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Chapter 5
Direct Current Generators
5.1 INTRODUCTION
The linear machine was introduced because of its simple construction and the fact that it served to
demonstrate clearly the principles of electromechanics. It also allowed us to establish a model to show
symbolically the relationships which exist with devices of this kind. At this point we must take a quantum
jump to see how these principles have been implemented in rotating machines, specifically generators and
motors. Historically, d-c machines came into being before a-c machines because the scientists of that time
(about the middle of the 19th century) were only familiar with battery sources and consequently strived to make
motors which operated from batteries, as well as generators to charge the batteries and operate arc lamps.
Although superior in many ways, a-c machines have not completely replaced d-c machines, and will not in the
foreseeable future, since the d-c motor offers a controllability not yet approached by a-c motors. The d-c
generator, on the other hand, is declining rapidly in use since its functions have been largely taken over by solid
state rectification of alternator outputs (in automobiles for example). Nevertheless a thorough study of d-c
generators is worthwhile because the construction of motors and generators is the same, and the bilateral nature
of the energy conversion process means their “inner workings” share much in common.
5.2 CONSTRUCTION AND TERMINOLOGY
No mention was made of the origin of the magnetic field in the linear machine; in the loudspeaker it
was furnished by a permanent magnet. In d-c machines the magnetic field is supplied by field coils” of wire
wound around “pole cores” which are part of the magnetic circuit. These terms and others relating to the
construction of a typical d-c generator or motor are explained in the following table.
Figure 5.1. Cross Section of a Typical d-c Machine. Numbers refer to items in Table 5.1.
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1. FIELD COILS. Coils of insulated copper wire which provide the mmf for the magnetic
field.
2. POLE CORES. Steel cores around which field coils are wound. Adjacent poles alternate in
polarity (N-S-N-S etc.)
3. POLE SHOE. Part of the pole structure (steel) which conforms to the curvature of the
armature core in order to provide a uniform air gap length.
4. YOKE. Steel frame providing mechanical rigidity and also providing a path of low
magnetic reluctance between poles.
5. ARMATURE A stack of steel laminations mounted on the shaft of the machine. CORE
CORE Copper armature conductors are placed in the slots. A major part of the
magnetic circuit.
6. ARMATURE. The collection of copper wires in which voltages are induced (BLu) and on
which forces are produced by current (BLi). In the linear machine the ‘bar’
is the armature.
7. SLOTS. Rectangular openings around the periphery of the armature core into which
armature conductors are placed.
8. TEETH. Rectangular area between slots, around periphery of the armature core, i.e.
material left after slots are cut out.
9. COMMUTATOR. A ring of copper segments surrounding the shaft, which are insulated from
each other by strips of mica. The ends of armature conductors are
connected to commutator segments.
10. SHAFT. A steel rod on which the armature core is mounted. The means by which
mechanical power is delivered from a prime mover.
11. BRUSHES. Stationary rectangular carbon and graphite blocks which make electrical
contact with the rotating commutator for the purpose of completing the
current path from the external terminals through the armature conductors
and return.
12. BRUSH RIG- Mechanical assembly which holds the brushes in place and which GING
provides for adjusting tension of the springs which push the brushes against the commutator.
13. END BELLS. Steel structures on both ends of the machine which provide support for the
bearings and brush rigging.
14. EXCITATION. A general term referring to the production of the magnetic field within the
machine. “Separate excitation” refers to the supplying of field-coil current
from an outside source such as a battery; whereas “self excitation” refers to
the generator supplying field-coil current from its own armature.
Table 5.1 MACHINE TERMINOLOGY
A typical armature assembly is shown below in Figure 5.2. Part (a) shows a lamination used in the core.
Multiple laminations are stacked to build the core. Laminations are used for ease of construction and to prevent
eddy currents. Part (b) shows the laminations stacked together on a shaft and part (c) includes the wires and
commutator segments.
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Figure 5.3: The Elementary Generator
(a) (b) (c)
Figure 5.2. Components of a Typical Armature Assembly
5.3 THE ELEMENTARY GENERATOR AND MECHANICAL COMMUTATION:
A very simple elementary AC generator is discussed first because of the similarity in concepts and
construction between AC and DC generators. Then the process of mechanical commutation is introduced.
Commutation is used to change an AC into a DC machine be it a generator or a motor.
Elementary Generator Construction
The elementary generator of Figure 5.3 consists of a loop of wire free to rotate in a stationary
magnetic field. Relative motion between the wire and the magnetic field will induce a potential difference
between the ends of the loop. Sliding contacts are used to connect the rotating loop to an external circuit in
order to use the induced voltage.
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Figure 5.4: How the Elementary Generator Works
The pole pieces are the north and south poles of the magnet which supply the magnetic field. The
loop of wire which rotates through the field is called the "armature." The ends of the armature loop are
connected to rings called "slip rings," which rotate with the armature. Brushes ride up against the slip rings
providing a sliding electrical contact to pick up the electricity generated in the armature and carry it to the
external circuit.
In the de******ion of the generator action which follows, visualize the loop rotating through the
magnetic field. As the sides of the loop cut through the magnetic field, they generate an induced voltage which
causes a current to flow through the loop, slip rings, brushes, zero-center current meter and load resistor all
connected in series. The induced voltage that is generated in the loop, and therefore the current that flows,
depends upon the position of the loop in relation to the magnetic field. Now, lets analyze the action of the loop
as it rotates through the field.
