DC Motor Earlier

D. C. Motor1. Highly versatile energy conversion device2. Flexible machine.3. Variety of purposes.From 3 mm stepper drawing a few A at 1.5 in a quartz crystalwatch.to the giant 75000MW or more rolling mill motor.4. It can produce high starting torque, accelerating torqueretarding torque5. It can give wide range of speed control6. Suitable for fast reversal of speed

D. C. Motor7. Characteristics can be modified by feedback circuit.8. In spite of high initial cost, widely used in industries for tough jobs.9. At present, annual production and sale are about 40% of rupee volume. This is due to the automobile industry.The batteries are used for storing electrical energy and for starting the vehicles by dc motor.It is also used for wipers, fans and for raising the windowglass.10. Due to this, almost millions of dc motors are manufactured every year.

The design of machines become a verywith improved magnetic, electrical andinsulating materials and the use of computer.Constructioninteresting and challenging topics

Air GapSTATORROTORStator CoreRotor CoreShaftRotor wdgStator wdgBearingsStator FrameBaseSTATORConstructionCommon constructional features for ALL rotating electrical machines

1. An outer stationary member is called as STATOR ROTOR3. The clearance between stator and rotor is called as4. The rotor is mounted on the shaft and bearings.2. The member is called asrotatingAIR GAP Bearings are fixed to stator frame.5. Both stator and rotor carry concentric cylindrical cores.Cores are laminated and made up of ferromagnetic materialsto give optimum electric and magnetic characteristics.6. When wdgs are excited, a common flux is set up.This flux crosses the air gap from one core to the other.N and S poles are produced on both stator and rotor.

Such a magnetic system with alternate N and S polesalong the air gap periphery is called Hetropolar system.7. This stator and rotor move relative to each other for Electro-Mechanical Energy Conversion.Construction of DC machineField winding is on STATORArmature winding is on ROTORThe clearance between stator and rotor is called asAIR - GAPThe energy transfer from stator to rotor is through Air Gap

Yoke or FramePolePole Shoe

Earth's Magnetic FieldEarthMagnetNSSNCompass

The magnetic pole model: two opposing poles, North (+) and South (-), separated by a distance d. Each arrow represents the direction of H at that point.

Yoke or FramePolePole ShoeArmature

Yoke or FramePolePole Shoe+ArmatureCommutator SegmentsMicaInsulation

Yoke or FramePolePole Shoe+ArmatureCommutator SegmentsMicaInsulationAAAFFF

Yoke or FramePolePole Shoe+ArmatureCommutator SegmentsMicaInsulationAAAFFFNS/2/2

Yoke or FramePolePole Shoe+ArmatureCommutator SegmentsMicaInsulationAAAFFFNS/2/2I/PI/PCompensating Winding

Yoke or FramePolePole Shoe+ArmatureCommutator SegmentsMicaInsulationAAAFFFNS/2/2I/PI/PCompensating WindingDirn of rotationM-AnticlockwiseGenr-ClockwiseDC i/p - MDC o/p - G

STATORSTATORROTORROTORSHAFT

STATORROTOR

1. STATOR: consists of i) Yoke or frameLaminated magnet material ii) Salient field polesLaminated magnet material,Concentrated field wdg. iii) Bearings iv) Brushes: Carbon, Graphite, copper-graphite v) Interpole wdg: Fixd to yoke in between main polesConnected in series with armature wdgCross-section Pole core is smaller than pole shoe vi) Compensating wdg: Housed in pole shoeConnected in series with armature wdgPole shoe is 70% of pole pitch

2. ROTOR: consists of i) Armature coreIt is made from 0.35 to 0.5 mm thick laminationof Cold Rolled Grain Oriented Silicon Steelto reduce iron loss.Provides the low reluctance path to the flux /2 ii) Armature WindingDistributed Lap or wave wdgs. Housed in slots, connected in series or parallelAn alternating emf is induced in the coil due to rotation iii) CommutatorConducting segments are seperated by insulation Arm wdg is connected to these segments

Carbon brushes are placed over commutatorIt converts AC to DC in case Gen (Rectifier)and DC to AC in case of Motor (Inverter)ShaftCommutatorInsulationBrushConductor

Carbon brushes are placed over commutatorIt converts AC to DC in case Gen (Rectifier)and DC to AC in case of Motor (Inverter)ShaftCommutatorArmature CoreArmature WdgInsulationBrushConductorCommutatorCommutator

iv) ShaftArmature core and commutator are fixed on the shaft by means of key.On both sides there are bearings which is connected to the yoke.The DC machine = AC machine + CommutaorThe DC machine = AC machine + Converter

Action of CommutatorTwo important Functions1. To convert AC to DC or vice-versa2. To keep armature mmf stationary in space, In case of Generator, commutator acts as aIn case of Motor, commutator acts as aDue to stationary mmf, electromagnetic torque Te is stationary or constant at particular position, whichis required for the rotation of rotor or armature.Rectifier, AC to DCInverter, DC to ACie cross & dot polarities in armature.

