried (in a special manner) in slotscarved in the laminated steel of the rotor.
Pole faces that are curved to provide uniform flux densityeverywhere under the faces.
The unprimed end of each loop is the outermost wirein each slot.
The primed end of each loop is the innermost wire inthe slot directly opposite.
The dc machine and its windings connection to the machines
commutator are shown below:
Notice that:
At t = 0, in the dc machine:
The 1, 2, 3 and 4 ends of the loops are under the northpole face and the voltage in each of the loop ends is given
by:
LoopStretches between
commutator
segments:
1 a and b
2 b and c
3 c and d
4 dand a
A fourloop twopole dc machine at
time t= 0.

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( ) pageofoutpositivevBllBveind ==
rr
r
The 1, 2, 3 and 4 ends of the loops are under the southpole face and the voltage in each of the loop ends is givenby:
( ) pagetheintopositivevBllBveind ==
rr
r
The overall result is shown below whereby each coilrepresents
one side (or
conductor)of a loop.
The total voltage at the brushes of the machine is:== 0henw4 teE
where e = vBl = induced voltage on one side of a loop.
Notice that there are two parallel paths for current throughthe machine. The existence oftwo or moreparallel paths
for rotor current is a common feature of all commutationschemes.
(8.5)
The voltage on the
rotor conductors of a
fourloop twopole
dc machine at timet= 0.

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Figure below shows the machine at time t = 45:
Loops 1 and 3 have rotated into the gap between thepoles, so the voltage across each of them is zero.
At this instant the brushes of the machine are shorting outcommutator segments ab and cd.
Only two loops 2 and 4 are under the pole faces, so theterminal voltageEis given by:
== 54henw2 teE (8.6)
The same four
loop twopole dc
machine at time
t= 45.
The voltage on the
rotor conductors of
the dc machine at
time t= 45.

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The rotor continues to turn another 45, i.e. t = 90:
The 1, 2, 3 and 4 ends of the loops are under the northpole face and the voltage in each of the loop ends is given
by:
( ) pageofoutpositivevBllBveind
==rr
r
The 1, 2, 3 and 4 ends of the loops are under the southpole face and the voltage in each of the loop ends is givenby:
( ) pagetheintopositivevBllBveind
==rr
r
The overall result is shown above on the right. Theterminal voltage of the machine is:
Note: The voltages on loops 1 and 3 have reversed between
t = 0 and t = 90, but their connections have also reversed.Hence, the total voltage is still built up in the same direction as
b f
(8.7)== 09when4 teE
The same twopole dc machine at time t= 90 with the voltages onthe conductors.