ELEMENTARY GENERATOR OPERATION
Assume that the armature loop is rotating in a clockwise direction and that its initial position is at A
(zero degrees) of Figure 5.4. In position A, the loop is perpendicular to the magnetic field and the black and
white conductors of the loop are moving parallel to the magnetic field. If a conductor is moving parallel to a
magnetic field, it does not cut through any lines of magnetic flux and no voltage is generated in the conductor.
This applies to the conductors of the loop at the instant they go through position A. No voltage is induced in
them and therefore no current flows through the circuit. The current meter registers zero.
As the
loop rotates from position A to position B, the conductors are cutting through more and more lines of flux until at 90
degrees (position B) they are cutting through a maximum number. In other words, between zero and 90 degrees, the
induced voltage in the conductors builds up from zero to a maximum value. Observe that from zero to 90 degrees the
black conductor cuts down through the field while at the same time the white conductor cuts up through the field. The
induced voltage in each of the conductors is therefore in series, and the resultant voltage across the brushes (the terminal
voltage) is the sum of the two induced emfs, or double that of one conductor since the induced voltages are equal to each
other. The current through the circuit will vary just as the induced emf varies, being zero at zero degrees and rising up
to a maximum at 90 degrees. The current meter deflects increasingly to the right between positions A and B, indicating
that the current through the load is flowing in the direction shown. (The reader should verify the polarity of the induced
voltage and the direction of resulting current by applying either the right hand rule or the cross product relationship.)
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Figure 5.5: How the Elementary Generator Works (cont’d)
The direction of current flow and polarity of the induced voltage depends upon the direction of the magnetic field and
the direction of rotation of the armature loop. The waveform shows how the terminal voltage of the elementary
generator varies from position A to position B. The simple generator drawing on the right is shown rotated by 900 to
illustrate the relationship between the loop position and the generated waveform.
As the loop continues rotating from position B (90 degrees) to position C (180 degrees), the
conductors, which are cutting through a maximum number of lines of flux at position B, cut through fewer
lines, until at position C they are again moving parallel to the magnetic field and no longer cut through any
lines of flux. The induced voltage, therefore, will decrease as the loop rotates from 90 to 180 degrees in the
same manner as it increased from zero to 90 degrees. The current flow will similarly follow the voltage
variations. The generator action at positions B and C is illustrated in Figure 5.5.
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From zero to 180 degrees the conductors of the loop have been moving in the same direction
through the magnetic field and, therefore, the polarity of the induced emf has remained the same. As the
loop starts rotating beyond 180 degrees back towards position A, the direction of motion of the conductors
through the magnetic field reverses. Now the black conductor cuts up through the field, and the white
conductor cuts down through the field. As a result, the polarity of the induced voltage and the current flow
will reverse. From positions C through D back to position A, the current flow will be in the opposite
direction than that from positions A through C. The generator terminal voltage will also have its polarity
reversed. The voltage output waveform for the complete revolution of the loop is as shown in Figure 5.6.
Figure 5.6: How the Elementary Generator Works (cont’d)
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RECTIFICATION BY MECHANICAL COMMUTATION
A simple AC generator was discussed in the section above. How can this machine be modified to
produce DC voltage and current which has the property that the polarity of the voltage does not change and
the direction of current flow is always the same, although it may not be constant? One method for
producing unidirectional voltage and current flow is shown below in Figure 5.7(a). The scheme is similar
to the elementary AC generator but the hardware associated with the connection of the rotating wire loop to
the outside world is modified. Instead of slip rings and brushes, the DC machine has commutator segments
and brushes. The slip ring of the AC machine has been divided into two halves insulated from each other
to form a commutator. Referring to Figure 5.7(a) , the right hand brush is always connected to the side of
the loop which is traveling downward next to the South Pole at the times of maximum induced voltage.
This means that the voltage at the left hand brush will always be “+” relative to that of the right hand brush.
A commutator is a rotating switch which reverses the connection between the armature (the rotating loop of
wire) and the outside world every half turn just at the point in the cycle the polarity would reverse in an
elementary AC generator. Note that the poles in Figure 5.7 are reversed relative to Figure 5.6. The
reader should again use the cross product or the right hand rule to verify the direction of current.
Consider it a homework assignment. The voltage at the left hand brush relative to that at the right hand
brush is shown in Figure 5.7(b).
(a) (b)
Figure 5.7 Mechanical Commutation Produces Unidirectional Output Voltage
Note that although the voltage across the brushes is unidirectional, it is not very constant. The
voltage pulsations are undesirable for many applications. How can the voltage generated be made
smoother? The answer is by including more loops of wire in the armature and correspondingly more
commutator segments. The armature in Figure 5.8(a) contains two mutually perpendicular loops with the
corresponding output shown in Figure 5.8(b) and a 3 loop armature machine and output are shown in
Figure 5.9. The generated DC voltage becomes more constant as more and more loops are included on the
armature.
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(a) (b)
Figure 5.8 A DC Generator With Two Loops On Its Armature and Corresponding Output
(a) (b)
Figure 5.9 A DC Generator With Three Loops On Its Armature and Corresponding Output
There are two primary methods of connecting the armature windings together. The turns are
connected together in parallel in the lap winding method. This provides for greater current capability at a
lower voltage. The turns are connected in series in the wave winding method providing for higher voltage
at reduced current. Combinations of these two methods can also be used. The details of these two
methods are beyond the scope of this course. If you are interested in learning more about either type of
winding, your instructor can help or direct you to an appropriate reference.