Action of CommutatorGen - Rectifier, AC to DCNS

Action of CommutatorGen - Rectifier, AC to DCNSFor Gen, direction of rotation of arm - Clockwise

Action of CommutatorGen - Rectifier, AC to DCNSFor Gen, direction of rotation of arm - ClockwiseObserve the movement of arm conductor and fluxThey move parallel to each otherNo change in flux w r t timeNo voltage is voltage is induced in the arm conductor if it is along brush axisSo conductors are having no polarities

Action of CommutatorGen - Rectifier, AC to DCNSA1 = 0 and A2 = 0B1 = 0 and B2 = 0VA1A2VB1B2A2A1B1B2 not connected to condr

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A2A1Find out the polarities of A1 and A2For gen apply RHRNow rotate the armature by 900.A2A1A1 = 0 and A2 = 0B1 = 0 and B2 = 0B1B2 not connected to condr

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A2A1A1 =A2 =B1 =B2 =Find out the polarities of A1 and A2For gen apply RHRNow rotate the armature by 900.A1 = 0 and A2 = 0B1 = 0 and B2 = 0B1B2 not connected to condrCrossCrossDotDot

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A2A1Find out the polarities of A1 and A2Again rotate the armature by 900.A1 and A2 move parallel to fluxNo voltageA1A2A1 =A2 =B1 =B2 =0000

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A1A2Again rotate the armature by 900.A1A2

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2Find out the polarities of A1 and A2For gen apply RHRAgain rotate the armature by 900.A2 = A1 =B1 =B2 =A1A2DotCrossCrossDot

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2Again rotate the armature by 900.A1A2Find out the polarities of A1 and A2A1 and A2 move parallel to fluxNo voltageA1A2A1 =A2 =B1 =B2 =0000

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A1A2

Action of CommutatorGen - Rectifier, AC to DCNSVA1A2VB1B2A1A2A1 under N pole is alwaysA1 under S pole is alwaysSo arm wdg is having AC emf.But B1 is always + and B2 is always -So output is DCRectifier actionThis is due to commutatorpositivenegative

If there is no CommutatorAs per AC genr action, o/p is ACAction of CommutatorGen - Rectifier, AC to DC

Action of CommutatorGen - Rectifier, AC to DCNB1VA1A2+SVB1B2B2

Action of CommutatorGen - Rectifier, AC to DCNB1B2VA1A2+SVB1B2

Action of CommutatorGen - Rectifier, AC to DCNB1B2+S

Action of Commutator - Inverter, DC to ACNB1B2+SB1 = + and B2 = -A1 = and A2 =VA1A2VB1B2 = I/p DCA1A2A1A2 conductors are not connected to brushes.0, 0 Input DC supply is given to motorMotor

Action of Commutator - Inverter, DC to ACNB1B2+SFor Motor, direction of rotation of arm Anti-clockwiseVA1A2VB1B2 = I/p DCA1A2MotorRotate the armature by 900 in anticlockwise dirN.

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCA1A2MotorA1A2For Motor, direction of rotation of arm Anti-clockwiseRotate the armature by 900 in anticlockwise dirN.

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCMotorA1A2A1 =A2 =+-Voltage VB1B2 =DCFor Motor, direction of rotation of arm Anti-clockwiseRotate the armature by 900 in anticlockwise dirN.

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCMotorAgain, rotate the armature by 900 in anticlockwise dirNA1A2A2A1A1 =A2 =00Voltage VB1B2 =DCSupply is not connected to A1A2

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCMotorAgain, rotate the armature by 900 in anticlockwise dirNA2A1A2A1Voltage VB1B2 =DCA1 =A2 =-+

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCMotorAgain, rotate the armature by 900 in anticlockwise dirNA2A1Voltage VB1B2 =DCA1 =A2 =00A2A1

Action of Commutator - Inverter, DC to ACNB1B2+SVA1A2VB1B2 = I/p DCMotorA2A1A1 under N pole is alwaysA1 under S pole is alwaysSo arm wdg is having AC Voltage even though input is DC.Due to DC supply is given, B1 is always + and B2 is always -Inverter actionThis is due to commutatorpositivenegative

Action of CommutatorNB1+S - Inverter, DC to ACMotorB2

Action of CommutatorNB1+S - Inverter, DC to ACMotorB2All conductors under N pole are having CROSS polarityAll conductors under S pole are having DOT polarityThis produces stationary armature mmf in space, Therefore stationary or constant FORCE applied to each conductor This is due to commutator

Generation of emfThe difference between AC and DC generator.Collector or Slip Rings

Generation of emfThe difference between AC and DC generator.Collector or Slip RingsN turn CoilNSAC O/PFig.: AC GenrNSDC OUTPUT+-CommutatorN turn CoilFig.: DC Genr

Generation of emfIn rotating machines, emf can be generated in armature wdg by1. Rotating these windings through a magnetic field2. Rotating magnetic field w r t these windings3. Designing magnetic circuit to have variable reluctance with rotor rotationConsider first case

Generation of emf1. Rotating these windings through a magnetic fieldNSA1A2Main Field AxisArmature Field AxisT turn arm coilr is angular velocity in radian per secondMain field flux is assumed to be sinusoidal in air gapr = rt = 0Both magnetic axes are COINCIDENTThe flux passing through or linking the coin A1A2 is MINIMUMMAXIMUMFAFlux linkage Cos or CostCos0 = 1 =MAXIMUM

NSA1A2Main Field AxisArmature Field Axisr = rt = 90Both magnetic axes are displaced by 900The flux passing through or linking the coin A1A2 is MINIMUMFACos90 = 0 = MINIMUMFlux linkage Cos or Cost

A1A2Main Field AxisArmature Field AxisT turn arm coilrThe flux linkage with the arm coil at any time t is = T Cos = T Cost = rtFor =1800, the flux passing through the coil is maximum but in reversed directionTherefore, in general, the diagram isBy Faradays Law, the emf induced in T turn coil ise = -d/dt = -d(T Cos)/dt = T r Sin rt - T {d/dt} Cos rt

