DC Generators

1
DC Generators
But first, AC an AC generator. This consists of
a coil rotating through a magnetic field, as
shown below. Electrical connection to the coil
is made through slip rings, so the same end of
the coil is always
connects to the terminal of the load. There
is a flux between the two magnets (which may
actually be opposite poles of the same magnet),
which the turns of the coil cut.
60 rpm
The two segments of each turn which pass close to
the magnet poles may be considered the active
portion of the turn, which determines the induced
voltage.
S
N
Slip rings
-

Load
2
DC Generators
The active lengths of the conductor are moving
with a velocity v in the counter-clockwise
direction as they cut the flux, and the total
active length (for a coil of n turns) is 2nlv
(because each turn has two active
segments, and there are n turns. A previous
result says that one active segment cutting a
flux of density B will have induced across itself
a voltage given by
60 rpm
S
N
So, for this generator,
Slip rings
-

Load
3
DC Generators
If the active conductors were merely moving with
constant velocity through a uniform magnetic
field (uniform flux density, and always the same
direction relative to each active conductor), the
induced voltage
would be constant. This would be a very nice DC
generator. However, the flux density is not
uniform. The density of the flux cut by the
conductors is greatest when the coil is in the
horizontal position, as shown. As the coil
rotates toward the vertical position, the flux
density diminishes, and reaches a minimum as the
coil passes through vertical. At this point,
those active conductors which were closer to the
south pole become closer to the north pole.
60 rpm
S
N
Slip rings
-

Load
4
DC Generators
The coil position can be represented by an angle,
q. Well say that q 0 when the coil is in the
position shown. After ¼ revolution, as it passes
through vertical, q p/2 (90 deg.) after ½
revolution, q p (180 deg.), and after one
complete revolution, q 2p (360 deg.)
60 rpm
S
N
Slip rings
-

Load
5
DC Generators
We can plot the flux density, B, cut by the
active conductors as a function of coil position,
q. Thus, B is a function of coil angle, B
B(q). By carefully designing the shape of the
magnets, we can make the shape of B(q)
approximately sinusoidal so, B(q) sinq.
Recalling that
60 rpm
We now have
Since v is constant, if B(q) is sinusoidal, E is
a sinusoidal function of q since the coil
rotates at a constant velocity, E is a sinusoidal
function of time.
S
N
Slip rings
-

Load
6
DC Generators
Thus, for the AC generator shown, rotating at 60
rpm, the load voltage would be given by
60 rpm
S
N
Slip rings
-

Load
7
DC Generators
The slip rings insure that the coil terminal
denoted a and b below always connect to the load
terminals marked by and -, respectively.
60 rpm
S
N
b
Slip rings
-

a
Load
8
DC Generators
Now well replace the slip rings with a
commutator. x and y are sliding contacts which
connect the coil to the brushes, a and b.
Consider the coil rotating as before, starting
from the horizontal position shown (q 0).
As the coil passes through vertical (q 90
deg.), the load voltage passes through zero. At
that moment, contact y leaves brush b and goes to
brush a contact x moves from a to b. This
reverses the connections between the coil and the
load, inverting the negative half cycle which we
could observe across contacts x and y.
Load
brushes

