DC Machinery Fundamentals

EEEB283 Electrical Machines and Drives Drives – DC Machinery Machinery Fundamentals

Chap. 8: DC Machinery Fundamentals This chapter will firstly explain the principles of dc machine operation using simple examples before considering some of the complications that occur in real dc machines. 8.1.

A simple rotating loop between curved pole faces

The simplest rotating dc machine is shown below.

It consists of: • A rotating single loop of wire in a slot carved in a cylindrical ferromagnetic ferromagnetic core c ore – rotor

• A pair of magnetic poles (north and south) – stator The 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. ℜ gap is the same everywhere under the pole faces). magnetic flux density is under the pole surface.

constant

everywhere

From Chapter 1, we know that ℜ gap >> ℜiron , hence magnetic flux takes shortest possible path through the air gap (i.e. perpendicular to the rotor surface). _____________________________________________________________________ 1 Dr. Ungku Anisa, UNITEN UNITEN © 2006


The voltage induced in a rotating 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 and and fron frontt view view of the the sim sim le dc mach machin inee

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 eind = v × B • l r

r

r

Hence, the total induced voltage in the loop:

eind = etot = eba + ecb + edc + ead

2vBl under the pole faces ∴ eind =  beyond the pole edges 0

(8.1)

When the loop rotates through 180 °, • segment ab is under the north pole instead of the south pole but magnitude • direction of e ba and e dc remains constant _____________________________________________________________________ 2 Dr. Ungku Anisa, UNITEN UNITEN © 2006


The voltage induced in a rotating 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 and and fron frontt view view of the the sim sim le dc mach machin inee

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 eind = v × B • l r

r

r

Hence, the total induced voltage in the loop:

eind = etot = eba + ecb + edc + ead

2vBl under the pole faces ∴ eind =  beyond the pole edges 0

(8.1)

When the loop rotates through 180 °, • segment ab is under the north pole instead of the south pole but magnitude • direction of e ba and e dc remains constant _____________________________________________________________________ 2 Dr. Ungku Anisa, UNITEN UNITEN © 2006


The resulting voltage etot is shown in the figure below.

Output voltage of the simple dc machine

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:

Derivation of an alternative form of induced voltage equation.

The tangential velocity v of the loop edges can be expressed as v ω. Therefore, = r ω

2r Bl under the pole face eind =  beyond the pole edges 0 _____________________________________________________________________ 3 Dr. Ungku Anisa, UNITEN UNITEN © 2006

EEEB283 Electrical Machines and Drives – DC Machinery Fundamentals

For a 2-pole 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

AP =

2π rl 2

= π rl

Therefore,

2  AP Bω under the pole face eind = π 0 beyond the pole edges Since the air gap flux density B is constant everywhere under the pole faces, the total flux under each pole is:

φ = AP B Thus, the final form of the voltage equation is:

2  φω under the pole face eind = π 0 beyond the pole edges

(8.2)

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? _____________________________________________________________________ 4 Dr. Ungku Anisa, UNITEN © 2006


By using a mechanism called commutator and brushes (see below)

• two semicircular conducting segments (commutator segments) are added to the end of the loop • two fixed contacts (brushes) are placed at an angle such that at the instant the voltage in the loop is zero, the contacts short-circuit 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).

Output voltage of a dc machine with commutator and brushes.

This connection-switching process is known as . _____________________________________________________________________ 5 Dr. Ungku Anisa, UNITEN © 2006


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?

The simple dc machine connection for derivation of induced torque equation.

Front view of the machine (the iron core is omitted for clarity).