5.4 THE EQUIVALENT CIRCUIT
Since the d-c generator is a practical form of the linear generator, the model introduced for the
latter may be used with only minor modifications. First, because control of the magnetic field strength by
adjusting current in the field coils is an important feature of the d-c generator, symbols for the field coils
and their resistance, RF , are included in the model. Second, since the generator is a rotating machine, the
model will show the armature symbol connected to a rotating shaft which is driven by a prime mover, as a
constant reminder that energy is being converted from one form to the other. Table 5.2 illustrates how the
basic relationships for the linear machine are modified to fit the rotating machine.
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V f
R f
Prime Mover
T a
T d
Ω
T f
R A
V T
+
-
IA
+
-
φ E = KφΩ
IF
Field Armature
Figure 5.10 Separately Excited DC Generator
Table 5.2.
Item Linear Machine Rotating Machine
Velocity/Speed u m/s 6 rad/s or n rpm
Current, Armature I amperes I A amperes
Current, Field ------------- I f amperes
Magnetic Field B Webers/m2 1 Webers (per pole)
Voltage, Induced e = BLu newtons E = K16 = k1n volts
Voltage, Terminal v Volts VT Volts
Force/Torque (Developed) fd = BLi newtons Td = K1IA N-m
Force/Torque (Applied) fa newtons Ta newton-meters
Resistance, Armature r ohms RA ohms
Mechanical Power Dev. fdu watts Td 6 watts
Electrical Power Dev. ei watts EIA watts
Power Equivalence ei = fdu EIA = Td 6
Machine Constants* ---------------- K,k
*The constant for a given machine lumps together all the effects of the unchanging quantities
such as the number of poles, dimensions of the armature core, number of armature coils and
the method of interconnecting them.
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A few comments are in order for clarifying information given in the table. First, the rotational
velocity of the machine is stated formally as 6 which is related to tangential velocity of u = radius x 6, but
tachometers in present use are calibrated in the more practical unit of revolutions per minute, symbol n. Thus
in the equation
volts (E=KφΩ=kφn 5-1)
The two constants K and k differ, but we need not be concerned with their actual numerical values. The
relationship between the two velocities is
Ω =2πn /60rad/s This is from (n rev/min)(2Œ rad/rev)(1 min/60sec) (5-2)
The magnetic field is traditionally represented as flux (1) instead of flux density (B) because of the size of the
average voltage E is arrived at by considering the total flux encountered by a single revolution of a conductor
rather than the instantaneous flux densities. Finally, it will be remembered that tangential force applied to a
rotating cylinder is interpreted as “torque”, which is the product of the tangential force and the radius of the
cylinder,
Td = Σfd x (radius) newton-meters (5-3)
whereΣ fd is the sum of the tangential forces developed by the individual conductors, and the radius is called
the “torque arm”. Consequently,
Td=KφIANewton-meters (5-4)
Where the constant “K” is the same as in E = K1:.
5.5 THE MAGNETIZATION CURVE
Magnetic conditions within the generator can be investigated by testing a separately excited generator
using a voltmeter connected to the output of Figure 5.10, where VT is indicated, and an ammeter in series with
the field circuit to measure field current. The test procedure is to adjust the speed of the prime mover to the
rated value as indicated on the generator name plate and then to adjust the variable resistor for increasing
values of field current (including zero) as read on the ammeter. For each setting of field current, the voltmeter
indicates the corresponding induced voltage. Data is usually taken for voltages up to 25 to 50 percent above
the rated value. The data is then plotted on a graph called the “saturation curve” or “magnetization curve”, a
typical curve is illustrated in Figure 5.11. If for any reason it should be desired to have curves for other
speeds, they can be obtained without actually running more experimental tests, by using direct proportionality.
For example if the curve is wanted for 1000 rpm a point whose voltage is 250 x 1000/1200 would be located
directly below the 250-volt ordinate on the 1200 rpm curve. By calculating a series of points in this manner
a new curve can be sketched below the 1200 rpm curve. The justification for this procedure is that for any
given fixed current the flux is the same for any speed and hence induced emf is proportional to speed only.
We shall next examine the magnetization curve in detail because its shape has a direct bearing on the operation
of generators and motors.
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50
100
150
200
250
300
0
0 0.5 1.0 1.5 2.0 2.5
Field Current in Amps
Generated Voltage in Volts
1200 rpm
Figure 5.11 Magnetization Curve for a Typical DC Generator.
Now this graph can be interpreted as a plot of air gap flux vs mmf as well as voltage vs current, since
voltage is proportional to flux when the speed is held constant E=kφn, and the current becomes mmf-perpole
when multiplied by the number of turns of wire in each field coil. Of course the numerical scales would
differ, but the point is that with this interpretation we can explain the behavior of the magnetic circuit within
the machine by referring to the shape of the curve. For this reason the curve has been redrawn with flux and
mmf coordinates in Figure 5.12.
Beginning in the lower left corner of the graph it can be seen that a flux exists for zero mmf. This
should not be surprising since we know from previous study of magnetic phenomena (hysteresis) that residual
magnetism is left in the steel structure from previous use of the generator. Although this flux is responsible
for only a small induced voltage, it fulfills an extremely important role in self-excited generators by providing
the “seed” flux which starts the process of “build-up” to normal flux levels — to be explained in detail later.