-
a
b
coil
Shaft
S
N
x
y
commutator
9
DC Generators
Here are plots of the voltage across the
terminals of our simple AC generator, and our
simple DC generator. The DC generator voltage is
not a constant voltage, but it is always
nonnegative. It looks like the AC generators
voltage waveform, full-wave rectified.
10
DC Generators
Can we reduce the amount of pulsation in the DC
generators voltage? Sure! Well use four
coils, and a four-section commutator, as shown
below. When the rotor is in the position shown,
coils B and D cut maximum flux, but A and C cut
no flux. Therefore, the voltage across the
series combination is Emax 0V. This is the
voltage across brushes x and y.
A
y
Shaft
S
N
D
B
x
C
11
DC Generators
After rotating 45 degrees, the coils are in the
position shown. Now all the coils are midway
between minimum and maximum flux, so the voltage
measured at the brushes is between zero and Emax.
After another 45 degrees, A and C will be
cutting maximum flux and B and D will cut none,
so the brush voltage will be Emax once again.
Thus, the brush voltage never falls to zero.
A
y
B
Shaft
S
N
D
C
x
12
DC Generators
Here are the output voltage waveforms for two DC
generators, one with 2 coils, and one with 4.
The 4-coil generators ripple amplitude is
reduced from 20 Volts to 6 Volts, and its ripple
frequency is doubled. Further improvement can be
had by adding more coils. The coil assembly
(everything that rotates) is called the armature.
13
DC Generators
In real life, the armature looks something like
the one shown below. Coil A is wound from the
top slot to the bottom slot, and coil C is wound
from the bottom slot to the top. Coils B and D
are wound using the two horizontal slots. Each
slot contains conductors of two of the four
coils. The commutator is connected as shown. A
real armature would probably have more than four
slots, and more than four coils. This type of
armature winding is called a lap winding.
Notice that when the armature is in the position
shown, coils A and C are short-circuited by
brushes x and y. However, these two coils cut no
flux in this position, so their induced voltage
is zero and the short circuit does no harm. The
brushes are said to be in neutral position.
a1
d1
b1
S
N
x
y
b2
d2
a2
c1
14
DC Generators
The armature shown previously had four coils. A
real DC generator would probably have many more
in order to reduce the ripple amplitude. Twelve
coils might be a reasonable number such an
armature is shown below. In the position shown,
the voltage induced across coil A (in slot 1) is
zero, and the voltage across coil D (in slot 4)
is at its maximum (e.g., 20 V). The coils in
slots 2 and 3 would then be at 7V and 18V
respectively. The coils in slots 4, 5, 6, and 7
would also be at 0V, 7V, 18V and 20V
respectively. These voltages are summed at the
brushes. The coils in slots 1, 2, 3, 4, 5, 6 and
7 are in series between brushes x and y, so these
voltages add, making the voltage
Coil A
Coil D
B
across the two brushes 0 7 18 20 18 7
0 70 V. Because of symmetry, the voltage
induced across the other coils is also 70V. Note
that, when the armature is in the position shown,
the brushes still short-circuit the two coils
whose individual voltages are zero.
1
D
2
12
3
11
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
15
DC Generators
The two coils which are shorted by the brushes
are at the top and bottom, in an area where the
flux density is zero. Therefore, no voltage is
induced across them while they are in these
neutral zones.
Coil A
Coil D
B
1
D
2
12
3
11
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
16
DC Generators
If the brushes were shifted by 30 degrees as
shown below, they would short-circuit two coils
which are not in the neutral zone. In fact,
these two coils have 7 volts across them, causing
large currents to flow through the short
circuits. This is not good, so the brushes are
always placed so they can only short the coils
passing through the neutral zones.
Coil A
Coil D
B
1
D
2
12
3
11
y
S
N
4