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 on each segment is given by F = i l × B . r

And

the

torque

on

the

segment

r

is

r

given

by

T = r × F = rF sin θ , where θ is the angle between r and F . r

r

r

r

r

When the loop is beyond the pole edges , T = 0 (since B = 0). _____________________________________________________________________ 6 Dr. Ungku Anisa, UNITEN © 2006


Hence, the total induced torque in the loop is:

T ind = T tot = T ab + T bc + T cd + T da

2rilB under the pole faces ∴ T ind =  beyond the pole edges 0

(8.3)

By employing the facts that AP ≈ π rl and φ = AP B , the torque expression can be reduced to:

2  φ i under the pole face T ind = π 0 beyond the pole edges

(8.4)

In general, the torque in any real machine will depend on the same three factors:

• the flux in the machine • the current in the machine • a constant representing the machine construction

_____________________________________________________________________ 7 Dr. Ungku Anisa, UNITEN © 2006




commutators



8.2. Commutation in a simple four-loop DC machine (see page 35 for the detailed explanation)

The resulting terminal voltage as a function of time is shown below:

The resulting output voltage of the four-loop two-pole dc machine.

Note: this is a better approximation to a constant dc level than that 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 of switching 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 constant dc 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. _____________________________________________________________________ 10 Dr. Ungku Anisa, UNITEN © 2006


8.3.

Commutation and armature construction in real DC machines

In real dc machines, there are several ways in which the loops on the rotor (also called the armature) can be connected to its commutator 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 the windings are connected to the commutator segments, the rotor windings consist of diamondThe shape of a shaped preformed coils which typical are inserted into the armature preformed slots as a unit (see figure). rotor coil. 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 machine’s armature is given by:

Z = 2CN c

(8.8)

where Z = number of conductors on rotor C = number of coils on rotor N c = 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, the other side is under the centre of a pole of opposite polarity. _____________________________________________________________________ 11 Dr. Ungku Anisa, UNITEN © 2006


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:

θ e =

P

2

(8.9)

θ m

where θ e = electrical angle, in degrees θ m = mechanical angle, in degrees P = number of magnetic poles on the machine If a coil spans 180 electrical degrees, the voltages in the conductors 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 full-pitch coil. A fractional-pitch coil spans less than 180 electrical degrees, and a rotor winding wound with fractional-pitch coils is called a chorded winding. The amount of chording in a winding is described by a pitch factor p, defined by:

p =

electrical angle of coil 180°

× 100%

(8.10)

Most rotor windings are two-layer windings, meaning that sides from 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.



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 y c = distance (in number of segments) between the commutator segments to which the two ends of a coil are connected.

A coil in a progressive winding. Progressive winding (yc =1) = the end of a coil (or a set number of coils) is connected to a commutator segment ahead of the one its beginning is connected to.

A coil in a retrogressive winding. Retrogressive winding ( yc =-1) = the end of a coil (or a set number of coils) is connected to a commutator segment behind the one its beginning is connected to.

If everything else is identical, the rotation direction of a progressive-wound rotor is opposite to that of a retrogressivewound rotor. Rotor (armature) windings are further classified according to the plex of their windings: 1. A simplex rotor winding is a single, complete, closed winding wound on a rotor. 2. A duplex rotor winding is a rotor with two complete and independent sets of rotor windings. Each of the windings will be associated with every other commutator segment: one winding will be connected to segments 1, 3, 5, etc., _____________________________________________________________________ 13 Dr. Ungku Anisa, UNITEN © 2006


and the other winding will be connected to segments 2, 4, 6, etc. 3. A triplex winding will have three complete and independent sets of windings, each winding connected to every third commutator segment on the rotor. 4. All armature with more than one set of windings are called multiplex windings. Finally, armature windings are classified according to the sequence of their connections to the commutator segments:

• Lap winding • Wave winding • Frog-leg 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 machines

In practice, the commutation process is not as simple as described theoretically in Sections 8.2 and 8.3. There are two major effects that disturb the commutation process : I. Armature reaction II. L di/dt voltages

This section explores the nature of these problems and the solutions employed to mitigate their effects.



I. Armature reaction

In 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 of the machine’s 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 machine’s 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: Neutral-plane shift Magnetic neutral plane = the plane within the machine where the velocity of the rotor is exactly parallel to the magnetic in the conductors in the plane. flux lines, so that

To understand the problem of neutral plane shift, the two-pole dc machine shown below is employed. _____________________________________________________________________ 15 Dr. Ungku Anisa, UNITEN © 2006


• The flux is distributed uniformly under the pole faces. • The rotor windings have voltages built up as shown on the left. • Hence, the neutral plane is exactly vertical .