= T r Sin rt - T {d/dt} Cos rtThe first term is term because it contains in itIIISpeed Voltagespeed rThe second term is term because it containsTransformer Voltagetime derivative of flux , i.e., d/dt,The second term is term because it contains in it.Here flux is sinusoidal, but is not changing with time due to DC supply to field wdg.Therefore, the flux is time-invariant, and d/dt = 0.e = T r Sin rtThe voltage appearing at the brushes is unidirectional or DCAverage or DC value is T r Sin rt d(rt ) T r

T rr= If z is the total number of conductors in one coil of T turns T Tthen, z =2 T, because single turn consists of two conductorsConductor 1Conductor 2 z n PAverage emf per conductors n PIn the armature wdg consider there are Z total number of condRsCoil 1Coil 2Coil 3Coil 12For increasing voltage, coils are connected in seriesFor increasing current, coils are connected in parallel

Suppose 4 coils are connected in seriesFor 3 parallel paths, total number of =Z/360V, 1A60V60V60Vand there are such 3 parallel pathsAAARating of arm wdg220V, 3AFor a parallel paths, total number of series conductors =Z/aseries conductorsThus the resultant voltage between A and AA, ie Ea is given byZ/aAverage emf per conductor xTotal number of series conductors in each parallel path= n P x Z n P = Ka n where, Ka=Z P/a = constant for a machine

where, N = Speed in r p mEa depends upon the flux per pole and not upon the shape of the flux density wave.

Flat Topped Sinusoidal Average value of Flat Topped Average value of Sinusoidal Thus for the same value of peak flux density Average value of armature voltage for flat topped is more than a Sinusoidal flux density wave. It is because for this reason the DC machines are designed with flat topped B wave. 0 B

ARMNSConductors move parallel with the flux.No change in flux. No induced voltage in the coils in this zone.Schematic Diagram

NSBrushes are positioned along Q axis or along MNA, which is also called as Commutation Axis.Schematic DiagramM. N. A. or G. N. A.or Commutation Axis The armature MMF is directed along brush axis, Interpolar Axis or Q axis (downwards).

NSSchematic DiagramM. N. A. or G. N. A.or Commutation Axis The armature MMF is directed along brush axis, Interpolar Axis or Q axis (If polarity of conductors are reversed then upwards).

NSSchematic Diagram + Two Pole DC machine

NSSchematic Diagram +DC machine with any Number of PolesFFF +

NSEMF Waveform 900 elect. displacement between 11 and 22FFF +1221

NSEMF Waveform If field winding is excitedFFF +1221V11V22ACand rotor rotates

NSEMF Waveform +Across armature we get DC voltage (due to Commutator-Rectifier)FFF +1221Va-aaDC

NSEMF Waveform +If the brushes are put along D axis. FFF +1221Va-aaQ axisD axisQ axisD axis

NSEMF Waveform +FFF +1221Va-aaQ axisD axisQ axisD axisIf the brushes are put along D axis. If rotor rotates. There is more spark along D axis.Due to more voltage wdg may burn.

NSEMF Waveform +FFF +1221Va-aaQ axisD axisQ axisD axisThats why the brushes are not installed along D axis.Due to more voltage wdg may burn.

NSEMF Waveform +Across armature we get DC voltage (due to Commutator-Rectifier)FFF +1221Va-aaDC Voltage

NSEMF Waveform +DC voltage with Two Coils.FFF +1221Va-aaDC Voltage with two coils

NSEMF Waveform +DC voltage with Three ColisFFF +122133

NSEMF Waveform +If field wdg is excited.If rotor rotates.Then three phase voltages are generatedFFF +1221Va-aaDC Voltage with Three coils33

EMF WaveformAcross armature we get DC voltage (due to Commutator-Rectifier)Va-aaDC Voltage with two coilsDC Voltage with Three coils

EMF WaveformTherefore, as the number of series coils between the brushes increases, 1. Magnitude of voltage increases. 2. Brush frequency increases. 3. Ripple content decreases.Va-aaDC Voltage with many coils in series

Distribution factor KdCosider the coil 11.Due to large number of coils, the phasor diagram for the emf polygon becomes almost a circle.Therefore, the distribution factor for a DC machine is

Working Principle(Principle of Operation)In 1819, it was discovered by a Danish Language Netherlands Physicist, Hans Christian Oersted that an electric current is always accomplished by a certain magnetic effects. He found that when a current is passed through a conductor placed above the magnetic needle, the needle turns in a certain direction. which is also called as Motor Rule.

Working Principle(Principle of Operation)Hans Christian OerstedRight Hand Rule

Working Principle(Principle of Operation)Whenever a current carrying conductor is placed in a magnetic field, it experiences a force whose direction is given by Flemings Left Hand Rule

Working Principle(Principle of Operation)Pole FieldNSConductor Field

Working Principle(Principle of Operation)NSOppositeAdditionNSForceNN

Working Principle(Principle of Operation)NSOppositeAdditionNSForceNSAdditionOppositeNSForce

Working Principle(Principle of Operation)NSSS

Working Principle(Principle of Operation)NSSSMNA is perpendicular to field flux.MNA

Working Principle(Principle of Operation)NSSSMNAWhen supply is given to armature, current flows through armature conductors.

