DC Motor Control Janet Klinkhachorn Jeannine Meyers Cheri Settell December 6, 2002.

DC Motor ControlDC Motor ControlJanet KlinkhachornJanet Klinkhachorn

Jeannine MeyersJeannine Meyers

Cheri SettellCheri Settell

December 6, 2002

OverviewOverviewIntroductionIntroductionTheoryTheoryDesign ApproachDesign ApproachProcedureProcedureResultsResultsAnalysisAnalysisConclusionConclusion

December 6, 2002Janet KlinkhachornJanet Klinkhachorn

IntroductionIntroductionTaskTaskDC MotorsDC Motors

Electrical power into Electrical power into mechanical powermechanical power

Used for heavy duty Used for heavy duty applicationsapplications

DC Motor ComponentsDC Motor ComponentsStatorStatorRotorRotor


http://www.prestoliteasia

TheoryTheoryVVff = R = RffIIff + L + LffdIdIff/dt/dt

VVtt = I = IaaRRaa + E + Eaa + +

LLaadIdIaa/dt/dt

Jdω / dt = TJdω / dt = Tee - T - TLL

EEaa = K = Kee * I * Iff *ω *ω

http://murry.newcastle.edu.au


OverviewOverviewIntroductionIntroductionTheoryTheoryDesign ApproachDesign ApproachProcedureProcedureResultsResultsAnalysisAnalysisConclusionConclusion

December 6, 2002

Design ApproachDesign ApproachInitial ConditionsInitial Conditions

Ke=0.33Ke=0.33Vf=150 VDC Vf=150 VDC Vt=100 VDCVt=100 VDCRa= 0.9 ΩRa= 0.9 Ω La= 0.01 HLa= 0.01 HRf= 75 ΩRf= 75 Ω Lf= 50 HLf= 50 HJ= 0.2 kgJ= 0.2 kgm2m2TL= 5 NTL= 5 Nmm


December 6, 2002Jeannine Meyers

Design ApproachDesign ApproachMeasure If, Ia, E, Measure If, Ia, E, ωωChange Rf and LfChange Rf and Lf

Rf= 100 ΩRf= 100 ΩLf= 75 HLf= 75 HBoth Rf and LfBoth Rf and LfRf= 50 ΩRf= 50 ΩLf= 25 HLf= 25 HBoth Rf and LfBoth Rf and Lf



ProcedureProcedureDefinition of integralDefinition of integral

[i(t + Δt) – i(t)] / Δt[i(t + Δt) – i(t)] / ΔtCT to DTCT to DT

VVff = R = RffIIff + L + LffdIdIff/dt/dt VVtt = I = IaaRRaa + E + Eaa + L + LaadIdIaa/dt/dtJdω / dt = TJdω / dt = Tee - T - TLL EEaa = K = Kee * I * Iff *ω *ω


ProcedureProcedureExampleExample

VVtt = I = IaaRRaa + E + Eaa + L + LaadIdIaa/dt/dt

dIdIa a / dt = [(V/ dt = [(Vtt – I – IaaRRaa – E – Eaa) / L) / Laa]]

[I[Iaa(t + Δt) – i(t + Δt) – iaa(t)] / Δt] = [(V(t)] / Δt] = [(Vtt – I – IaaRRaa – E – Eaa) / L) / Laa]]

IIaa(t + Δt) = i(t + Δt) = iaa(t) + Δt * [(V(t) + Δt * [(Vtt – I – IaaRRaa – E – Eaa) / L) / Laa]]

MatLab CodeMatLab CodeIa = I1 + dt*(Vt – (Ra*I1) – ea) / La Ia = I1 + dt*(Vt – (Ra*I1) – ea) / La


OverviewOverview

IntroductionIntroductionTheoryTheoryDesign ApproachDesign ApproachProcedureProcedureResultsResultsAnalysisAnalysisConclusionConclusion

December 6, 2002Cheri Settell

ResultsResultsChange the value of field resistance Change the value of field resistance

and inductance by 25and inductance by 25Increase Rf Increase Rf Increase LfIncrease LfIncrease both Rf and LfIncrease both Rf and LfDecrease RfDecrease RfDecrease LfDecrease LfDecrease both Rf and LfDecrease both Rf and Lf

Cheri Settell December 6, 2002

Rf= 75 ΩRf= 75 Ω Lf= 50 H Lf= 50 H


Rf= 100 Ω Lf= 50 HRf= 100 Ω Lf= 50 H


Rf= 75 Ω Lf= 75 HRf= 75 Ω Lf= 75 H


Rf= 100 Ω Lf= 75 HRf= 100 Ω Lf= 75 H


Rf= 50 Ω Lf= 50 HRf= 50 Ω Lf= 50 H


Rf= 75 Ω Lf= 25 HRf= 75 Ω Lf= 25 H


Rf= 50 Ω Lf= 25 HRf= 50 Ω Lf= 25 H


Analysis of RfAnalysis of RfTime IC Rf inc Rf dec

Back emf 2.5 3.75 1.25

Field Current 3.75 2.5 5

Armature Current 2.5 3.75 2.5

Speed 1.9 3.75 1.25

Back emf 92 92 95

Field Current 2 1.5 3

Armature Current 115 115 115

Speed 140 180 99


Analysis of LfAnalysis of LfTime IC Lf inc Lf dec

Back emf 2.5 2.5 1.9

Field Current 3.75 5 1.9


Speed 1.9 1.9 1.9

Back emf 92 95 95

Field Current 2 2 2


Speed 140 140 140


Analysis of BothAnalysis of BothTime IC B inc B dec

Back emf 2.5 3.75 1.25

Field Current 3.75 3.75 2.5


Speed 1.9 3.75 1.25

Back emf 92 90 95

Field Current 2 1.5 3


Speed 140 185 98


ConclusionConclusionSolutionsSolutions

Increase Rf to 100 ohmsIncrease Rf to 100 ohmsDecrease Lf to 25 ohmsDecrease Lf to 25 ohms

Team’s DecisionTeam’s DecisionLf = 25 ohms and Rf = 75 ohmsLf = 25 ohms and Rf = 75 ohms


Questions?Questions?

DC Motor Control Janet Klinkhachorn Jeannine Meyers Cheri Settell December 6, 2002.

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