• Now suppose a load is connected such that the machine acts as a generator (i.e. no power supplied to rotor windings). • Current flows out of the positive terminal of the generator, as shown on the left.

• This current flow produces a magnetic field from the rotor windings, as shown in picture (c). (direction: right-hand rule).. • This rotor magnetic field affects the original magnetic field from the poles that produced the generator’s voltage in the first place. • In some places under the pole surfaces, it subtracts from the pole flux. In other places, it adds to the pole flux.



• The overall result is that the magnetic 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 a conductor 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 current direction, 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 neutral-plane shift?

• The brushes must short out commutator segments just at the moment when the voltage across them is zero. • If the brushes are set to short out conductors in the vertical plane, then the voltage between the segments is zero until the machine is loaded . • When the machine is loaded, the neutral plane shifts and the brushes short out commutator segments with finite voltage across them.



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 at the

End result: 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 full-load neutral plane , as it will then cause sparks at no load. In extreme cases, the neutral-plane shift can lead to flashover in the commutator segments near the brushes. (The air near the burshes in a machine is normally ionized as a result 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 commutator’s surface.) Problem 2: Flux weakening To understand flux weakening, refer to the magnetisation curve shown below:



• Most machines operate near the saturation point (knee of curve). • At locations on the pole surface where rotor mmf adds to the pole mmf, only a small increase in flux occurs. • But at locations on the pole surface where rotor mmf subtracts from the pole mmf, a decrease in flux occurs.

The net result: The total average flux under the entire pole face is decreased (see figure below).



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 flux-weakening 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 reach condition, where motor speed just keeps on increasing until the machine is disconnected from the power line or until it destroys itself. _____________________________________________________________________ 20 Dr. Ungku Anisa, UNITEN © 2006


II. L di/dt voltages

The 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 below representing 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 are 50 commutator segments (a reasonable number for a typical motor), each commutator segment moves under a brush and clears it again in t = 0.0015s. Calculations: 800 rpm = 83.78 rad/s 1 segment takes up 0.126 rad ( = 50 / 2π ) Time to clear 1 segment = 0.126 / 83.78 = 0.0015s _____________________________________________________________________ 21 Dr. Ungku Anisa, UNITEN © 2006


Therefore, the rate of change in current with respect to time in the shorted loop must average: real commutation

di dt

=

200 − ( −200) 0.0015 s

= 266,667 A/s

ideal commutation

- 200 A

The current reversal in the coil undergoing commutation as a function of time for both ideal and real commutation, with the coil inductance taken into account.

With even a tiny inductance L in the loop, a very significant inductive voltage kick v = L di/dt will be induced in the shorted commutator segment. This high voltage naturally causes sparking in the brushes of the machine, resulting in the same arcing problems as caused by the neutral-plane shift. Solution to the problems with commutation

The following three approaches have been developed to partially or completely correct the problems of armature reaction and L di/dt voltages: I.

Brush shifting

This method attempts to stop the sparking at the brushes caused by the neutral-plane shifts and L di/dt effects. _____________________________________________________________________ 22 Dr. Ungku Anisa, UNITEN © 2006


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 shift direction reverses when going from motor to generator operation.

• Even though brush sparking is stopped, brush shifting aggravates the flux-weakening effect of armature reaction in the machine. This is true because of two effects: i. The rotor mmf now has a vector component that opposes the mmf from the poles (see figure below). ii. The change in armature current distribution causes the flux to bunch up even more at the saturated parts of the pole faces.

Before brush shifting (i.e. brushes over the vertical lane)

After brush shifting (i.e. brushes over the shifted neutral lane)

However, this method is obsolete . _____________________________________________________________________ 23 Dr. Ungku Anisa, UNITEN © 2006


It is only used in very small motors where other better solutions are not economical. II. Commutating poles or interpoles

Basic 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 called commutating poles or interpoles are placed midway between the main poles and directly 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 the machine 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 is not 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 . _____________________________________________________________________ 24 Dr. Ungku Anisa, UNITEN © 2006


A dc machine with interpoles

As the load increases The rotor current ( I A) increases Magnitude of neutral plane shift and size of L 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. I A↑ , I interpole ↑) Gives larger voltage in the conductor which opposes voltage due to the neutral-plane shift The net result is: Commutating pole cancels the neutral-plane shift and L di/dt effects over a broad range of loads . Note: Interpoles work for both motor and generator operation. _____________________________________________________________________ 25 Dr. Ungku Anisa, UNITEN © 2006


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 neutral-plane shift and L di/dt effects. For a generator, • neutral-plane 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 same polarity as the next upcoming main pole in a generator.

