ElectroMSystem - ME375 Handouts ElectroElectro-Mechanical...

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Unformatted text preview: ME375 Handouts ElectroElectro-Mechanical Systems • DC Motors – Principles of Operation – Modeling (Derivation of Governing Equations (EOM)) • Block Diagram Representations Di – Using Block Diagrams to Represent Equations in s - Domain – Block Diagram Representation of DC Motors • Example School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 1 1 ME375 Handouts DC Motors Motors are actuation devices (actuators) that that generate torque as actuation. torque • Terminology – Rotor : the rotating part of the motor. – Stator : the stationary part of the motor. – Field System : the part of the motor that provides the magnetic flux. – Armature : the part of the motor th th which carries current that interacts with the magnetic flux to produce torque. – Brushes : the part of the electrical circuit through which the current is supplied to the armature. to the armat – Commutator : the part of the rotor that is in contact with the brushes. School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 2 2 ME375 Handouts DC Motors - Principles of Operation • Torque Generation B B df i dL Force will act on a conductor in a magnetic field with current flowing magnetic field with current flowing through through the conductor. d f = ia ⋅ d L × B Integrate over the entire length: i f= Total torque generated: torque generated: B τ Coil = School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 3 3 ME375 Handouts DC Motors - Principles of Operation Let N be the number of coils in the motor. The total torque generated generated from the N coils is: τ m = N ⋅ (2 ⋅ ia ⋅ B ⋅ L ⋅ R) = For a given motor, (N, B, L, R) are fixed. We can define fi KT = 2 ⋅ N ⋅ B ⋅ L ⋅ R Nm / A as the Torque Constant of the motor. Torque The torque generated by a DC motor is proportional to the armature current ia : τ m = KT ⋅ ia Large KT : – Large (N, L, R). (N, L, R) is limited by the size and weight of the motor. For a DC motor, it is desirable to have a large KT . However, size and other physical limitations often limits the achievable KT . School of Mechanical Engineering Purdue University – Large B: Need to understand the methods of generating flux ... ME375 ElectroMechanical - 4 4 ME375 Handouts DC Motors - Principles of Operation • Back-EMF Generation BackElectromotive force (EMF) is generated in a conductor moving in a magnetic field: B v ×B de = ( v × B ) ⋅ dL emf Integrate over the entire length L: e em f = v Si Since the N armature coils are in series, the total EMF is: EMF E em f = 2 N ( R ω ) B L = v Define the Back-EMF Constant Kb : Back- Kb = 2 ⋅ N ⋅ R ⋅ B ⋅ L V / (rad / sec) The Back-EMF generated due to the rotation of the motor armature is opposing the applied Backvoltage and is proportional to the angular speed ω of the motor: E em f = K b ⋅ ω Note: KT = Kb is true only if consistent SI units are used ! School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 5 5 ME375 Handouts DC Motors - Modeling Schematic + eRa − + ei(t) _ RA + eLa − iA LA + Eemf _ τm θ, ω τL Element Laws: Electrical Subsystem JA B Mechanical Subsystem FBD: Interconnection Laws: Laws School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 6 6 ME375 Handouts DC Motors - Modeling Derive I/O Model: I/O Model from ei(t) and τL to angular speed ω : and FG H FL B+ R J θ +G K HK IJ FG KH IJ θ + FG R B + K K HK IJ b K IJ θ = e ( t ) − b L K LA τ L + RA τ LA J A LA B RJ RA B ω+ + A A ω+ + K b ω = ei ( t ) − KT