Working Principle(Principle of Operation)NSSSSMNADue to + and polarities, flux direction changes .

Working Principle(Principle of Operation)NSSSSMNANew MNA shifts towards the direction opposite to the direction of rotation.So shift the brush axis in a direction opposite to the direction of rotation due to which there is less spark.New MNA

Working Principle(Principle of Operation)NSSSSMNANew MNA shifts towards the direction opposite to the direction of rotation. New MNA

Torque DevelopedNSSSSMNAConductors under N pole with + polarity, when reacted with field flux, produces downwards force. New MNA Conductors under S pole with polarity, when reacted with field flux, produces upwards force. Commutator maintains these polarities constant. So there is continuous torque and continuous rotation.

Torque DevelopedSForce on a conductor, F=B i l.If motor contains Z conductors, then total torque Ta = B i l r.ZIf Ia is the total armature current, then I = Ia/a.Torque developed by each conductor, T=B i l.rWhere B = Flux density wb/m2 i = Armature current in a conductor, Amp l = Active length of conductor, m r = Average radius at which conductors are placed, m

Torque DevelopedSIf is the total flux and A is the cross sectional area of flux path at radius r, then B = /A, Where A = (2 r l )/ p. Therefore, the total torque Ta = B i l r.Z =( /A) x (Ia/a) x l r Z = ( Ia l r Z ) / (aA)Thus the torque is directly proportional to the field flux or field current and armature current. = ( Ia l r Z ) / (a x (2 r l )/ p) = ( Ia Z p) / (2 a ) = KT Ia Nm, where KT = Z p / 2 a = K If Ia Nm, where K = KKT

Torque DevelopedSIaBN PoleS PoleForce or TorqueAverage Force or Torque+

Torque DevelopedSGross Torque TG: The torque developed by the armature Ta is called as Gross Torque. Shaft Torque Tsh: Some torque developed by the armature is lost to overcome the Iron and Friction and Windage loss, Tsh=Ta-TLost Torque. The torque available at the shaft is called as Shaft Torque Tsh (Useful Torque).B. H. P.= Brake Horse PowerTa-Tsh=Lost Torque

Back EMFNSSSSMNAAs the armature rotates, the armature conductors (with + and dot) cut the lines of field flux.So that an EMF is induced in the armature whose direction is given by Right Hand Rule.

Back EMFNSSSSMNAApply Right Hand Rule under N pole.Apply Right Hand Rule under S pole.Eb

Back EMFNSSSMNAApply Right Hand Rule under N pole.Apply Right Hand Rule under S pole.ROTATIONROTATIONEb

Back EMFNSSSMNATherefore it is called as Back EMF or Counter EMF.Again it is Back EMF as per Lenzs Law, which states that the direction of induced emf is such to oppose the cause of it and which is the applied voltage.Eb

Back EMFNSSSMNAThis is the voltage due to generator action.Therefore it is given by Eb=ZnP/aThis voltage is always less than applied voltage V. Due to the difference between these two armature current flows through the armature resistance.Eb

Back EMFNSSMNA Voltage Equation for DC Motor, V=Eb+Iara The resistance ra is always present in the machine. Eb is induced due to rotation.Therefore Eb is also called as Voltage Behind Armature Resistance, ra.Eb

Back EMFS This Back EMF makes DC motor as Self-Regulating Machine. This makes to draw required Ia to match the Load Torque. Under no load, torque is required to overcome F & W losses. Therefore Ia is very low. V and Eb are nearly equal. When load increases, speeddecreases, Ebdecreases, increases to drive the increased load.Ia increases, Ta When load decreases, speeddecreases to drive the decreased load.Ia decreases, Ta increases, Ebincreases, The steady state conditions are attained when Ta matches load torque TL

Back EMFThus back EMF Eb acts as a GOVERNER i. e, it makes the motor SELF REGULATING so that it draws as much current as just required.

Equivalent CircuitIn DC machine armature have parallel paths equal to a.Each path have resistance rp and generated EMF Eb.rpEb

Equivalent CircuitIn DC machine armature have parallel paths equal to a.Each path have resistance rp and generated EMF Eb.

Equivalent CircuitCircuit Model of DC MotorAAAV raIaSeparately Excited DC Motor

Equivalent CircuitCircuit Model of DC MotorSeparately Excited DC Motor

Equivalent CircuitCircuit Model of DC MotorFFFDC Shunt Motor

Equivalent CircuitCircuit Model of DC MotorFFFDC Shunt MotorOR

Equivalent CircuitCircuit Model of DC MotorAAAV raFFFDC Shunt MotorOR

PowerAAAVFFFDC Shunt Motor Voltage Equation for DC Motor, V=Eb+Iara+Vb Vb = Brush Contact Voltage Drop =1-2 volts. Hence neglected. Multiplying by Ia on both sides.VIa=EbIa+Ia2raThis is a Power EquationVIa=Electrical Input to the armatureIa2ra= Copper Loss in the armatureEbIa= Output Power Electromagnetic Power

EbIa= Output Power=Electromagnetic Poweror Electrical Equivalent of Mechanical Power Pm developed in the Armature causing Rotation of the Armature. Psh=Pm - Constant Loss. =EbIa - (Iron Loss +F & W Loss).If brush contact loss is givenVIa=EbIa+Ia2raEbIa+Ia2raPower