Determining the required polarity of an interpole. The flux from the interpole must produce a voltage that opposes the existing voltage in the conductor.

For a motor, • reverse in rotor current direction compared to generator mode • neutral-plane 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. _____________________________________________________________________ 26 Dr. Ungku Anisa, UNITEN © 2006


Use of commutating poles: • very common • they correct sparking problems at low cost BUT flux-weakening problem is till present! III. Compensating windings

Flux-weakening 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 parallel to the rotor conductors . • Connected in series with the rotor windings such that load changes will change the currents in the compensating windings. Figures below shows the basic concept of compensating windings. 1. Here, the pole flux is shown 2. The rotor flux and compensating winding flux is shown. by itself.

3. The sum of the fluxes is just equal to the original pole flux by itself .



A more careful development of the compensating winding effect is shown below:

The flux and magnetomotive forces in a dc machine with compensating windings.

• Mmf due to compensating windings is equal and opposite to the mmf due to the rotor at every point under the pole surface. • The resulting net mmf is just the mmf due to the poles. Flux in the machine is unchanged regardless of load .



The major disadvantage of compensating windings: • expensive – must be machined into pole faces • does not cancel L di/dt effects – also requires interpoles 8.5.

Internal generated voltage and induced torque equations for real DC machines

Internal generated voltage equation

The voltage out of the armature of a real machine is:

 number of conductors  E A =   × (voltage on each conductor)  per current path  ∴ E A =

Z a

(vBl )

(8.11)

where Z = total number of conductor a = number of current paths The velocity of each conductor in the rotor can be expressed as v = r ω , where r is the radius of the rotor. Therefore,

E A =

Z a

(r ω Bl )

Since the area per pole AP = 2π rl P , hence the total flux per pole in a P pole machine is:

φ = AP B =

2π rl P

B

(8.12)

The voltage out of the armature of a real dc machine is:

ZP  2π rlB  ZP φω E A = (rlB )ω =  ω = a a 2π  P  2π a Z

∴ E A = K φω

(8.13)



Induced torque equation

The torque on the armature of a real machine is:

T ind = (number of conductors ) × (torque on each conductor )

 rI AlB  ∴T ind = Z   a  

(8.14)

The flux per pole in the machine can be expressed as:

φ = AP B =

B(2π rl ) P

Therefore, torque on the armature of a real dc machine is:

T ind =

Z a

(rlB ) I A =

ZP

2π a

φ I A

∴T ind = K φ I A 8.6.

(8.15)

The construction of DC machines

A simplified diagram of a dc machine is shown below.



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

Rotor

Air gap = distance between stator and rotor.

There are two principal windings on a dc machine: o

o

Armature windings – windings in which voltage is induced (i.e. rotor windings in the dc machine ) Field windings – windings that produce the main magnetic flux in the machine (i.e. stator windings in the dc machine)

Note: Since the armature windings are on the rotor, a dc machine’s rotor is sometimes called an armature. 8.7.

Power flow and losses in DC machines

The efficiency of a dc machine:

η =

Pout Pin

× 100% =

Pout Pout + Ploss

× 100% =

Pin − Ploss Pin

× 100%



Losses in DC machines 2

1. Electrical or copper losses (I R loss): Armature loss: P A = I A R A 2

Field loss

: PF = I F RF 2

2. Brush losses – power loss across the contact potential at the machine brushes. Brush drop loss: P BD = V BD I A 3. Core losses – hysteresis and eddy current losses occurring in the metal of the motor. 4. Mechanical losses – losses associated with mechanical effects, i.e. friction and windage losses. 5. Stray losses (or miscellaneous losses) – losses that cannot be placed in any of the above categories. The power flow diagram

