Concordia University
Department of Electrical and
Computer Engineering
ELEC-331
Lecture 12
Dc Machines (8.1, 9.1-9.5)
Outline of the lecture
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Principles of DC machinery
The equivalent model of a DC motor
Characteristics of shunt DC motors
Speed control of DC motors
Characteristics of permanent magnet DC
motors
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Rotating loop between curved pole faces
• The rotor rotates with constant speed (ωm). Due to the curved pole faces, the magnetic field is always perpendicular to
the rotor surface under a pole face. What is the voltage induced in the loop?
etot = eda = edc + eba
e = (v × B ) • l
→ v perpendicular to B
under a pole face.
→ B = 0 beyond the pole edge.
edc = v B l , when dc under NP
edc = −v B l , when dc under SP
eba = edc , for all cases.
Voltage induced in the rotating loop
• By definition: eind = (v × B) • l
v perpendicular to B under a pole face. B = 0 beyond the pole edge.
edc = v B l , when dc under NP
edc = −v B l , when dc under SP
edc = 0 when dc beyond pole edges
Since v is always perpendicular to B,
the magnitude of the induced voltage
(e) is constant!
eba = edc , for all cases.
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⎧
⎪2vBl = φω under the pole faces
etot = ⎨
π
⎪⎩0
beyond the pole edges
2
Getting DC voltage (commutator & brushes)
• Whenever segment dc is under NP, eout = etot = eda. If segment dc is under SP, eout = -etot = ead. Commutator does it!
• The commutator (two isolated semicircular conductors) is
fixed on the rotor, is connected to the loop and rotates. The
brushes do not rotate and alternately connect etot and - etot
to the output terminals of the dc
generator.
Dc machine as a motor: Induced torque.
• Flow of current in the loop is required. F = i(l × B) τ = rF sin θ
Due to the curved pole faces, B is always perpendicular to l, the magnitude of the Force (F)
is constant! F is always perpendicular to r, the magnitude of the torque (τ) is constant!
τ ab = τ cd = rilB, τ bc = τ da = 0, τ ind = 2rilB
Verify the direction of F!
2
⎧
⎪2rliB = φ i A under the pole faces
τ ind = ⎨
⎪⎩0
π
beyond the pole edges
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How to keep the rotor spinning?
• Note that there is no rotating magnetic field for the rotor magnetic field to follow! Therefore, whenever segment dc goes
beyond the lower edge of the NP the direction of the current
that flows through it must be reversed to reverse the direction
of the rotor’s magnetic field. The commutator does it!
Example 8-1
• Fig. 8.6 shows a simple rotating loop between curved poles
connected to a battery and a resistor through a switch.
Values for r, l, B, R and Vb are given.
• A) What happens when the switch is closed?
• B) Starting current and steady-state speed at no load?
• C) A 10N.m load is connected to the shaft. New steady
state shaft speed? How much power is supplied to the
machine shaft? How much power does the battery supply?
• D) A 7.5 N.m torque is applied in the direction of rotation.
New steady-state speed? Motor or generator operation?
• E) What happens to the steady-state speed of the unloaded
machine when B decreases?
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Characteristics of DC motors
Chapter 9
• Great performance for applications that require
wide variations in speed.
• Main types of DC motors:
- Separately excited DC motor
- Shunt DC motor
- Permanent magnet DC motor
- Series DC motor
- Compounded DC motor
The equivalent circuit of a DC motor
• Armature and field circuits.
Stator circuit
Field
Ea = Kφω
τ ind = KφI a
Rotor circuit
Armature
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The magnetization curve of a DC motor
• Variation of the internal voltage with the field
winding current (that produces B).
E A = K ′φ n
Separately excited and shunt DC motors
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Terminal characteristics of a shunt motor
• Plot of the output quantities (torque x speed).
VT = E A + R A I A
E A = Kφω τ ind = KφI A
ω=
VT
RA
−
τ ind
Kφ (Kφ )2
• Response to a load variation (increase):
• If: τload↑, ω↓, EA↓, IA↑, τind↑, ω↑ (ω2< ω1)
Speed control of shunt DC motors
• Changing the field resistance (and the field flux)
• If: RF↑, IF↓, φ ↓, EA↓, IA↑ ↑, τind↑, ω↑ (ω2>ω1)
Ia =
ω=
VT − E A
RA
τ ind = KφI a
VT
RA
−
τ ind
Kφ (Kφ )2
• Slope variation changes the “control logic” at low speeds!
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Speed control of separately excited DC
motors
• Changing the armature voltage (VA) with IF fixed
• If: VA↑, IA↑, τind↑, ω↑ (ω2> ω1) I = V R− E τ = KφI
A
A
a
ind
a
A
ω=
Separately excited machine
VA
RA
−
τ ind
Kφ (Kφ )2
Notice that the no-load speed of the
motor changes and the slope of the
curve is kept constant
Dual (IF and VA) speed control
• If: RF↑, IF↓, ω↑ and if: RF↓, IF ↑, ω ↓. Since IF< IFbase this method is
good for ω> ωbase!
• If: VA↑, ω↑ and if VA ↓, ω ↓. Since VA< VAbase this method is good for
ω< ωbase!
τ max = KφI A,max
Pmax = τ maxω
• Risk of overspeeding if the field “opens” due to a blown fuse!
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The permanent magnet DC motor
• No external field circuit: Permanent magnets generate the magnetic
field.
• Lower size and lower torque than a shunt motor. Can be demagnetized
when overloaded.
• Speed control by armature voltage control.
Power & torque in a dc machine
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Armature input power:
Field input power:
Total input power:
Armature copper losses:
Field copper losses:
Converted power:
Induced torque:
Output power:
Output torque
Pin _ A = V A I A
Pin _ F = VF I F
Pin _ Tot = VA I A + VF I F
Ploss _ A = I A R A
2
Ploss _ F = I F RF = Pin _ F
2
Pconv = E A I A
P
τ ind = conv
ωm
Pout = Pconv − PF &W − Pcore
τ out =
Pout
ωm
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Power & torque in a dc machine
• Mechanical power losses:
• Core losses:
• Efficiency:
PMech = f (n)
Pcore = f ( B, n)
η=
Pout
x100%
Pin _ Tot
Example
• A separately excited dc motor has the following
characteristics: armature resistance, 0.20 Ω, induced
voltage at 1800 rpm and rated field current: 230 V,
mechanical losses: 750 W at 1800 rpm and proportional to
speed, core losses at 1800 rpm: 500 W proportional to the
square of the speed. The machine is not saturated. For a
120 V armature voltage, the recorded armature current is
80 A. Find: (a)The motor speed, (b) the electromagnetic
(converted) power, (c) the electromagnetic (induced)
torque and (d) the shaft (output) power and torque.
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End!
Voltage induced in a rotating loop
(elementary ac machines lecture)
eind = (v × B ) • l
• Basic equation (section 1.7):
• Polarity of the induced voltage: Right hand rule. v → index
finger, B → middle finger and thumb → eind)
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Torque induced in a current-carrying loop
(elementary machine with flat poles)
F = i (l × B ) τ = F (r sin θ )
τ ind = 2rilB sin θ
(clockwise), since θ ab = θ cd
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