10
x
5
9
C
8
6
7
A
Coil C
Coil B
17
DC Generators
If the brushes are in the neutral position, the
terminal voltage of a DC generator with a lap
winding is given by
Coil A
Coil D
B
1
D
2
12
3
11
This is the no-load voltage, the voltage between
the brushes when no current (and therefore no
power) is being supplied by the generator.
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
18
DC Generators
Suppose the generator is supplying current to a
load. In the figure below, terminals x and y are
connected to brushes x and y, so the load is
connected across the brushes. For example,
assume this generator is rotating
counterclockwise, with current I as shown flowing
through the load. Because
the armature coils are connected in series, I
flows through every conductor in the armature.
The current always flows in the same direction in
the conductors that happen to be momentarily
under the N pole, and in the opposite direction
in those conductors momentarily under the S pole.
For example, the current in the conductors under
the S pole flows into the page, the current in
the conductors under the N pole flow out of the
page.
Coil A
Coil D
B
1
D
2
12
3
11
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
x
y
I
Load
19
DC Generators
Remember that a conductor with a current flowing
through it in a magnetic flux is subject to the
Lorentz force. Close examination reveals that,
when the armature rotates counterclocwise, the
direction of the force on the conductors under
the N pole is upward, and the force on the
conductors under the S pole
acts in the downward direction. Both forces
produce torque in the direction opposite that of
the armature rotation. If no current is drawn,
there is no opposing torque increasing the load
current (and therefore the power supplied by the
generator) increases the opposing torque.
Mechanical power must be supplied in the form of
torque to the shaft in order to keep the armature
rotating this is the power which is converted to
electrical power.
Coil A
Coil D
B
1
D
2
12
3
11
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
I
Load
20
Armature Reaction
As long as no load current flowed through the
armature winding, the only source of
magnetomotive force in the generator was the the
magnet whose N and S poles are shown. As soon as
a load is connected and current flows through the
armature winding, each conductor has a magnetic
field
Surrounding it. In other words, the armature
is also a source of MMF, which weakens and
distorts the shape of the pole flux. This is
called armature reaction. One consequence of
this is that, due to the armature mmf, the two
regions which were neutral zones with no load
connected now have nonzero flux density. In
other words, they arent neutral zones anymore.
Coil A
Coil D
B
1
D
2
12
3
11
S
N
4
10
x
y
5
9
C
8
6
7
A
Coil C
Coil B
I
Load
21
Armature Reaction
Under load, the neutral zones are actually
shifted in the direction of rotation by an angle
which depends on the load current. We could
compensate for this by also shifting the brushes
counterclockwise, but the shift angle required to
compensate for the neutral zone shift would also
vary according to the load
current. This is not usually practical. The
armature reaction also distorts the field in such
a way that, due to saturation, the total flux due
to the N and S poles is reduced. This reduces
the terminal voltage under load.
Coil A
Coil D
B
1
D
2
12
1
3
3
11
S
N
4
10
x
y
5
9
2
4
C
8
6
7
A
Coil C
Coil B
I
Load
22
Armature Reaction
Heres another way to compensate for armature
reaction. Its usually used in
Medium and large generators. A set of
commutating poles is placed between the main
poles, and the commutating pole windings are
placed in series with the armature windings.
Thus, the load current flows through the
commutating pole windings and produces mmfc. The
direction of this mmf is opposite the armature
mmf, and like the armature mmf, its strength
depends on the load current. Careful design will
compensate to a degree for the
I