For a dc motor:

Pout = Tappωm T



8.1

A simple rotating loop between curved pole faces

Detailed segment analysis to obtain the voltage induced in a simple rotating loop

Segment ab: • Velocity vab tangential to rotation path , under the pole  ⊥ to surface • B =  , beyond the pole edges 0 vBl positive into page , under the pole face • eba =  , beyond the pole edges 0 r

Segment bc: • direction of (v × B ) either into or out of page • (v × B ) ⊥ l • ecb = 0 r

r

r

r

r

Segment cd : • Velocity vcd tangential to rotation path , under the pole  ⊥ to surface • B =  , beyond the pole edges 0 vBl positive out of page , under the pole face • edc =  , beyond the pole edges 0 r

Segment da: • (v × B ) ⊥ l • ead = 0 r

r

r



Detailed segment analysis to obtain the torque induced in a simple rotating loop

Segment ab: • Current directed out of the page , under the pole  ⊥ to surface • B =  , beyond the pole edges 0 ilB tangent t o motion direction , under the pole face • F ab =  , beyond the pole edges 0 rilB counterclo ckwise , under the pole face • T ab =  , beyond the pole edges 0 r

r

Segment bc: • Current flowing from upper left to lower right • F bc = i (l × B ) = 0 since l is parallel to B r

r

r

r

r

• T bc = 0 Segment cd : • Current directed into the page , under the pole  ⊥ to surface • B =  , beyond the pole edges 0 ilB tangent t o motion direction , under the pole face • F cd =  , beyond the pole edges 0 rilB counterclo ckwise , under the pole face • T cd =  , beyond the pole edges 0 r

r

Segment da: • Current flowing from upper left to lower right • F da = i l × B ) = 0 since l is parallel to B r

r

r

r

r

• T da = 0



8.2

Commutation in a simple four-loop DC machine

Commutation = process of converting the ac voltages and currents in the rotor of a dc machine to dc voltages and currents in its terminals.

A simple four-loop, two-pole dc machine (see figure below) has:

• Four complete loops buried (in a special manner) in slots carved in the laminated steel of the rotor. • Pole faces that are curved to provide uniform flux density everywhere under the faces. • The “unprimed” end of each loop is the outermost wire in each slot. • The “primed” end of each loop is the innermost wire in the slot directly opposite. The dc machine and its winding’s connection to the machine’s commutator are shown below: Notice that:

Loop 1 2 3 4

Stretches between commutator segments: a and b b and c c and d d and a

A four-loop two-pole dc machine at time ω t = 0°.

At ωt = 0°, in the dc machine:

• The 1, 2, 3’ and 4’ ends of the loops are under the north pole face and the voltage in each of the loop ends is given by: _____________________________________________________________________ 35 Dr. Ungku Anisa, UNITEN © 2006


eind = (v × B ) • l = vBl r

r

r

positive out of page

• The 1’, 2’, 3 and 4 ends of the loops are under the south pole face and the voltage in each of the loop ends is given by:

eind = (v × B ) • l = vBl r

r

r

positive into the page

• The overall result is shown below whereby each coil represents one side (or conductor) of a loop.

The voltage on the rotor conductors of a four-loop two-pole dc machine at time ω t = 0°.

• The total voltage at the brushes of the machine is:

E = 4e

when t = 0°

(8.5)

where e = vBl = induced voltage on one side of a loop.

• Notice that there are two parallel paths for current through the machine. The existence of two or more parallel paths for rotor current is a common feature of all commutation schemes.



Figure below shows the machine at time ωt = 45°:

The same fourloop two-pole dc machine at time ω t = 45°.

• Loops 1 and 3 have rotated into the gap between the poles, so the voltage across each of them is zero. • At this instant the brushes of the machine are shorting out commutator segments ab and cd .

The voltage on the rotor conductors of the dc machine at time ω t = 45°.

• Only two loops 2 and 4 are under the pole faces, so the terminal voltage E is given by:

E = 2e

when t = 45°

(8.6)


DC Machinery Fundamentals

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