Direct Current Motors
MTE 320
Spring 2006
E.F. EL-Saadany
Direct Current Motors
1. DC Machines Constructions
Direct current machines can be considered consisting of four main parts namely; Stator, Rotor, Air
Gap, and Commutators and Brushes. Figs. 1 and 2 present an illustrative diagram to the main
mechanical components of DC machines.
Fig. 1 Main components of 2-pole DC machines [1]
Fig. 2 Illustrative view of the main components of DC machines [1]
Stator: This is the stationary part in DC machines. It is responsible for generating a stationary
electromagnetic field in the air gap. It usually houses the field winding and known as woundfield DC motor. It is usually made of solid cast steel frame mounted to which a set of salient
poles from the inner layer. The poles are usually made of stacked iron laminations. The field
windings (coils) are mounted on these poles. These field coils carry the DC excitation current
when connected to an external DC source. This excitation current is responsible for generating
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Direct Current Motors
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the magnetic flux in the machine. In some cases, the stator may be constructed from permanent
magnets instead of the field windings and referred to as permanent magnet DC motor.
Rotor: The rotor and the commutators represent the main rotating parts of the DC machine as
shown in Fig. 3. The rotor is usually made of slotted, insulated iron laminations cylinder as
shown in Fig. 4. It carries the armature winding that usually arranged in slots as shown in Fig. 5.
These armature conductors are insulated from each other and from the iron core by paper or mica
layers.
Fig. 3 DC machines rotating parts [1]
Fig. 4 Slotted, iron laminations rotor of DC machines [1]
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Air gap: It is the air section separating between the stator and the rotor. The energy conversion
phenomenon occurs in the air gap.
Commutators and Brushes: The commutator is made of tapered copper segments mounted on
the rotating shaft. These segments are insulated from each other through mica sheets and are
connected to the armature windings as shown in Fig. 3. On the other hand, the brushes are made
of carbon (a mixture of carbon and small amount of copper is sometimes used to increase
conductivity). They are mounted on the commutator as shown in Fig. 6. Springs are used to
provide suitable pressure for the brushes to be in good contacts with the commutators.
Fig. 5 Crosssection of DC machines rotor slots carrying four conductors [1]
Fig. 6 Commutators and brushes [1]
2. Principle of Operation
2.1 Generator Mode of Operation
Fig. 7 presents a schematic diagram of a single coil DC generator where the coil rotates using an
external prime-mover. During the coil rotation, an emf (voltage) is induced between its terminal a and
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d. This induced emf depends on the coil angular position with respect to the two magnet poles. This
voltage is at its maximum positive value whenever side ab faces the north (N) pole as shown in Fig. 8
for one revaluation of the coil. On the other hand, the induced voltage is at its maximum negative value
whenever side ab faces the south (S) pole. Moreover, whenever the coil is at a vertical position the
generated emf is zero because there are no cut fluxes. Therefore, the generated voltage at the coil
terminal is an AC voltage as indicated by Fig. 8.
Fig. 7 Schematic diagram of a single coil DC generator.
Fig. 8 Induced emf in the armature windings.
The DC generator is equipped with a pair of commutators and brushes as shown in Fig. 7. Each half
(segment) of the commutator slip ring rotates with the coil while the brushes remain stationary. Each
segment is connected to one of the coil-end terminals. The generated voltage across the segments is
picked up by the stationary brushes and it is alternating as descried previously. The segments’ design
allows synchronization between changing the polarity of the generated voltage at each segment and
changing its contact with the brushes. Therefore, the generated load voltage will be rectified (with
constant polarity). Consequently, the load current will always be unidirectional current.
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Spring 2006
E.F. EL-Saadany
Fig. 9 Induced emf in the external load.
2.2 Motor Mode of Operation
A DC magnetic field is generated in the air gap by either the permanent magnet or the DC current
passing through the field windings (coils). Another DC current is forced to flow through the armature
windings (coils) from an external DC source passing through the commutators and brushes. A
magnetic force is therefore, generated as a result of the interaction between the field circuit magnetic
field and the armature current. This force acts on the side conductors of the same loop in an opposite
direction (because of the opposite direction of the currents as shown in Fig. 10) to create a torque that
starts to rotate the DC motor’s armature (in this case in clockwise direction).
Fig. 10 Principle of operation of DC motors. [2]
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Direct Current Motors
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Spring 2006
E.F. EL-Saadany
As the armature starts to rotate a counter (back) emf is induced in the armature windings that opposes
the armature voltage and, therefore, reduces the armature current. During the motor’s rotation, each
coil (turn) reaches a point in-between the field poles (neutral point) where the conductors are no longer
cuting the field’s magnetic field. Therefore, the force acting on the conductors at this point is zero. Due
to the machine’s inertia and because there are other armature coils being cut by the field at the same
instant, the rotor continuous its clockwise rotation.