As current (mmf) is increased, it is seen that the curve is straight for an appreciable portion, i.e. flux is directly
proportional to the mmf. In this region the steel parts of the magnetic circuit account for only a small portion
of the total mmf, most of the mmf drop occurring in the air gap between the armature core and the pole shoe.
As current is increased beyond the linear region, rotation of magnetic domains and finally magnetic saturation
take place in the steel parts resulting in a marked departure from the straight-line relationship of the curve.
Since this is a gradual process rather than an abrupt one, there is a transition region between the linear part and
the saturated part, which is referred to as the “knee” of the curve.
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0 MMF, Magnetomotive Force which
is proportional to Field Current
FLUX
0
LINEAR
SATURATION
Figure 5.12 Magnetization curve with modified scales.
Operating the generator within the linear portion of the curve is recommended for applications where
terminal voltage is to be varied over a wide range, as in situations where the generator has direct control of a
single load. On the other hand, if an application requires the voltage not change much once it is set, operation
in the saturated region is preferred. In the former case the armature is more responsive to control since a given
change in field current causes a larger change in voltage than in the saturated region.
5.6 SEPARATELY-EXCITED GENERATORS
The simplest generator and one which is used for a wide variety of applications is “separately
excited”. The modifier “separately” is used to distinguish it from another class of generators which are “selfexcited”,
referring to the manner in which electrical energy is supplied to the field coils. In generators using
self excitation (to be treated later) the field is connected to the armature; whereas with separate excitation the
field is connected to a separate source. The obvious advantage of self excitation is the cost savings realized
by not requiring additional capital investment for a separate source. The advantage of separate excitation
which offsets the extra cost is the much greater range of control available, including the ability to change
polarity at will. The principle application for this type of generator is in motor control. The method of wiring
was indicated in the test for acquiring data for the magnetization curve, but in Figure 5.13 a load resistor has
been added and the field control potentiometer has been modified to allow reversal of field current. We shall
next investigate how the generator reacts under load conditions, but first a few words about ratings.
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V f
R f
Prime Mover
T a
T d
Ω
T f
R A
VM
IA
+
-
φ E = KφΩ
IF R L
AM
+
+
-
-
IL
Figure 5.13 Separately excited generator.
The chief parameters for rating rotating machines are power output, voltage, current and speed. These
ratings assume continuous operation, 24 hours a day and 7 days a week unless otherwise stated. Operation of
a machine above its current rating for an appreciable length of time could result in overheating and to
deterioration of insulation or complete burnout. The current rating is the most likely rating to be exceeded
since it depends directly on the loading of the machine. In the graphs of external characteristics which follow,
rated (i.e. “full-load) values will be indicated for reference. Operation beyond the rated limits is shown for
completeness, for the sake of perspective, not as recommended practice.
The load characteristic for a separately-excited generator is displayed in Figure 5.14. This graph is
the I-v relationship or terminal characteristic of the generator. The drop in terminal voltage as the current
increases is caused by the IARAvoltage drop within the armature in accordance with
VT=E−IARA volts (5-5)
Just as it was in the linear generator. The voltage regulation of the generator is defined as:
V.R. = Voltage Regulation = VNL −VFL  /VFL (5-6)
Suppose it is desired to investigate the effect on the external characteristics of changing the field
current by readjusting the variable resistor in the field circuit, without changing the load resistance. First, we
can draw a family of curves for several settings of the field current, as shown in Figure 5.15. The spacing of
the curves will be equidistant for equal field current increments provided that operation is limited to the linear
part of magnetization curve, otherwise not. Second, a “load line” is drawn from the origin through the point
where the coordinates are rated current and voltage. The slope of this line is the resistance of the load RL. The
load line is the i-v characteristic of RL. The intersection of this load line with the external characteristic of the
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generator is the point at which the generator will operate. For field current IF4 and the load line shown in
Figure 5.15, rated current and voltage will result. Now it can be seen that increasing the field current above
the IF4 value results in current and voltage overloads, but reducing it below IF4 affords full control of current
without overload.
V FL
V NL
V T ,Terminal Voltage
Rated Voltage
Rated Current
Load Current
IL = IA
Load Line Slope = VL /IL = RL
This Load Intersects Machine
Characteristic at Greater Than
Rated Load Current Causing
Overheating
Figure 5.14 External Characteristic of a Separately-Excited Generator.
V FL
V T ,Terminal Voltage
Rated Voltage
Rated Current
Load Current
IL = IA
IF1
IF2
IF3
IF4
IF5
Figure 5.15 External Characteristics of a Separately-Excited Generator with
Various Settings of Field Current
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To further demonstrate the versatility of the separately-excited generator, note that moving the
“slider” of the field control resistance to the left of the centerpoint (Figure 5.13) reverses the polarity of the
voltage applied to the field coils, which reverses the direction of field current and eventually the magnetic field.
This extends operation of the generator to negative terminal voltage and negative line current with respect to
Figures 5.14 and 5.15 or into the third quadrants of these plots.
 EXAMPLE 5-1 
A separately-excited generator with the following nameplate ratings is delivering energy at rated
voltage to 8 parallel loads with resistances of 25 ohms each:
5 Kilowatts 1800 rpm
125 Volts 40 Amperes
When the loads are disconnected the voltage rises to 130 volts.
a. Calculate the voltage regulation.
b. Calculate the armature resistance.
c. Calculate the internal torque when supplying the original loads.
