DC MOTOR
• ADC motor is a mechanical rotating device
which converts electrical energy into
mechanical energy. It is based on the principle
of force acting on a current carrying wire in a
magnetic field.
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DC Motor Principle
•DC motor working is based on the principle that
when a current carrying conductor is placed in a
magnetic field, the conductor experiences a
mechanical force.
• The direction of this force is given by Fleming’s
left-hand rule and magnitude is given by
• F = BIL Newtons
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Working of DCMotor
• When the terminals of the motor are connected
to an external source of DC supply:
• the field magnets are excited developing
alternate North and South poles.
• the armature conductors carry currents.
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• All conductorsunder North-pole carry currents in one
direction while all the conductors under South-pole
carry currents in the opposite direction.
• The armature conductors under N-pole carry currents
into the plane of the paper (denoted as ⊗ in the
figure). And the conductors under S-pole carry
currents out of the plane of the paper (denoted as ⨀
in the figure).
• Since each armature conductor is carrying current and
is placed in the magnetic field, a mechanical force acts
on it.
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7.
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• On applyingFleming’s left-hand rule, it is clear
that force on each conductor is tending to rotate
the armature in the anticlockwise direction. All
these forces add together to produce a driving
torque which sets the armature rotates.
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Back EMF inDC Motor
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• When the current carrying conductor placed in a
magnetic field, the torque induces on the conductor.
The torque rotates the conductor which cuts the flux
of the magnetic field. According to the
Electromagnetic Induction Phenomenon when the
conductor cuts the magnetic field, EMF induces in
the conductor.
• On applying the right-hand rule it is seen that the
direction of the induces emf is opposite to the
applied voltage. Thereby the emf is known as the
counter emf or back EMF.
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Advantages of BackEmf in DC Motor
• 1. The back emf opposes the supply voltage. The supply voltage
induces the current in the coil which rotates the armature. The
electrical work required by the motor for causing the current
against the back emf is converted into the mechanical energy. And
that energy is induced in the armature of the motor. Thus, we can
say that energy conversion in DC motor is possible only because of
the back emf.
• The back emf makes the DC motor self-regulating machine, i.e., the
back emf develops the armature current according to the need of
the motor.
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Len’s Law
• Thedirection of induced emf is given by Lenz’s
law.According to this law, the induced emf will
be acting in such a way so as to oppose the
very cause of production of it .
• e = -N (dØ/dt) volt
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Torque equation ofa DC motor
• When armature conductors of DC motor carry current
in the presence of stator field flux, a mechanical
torque is developed between the armature and the
stator. Torque is given by the product of the force and
the radius at which this force acts.
• Torque T = F × r (N-m) …where, F = force and r = radius
of the armature
• Work done by this force in once revolution = Force ×
distance = F × 2πr (where, 2πr = circumference of
the armature)
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• Net powerdeveloped in the armature = word done /
time
= (force × circumference × no. of revolutions) / time
= (F × 2πr × N) / 60 (Joules per second)
• But, F × r = T and 2πN/60 = angular velocity ω in
radians per second. Putting these in the above
equation
Net power developed in the armature = P = T ×
ω (Joules per second)
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Armature torque (Ta)
•The power developed in the armature can be given
as, Pa = Ta × ω = Ta × 2πN/60
• The mechanical power developed in the armature is
converted from the electrical power,
Therefore, mechanical power = electrical power
That means, Ta × 2πN/60 = Eb.Ia
• We know, Eb = PΦNZ / 60A
• Therefore, Ta × 2πN/60 = (PΦNZ / 60A) × Ia
• Rearranging the above equation,
Ta = (PZ / 2πA) × Φ.Ia(N-m)
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Shaft Torque (Tsh)
•Due to iron and friction losses in dc machines, the
total developed armature torque is not available at
the shaft of the machine. Some torque is lost, and
therefore, shaft torque is always less than the
armature torque.