Moreover, the commutators and brusches sets are designed in such a way to reverse the current in the
armature coils after passing this neutral point to maintain the clockwise rotation direction. Thus, the
current in the armature windings is an AC current while the current in the armature external circuit
(before the brushes) maintain its direction, i.e a DC current.
3. Induced Voltage and Developed Torque of DC Machines
The induced voltage in the armature winding Ea is known as the generated voltage in the case of
generator and the back emf if the machine is operated as a motor. The mathematical expression for this
back emf can be derived as follows;
Consider a single turn (coil) rotating between two curved poles as shown in Fig. 11. The total induced
emf in the turn (eind) can be expressed, in the terms of the induced emf in each segment, by;
eind = eba + ecb + edc + ead
where:
eba
is the induced emf in segment ba and is given, in terms of the magnetic field
density vector ( B ), the tangential velocity of the segment ba ( V ), and the
segment length ( L ), by,
(
)
eba = V × B • L
Under the pole face the vectors B and V are perpendicular to each other and
their cross product vector aligned with the vector L . On the other hand, beyond
the pole face edges the vector B is zero. Therefore,
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⎧v B l
eba = ⎨
⎩ 0
E.F. EL-Saadany
under pole faces (directed into the page)
beyound pole edges
Fig. 11 Single turn (coil) rotating between two curved poles. [3]
ecb
is the induced emf in segment cb. The vectors B and V are perpendicular to
each other and their cross product vector is always perpendicular to the vector
L . Therefore,
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E.F. EL-Saadany
ecb = 0
edc
is the induced emf in segment dc. Under the pole face the vectors B and V are
perpendicular to each other and their cross product vector aligned with the
vector L . On the other hand, beyond the pole face edges the vector B is zero.
Therefore,
⎧v B l
edc = ⎨
⎩ 0
ead
under pole faces (directed out of the page)
beyound pole edges
is the induced emf in segment ad. The vectors B and V are perpendicular to
each other and their cross product vector is always perpendicular to the vector
L . Therefore,
ead = 0
Therefore, the total induced emf in the turn (eind) can be expressed by;
⎧2 v B l
eind = eba + ecb + edc + ead = ⎨
⎩ 0
under pole faces
beyound pole edges
Since the tangential velocity ( v ) can be related to the rotational speed ( ω ) and the coil radius ( r ) by;
v=rω
Therefore,
⎧2 r B l ω
eind = ⎨
0
⎩
under pole faces
beyound pole edges
Since there are two poles, the area under each pole can be expressed approximately by;
AP =
2π r l
=π r l
2
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Therefore,
eind
⎧2
⎪ B AP ω
= ⎨π
⎪⎩
0
under pole faces
beyound pole edges
Since the flux density is always constant under the pole faces, then the total flux under each pole can
be expressed by;
φ = AP B
Therefore,
⎧2
⎪ φω
eind = ⎨ π
⎪⎩ 0
under pole faces
beyound pole edges
For DC motors with N turns, P poles, and a parallel paths, the total induced emf in the armature (Ea)
can be expressed by;
Ea = K a φ ωm
where:
Ka
The machine or the armature constant and is given by K a =
φ
The total flux under the pole (wb),
ω
m
NP
,
πa
The speed in radians/second,
Ea
The induced voltage/or back emf(v),
N
the total no of turns in the armature winding.
P
the number of poles in the machine.
a
the number of parallel path in the machine.
Once again, consider the single turn (coil) rotating between two curved poles as shown in Fig. 11. The
total induced (developed) torque in the turn (τind) can be expressed, in the terms of the induced torque
by each segment, by;
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Direct Current Motors
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τ ind = τ ba + τ cb + τ dc + τ ad
where:
τ ba
is the induced torque on the rotor by segment ba. The force on the wire segment
ba can be expressed, in terms of the magnetic field density vector ( B ), the
flowing current magnitude (i), and the segment length ( L ), by,
(
Fba = i L × B
)
Under the pole faces, the vectors B and L are perpendicular to each other
while, beyond the pole face edges the vector B is zero. Therefore,
⎧i l B
Fba = ⎨
⎩ 0
under pole faces
beyound pole edges
The induced torque caused by this force is given by;
⎧r i l B
⎩ 0
τ ba = r Fba sin (90°) = ⎨
τ cb
under pole faces
beyound pole edges
is the induced torque on the rotor by segment cb. The vectors B and L are
parallel to each other. Therefore,
τ cb = 0
τ dc
is the induced torque on the rotor by segment dc. Under the pole faces, the
vectors B and L are perpendicular to each other while, beyond the pole face
edges the vector B is zero. Therefore,
⎧i l B
Fdc = ⎨
⎩ 0
under pole faces
beyound pole edges
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Direct Current Motors
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Spring 2006
E.F. EL-Saadany
The induced torque caused by this force is given by;
⎧r i l B
⎩ 0
τ dc = r Fdc sin (90°) = ⎨
τ ad
under pole faces
beyound pole edges
is the induced torque on the rotor by segment ad. The vectors B and L are
parallel to each other. Therefore,
τ ad = 0
Therefore, the total induced torque on the rotor by the turn (τind) can be expressed by;
⎧2 r i l B
⎩ 0
τ ind = ⎨
Once again, since v = r ω , AP =
τ ind
under pole faces
beyound pole edges
2π r l
= π r l , and φ = AP B , therefore,
2
⎧2
⎪ φi
= ⎨π
⎪⎩ 0
under pole faces
beyound pole edges
For DC motors with N turns, P poles, and a parallel paths, the total induced (developed) torque (T) can
be expressed by;
T = Ka φ Ia
where
T
The developed torque in Nm.