While the loads are still connected, the speed governor malfunctions, reducing the speed to 1400 rpm.
d. Calculate the new values of IL and V.
e. Calculate the new value of torque.
f. To what value will the voltage now rise when the loads are disconnected?
Solution
a. V.R. = VNL−VFL/VFL=130 −125/125 =.040
b. RL=25/8=3.125 O h m s ; IA=IL=V/RL=125/3.125=40 a m p e r e s ,
RA=E−V/IA=130−125/40=0.125 Ohms
c. Td=EIA/Ω=130x40/2π 1800 / 60=27.6 N-m.
d. By proportionality, E2=E1(n2/n1)for a given setting of field current, therefore,
E=130(1400/1800)=101.1 volts
IL=IA=E/RA+RL=101.1 /0.125+3.125=31.1 amperes
V=E−IARA=101.1−31.1x0.125=97.2 volts.
e. Td=EIA/Ω=101.1x31.1 / 2π 1400 / 60=21.5 N-m.
f. E = 101.1volts.
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5.7 SELF EXCITED SHUNT GENERATORS
A large savings in the initial cost and complexity of a DC generator can be realized if the field circuit
is connected across the armature terminals instead of to a separate source. Moreover, a great amount of control
over the terminal voltage is still available, with the aid of a “field rheostat” (RC) connected in series with the
field coils.
R A
IA
+
-
E = KφΩ
φ
IF
+
-
V T
R A
IA
+
-
E = KφΩ
+
-
V T
IL
φ
R F
R C
FIELD FIELD
IF
V F
R F
(a) Separately-excited Generator (b) Self-excited Generator
Figure 5.16. Comparison of generator connections.
The term “shunt” is synonymous with “parallel”. Since the field circuit is wired in parallel with the load, as
far as the armature is concerned, the field circuit is just another load to be supplied with current. From KCL:
IA=IL+IF (5-7)
and then using Ohm’s Law:
IF=VTRF+RC. (5-8)
There is an interdependence between the field circuit and the armature circuit since the field current supplies
the magnetic field which produces the armature induced voltage but at the same time the armature supplies
current to the field current, a feedback situation. This can be expressed in mathematical terms as
(5-9) E f IF =  
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which describes the magnetization curve, and Equation (5-8) which describes the field circuit. Since Equation
(5-9) describes a non-linear, non-analytic function it is necessary to resort to a graphical solution of these two
simultaneous equations.
Since Equation (5-8) is a linear equation (Ohm’s law) it can be plotted as a straight line on a graph.
Specifically, we choose to plot it on the same graph as the magnetization curve. Since it is a straight line
through the origin and has a slope of , one need only pick a single current at random RF+Rc and calculate
the corresponding voltage to find one point on the line, then drawing a line through that point and the origin
to complete the job, as illustrated in Figure 5.17. Since the magnetization curve indicates values of E and the
field resistance line indicates values of VT, the vertical difference between the curves must correspond to the I R A A
drop in the armature.
0
0
IAR A
E
V T
IF
Field Resistance Line
E, VT
Figure 5.17 Field resistance line drawn on magnetization curve showing
voltage division in armature.
Voltage Build-Up
The build-up of voltage in a self-excited generator can be visualized with the aid of this graph. First
assume there is no load resistance connected to the terminals of the generator except the shunt field, and since
the field current is small compared with the rated load current (to keep losses low) it can also be assumed that
theIARA = IFRA voltage drop is negligible, so VT must equal E in the steady state. Second, assume the prime
mover has been brought up to rated speed. The small residual magnetic field induces a small voltage in the
armature which in turn causes a small field current. This field current increases the strength of the magnetic
field which then causes an increase in induced voltage, which causes an increase in field current, and so on.
The growth of the magnetic field, with concurrent growth in the magnitude of induced voltage, is thus seen to
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be an automatic, positive feedback situation. Now, what prevents the build-up, the positive feedback cycle,
from proceeding ad infinitum? The non-linearity of the magnetization curve gives us a clue: saturation. An
increase in flux for a given increase in field current becomes smaller beyond the knee of the curve, until finally
an equilibrium condition exists. Then the field current is precisely enough to keep the induced voltage at the
level required by Ohm’s law to maintain that amount of current in the field circuit.
Remembering our assumption of negligible drop under no-load conditions, the establishmIARA ent
of an equilibrium condition must result in E = VT, and the only place on the graph where this is true is at the
intersection of the resistance line and the magnetization curve. This point can also be interpreted as the
graphical solution of simultaneous Equations (5-8) and (5-9). We can make use of this knowledge by coupling
it with the fact that the slope of the field resistance line is equal to the combined field circuit resistances
RF+RC. To be specific, we can control the voltage at which the intersection occurs by adjusting the field
rheostat Rc, as illustrated in Figure 5.18. Beginning with the original setting Rc1, which produces a voltage E1
an increase to RC2 (steeper slope = larger resistance) drops the voltage to E2. Finally, an increase to RC3 drops
the induced voltage drastically, where build-up is practically nonexistent. This brings up a point: if a selfexcited
generator fails to build up, what are the usual causes?
0
0
IF
E, VT R C3 R C2 R C1
E 1
E 2
E 3
Figure 5.18. Effect of varying field rheostat on no-load voltage build-up.
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Table 5.3. Failure of a Generator to Build Up*
Symptom Reason Remedy
1. Voltage low, drops when field
is disconnected.