S
N
x
y
armature mmf in the neutral zone, reducing the
short-circuit current caused by the brushes.
However, the load voltage is still reduced
because the field is still weakened.
-
23
Separately Excited DC Generator
Up to this point, we havent considered the
magnets which create the field. They could be a
permanent magnets, but in most cases theyre a
pair of electromagnets. These are called the
field poles. A DC current, called the excitation
current, must be supplied to them, possibly by an
external source such as a battery. When this is
the case, the generator is said to be separately
excited. When the armature is driven to rotate
by a source of mechanical power (motor, diesel
engine, etc.) a voltage appears between the brush
terminals x and y. Under no-load conditions, a
change in excitation
Current results in a change in the induced
voltage. A saturation curve, plotting Eo vs.
excitation current, will illustrate this.
Ix
x

a
S
N
Eo
b
-
y
24
Separately Excited DC Generator
If we plot the induced (generated) voltage vs.
excitation current for generator turning at a
constant speed, well get a curve similar to the
one shown below. Notice that, as Ix is increased
from 0A, Eo increases linearly at first. This
portion of the curve is the linear region. At
some point, the increase in Eo for a given
increase in Ix starts to get smaller. This
region is called the knee. Above the knee, a
large increase in Ix will result in only a small
increase in Eo.
Why does this curve exhibit saturation? Remember
that the induced voltage is proportional to the
flux density B in the air gap (the region between
the poles, where the armature is)
150
120
So if the flux density vs. excitation curve
exhibits saturation, so does the Eo vs.
excitation curve.
90
Eo (V)
60
30
1.00
2.00
3.00
Ix (A)
25
Separately Excited DC Generator
Shown below is a plot of flux in the air gap vs.
excitation current for our hypothetical
generator. Its shape is the same as the plot of
Eo vs. Ix we just saw, and this is the reason
that the Eo vs. Ix curve exhibits saturation.
The ferromagnetic material the poles are made of
is not saturated as small values of Ix, so a
small increase in Ix results in a corresponding
increase in flux. This is the linear region. At
some point, the flux is strong enough to have
lined up most of the magnetic domains in the pole
material. At this point the poles start to
saturate, so a large increase in Ix results in
only a small increase in flux.
Because Eo is proportional to flux density in the
air gap, when the pole flux saturates, so does
Eo. However, increasing the speed of the
generator still increases Eo.
1.25
1.00
0.75
f (per unit)
0.50
0.25
1.00
2.00
3.00
Ix (A)
26
Separately Excited DC Generator
The plot of Eo vs. Ix is called the no-load
saturation curve. A DC generator are normally
designed to produce its rated voltage with the
excitation current just above the knee region of
the saturation curve (at its rated speed). The
excitation current provides a mechanism for
controlling or regulating the induced voltage to
a degree. If the speed of the generator drops
for some reason (possibly because of load
conditions), resulting in a drop in voltage,
It may be possible to compensate for a small
reduction in voltage by increasing Ix. Reversing
Ix would reverse the polarity of Eo.
rated voltage
150
120
90
Eo (V)
60
30
1.00
2.00
3.00
Ix (A)
27
Shunt DC Generator
A shunt generator is basically the same as a
separately excited generator, but its field
winding (called a shunt field winding in this
case) is connected in parallel (shunt) with the
armature terminals. A shunt generator doesnt
require a separate source of excitation current.
Its pretty obvious how this works once the
generator is running, but it would appear that
starting the generator could be a problem. With
no excitation current (as the generator starts),
Eo is zero, so no current flows in the excitation
winding. As the generator speed
Increases, there is still no induced voltage, so
Ix remains at zero. How does Ix ever increase
above zero, to cause a nonzero induced voltage?
This looks like catch-22.
Ix
x

a
S
N
Eo
b
-
y
28
Shunt DC Generator
This situation is resolved because of a small
residual flux in the poles. With zero
excitation, the poles act like a weak permanent
magnet. This residual flux is sufficient to
produce a small voltage and excitation current,
which increases the field flux, increasing the
induced voltage, which increases the excitation
and the field flux, and so on. Eventually, the
resistance of the shunt field winding, together
with saturation of the poles, stops the increase
in excitation current and equilibrium is achieved.
Ix
x

a
S
N
Eo
b
-
y
29
Shunt DC Generator
The excitation current can be controlled by
adding a field rheostat in series with the field
winding, as shown below. The field rheostat can
be used to control the induced voltage. Ix is
related to Eo and Rt (the total series resistance
of the field winding and the field rheostat)
according to Ohms law
Ix
x