Ka
The machine constant.
φ
The total flux under the pole (wb).
Ia
The armature current (A).
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E.F. EL-Saadany
For loss- free system,
mechanical power = electric power
T × ωm =
n=
Note:
where
Ea I a
ωm
× 60
2π
n is the machine speed in rpm and is given by
4. Types of DC Motors
In general, DC motors can be classified into Four main categories; based on its field winding
connection; namely; Separately Excited Motors, Shunt Motors, Series Motors, and Compound Motors.
4.1 Separately Excited DC Motors
In this type, the armature and the field winding are separated from each other and connected to
different DC sources as shown in Fig. 12.
Fig. 12 Separately excited DC motor equivalent circuit
The mathematical equations that illustrate the operation of separately excited DC motors can be
expressed as follows;
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Direct Current Motors
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E.F. EL-Saadany
E a = Vt − I a Ra
It = Ia
Vf = I f Rf
E a = K aφ ω m
T = K aφ I a
The equations that represent the terminal characteristic (speed – torque characteristic) for separately
excited DC motors, presented in Fig. 13, can be derived as follows;
Q E a = Vt − I a Ra
∴ Ia =
Vt − E a
Ra
Q E a = K aφ ω m
∴ Ia =
Vt − K aφ ω m
Ra
Q T = K aφ I a
⎛ V − K aφ ω m
∴ T = K aφ ⎜⎜ t
Ra
⎝
∴ωm =
⎞ K aφ Vt (K aφ )2 ω m
⎟⎟ =
−
Ra
Ra
⎠
Vt
Ra
−
T
K aφ (K aφ )2
∴ ω m = K 1 − K 2T
Fig. 13 Separately excited DC motor terminal characteristic
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where
K1 =
K2 =
Vt
K aφ
Ra
( K a φ )2
4.2 Shunt DC Motors
In this type, the armature and the field winding are connected in parallel and they are connected to the
same DC source as shown in Fig. 14. Shunt DC motors are usually used with applications requiring
high speed and low torque.
Fig. 14 Shunt DC motor equivalent circuit
The mathematical equations that illustrate the operation of shunt DC motors can be expressed as
follows;
Vt = I a Ra + E a
It = I a + I f
E a = K aφ ω m
Vt = I f R f
T = K aφ I a
The equations that represent the terminal characteristic for shunt DC motors, presented in Fig. 15, can
be expressed by (similar to the case of separately excited DC motors);
ω m = K1 − K 2T
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Direct Current Motors
where
K1 =
K2 =
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Spring 2006
E.F. EL-Saadany
Vt
K aφ
Ra
( K a φ )2
Fig. 15 Shunt DC motor terminal characteristic
4.3 Series DC Motors
In this type, the armature and the field winding are connected in series and they are connected to the
same DC source as shown in Fig. 16.
Fig. 16 Shunt DC motor equivalent circuit
The mathematical equations that illustrate the operation of series DC motors can be expressed as
follows;
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E.F. EL-Saadany
E a = Vt − I a ( Ra + R f )
E a = K aφ ω m
T = K aφ I a
φ =k If
It = Ia = I f
The equations that represent the terminal characteristic for series DC motors, presented in Fig. 17, can
be derived as follows;
(
Q E a = Vt − I a Ra + R f
)
Q It = Ia = I f = I
(
)
∴ I Ra + R f = Vt − E a
Q E a = K aφ ω m = K a k I f ω m = K I ω m
(
)
∴ I Ra + R f = Vt − K I ω m
∴I =
Vt
Ra + R f + K ω m
(
)
Q T = K aφ I a = K a k I f I a = K I 2
⎡
⎤
Vt
K (Vt )2
∴T = K ⎢
⎥ =
Ra + R f + K ω m
⎣⎢ Ra + R f + K ω m ⎥⎦
2
(
)
(
∴ Ra + R f + K ω m
)
2
∴ Ra + R f + K ω m =
∴ωm =
∴ωm ∝
(
(
)2
K (Vt )2
=
T
K
Vt
T
)
K
Vt − R a + R f
Ra + R f
V
1
T
= t ⋅
−
K
K
K
T
1
T
Note: at T = 0, ω m = ∞
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Fig. 17 Series DC motor terminal characteristic
4.4 Compound DC Motors
In this type, the field windings are divided into two parts, a series part connected to in series to the
armature winding and a parallel part as shown in Fig. 18.