Field rheostat set for too
much resistance
Readjust field rheostat for
lower resistance
2. Voltage low, rises slightly
when field is disconnected
Field mmf is opposing
residual magnetism
Switch shunt field
connections
3. No voltage No residual magnetism Disconnect shunt field and
reconnect it to a d-c source
(“flasher”) for a few
seconds. Disengage field
from source and reconnect to
armature.
4. No voltage Open circuit Shut down and check circuits
for continuity
5. Wrong Polarity Residual magnetism in
wrong direction
“Flash” field as described
above
*Assuming direction of rotation is correct.
Load Characteristic
If a self-excited shunt generator is connected to a load and a test is run for obtaining the load
characteristics, it is found that the voltage drops more, for a given load current than it does when connected
for separate excitation, as illustrated in Figure 5.19.
V NL
V T ,Terminal Voltage
IL
Self-excited
Separately-excited
Rated Current
Figure 5.19 Comparison of External Characteristics For The Same
Generator Connected Two Ways, With The Same No-load Voltage
20
The additional armature current required by the field does not sufficiently increase the IARIA=IL+IF A drop to
account for the entire additional drop — we must look elsewhere for a reasonable explanation. The answer is found
in Equation (5-8) where the field current is seen to be directly proportional to the terminal voltage VT. The field current
does not remain constant as it would in a separately excited generator. A drop in VT causes a drop in IF, which in turn
results in a decreased induced voltage E. This can be seen graphically in Figure 5.20, where E has dropped from the
no-load value at the intersection (E1) to the loaded value, E2, with the accompanying change in VT from VT1 to VT2.
Correspondingly, the drop in VT has reduced field current from IF1 to IF2. In other words, the difference between the two
curves in Figure 5.19 is accounted for by the difference between E1 and E2 in Figure 5.20.
0
0
IAR A
E 2 V T2
IF2
E, VT
IF1
E 1 = VT1
IF
Figure 5.20 Effect of loading on location of operating point, in self-excited
generator. Condition 1 is no-load condition 2 is with load.
EXAMPLE 5-2
The generator of example 5-1 is rigged for self-excitation and adjusted for rated voltage and load
current, with the same load as before. The field current is noted to be 1.85 amperes. When the load is
disconnected the terminal voltage rises to 138 volts.
a. Calculate voltage regulation.
b. Calculate total field resistance.
c. Calculate generated emf under full-load conditions.
d. Calculate field current under no-load conditions.
21
Solution
a. V.R. = (138 - 125)/125 = .104
b. RF+RC=V/IF=125/1.85=67.6 ohms
c. E=VT+IL+IFRA=125+41.850.125+130.2 volts
d. IFno−load=VT/RF+RC =138/67.6=2.0 amperes
5.8 COMPOUND GENERATORS
The “droop” in the external characteristic of the shunt generator is a disadvantage when it is serving
a multiplicity of loads, since the voltage at one load location will fluctuate up and down as other loads are
turned on and off. This effect is rendered all the worse by the resistance of wiring between the generator and
the loads, which causes additional voltage drop. An ingenious but simple method of overcoming this difficulty
was thought of in the early days of electrical engineering, consisting of a means of incorporating automatic
positive feedback proportional to load current — the “series’ field
The key to offsetting voltage drop is in increasing the induced voltage by adding extra mmf. Although
this could be done by someone continually adjusting the shunt field rheostat, a better way is to add another
field winding called a “series field”, as shown in Figure 5.19(a). The series field is wired in series with the
load and thus carries load current. The physical location of the shunt and series field windings are the same,
but because the full-load current is typically many times greater than the shunt field current, the series field
must be wound with much larger wire. Increased wire size is offset, however, by the small number of turns
needed by the series field to achieve a significant amount of mmf. For example, a 500-turn per-pole shunt field
with a current of 2 amperes produces an mmf of 1000 ampere-turns, but a series field of only 5 turns per pole
with a current of 40 amperes produces an mmf of 200 ampere-turns.
IF R L
IL
R C
SHUNT SERIES
+
-
E
+
-
V T
V NL
V T ,Terminal Voltage
Rated Current
IL
Compound
Separately
Excited
Shunt
(a) Wiring Diagram (b) Comparative External Characteristics
Figure 5.21. Compound generator
22
The external characteristic illustrated in Figure 5.21(b) shows a rising terminal voltage as load current
increases. This is often more desirable than a horizontal curve because the additional voltage above the noload
value can be used to cancel the effect of IR drop in the wires leading to the loads, If the voltage rise is
greater than needed for a given situation, a variable resistor called a “diverter” can be added in parallel with
the series field to bypass a portion of the load current around the field, thus, lowering the curve. If the curve
is horizontal, the generator is said to be “flat-compound” because the full-load and no-load voltages are the
same (voltage regulation is zero). “Overcompound” and “undercompound” generators have characteristic
curves above or below the flat curve, respectively. Generators are normally furnished by the manufacturer as
overcompound so the user can adjust the compounding to suit his or her application, by adding a diverter.
An interesting possibility arises when one considers that the series field might also be connected so
its mmf opposes the mmf of the shunt field. Under this circumstance the terminal voltage would drop quite
drastically when a load is connected. Although this defeats the purpose of adding the series field in the
beginning, there is one application which uses the series field in this way — the d-c arc welding generator.
The sharply drooping voltage characteristic fits the need in welding for a large voltage to start an arc but once
the arc begins the voltage needed to sustain it is considerably less.
Figure 5.22 Physical Location of Shunt And Series Field Windings Shown On One Pole.