Shaft torque of a DC motor is given as,
Tsh = output in watts / (2πN/60) ....(where, N is
speed in RPM)
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Characteristics of DCmotors
• Generally, three characteristic curves are
considered important for dc motors which are,
(i) Torque vs. armature current,
• (ii) Speed vs. armature current and
• (iii) Speed vs. torque.
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Aplications of dcdc motors
D.C Shunt Motors:
It is a constant speed motor.Where the speed is required
to remain almost constant from noload to full load.Where
the load has to be driven at a number of speeds and any
one of which is nearly constant.
• Lathes
• Drills
• Boring mills
• Shapers
• Spinning and Weaving machines.
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• D.C Seriesmotor:
It is a variable speed motor.The speed is low at high
torque at light or no load ,the motor speed attains
dangerously high speed.The motor has a high starting
torque.
• Electric traction
• Cranes
• Elevators
• Air compressor
• Vacuum cleaner
• Hair drier, Sewing machine
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• D.C Compoundmotor:
• Differential compound motors are rarely used
because of its poor torque characteristics.
• Presses
• Shears
• Reciprocating machine.
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Speed control ofshunt motors
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• Armature Control Method
• This armature control method is based on the fact
that by varying the voltage across the required
voltage. The motor back EMF (Eb) and Speed of the
motor can be changed. This method is done by
inserting the variable resistance in series with the
armature.
Flux Control Method
•In this flux control method, speed of the
motor is inversely proportional to the flux.
Thus, by decreasing flux and speed can be
increased vice versa. To control the flux , he
rheostat is added in series with the field
winding will increase the speed (N), because
of this flux will decrease. So, the field current
is relatively small and hence I2R loss
is decreased
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Speed Control OfSeries Motor
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• Flux Control Method
• Field Diverter : A Rheostart is connected
parallel to the series field .This variable
resistor is also called as a diverter, as desired
value of the current can be diverted through
this resistor and hence current through field
coil can be decreased. Hence flux can be
decreased to desired amount and speed(N)
can be increased.
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• Armature Diverter: Rheostat (Divider) is
connected across the armature of the coil. For
a given constant load torque, if armature
current is reduced, then flux must increase. As
armature torque Ta α ØIa. This will result in an
increase in current taken from the supply and
hence flux Ø will increase and subsequently
speed of the motor will decrease.
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• In thismethod, field coil is tapped dividing the
number of turns. Thus we can select different value
of Ø by selecting a different number of turns. In this
method flux is reduced and speed is increased by
decreasing the number of the turns of the series field
winding. The switch S can be short circuit any part of
the field winding, thus decreasing the flux and raising
the speed (N) with full turns of coil.
Construction of 3Point Starter
• Construction wise a starter is a variable resistance
integrated into the number of sections . The contact
points of these sections are called studs and are
shown separately as OFF, 1, 2, 3, 4, 5, RUN. Other
than that there are three main points, referred to as
• ‘L’ Line terminal (Connected to positive of supply)
• ‘A’ Armature terminal (Connected to the armature
winding)
• ‘F’ Field terminal (Connected to the field winding)
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• The point‘L’ is connected to an electromagnet called overload
release (OLR) as shown in the figure. The other end of OLR is
connected to the lower end of conducting lever of starter
handle where spring is also attached with it, and the starter
handle also contains a soft iron piece housed on it. This
handle is free to move to the other side RUN against the force
of the spring. This spring brings back the handle to its original
OFF position under the influence of its own force. Another
parallel path is derived from the stud ‘1’, given to another
electromagnet called No Volt Coil (NVC) which is further
connected to terminal ‘F.’ The starting resistance at starting is
entirely in series with the armature. The OLR and NVC act as
the two protecting devices of the starter.
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• To start, keep starting handle in OFF position and
switch on the power supply.
• The handle is now moved clockwise to the first stud.