Fig. 18 Compound DC motor equivalent circuit
Compound DC motors can be classified into two main categories.
•
Cumulative Compound where the mmfs generated from both field windings are add
together to strengthen the net mmf. The total mmf increases with the increase of the load.
In this type, there is one components of the field flux that is constant while the other is
proportional to the armature current. At light loads, it behaves very close to the shunt
motor characteristic as the effect of the series field flux will be very small. At higher
loads, it behaves very close to the series motor characteristic as the effect of the series
field flux increases with the increase in loading conditions will be very small. This type of
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connections combines the best features of both the series type (represented by high
starting torque) and the shunt type (represented by no over-speed occurrence at no-load).
•
Differential Compound: where the mmfs generated from both field windings opposes
each other. The total mmf decreases with the increase of the load. As the load motor
increases, the armature current increases. Thus the net motor flux decreases that leads to
an increase in the motor rotational speed. This increase in the motor speed increase the
motor loading that result in further increase in the armature current, further decrease in
the net flux, further increase in the rotor speed, and so on. Therefore, these types of
motors are unstable and tend to run away.
The mathematical equations that illustrate the operation of compound DC motors can be expressed as
follows;
E a = Vt − I a ( Ra + R fs )
E a = K aφ ω m
V f = I f R fp
T = K aφ I a
It = Ia + I f
The terminal characteristic for compound DC motors are presented in Fig. 19. Fig. 20 presents the
terminal characteristics for most DC motors types.
Fig. 19 Compound DC motor terminal characteristic
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Fig. 20 DC motors terminal characteristic
5. Power Flow
In general, the power flow equations for DC motors can be expressed by;
Pin = Pout + Losses
Losses = I a2 Ra + I 2f R f
Pin = Vt I t
Pout = Pout , net + Protational
where
Pout = E a I a = T * ω m ,
Protational = mechanical losses,
Pout , net is called the shaft power,
η = efficiency and is given by η =
moreover, η net =
Ea I a T ωm
=
,
Vt I t
Vt I t
Pout net
Pin
Fig. 21 presents the power flow diagram for DC motors.
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E.F. EL-Saadany
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E.F. EL-Saadany
Fig. 21 Power flow diagram for DC motors
6. Starting of DC Motors
The main problem with the starting of DC motors is that at starting ω m = 0 and hence E a = 0 .
Since, I a =
Vt − E a
Vt
, therefore, at starting I a =
.
Ra
Ra
Since R a is very small, then I a is very large at starting.
The solution for this problem is to add a resistance to the armature at starting such that R a new is
high to limit the current and then cut it in steps as E a builds up (this technique is called starter). Fig.
22 presents a schematic diagram for a manual face-plate starter. When using this starter, the contact
arm starts at position N where all starter resistances (R1, R2, R3, and R4) are being connected in series to
the armature windings. The armature current will be limited to its maximum permissible value Imax
(usually ranging from 1.5 to 2 times the armature full-load current) as shown in Fig. 23. The induced
emf starts to build up and the motor starts to rotate. With the increase in the rotor’s rotational speed,
the induced emf increases and the armature current decreases. This process is continued until the
rotational speed can’t further increased with such connection and the corresponding armature current
will be Imin. At this operating point the contact arm is moved to the next step to disconnect the starter
resistance R1. The armature current will increase again to its permissible maximum value and the
process is repeated with the new starter resistance. This operation continues until the starter’s
resistances are disconnected. It is worth mentioning that electronic semiconductor devices are currently
used in the starting of DC motors.
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Fig. 22 Manual face-plate starter for shunt DC motors.
1: Contact arm,
2: Insulated handle to move the arm.
3: Spring to pull the arm to its original dead position when the supply voltage is interrupted.
4: Eelctromagnet to hold the arm magnetically in its last contact position
Fig. 23 Armature current variation during starting
7. Speed Control of DC Motors
Since E a = Vt − I a Ra and E a = K aφ ω m ,
Therefore, ω m =
Vt − I a Ra
, i.e. ω m = K1 Vt − K 2T
K aφ
To control the motor speed one of the following parameters has to be changed:
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1. The armature resistance Ra: Increasing the armature resistance reduces the output torque for
the same rotational speed as shown in Fig. 24. This is governed by the speed equation
ω m = K1 Vt − K 2T .