PowerEbIa= Output Power=Electromagnetic Poweror Electrical Equivalent of Mechanical Power Pm developed in the Armature causing Rotation of the Armature. Psh=Pm - Constant Loss. =EbIa - (Iron Loss +F & W Loss).If brush contact loss is givenVIa=EbIa+Ia2raEbIa+Ia2ra+Ia2raEbIa +Constant Loss. Psh

PowerEbIa= Output Power=Electromagnetic Poweror Electrical Equivalent of Mechanical Power Pm developed in the Armature causing Rotation of the Armature. Psh=Pm - Constant Loss. =EbIa - (Iron Loss +F & W Loss).If brush contact loss is givenVIa=EbIa+Ia2raEbIa+Ia2raEbIa Psh Constant Loss. Ia2ra( Tsh)

PowerEbIa= Output Power=Electromagnetic Poweror Electrical Equivalent of Mechanical Power Pm developed in the Armature causing Rotation of the Armature. Psh=Pm - Constant Loss. =EbIa - (Iron Loss +F & W Loss).If brush contact loss is givenVIa=EbIa+Ia2raEbIa+Ia2raEbIa Psh Constant Loss. Ia2ra( Tsh)PiPmPo

PowerCondition for Maximum PowerWe know that Pm=EbIa=Via - Ia2raDifferentiating both sides with respect to IaThus mechanical power developed is maximum when back EMF is equal to half the applied voltage.Under this condition, Ia would be more than rated current.Half the input current Ia is wasted in copper loss.Considering constant losses, the efficiency will be below 50%. Some Calculations.

The EFFECT or ACTION of armature mmf on the field flux is called armature reaction. There is a ACTION of the field flux and under operating condition there is a REACTION of armature flux on main flux or field flux. Thats why the word REACTION.

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNAFig. Main Pole Flux due to Field Current only

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNAif supply is given to Armature wdg.With Field wdg unexcited,

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNACurrent flows through the armature wdg with cross under N pole and dot under S pole.Cross produces clockwise fluxand dot produces anticlockwise flux.if supply is given to Armature wdg.With Field wdg unexcited,

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNADue to all conductors, a Fig. Armature flux only.Armature flux is perpendicular to main flux or CROSSES the main fluxor armature flux is entirelyarmature sets up vertical downward flux.Cross-Magnetizing (1st effect).

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNAWhen the current flows in the FieldaArm flux adds at the upper end of N poleand lower end of S pole. This strengthens the main field in the said area.and Armature windings.

NSQ axis or GNA or MNAD axisField FluxD axisQ axis or GNA or MNAaArm flux opposes at the lower end of N poleand upper end of S pole. This weakens the main field at this ends.r

NSQ axis or GNA D axisField FluxD axisaIn the phasor r is the resultant flux.rMNA is always perpendicular to the axis of resultant flux. MNA MNA or MNAQ axis or GNA or MNArMNA is shifted from GNA (2nd effect). This shift is dependant on armature current. If Ia is more shift is more.

NSQ axis or GNA D axisField FluxD axisar MNAQ axis or GNA rUnder heavy load condition flux is set up over the brush, so there is more spark. MNA

We have seen that there strengthening and weakening effect on main field due to armature current.Strengthening and weakning of main field are equal and resultant flux remainsunchanged due to same SLOP.If there is no saturation, B-H curve is a straight line.HBut for a practical machine, there is magnetic saturation because machines are designed at knee point for ecomomy.High B Low B Below knee point, Constant Flux= Low B/More Area (more cost)Above knee point, Constant Flux= High B/Less Area (less cost)B

HBmore slop Strengthening effect means more flux i. e., saturation and with less slop strengthening effect is lessas compared to weakening effect with more slop. So resultant flux r1 decreases from its no load value.This is called as effect (3rd effect).less slop H0H2H1B0B1B2For constant H0H2=H0H1, B0B2 < B0B1Under saturation, slop is less and under no saturation i. e, for straight line slop is more.De-Magnetizing

HBmore slop Strengthening effect means more flux i. e., saturation and with less slop strengthening effect is lessas compared to weakening effect with more slop. So resultant flux r1 decreases from its no load value.This is called as effect (3rd effect).less slop H0H2H1B0B1B2For constant H0H2=H0H1, B0B2 < B0B1Under saturation, slop is less and under no saturation i. e, for straight line slop is more.Sr1De-Magnetizingar

Thus the net effect of armature reaction or the net effect of armature flux on the main flux is

Field wdg is excitedStatorRotor

StatorRotorField wdg is excitedFlux Density B Waveform for Main Field alone

StatorRotorFlux Density B Waveform for Main Field aloneField wdg is un-excited

StatorRotorVoltage V is given to Armature wdg

StatorRotorVoltage V is given to Armature wdgArmature MMF Waveform

StatorRotorWhen Field and Arm are Excited Armature MMF Waveform

StatorRotorAddition SubstractionArmature MMF Waveform

StatorRotorMore SparkArmature MMF Waveform

StatorRotorNow draw armature flux waveform.Under poles air gap is uniform.So arm flux varies uniformly.Along Q axis air gap is maximum.So flux is minimum.Saddle shaped arm fluxArmature MMF Waveform

StatorRotorNow add field flux and armature flux waveform.