a
S
N
Eo
b
-
y
30
Shunt DC Generator
Given this relationship between Ix, Eo and Rt, we
can plot Eo vs. Ix on the same axes as the
no-load saturation curve, as shown below. For a
given field rheostat setting, this is a straight
line with slope equal to Rt. This line must have
a slope less than that of the linear region of
the saturation curve, so it intersects the
saturation curve at some point. The value of Eo
at the point of intersection is the induced
voltage that will appear at the armature
terminals For that setting of the field rheostat.
Reducing the resistance of the field
rheostat reduces the slope, moving the point of
intersection to the right and increasing Eo.
Conversely, increasing the resistance increases
the slope. If the resistance is increased enough
that the slope of the line exceeds the slope of
the linear portion of the saturation curve (to
the left of the knee), the two curves only
intersect at Eo 0V. In this situation, the
induced voltage drops to zero.
Rt 75W
Rt 120W
150
120
90
Eo (V)
60
30
1.00
2.00
3.00
Ix (A)
31
Equivalent Circuit
Like any practical DC voltage source, a DC
generator may be modeled using its equivalent
circuit shown below. The ideal voltage source Eo
represents the no-load induced voltage. The
resistance Ro represents the series resistance of
the armature windings, comutator, and brushes. Ro
usually has a very small value, much less than
one Ohm.
1
F1
Eo
F2
2
32
Equivalent Circuit
Consider a separately excited generator under
load. The equivalent circuit is shown below
Assume that the speed and excitation current are
held constant, so Eo is constant. Iload causes a
voltage drop across Ro, reducing the voltage
which appears across the armature terminals , 1
and 2. In other words, drawing load current
reduces the load voltage, like any practical DC
voltage source. In the case of a shunt
generator, the load voltage decreases more
rapidly with increasing load current, because the
field current drops as load voltage drops. The
shunt generator appears to have greater series
Resistance Ro than the separately excited
generator.
Iload
F1
1
Rload
Eo
2
F2
33
Compound Generator
The compound generator includes a clever measure
to reduce the load voltage drop due to increasing
load current. There is a second field winding in
series with the load, so increasing load current
effectively increases the field flux. This
causes an increase in induced voltage, which
counteracts the induced voltage reduction due to
the voltage drop across Ro. Connection of the
series field winding is shown below
Series Field
1
Iload
Rload
Shunt Field
Eo
2
34
Differential Compound Generator
The differential compound generator is a compound
generator with the series field winding polarized
to oppose the mmf produced by the shunt field
winding. This means that increasing load current
reduces the flux in the air gap, causing the
armature terminal voltage to decrease with
increasing load current even more rapidly than
would be the case with a simple shunt generator.
This has the effect of increasing Ro, and
provides a degree of current limiting under
overload conditions.
Series Field
1
Iload
Rload
Shunt Field
Eo
2
35
Generator Specifications
A generator normally has a nameplate attached to
it with some useful information about the
generators specifications. A typical nameplate
might read like this Power 100 kW Speed 1200
rpm Voltage 250 V Type Compound Exciting
current 20 A Class B Temperature rise 50
deg. C This indicates that the generator will
develop the rated power and voltage at its rated
speed and excitation current, and will not rise
in temperature by more than the rated amount.
Class B refers to the type of insulation. The
rated current can be obtained by dividing the
rated power by the rated voltage, resulting in a
rated current of 400 A for the generator whose
nameplate is shown here.
36
Multipole Generators
Up to this point, weve talked about motors that
have two poles one pair of poles. Many
generators, especially larger ones, have more
than one pole pair arranged around the armature.
As an example, a 6-pole generator is shown below.
Note that there are also 3 sets of brushes,
wired as shown. A multipole generator has as
many sets of brushes as it has pole pairs. The
advantage of a multipole generator is primarily
reduced size and cost, along
With some improvement in performance.
I
37
Generator Construction
Sections 4.23 through 4.26 of the textbook are a
very good description of the construction of
actual DC generators. The student is urged to
read these sections.
38
Ideal Commutation
The figure below is a representation of a portion
of a 12-coil armature. Note that each coil
carries the same current since they are in