Fig. 24 Speed control by varying the armature resistance.
2. The flux φ by varying the field terminal voltage or the field resistance. In this case, increasing
the field current, or reducing the field resistance, reduces the output torque for the same
rotational speed as shown in Fig. 25. This is governed by the speed equation
ωm =
Vt
Ra
−
K If
K If
(
)2
T.
Fig. 25 Speed control by varying the field current or resistance.
3. The armature terminal voltage Vt: Increasing the armature terminal voltage increases the
output torque for the same rotational speed as shown in Fig. 26. This is governed by the speed
equation ω m = K1 Vt − K 2T .
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Fig. 26 Speed control by varying the armature terminal voltage.
4. Solid state devices speed control: Speed control of DC motors can be also performed using
solid state switches that are used to vary the armature voltage and / or the field current. This
will be discussed in later lectures.
8. Numerical Examples for DC Motors
Example 1: A DC shunt motor running at 1150 rpm, takes 38.5 A from a 230 V DC voltage
supply. What value of extra resistance inserted into the rotor circuit (armature
circuit) would provide a speed of 450 rpm at 120 % of rated torque. The armature
resistance is 0.3 Ω and the field resistance is 128 Ω.
Given:
n1 = 1150 rpm,
It1 = 38.5 A,
Ra1 = 0.3 Ω,
Rf = 128 Ω,
Vt = 230 V.
Solution:
For shunt DC motors shown in Fig. 27.
Fig. 27 Shunt DC motor equivalent circuit
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T2 = 1.2 T1 and
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The field current is,
If =
Vt
230
=
= 1.8 A
Rf
128
This current is constant since Vt and Rf are constants. Accordingly φ will be
constant.
The armature current is,
I a1 = I t1 − I f = 38.5 − 1.8 = 36.7
A
The back emf is,
E a1 = V1 − I a1 ⋅ Ra1 = 230 − 36.7 x 0.3 = 219 V
Since T = K aφ I a and φ is constant, therefore,
T2 K aφ I a 2 I a 2
=
=
T1
K aφ I a1
I a1
Since T2 = 1.2 T1, therefore
Ia2 = 1.2 Ia1 = 1.2 x 36.7 = 44.04 A.
Since E a = K aφ ω m and φ is constant, therefore,
E a 2 K aφ ω m 2 ω m 2 n2
=
=
=
E a1
K aφ ω m1 ω m1
n1
Therefore
E a 2 = E a1 x
n2
450
= 219 x
= 85.7 V .
1150
n1
Since Vt = I a 2 Ra 2 + E a 2 , therefore,
Ra 2 =
Vt − E a 2 230 − 85.7
=
= 3.28 Ω .
44.04
I a2
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Therefore, the extra resistance to be added to the rotor is then ∆R and is equal to,
∆R = Ra 2 − Ra1 = 3.28 − 0.3 = 2.98 Ω .
Example 2: 100 V, 10 hP, 1100 rpm, DC shunt motor has an armature resistance of 0.05 Ω. The
field winding resistance is 50 Ω. Friction and windage losses (rotational losses) are
250 W. When the motor is delivering rated load at 1100 rpm, calculate:
1. The armature current,
2. The back emf.,
3. The no-load speed,
4. The efficiency of the motor.
Given:
n = 1100 rpm,
Pout, net = 10 hP,
Ra = 0.05 Ω, Rf = 50 Ω,
and Protational losses = 250 W.
Solution:
For the shunt DC motor circuit shown in Fig. 28.
Fig. 28 Shunt DC motor equivalent circuit
1. The output mechanical power is,
Pout = E a I a = Pout , net + Protational
∴ Pout = E a I a = 10 x 746 + 250 = 7710 W
25
Vt = 100 V,
Direct Current Motors
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Spring 2006
E.F. EL-Saadany
Multiplying both sides of Vt = I a Ra + E a by Ia results in
Vt I a = (I a )2 Ra + I a E a
∴ (I a )2 Ra − Vt I a + I a E a = 0
∴ (I a )2 x 0.05 − 100 x I a + 7710 = 0
∴I a =
100 ±
(100)2 − 4 x 0.05 x 7710
2 x 0.05
⎧1919.7 A Not Realistic Solution
∴I a = ⎨
A
⎩ 80.3
Therefore, the armature current is
∴I a = 80.3
A
2. The back emf can be calculated from the output power as follows,
E a I a = Pout
∴ Ea =
Pout 7710
=
= 96 V
Ia
80.3
3. The field current can be calculated from,
If =
Vt
100
=
=2
Rf
50
A
This current is constant since Vt and Rf are constants. Accordingly, φ will be
constant.