StatorRotorNow add field flux and armature flux waveform.Un-saturated B-H CurveIf saturation is considered thenthere is DE-MAGNETIZING actionFlux decreases

StatorRotorThis is resultant flux r waveform.MNA moves towards right Rotation of Motor towards leftSo this is Leading Edge of PoleMore FluxThis is Trailing Edge of PoleLess FluxMove brushes towards right for less spark

Detrimental Effects of Armature Reaction:1. Distortion of the main field flux. a) This increases iron loss. The flux and flux density increase in teeth and pole shoe. As a result iron loss increases which is more than no load condition. b) Poor commutation. The zero crossing of the flux density wave is shifted by an angle from GNA or Q axis. The coils undergoing commutation have some voltage induced in them. Due to which there is current which resulted in detrimental sparking or poor commutation at the brushes.

c) Sparking. Under heavy load condition, the flux density waveform is distorted considerably. So more voltage may be generated in the coils. If adjacent commutator segments voltage exceeds 30 or 40V, a spark may occur between these segments. Some times this spark may spread around the commutator in the form of ring fire.

2. Net reduction of the main field flux.Induced voltage in the armature decreases in case of generator. b) Due to lower voltage, current decreases and torque decreases in case of motor. c) In order to compensate the above, it is required to increase the field mmf which is possible by increasing the number of turns and diameter of field winding. Therefore cost increases.

Methods for limiting the effects of Armature Reaction.1. Chamfered or Eccentric Pole. Chamfered or Eccentric Pole.Long Air GapShort Ai GapThe air gap at the pole tips can be increased to increase the reluctance for arm flux.

Methods for limiting the effects of Armature Reaction.2. Staking of Laminations Alternately.2 1 1 2 1 2 2 1 Staking of Laminations Alternately.1 2 1 2 1 2 Air Gap High Reluctance to Arm Flux

Methods for limiting the effects of Armature Reaction.3. Reduction in Cross Armature Flux. Rectangular Holes for Air Gap High Reluctance to Arm Flux Cross Arm Flux decreases

Methods for limiting the effects of Armature Reaction.4. Interpoles.Q axisInterpoleInterpoleInterpoles are placed in between main poles.They are Smaller in size.Interpolar magnetic axis is along Q axisInterpole wdg is connected in series with arm wdg.Interpole mmf neutralizes the effect of arm mmf along Q axis at all load current.

Methods for limiting the effects of Armature Reaction.4. Interpoles.Q axisInterpoleInterpoleFor the polarities + and dot for arm, arm mmf is DOWNWARDS. To neutralizes the effect of arm mmf along Q axis lower interpole must be N and upper interpole must be SNSSo for motor in the direction of rotation main pole comes first and then interpole for same type i. e. N or S

Methods for limiting the effects of Armature Reaction.5. Compensating Winding.Q axisInterpoleInterpoleCompensating wdg is housed in the slots which are at the pole face or pole shoe. NSCompensating wdg polarities are opposite to that of arm wdg polarities in order to reduce the armature reaction.Compensating wdg is connected in series with armature wdg

NSSMNADue to + and polarities of armature, flux direction changes .

NSSMNANew MNA shifts towards the direction opposite to the direction of rotation.So shift the brush axis in a direction opposite to the direction of rotation due to which there is less spark.New MNA

NSSMNA shifts in the direction of rotation in case of generator and against the direction of rotation in case of motor. New MNA If brushes are given forward shift in case of generator and backward shift in case of motor, then there is GOOD commutation (Less Spark).

NSSQ Axis

NSSQ AxisResolve aadaq1. aq is right angle to main flux f .Cross-magnetizing or Distorting component of arm flux.2. ad is opposing the main field This reduces the main field. aq = a CosDemagnetizing Component of arm flux ad= a Sin

NSSrQ AxisadaqResultant flux r is reduced due to brush shift even if there is no saturation aq and ad will increse with the increase in armature current.

SrQ Axisadaq

180-2 180-2SrQ Axisadaq

180-2 180-2SrQ AxisadaqThis gives adThis ad is demagnetizing arm flux.Conductors in 4 gives demagnetizing ampere turns.Demagnetizing ampere turns per pole ATd= angle in electrical degreesConsider 2

180-2 180-2SrQ AxisadaqThis gives aqThis aq is crossmagnetizing arm flux.Conductors in 360-4 gives crossmagnetizing ampere turns.Crossmagnetizing ampere turns per pole ATc= angle in electrical degreesConsider 180-2

If =0, then brushes are on GNA and ATd=0, and entire AT=ATc.Cross Magnetizing Conductors,Cross Magnetizing Conductors, (180-2)0 electrical (180-2)0 electricalDemagnetizing Conductors, 20 electricalDemagnetizing Conductors, 20 electricalSome Calculations.

Armature conductor carries + current under N pole and current under S pole. COMMUTATIONSo current is reversed. This reversal takes place along brush axis.During reversal armature coil is short circuited.If coil is having low induced voltage along brush axis, large current flows in short circuited coil.As you know, under short circuit test of transformer, low voltage is required for the flow of rated current.The process by which current is a coil is reversed during the short circuit is known as COMMUTATIONThe period during which the coil remains short circuited is called commutation period and is represented by Tc.

Good commutation means no sparking at the brushes.COMMUTATIONPoor commutation means sparking at the brushes.1. An elliptical form of the commutator.2. Unbalancing of the commutator.4. Vibration of the brush holders.3. Uneven projection of the commutator segments.The causes of sparkingPoor commutation may be due to1. Increase in voltage between commutator segments, which may result in flash over.2. Increase in current density at the trailing edge of the brush.

NSS COMMUTATIONArmature WindingArmatureCommutatorCommutator segmentsConnection between Commutator and arm wdg.