series, the current in coil 12 is left to right,
the current in coil 1 is right to left. Both
coil currents flow into commutator section 1, and
combine according to KCL to make the brush
current 80 A.
40 A
40 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
39
Ideal Commutation
A moment later, the armature has moved so the
brush contacts section 2 instead of section 1.
The 40 A currents in coils 1 and 2 add to make
the brush current 80 A as in the previous figure,
but notice that the direction of the current in
coil 1 has reversed in the fraction of a second
that elapsed between the two figures. This
reversal is called commutation. Lets take a
closer look at this process.
40 A
40 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
40
Ideal Commutation
Here again, the brush contacts section 1, and 80
A flows from section 1 into the brush. All coils
carry a 40 A current. In the lower figure, due
to armature rotation, the brush momentarily
contacts sections 1 and 2.
40 A
40 A
40 A
40 A
40 A
1
2
3
4
5
12
40 A
20 A
40 A
40 A
40 A
80 A
1
2
3
4
5
12
80 A
41
Ideal Commutation
The brush current is still 80 A, but part of it
comes from section 1 and part from section 2.
The current from section 1 is greater, because in
the position shown more of the surface area of
the brush contacts section 1 than section 2. 20
A flows through section 2, from coil 2 to the
brush, but 20 amps flows through section 2 from
coil 2 to coil 1. This 20 A combines with the 40
A from coil 12 in section 1, so 60 A flows from
section 1 into the brush. The sum of the
currents flowing from sections 1 and 2 into the
brush is 80 A, so the brush current is unchanged.
40 A
20 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
42
Ideal Commutation
Now the armature has moved a bit more, so that
equal portions of its surface area contact
sections 1 and 2. Coils 2 and 12 still carry 40
A of current, but the current in section 1 has
dropped to 0 A. 40 A flows from coil 2, through
section 2, into the brush. 40 A flows from coil
12, through section 1, into the brush, so the
total brush current is still 80 A.
40 A
0 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
43
Ideal Commutation
Further rotation results in the more of the brush
contacting section 2 and less contacting section
1, so now 60 A flows from section 2 into the
brush. 20 A comes from coil 1, whose current has
now reversed, and 40 A comes from coil 2. Of
coil 12s current of 40 A, 20 A flows through
section 1 into the brush, and 20 A flows through
section 1 into coil 1. The total brush current
is still 80 A.
40 A
20 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
44
Ideal Commutation
Finally, the armature has rotated so the brush
entirely contacts section 2. Now all the coils
carry 40 A, and the direction of the current in
coil 1 has completely reversed.
40 A
40 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
45
Practical Commutation
The reference to this process as ideal
commutation suggests that there is also
practical (i.e., real world) commutation, which
is somewhat different. This is the case. Ideal
commutation involves reversal of large coil
currents in short intervals of time, but the
coils are inductive, so rapid changes in current
induce large voltages which oppose the change in
current. To make a long story short, the result
would be similar to the following. Ideally, at
the moment shown, the currents flowing into the
brush from sections 1 and 2 are each 40 A. The
current density is uniform across the brush
surface, and the current in coil 1 is momentarily
0 A.
40 A
0 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
46
Practical Commutation
In the case of practical commutation, due to the
inductance of the coils, the current in coil 1
has been prevented from falling to 0 A, its
value is 30 A. KCL tells us that 70 A flows into
section 1 (40 A from coil 12 and 30 A from coil
1) and from section 1 into the brush. KCL also
tells us the current flowing from section 2 into
the brush is 10 A. The total brush current is
still 80 A. Equal portions of the brush surface
contact both sections, but because of the coil
inductance, more current flows through the
portion of the brush surface that contacts
section 1 and less through the portion that
contacts section 2. The current density is no
longer uniform across the brush surface.
40 A
30 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A
47
Practical Commutation
This nonuniform current density results in
nonuniform heating across the surface of the
brush, which can lead to overheating,
incandescence, and sparking. This can be
minimized by reducing the coil inductance, and by
minimizing the resistance of the brush surface.
40 A
30 A
40 A
40 A
40 A
1
2
3
4
5
12
80 A