5.9 TACHOMETER GENERATORS
Aside from its in energy conversion, the d-c generator finds application as a transducer which
provides an output voltage proportional to shaft speed. This is accomplished quite simply by using a
permanent magnet to provide a magnetic field of constant strength and coupling its shaft directly to the rotating
shaft whose speed is to be measured. The armature terminals are then connected to a voltmeter calibrated
directly in rpm. Another use for tachometer generators is as a link in the feedback path of a servo system, for
example in a system to maintain motor speed at a steady value in spite of fluctuating load.
23
5.10 SUMMARY
 Knowledge of the terminology and construction of d-c machines is essential to understanding their
operation
 The induced emf and current emf direction in each armature coil undergoes a reversal in traveling the
distance of one pole pitch. The brush-commutator system serves as a means of rectifying the voltage and
current so they appear at the brushes as unidirectional.
 The magnetization curve is a plot of induced voltage vs. Shunt field current, determined experimentally by
separately exciting the generator under no-load conditions. It is useful in explaining and predicting machine
performance. Curves for speeds other than the test speed can be constructed by direct proportionality.
 The separately-excited generator requires a voltage source to supply power to the field circuit. This allows
complete control of the generator’s armature voltage from zero up to rated, for either polarity.
 The self-excited generator saves the cost of providing a source for the field by wiring the field in parallel
with the armature; field current is then supplied by the armature. Initial flux to allow the system to build
up voltage is supplied by the small residual magnetism in the magnetic circuit. Indefinite build-up of
voltage is prevented by saturation of the magnetic circuit.
 Reasons for the failure of a self-excited generator to build up voltage are enumerated in Table 5.3.
 Addition of a series field to a self-excited shunt generator, wired in series with the armature*, provides
additional mmf as the generator load current increases, thus offsetting the voltage “droop” which normally
occurs with the shunt field only. A generator having both fields, shunt and series, is called a compound
generator.
*or the total
5.11 QUESTIONS
Q5.1 Describe the function of the commutator in a DC generator.
Q5.2 What is the advantage of multiple pairs of commutator segments in a DC generator?
Q5.3 Describe the source of the magnetic field in which the rotor spins for (a) separately excited DC
generator and (b) a self-excited DC generator.
Q5.4 What allows the approximation to be made that the no load terminal voltage is about equal to be armture
induced voltage for a self-excited DC shunt generator?
Q5.5 What would be the effect of zero residual magnetic field in a self-excited, DC shunt generator upon
startup? How would you correct this problem?
Q5.6 What is the major advantage of a compound DC generator?
24
5.12 PROBLEMS
P5.1 Assuming that Figure 5.8 of the **** applies to a particular DC generator, find the open circuit
voltage of that generator if (a) the field current is 0.5A and the speed is 1800 rpm and (b) the field
current is 2.0A and the speed is 1000 rpm.
P52. Repeat Example 5-1 of the **** if 10 parallel loads are supplied, each having a resistance of 20
Ohms and the generator is rated at 120 Volts, 1800 rpm, 60 Amps and 7.2 kW.
P5.3 The generator of P5.2 is reconnected for self-excitation and adjusted for rated voltage and load
current, with the same load as before. The field current is noted to be 1.8 Amps. When the load
is disconnected, the terminal voltage rises to 140 V.
a. Calculate the voltage regulation.
b. Calculate the total field resistance.
c. Calculate the generated emf under full-load conditions.
d. Calculate the field current under no-load conditions.
P5.4 A separately excited DC generator is rated at 125 V, and 1 Amp at 1200 rpm. When the load is
disconnected, the terminal voltage rises to 130 Volts. Calculate (at rated conditions):
a. the armature current,
b. the voltage regulation,
c. The armature resistance,
d. And the internal torque when supplying rated load.
P5.5 If the speed of the generator of P5.4 is increased to 2000 rpm, calculate: (Assume linear operation.)
a. The load current,
b. The terminal voltage,
c. And the mechanical power converted to electrical power.
P5.6 A self-excited DC shunt generator is rated at 125V, and 1 Amp at 1200 rpm. The field resistance is
125 Ohms and the armature resistance is 5 Ohms. Calculate:
a. The field current.
b. The armature current,
c. The no-load terminal voltage (the behavior is non-linear, and a graphical solution using the
curve is required),
d. The voltage regulation,
e. And the internal torque when supplying rated load.
25
P5.7 A separately excited DC generator rated at 22 kW, 220 V and 100 A at 1200 rpm has the
magnetization curve shown in Figure 5.8 of the ****.
a. Given the field resistance of 150 Ohms, estimate the field current if the generator has a no-load
voltage of 250 Volts.
b. Determine the armature resistance, RA.
If the generator is supplying rated voltage and current to the load, calculate:
c. the developed torque,
d. the total input power if rotational losses are 0.1 Hp.
e. the total power out,
f. the efficiency,
g. and the % voltage regulation.
P5.8 A self-excited DC generator rated at 22 kW, 220 V and 100 A at 1200 rpm has the magnetization
curve shown in Figure 5.8 of the ****. If the no-load voltage is 250 V, calculate:
a. the field resistance, Rf.
b. the armature resistance, RA.
c. the rated load resistance,
d. and the developed torque.
If mechanical losses are 0.1 Hp, calculate:
e. the total power,
f. the efficiency,
g. and the % voltage regulation.
P5.9 If the speed governor for the generator of P5.7 were to fail and the speed increased to 1500 rpm,
what would the new no-load voltage be?