COMMUTATION PROCESSQ axisBrushCommutator SegmentsStatorRotorWidth of brush=width of segment

COMMUTATION PROCESSQ axisBrushCommutator SegmentsStatorRotorCoils2IcConsider this portion

COMMUTATION PROCESSQ axis2IcIcIcIcIcN PoleS PoleConsider coil 3

COMMUTATION PROCESSQ axis2IcMove rotor towards leftN PoleS PoleWhich is the direction of rotation?

COMMUTATION PROCESSQ axis2IcBrush comes in contact with another segment 5So current flows from 5 to brushI5Ic-I5Ic2Ic-I5

COMMUTATION PROCESSQ axis2IcAgain move rotor towards left

COMMUTATION PROCESSQ axisBrush comes in contact with 50% segment 4 & 5So current flows from 4 & 5 to brush is Ic.IcIcIcIc=02Ic

COMMUTATION PROCESSQ axis2IcBrush comes in contact with 50% segment 4 & 5So current flows from 4 & 5 to brush is Ic.IcIcIcIc=0Coil is ShortedTotal current remains constant =If induced low voltage is present,high current flows.

COMMUTATION PROCESSQ axis2IcBrush comes in contact with 50% segment 4 & 5So current flows from 4 & 5 to brush is Ic.IcIcIcIc=0Coil is ShortedIf induced low voltage is present,high current flows.Coil may burn.Total current remains constant =

COMMUTATION PROCESSQ axis2IcBrush comes in contact with 50% segment 4 & 5So current flows from 4 & 5 to brush is Ic.IcIcAgain move rotor towards left

COMMUTATION PROCESSQ axisBrush comes in contact with 75% segment 5Current flows from 5 to brush is 2Ic - I4.2Ic-I4I4So current flows from segment 4 to brush is I4.2IcTotal current remains constant =Ic-I4Ic

COMMUTATION PROCESSQ axisBrush comes in contact with 75% segment 5Current flows from 5 to brush is 2Ic - I4.2Ic-I4I4So current flows from segment 4 to brush is I4.2IcTotal current remains constant =Again move rotor towards leftAs soon as rotor moves I4 is interrupted, so there is spark.

COMMUTATION PROCESSQ axisBrush comes in contact with 100% of segment 5Current flows from 5 to brush is 2Ic.So current flows from segment 4 to brush is zero.2IcIcTotal current remains constant =

COMMUTATION PROCESSQ axis2IcIcIcIcIcCoil 3Initially current polarity is dot, with current = +Ic +Ic0-Ic

COMMUTATION PROCESSQ axis2IcI5Ic-I5Ic2Ic-I5Coil 3Then current polarity is dot, and current =+ (Ic- I5) +Ic0-Ic

COMMUTATION PROCESSQ axis2IcIcIcIcIc=0Coil 3Now no polarity, and current =0+Ic0-Ic

COMMUTATION PROCESSQ axis2Ic-I4I42IcIc-I4IcCoil 3Then current polarity is cross, and current =- (Ic- I4) +Ic0-Ic

COMMUTATION PROCESSQ axis2IcIcCoil 3Current polarity is cross, and current = -Ic +Ic0-IcSo current is reversed from +Ic to -Ic There is spark due to interruption of I4.

COMMUTATION PROCESS+Ic0-IcThe time required by a coil to change the current from +Ic to Ic, is Tc, commutation period. TcThe time Tc, can be calculated from the relation

COMMUTATION PROCESS+Ic0-IcIf the current changes from +Ic to Ic linearly in the commutation period Tc, then the commutation is called asTcIdeal Commutation orIdeal CommutationStraight Line Commutation.If the current changes from +Ic to Ic in the time more than Tc, then the commutation is called asDelayed or Under Commutation.Delayed or Under CommutationCurrent in SparkAfter time Tc is over there is some current I4 in the coil which is to be disconnected, therefore due to this current there is spark.

COMMUTATION PROCESS+Ic0-IcTcIdeal CommutationIf the current changes from +Ic to Ic in the time less than Tc, then the commutation is called as Over or Accelerated Commutation.Delayed or Under CommutationCurrent in SparkThis gives satisfactory commutation.Over or Accelerated Commutation

COMMUTATION PROCESSThe nature of current in the coil being commutated depends on:1. The resistance of copper-carbon contact.2. The resistance of the coil being commutated.2. The emfs induced in the coil due toa) Self inductance andb) Mutual inductance with another coil.

IMPROVING COMMUTATIONThere are three methods of obtaining sparkless commutation.1. Resistance Commutation.2. Voltage Commutation.3. Compensating Commutation.

IMPROVING COMMUTATION1. Resistance Commutation If the time constant of coil i. e. L/R is reduced, then the current change is faster.So metal brush is replaced by high resistance carbon brush.Resistance CommutationIf the resistance of the coil and brush is considered then variation of current is as shown in figure.Such type of commutation is called as resistance commutation.

IMPROVING COMMUTATION1. Resistance Commutation Due to resistance, current Ic also decreases. For small rating or fractional machines, thin wires are used due to which there is more resistance. This gives good commutation.2. Voltage (emf) CommutationCurrent changes from +Ic to Ic with time interval of Tc (1 millisecond or less).So emf is induced as per equation Lc is self inductance, Lc is self reactance. Similarly some voltage is also induced due to mutual flux of the other coils. ( Flux Linkage per amp=Inductance, Lm and Xm=Lm.)These voltages are called as REACTANCE voltage.