For no-load conditions, Ia = 0 and Ea = Ea NL
∴ E a NL = Vt − I a ⋅ Ra = Vt = 100 V
Since E a = K aφ ω m and φ is constant, therefore,
E a NL
Ea
=
K aφ ω m NL
K aφ ω m
=
26
ω m NL
ωm
=
n NL
n
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Spring 2006
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Therefore, the no-load rotational speed is,
n NL = n x
E a NL
Ea
= 1100 x
100
= 1145.7 rpm .
96
4. The net output mechanical power is,
∴ Pout , net = 10 x 746 = 7460 W
I t = I a + I f = 80.3 + 2 = 82.3 A
The input electrical power can be calculated from,
Pin = Vt x I t = 100 x 82.3 = 8230 W .
Therefore, the motor efficiency is,
η=
Pout , net
Pin
=
7460
= 0.906 = 90.6 %
8230
Example 3: A 200 V, 20 hP, series DC motor has an armature resistance of 0.02 Ω. The field
winding resistance is 0.03 Ω. The rotational speed is 900 rpm when the armature
current is 50 A. Neglect friction and windage losses and assume that the machine is
unsaturated. Calculate:
1. The back emf.,
2. The input power in kW,
3. The developed torque,
If the load torque is halved, calculate:
4. The armature current,
5. The speed,
6. The output power.
Given:
n1 = 900 rpm,
Ia1 = 50 A,
Ra = 0.02 Ω, Rf = 0.03 Ω,
Protational losses = 0 W.
27
Vt = 200 V, and
Direct Current Motors
Solution:
MTE 320
Spring 2006
E.F. EL-Saadany
For the series DC motor circuit shown in Fig. 29.
Fig. 29 Series DC motor equivalent circuit
For series DC motors,
E a = K aφ ω m
T = K aφ I a
It = Ia = I f
φ = K I f since the machine is assumed to be unsaturated.
∴ E a = K a K I a ω m = K sr I a ω m
(1)
∴ T = K a K (I a )2 = K sr (I a )2
(2)
where K sr = K a K = Constant
1. The back emf can be calculated from,
E a = Vt − I a1 ( Ra + R f )
∴ E a = 200 − 50(0.02 + 0.03) = 197.5 V
2. The input electrical power can be calculated from,
Pin = Vt x I t = 200 x 50 = 10,000 W = 10 kW .
3. The output power can be related to the rotational speed and torque as follows,
Pout = E a1 I a1 = T1 * ω m1
∴ T1 =
E a1 I a1
ω m1
The rotational speed can be calculated from the following equation,
28
Direct Current Motors
MTE 320
ω m1 = n1 x
∴ T1 =
Spring 2006
E.F. EL-Saadany
2π
2π
= 900 x
= 94.25 rad / s
60
60
E a1 I a1
ω m1
=
197.5 x 50
= 104.78
94.25
Nm
For the new loading conditions, T2 = 0.5 T1 = 0.5 x 104.78 = 52.39 Nm
4. From equation (2) we can get,
2
(I a 2 )2 = ⎛⎜ I a 2
T2 K sr (I a 2 )
=
=
T1
(I a1 )2 ⎜⎝ I a1
K sr (I a1 )2
∴ I a 2 = I a1
T2
= 50 x
T1
⎞
⎟
⎟
⎠
2
1
= 35.36
2
A
Consequently,
E a 2 = Vt − I a 2 ( Ra + R f ) = 200 − 35.36(0.02 + 0.03) = 198.23 V
5. From equation (1) we can get,
E a 2 K sr I a 2 ω m 2 I a 2 n 2
=
=
E a1
K sr I a1 ω m1
I a1 n1
n 2 = n1 x
E a 2 I a1
198.23
50
x
= 900 x
x
= 1277.3 rpm
197.5 35.36
E a1 I a 2
6. Since Protational losses = 0 W.
∴ Pout = E a I a = Pout , net + Protational = Pout , net
∴ Pout = Pout , net = E a 2 I a 2 = 198.23 x 35.36 = 7009.4 W
Example 4: A 230 V, 50 hP, DC shunt motor is connected to a 230 V DC supply and delivers
power to a load drawing an armature current of 200 A while running at a speed of
1200 rpm. It has an armature resistance of 0.2 Ω. The rotational losses are 500 W.
Determine:
1. The value of the generated voltage at this loading conditions,
2. The value of the load torque,
3. The efficiency of the motor if the field resistance is 115 Ω.
29
Direct Current Motors
Given:
MTE 320
n = 1200 rpm,
Spring 2006
Ra = 0.2 Ω,
Ia = 200 A,
Rf = 115 Ω,
E.F. EL-Saadany
Vt = 230 V, and
Protational losses = 500 W.
Solution:
For the shunt DC motor circuit shown in Fig. 30.