The shunt field winding is now directly connected
across the supply while the whole starting resistance
is inserted in series with the armature circuit.
• As the motor gains speed , the handle is gradually
moved over to the final stud , the starting resistance
is cut off in steps. The handle is now held
magnetically by the no-volt release coil which is
energised by the field current.
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• Now ifsupply is interuppted or field excitation is
accidentally cut , the no-volt coil is demagnetised
and the handle moves back to OFF position under
the pull of the spring. If the spring was not used ,
then the handle would have left on the ON position
and on restoration of supply the motor would have
directly connected to the supply , thus damaging the
motor.
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• If themotor is over loaded, it will draw excessive
current from the supply . This large current is
sufficient to magnetically pull the magnetic rod
which will short the no-volt release coil . The no-volt
coil is demagnetised and starting handle is pulled to
OFF position under the pull of the spring . Thus , the
motor is automatically disconnected on
overloading.
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Types of Lossesin DC machine
• Electrical or Copper losses (I2R Losses)
• Core Losses or Iron Looses
• Mechanical losses
• Brush Losses
• Stray Load Losses
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Electrical or Copperlosses
• Resistive losses in the armature and field
windings of the machine are called electrical
or copper losses or Ohmic losses.
• This loss varies with the variation of load on
the machine and is called variable loss. When
a motor is loaded, its armature current
increases.
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The various cooperlosses in the DC machine are
as follows.
• Armature copper loss = I2
aRa
• Shunt field copper loss = I2
shRsh
• Series field copper loss = I2
seRse
• Interpole winding copper loss = I2
iRi
• Brush contact loss = I2
aRb = 2IaVb
• Compensating winding copper loss =I2
aRc
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Core Losses orIron Looses
• Core losses are also known as iron losses or
Magnetic losses. Core losses can be classified
into two types:-
• Hysteresis Losses
• Eddy current Losses
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Hysteresis Losses
• Thearmature core is made up of magnetic
material and is subjected to variations in
magnetic flux. When the armature rotates it
comes under North and South poles
alternately.
• Hysteresis loss occurs due to the alternate
magnetization of the atoms, forming domains
in the magnetic material of the core
• Hysteresis Loss = Kh.V.f.Bm
1.6
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Eddy current loss
•Eddy current loss is due to the presence of
circulating current in the core material. The
armature core cuts the magnetic flux during
its rotation and EMF is induced in the body of
the core according to the laws of
electromagnetic induction. This emf is very
small but it sets up a large current in the body
of the core.
• Eddy current losses = Ke × Bm
2 × f2 × t2
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Mechanical Losse
• Themechanical losses in a dc machine are the
losses associated with mechanical effects.
Mechanical losses occur at the bearing and
shaft and air friction (windage loss) due to
rotation of the armature.
• There are two basic types of mechanical
losses:
• Friction
• Windage.
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Stray Load Looses
•The magnetic and mechanical losses are
known as stray losses. Stray losses are also
known as rotational losses.
• Total Losses = Constant losses + Variable
losses
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Swinburne’s Test
• Itis an indirect method of testing of DC
machines. In this method the losses are
measured separately and the efficiency at any
desired load is predetermined. Machines are
tested for finding out losses, efficiency and
temperature rise. For small machines direct
loading test is performed. For large shunt
machines, indirect methods are used like
Swinburne’s or Hopkinson’s test.
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• Efficiency whenrunning as a motor
• Let IL be current drawn from supply mains by
the motor and Ia be the armature current
when efficiency is to be calculated.
• Ia= IL-If
• ηm = (Input - Losses)/Input = [Vt IL - Wo - Vt If -
Ia
2 Ra] /Vt IL
•
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• Efficiency whenrunning as a generator
• Let IL be current drawn by the load and Ia be
the armature current when efficiency is to be
calculated
• Ia= IL+If
• ηg = Output/Input = Vt IL/[Vt IL + Wo + Vt If +
Ia
2 Ra]
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