IMPROVING COMMUTATION As per Lenzs Law, this reactance voltage opposes the cause of it, and the cause is change in current.The change in current is delayed. Thats why it is also called as Delayed Commutation.

IMPROVING COMMUTATION The spark due to voltage commutation is minimised by the use of Interpoles.Interpoles are also called as Compensating Poles or Compoles.Q axisInterpoleInterpoleInterpoles are placed in between main poles.Interpolar magnetic axis is along Q axisThey are narrow and Smaller in size. Interpole wdg is connected in series with arm wdg.Interpole mmf neutralizes the effect of arm mmf along Q axis at all load currents.

Q axisFor the polarities + and dot for arm, arm mmf is DOWNWARDS. To neutralizes the effect of arm mmf along Q axis lower interpole must be N and upper interpole must be SNSSo for motor in the direction of rotation main pole comes first and then interpole for same type i. e. N or SInterpoleInterpoleIMPROVING COMMUTATION

Circuit Diagram of DC Shunt Motor with Interpole Winding.Interpole Winding.

Q axisFunctions of interpoles.1. To neutralizes the cross magnetizing effect of arm reaction.NSInterpoleInterpoleIMPROVING COMMUTATIONadaq cross magnetizingDue to this opposition, there is no flux in the commutating zone.

Q axisFunctions of interpoles.2. To neutralizes the reactance voltage and to make sparkless commutation.NSInterpoleInterpoleIMPROVING COMMUTATIONSome of the interpole flux neutralizes the cross mangetizing flux andremaining interpole flux induces rotational emf in the coil which opposes the reactance voltage. So reactance voltage decreases. Delaying of current decreases. That resulted in to less spark.

IMPROVING COMMUTATION+Ic0-IcTcIdeal CommutationDelayed or Under CommutationReactance Voltage due to linear commutation=Reactance Voltage due to sinusoidal commutation=

SStatorRotor 1NSNow draw flux under InterpoleUnder main S pole flux is positive So under S interpole flux should be positive Under main N pole flux is negative So under N interpole flux should be negative This is main field fluxThis is resultant flux due to arm reaction.

SStatorRotor 1NSNow add these two fluxes

SStatorRotor 1NS

SStatorRotor 1NSResultant flux due to interpole wdg

IMPROVING COMMUTATION3. Compensating CommutationThe interpole winding is effective only in the commutating zone.With interpole winding the resulted flux waveform isConsider coil aa at the peak position of flux, where flux is rising of falling suddenly.aaUnder this condition large voltage may be inducedIf this voltage is more than 40V between the commutator segments, then nearby air is ionized.This may result in to the flashover, damaging segments, mica sheet and sometimes armature winding.

IMPROVING COMMUTATION3. Compensating CommutationThe induction of voltage also depends on sudden change in heavy load.If load increases, current increases, flux increases or flux changes from lower to higher value, so induced voltage increases.If load decreases, current decreases, flux decreases or flux changes from higher to lower value, so induced voltage increases.The induction of voltage can be neutralized or compensated by providing winding in the pole shoe or pole face and connecting that winding in series with armature winding.Because load current flows through the armature winding.

Q axisThis wdg is called as compensating wdg which is housed in the slots which are at the pole face or pole shoe. Compensating wdg polarities are oppositeCompensating wdg is connected in series with armature wdg.3. Compensating CommutationIMPROVING COMMUTATIONto that of arm wdg polarities in order to reduce the induced voltage in the coilto have better commutation.

Circuit Diagram of DC Shunt Motor with Interpole and compensating Winding.Interpole Winding.Compensating Winding.

SStatorRotor NSArmature FluxResultant Flux due to Interpole winding

SStatorRotor NSFor cross arm current So for cross compensating wdg current Armature Flux

SStatorRotor NSArmature FluxFor dot arm current So for dot compensating wdg current

SStatorRotor NSArmature FluxFlux due to compensating winding in the air gapFor air gap Flux due to compensating winding

SStatorRotor NSArmature FluxNow add these two fluxes

SStatorRotor NSNow add these two fluxes

SStatorRotor NSResultant Flux due to Interpole and Compensating winding

SStatorRotor NSThis is Main field Flux

SStatorRotor NSNote the difference between two waveforms.Thus the detrimental effect of Armature Reaction on Main field is minimised due to Interpole and Compensating windings.The difference is negligible.

SStatorRotor NSMNA due to armature reaction shifted towards rightNow it is toward leftSo MNA is very near Q axis.Good CommutationLess spark

1. To neutralize the cross magnetizing effect of arm reaction.2. To shift MNA near to Q axis from angle .3. To reduce armature inductance by decreasing flux. Thus armature time constant (L/R) is reduced, which resulted into quick dynamic response.4. To have better commutation and less spark under heavy load condition.responseTimeSLOWFASTThe Functions of Compensating Winding

Compensating Winding MMF per Pole, ATcomp.Since compensating wdg mmf neutralizes arm mmf under pole face.Pole arc or pole shoe portion is approximately equal to 70% of pole pitch.

100%70%

IMPROVING COMMUTATIONThe methods of obtaining sparkless commutation are.Resistance Commutation: by increasing brush contact resistance.2. Voltage Commutation: by providing interpole winding.3. Compensating Commutation: by providing compensating winding.

CLASSIFICATIONTYPES OF MOTORS(METHODS OF EXCITATIONIn general, there are two methods of exciting the field winding.1. Separate excitationOR

DC Motor Earlier

Documents

Transcript of DC Motor Earlier