Fig. 30 Shunt DC motor equivalent circuit
1. The back emf can be calculated from
E a = Vt − I a Ra = 230 − 200 x 0.2 = 190 V
2. The output mechanical power is,
Pout = E a I a = Pout , net + Protational
∴ Pout , net = E a I a − Protational = 190 x 200 − 500 = 37,500 W
The rotational speed can be calculated from
ωm = n x
2π
2π
= 1200 x
= 125.66 rad / s
60
60
∴ Tout , net =
Pout , net
ωm
=
37500
= 298.4 Nm
125.66
3. The field current can be calculated from,
If =
Vt
230
=
=2
Rf
115
A
I t = I a + I f = 200 + 2 = 202
30
A
Direct Current Motors
MTE 320
Spring 2006
E.F. EL-Saadany
The input electrical power can be calculated from,
Pin = Vt x I t = 230 x 202 = 46,460 W .
Therefore, the net motor efficiency is,
η net =
Pout , net
=
Pin
37,500
= 0.807 = 80.7 %
46,460
Example 5: A 230 V, 12 hP, 1200 rpm DC series motor is connected to a 230 V DC supply and
draws a current of 40 A while rotating at 1200 rpm. It has an armature resistance of
0.25 Ω and a field winding resistance of 0.1 Ω. Assume magnetic linearity.
Determine:
1. The power and the torque delivered by the motor,
2. The speed, torque and delivered power if the motor draws 20 A.
Ia1 = 40 A,
Ra = 0.25 Ω, Rf = 0.1 Ω, and
Given:
n1 = 1200 rpm,
Solution:
For the series DC motor circuit shown in Fig. 31.
Fig. 31 Series DC motor equivalent circuit
1. The back emf can be calculated from
(
)
E a1 = Vt − I a1 Ra + R f = 230 − 40 x (0.25 + 0.1) = 216 V
The output mechanical power is,
Pout1 = E a1 I a1 = 216 x 40 = 8640 W
31
Vt = 230 V.
Direct Current Motors
MTE 320
Spring 2006
E.F. EL-Saadany
The rotational speed can be calculated from
ω m1 = n1 x
∴ Tout1 =
2π
2π
= 1200 x
= 125.66 rad / s
60
60
Pout1
ω m1
=
8640
= 68.75 Nm
125.66
2. If the motor draws 20 A and assuming linear relationship between φ and If, then
T = K a K (I a )2 = K sr (I a )2
K (I )2 (I )2 ⎛ I
T
∴ 2 = sr a 2 2 = a 2 2 = ⎜⎜ a 2
T1
(I a1 ) ⎝ I a1
K sr (I a1 )
⎛I
∴ T2 = T1 x ⎜⎜ a 2
⎝ I a1
The new back emf is
2
⎞
⎟
⎟
⎠
2
⎞
20
⎟ = 68.75 x ⎛⎜ ⎞⎟ = 17.19
⎟
⎝ 40 ⎠
⎠
2
(
Nm
)
E a 2 = Vt − I a 2 Ra + R f = 230 − 20 x (0.25 + 0.1) = 223 V
The new output mechanical power is,
Pout 2 = E a 2 I a 2 = 223 x 20 = 4460 W
The rotational speed can be calculated from
ωm2 =
Pout 2 4460
=
= 259.45 rad / s
T2
17.19
n2 = ω m 2 x
60
60
= 259.45 x
= 2477.6 rpm
2π
2π
Example 6: A 240 V, 2 hP, 1200 rpm, DC shunt motor drives a load whose torque varies linearly
with the rotational speed. The armature resistance of 0.75 Ω. With a field current of
1 A, the rotor draws a line current of 7 A and rotates at a speed of 1200 rpm.
Assume magnetic linearity. If the field current is now reduced to 0.7 A and the
rotational losses are 150 W., determine:
32
Direct Current Motors
MTE 320
Spring 2006
E.F. EL-Saadany
1. The operating speed of the motor,
2. The line current, mechanical developed power, and the motor efficiency.
Given:
n1 = 1200 rpm,
It1 = 7 A,
If1 = 1 A,
If2 = 0.7 A,
Ra = 0.75 Ω,
Vt = 240 V, and Protational losses = 150 W.
Solution:
For the shunt DC motor circuit shown in Fig. 32.
Fig. 32 Shunt DC motor equivalent circuit
1. Since the load torque is assumed to vary linearly with the rotational speed,
therefore,
TLoad = K Load n
At the initial operating conditions; n1 = 1200 rpm, It1 = 7 A, and If1 = 1 A
I a1 = I t1 − I f 1 = 7 − 1 = 6
A
E a1 = Vt − I a1 Ra = 240 − 6 x 0.75 = 235.5 V
Assuming magnetic linearity (unsaturated machine operating conditions);
T = K aφ I a
φ = K I f since the machine is assumed to be unsaturated.