P5.10 If the speed governor for the generator of P5.8 were to fail and the speed decreased to 1000 rpm,
what would be the new no-load voltage?
P5.11 A separately-excited DC generator is rated at 130 V, and 45 A at 1200 rpm. At rated conditions, the
mechanical power losses are 250 W. When the load is disconnected, the terminal voltage rises to
136 V. Sketch and completely label a circuit model for this generator showing the following Ta, Td,
Tf, 6, Prime mover, E, IL, Rl, V, IA, RA, 1, VF, RF, RC, and IP. At rated conditions calculate:
a. the power supplied to the load, POUT,
b. the induced voltage, E,
c. the percent voltage regulation,
26
d. the armature resistance, RA,
e. the load resistance, Rl,
f. the angular velocity, 6,
g. the power converted from mechanical to electrical power, POUT,
h. the developed torque Td,
i. the mechanical power in PIN,
j. and the torque due to friction Tf.
k. Sketch the power flow diagram for this generator, including numeric values for each power.
P5.12 A separately-excited DC generator is running at rated conditions of 215 v and 155 A at 1800 rpm.
It has an armature resistance of 0.12 Ohms and the magnetization curve shown in Figure P5.12
applies.
50
100
150
200
250
300
0
0 1 2 3 4 5
Field Current in Amps
Generated Voltage in Volts
1800 rpm
Figure P5.12
a. Sketch a circuit diagram showing all the parameters listed in P5.11 and then determine the value
of the equivalent load resistance.
b. Predict the terminal voltage if the load is disconnected.
c. Determine the amount of field current required at full-load to produce the correct terminal
voltage.
27
d. Predict the terminal voltage for the load found in (a) if the field current is reduced by 20%.
e. If the field current is restored to its original value and the speed of rotation is slowed to 1000
rpm, determine the terminal voltage.
P5.13 A separately-excited DC generator has the magnetization curve shown in Figure P5.13. The rated
output is 5.5 kW and 210 V at 1200 rpm. The input torque to sustain these rated conditions is 49.5
N-m.
50
150
200
250
300
350
0
0 1 2 3 4 5
Field Current in Amps
Generated Voltage in Volts
1200 rpm
100
Figure P5.13
a. Determine the rated load ILD.
To produce rated output it is found necessary to set the field current IF, equal to 1.6A. This current
is supplied from a 200 V DC source. The resistance, RF, in the field winding is 100 Ohms.
Determine:
b. The value of the adjustable resistance, RC,
c. And the equivalent load resistor, RLD, which must be connected to the generator output terminals
to receive this amount of power.
d. And the equivalent load resistor, RLD, which must be connected to the generator output terminals
to receive this amount of power.
With the machine delivering rated output, the load is disconnected but no other changes are made.
Study the system carefully and determine:
28
e. the no-load terminal voltage, VT,
f. the value of RA, the armature resistance,
g. the percent voltage regulation of this machine,
h. the power converted from mechanical to electrical, PCONV,
i. the required mechanical input power, PIN,
j. the friction and windage loss of the machine,
k. and the efficiency of operation at rated output.
P5.14 A self-excited, shunt DC generator is rated at 130 V and 95 A at 1200 rpm. The total field
resistance is 120 Ohms and the armature resistance is 0.13 Ohms. Mechanical losses at rated speed
are 480 W. Under rated conditions, sketch a complete model of the generator including all
appropriate labels. Then determine:
a. the field current,
b. the armature current,
c. the induced voltage,
d. the angular velocity,
e. the developed torque,
f. the applied torque,
g. and the torque due to mechanical losses.
h. construct a power flow diagram for this generator under rated condition, including the numeric
value for each power.
The load is now disconnected. Under unloaded conditions both the induced and terminal voltages
will increase. If E increases to 145 V, what will be the no load terminal voltage?
(Note: the IARA voltage drop under unloaded conditions is very small compared with E and VNL.
Therefore, the assumption that under no load conditions the voltage drop due to the armature
resistance can be ignored is a good approximation.)
P5.15 The rated output for a self-excited, DC generator is 5.2 kW. The terminal voltage is 195 V at 1200
rpm. The following resistances were measured: RA = 1.13 Ohms, and the total field resistance - 116
Ohms. The friction and windage losses are 250 W. Carefully sketch and label a circuit diagram for
this generator and then determine:
a. the terminal voltage and shunt field current for this fully loaded machine,
b. the value of the generated voltage,
c. the efficiency of the machine.
29
P5.16 A self-excited, DC generator is rated at 22 kW, 220 V and 100 A at 1800 rpm. It has the same
magnetization curve as shown in Figure P5.12 and its no load voltage is measured at 250 V. At
rated output, the mechanical losses are 1 Hp. Calculate:
a. the total field resistance (Hint: plot the field resistance line and make appropriate
approximations.),
b. the developed torque, the total input power and the efficiency at rated output.
P5.17 A DC shunt generator is operating at rated condition, which are: 18 kW, 150 V, and 1500 rpm. The
total field resistance is 30 Ohms and the armature resistance is 0.08 Ohms. Mechanical losses are
142 W. Draw an equivalent circuit and a power flow diagram and then calculate:
a. the field current,
b. the armature current,
c. the induced voltage
d. the power converted,
e. the angular velocity,
f. the developed torque,
g. the total torque due to mechanical losses,
h. the torque applied to the shaft by the prime mover,
i. and the input horsepower required to turn the shaft.