∴ T = K a K I a I f = K sh I a I f
where K sh = K a K = Constant
33
Direct Current Motors
MTE 320
Spring 2006
E.F. EL-Saadany
For the new operating conditions
T2 K sh I a 2 I f 2 I a 2 I f 2
=
=
T1
K sh I a1 I f 1
I a1 I f 1
(1)
Assuming a linear load torque-speed characteristics,
T2 TLoad 2 K Load n2 n2
=
=
=
T1 TLoad 1
K Load n1 n1
(2)
The armature current, in general, can be related to the terminal voltage and the
back emf using the following equation,
Ia =
∴
Vt − E a
Ra
I a 2 Vt − E a 2
=
I a1 Vt − E a1
(3)
Moreover, the back emf can be related to the field current and the rotational
speed as follows,
E a = K aφ ω m
φ = K I f since the machine is assumed to be unsaturated.
∴ E a = K a K ω m I f = K sh ω m I f
∴
E a 2 K sh ω m 2 I f 2 ω m 2 I f 2 I f 2 n2
=
=
=
⋅
E a1
K sh ω m1 I f 1
ω m1 I f 1 I f 1 n1
(4)
Dividing equation (3) by Ea1 we can get,
∴
I a2
I a1
Vt
E
− a2
E
E a1
= a1
Vt
−1
E a1
(5)
Substituting equation (4) into equation (5) we can get,
34
Direct Current Motors
MTE 320
∴
I a2
I a1
Spring 2006
I f 2 n2
Vt
−
⋅
E a1 I f 1 n1
=
Vt
−1
E a1
E.F. EL-Saadany
(6)
Substituting equation (6) into equation (1) and then equation (2) we can get,
∴
T2 I f 2
=
T1
I f1
I f 2 n2
Vt
−
⋅
E a1 I f 1 n1
n
⋅
= 2
Vt
n1
−1
E a1
(7)
Equation (7) can be rewritten as follows,
∴
If2
I f1
⎛If2
V
⋅ t −⎜
E a1 ⎜⎝ I f 1
2
⎞ n 2 n2 ⎛ Vt
⎞
⎟ ⋅
⎜⎜
=
− 1⎟⎟
⎟ n1 n1 ⎝ E a1
⎠
⎠
(8)
By solving equation (8) we can get,
If2
∴
⋅
Vt
E a1
I f1
n2
=
n1 ⎛
⎞ ⎛I
V
⎜⎜ t − 1⎟⎟ + ⎜ f 2
⎠ ⎜⎝ I f 1
⎝ E a1
⎞
⎟
⎟
⎠
2
(9)
Substituting n1 = 1200 rpm, If1 = 1 A, If2 = 0.7 A, Vt = 240 V, and Ea1 = 235.5 V
in equation (9)
0.7 240
⋅
n2
1
235.5
= 1.401
∴
=
2
n1 ⎛ 240
⎞ ⎛ 0.7 ⎞
− 1⎟ + ⎜
⎟
⎜
⎝ 235.5 ⎠ ⎝ 1 ⎠
∴ n 2 = 1.401 x n1 = 1.401 x 1200 = 1681.2 rpm
2. From equation (4)
∴ Ea2 =
E a 2 I f 2 n2
=
⋅
E a1
I f 1 n1
I f 2 n2
0.7
⋅
⋅ E a1 =
⋅ 1.401 ⋅ 235.5 = 230.95 V
1
I f 1 n1
35
Direct Current Motors
MTE 320
From equation (3)
∴ I a2 =
Spring 2006
E.F. EL-Saadany
I a 2 Vt − E a 2
=
I a1 Vt − E a1
Vt − E a 2
240 − 230.95
⋅ I a1 =
⋅ 6 = 12.07
240 − 235.5
Vt − E a1
∴ I t 2 = I a 2 + I f 2 = 12.07 + 0.7 = 12.77
A
A
The output mechanical power is,
Pout 2 = E a 2 I a 2 = Pout , net 2 + Protational
∴ Pout , net 2 = E a 2 I a 2 − Protational = 230.95 x 12.07 − 150 = 2,637.6 W
The input electrical power can be calculated from,
Pin 2 = Vt x I t 2 = 240 x 12.77 = 3,064.8 W .
Therefore, the net motor efficiency is,
η net 2 =
Pout , net 2
Pin 2
=
2,637.6
= 0.861 = 86.1 %
3,064.8
References
[1] Theodore Wildi, "Electrical Machines Drives, and Power Systems," Prentice Hall, Ohio, 2006.
[2] PC in Control, http://www.pc-control.co.uk/dc-motors.htm
[3] Stephen J. Chapman, “Electrical Machinery Fundamentals”, Fourth Edition, Mc Graw Hill, NY,
2005.
36