KT KT KT KT I/O Model from ei(t) and τL to angular position θ : and LA J A KT A A T T A A b i T School of Mechanical Engineering Purdue University A τ L + RA τ KT L L g g ME375 ElectroMechanical - 7 7 ME375 Handouts DC Motors - Modeling Transfer Function: Ω (s) = ⋅ Ei ( s ) − ⋅Τ L ( s ) θ (s) = ⋅ Ei ( s ) − ⋅Τ L ( s ) Q: Let the load torque be zero (No Load), what is the steady state speed (No-Load Speed) of the (Nomotor for a constant input voltage V ? Q: Let the load torque τL = T, what is the steady state speed of the motor for a constant input the load to what is the steady state speed of the moto fo constant input voltage V ? School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 8 8 ME375 Handouts Block Diagram Representation • Differential Equation → Transfer Transfer Function (System & Signals) Function Y (s) = G (s) ⋅ U (s) U(s) Input Signal Y (s) = U1(s) ± U 2 (s) U1(s) Y(s) G(s) • Signal Addition/Subtraction Output Signal Y(s) + ± U2(s) Ex: Draw the block diagram for the following DE: Ex: Draw the block diagram for the following DE: Jω = τ School of Mechanical Engineering Purdue University Jω + Bω = τ ME375 ElectroMechanical - 9 9 ME375 Handouts Block Diagram Representation • Transfer Function in Series • Multiple Inputs Y ( s ) = G 2 ( s ) ⋅ Y1 ( s ) , Y1 ( s ) = G1 ( s ) ⋅ U ( s ) b g Y ( s ) = Y1 ( s ) ± Y2 ( s ) Y ( s ) = G 2 ( s ) ⋅ G1 ( s ) ⋅ U ( s ) U(s) G1 (s) Y1(s) G2 (s) Input Signal = G1 ( s ) ⋅ U 1 ( s ) ± G 2 ( s ) ⋅ U 2 ( s ) Y (s) Output Signal • Transfer Function in Parallel X 1 ( s ) = G1 ( s ) ⋅ U ( s ) , X 2 ( s ) = G 2 ( s ) ⋅ U ( s ) Y ( s ) = X1 ( s ) ± X 2 ( s ) b Y1 ( s ) = G 1 ( s ) ⋅ U 1 ( s ) , Y2 ( s ) = G 2 ( s ) ⋅ U 2 ( s ) U1(s) U2(s) Input Signals G1 (s) G2 (s) Y1(s) + Y (s) ± Output Y2(s) Signal g = G1 ( s ) ± G 2 ( s ) ⋅ U ( s ) U(s) Input Signal G1 (s) G2 (s) X1(s) + Y (s) ± Output X2(s) Signal School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 10 10 ME375 Handouts Block Diagram Representation • Feedback Loop Loop U(s) + Input − Signal X (s) Y (s) G (s) Output Signal H(s) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 11 11 ME375 Handouts Block Diagram For Feedback Loop Negative feedback loop feedback loop U(s) + Input − Signal X (s) Positive feedback loop Positi feedback loop Y (s) G (s) U(s) + Output Signal Input + Signal H(s) X (s) Y (s) G (s) Output Signal H(s) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 12 12 ME375 Handouts Block Diagram Representation of DC Motors Schematic + eRa − + ei(t) _ RA ElectroElectro-Mechanical Coupling: + eLa − iA LA + Eemf _ θ, ω Take Laplace Transform of the Eqs. τm τL JA B Governing Equations: d LA iA + RA iA + Eemf = ei (t ) ( dt J A ω + Bω = τ m − τ L τ m = KT ⋅ i A ⎫ ⎬ Eemf = Kb ⋅ ω ⎭ ( ⎛ ⎜ ⎝ ) ) ⎞ ⎟ ⎠ School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 13 13 ME375 Handouts Block Diagram Representation of DC Motors Electrical Subsystem: Mechanical Subsystem: Take Laplace Transform of the Eqs. Take Laplace Transform of the Eqs. Q: Now that we generated a block diagram of a voltage driven DC Motor, can we derive the transfer function of this motor from its block diagram ? ( This is the same as asking you to reduce the multiblock diagram to a simpler form just relating inputs ei(t) and τL to the output, either ω or θ .) School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 14 14 ME375 Handouts Block Diagram Reduction From Block Diagram to Transfer Function • Label each signal and block + − − 1 LA s + RA KT + 1 JA s + B Kb • Write down the relationships between signals School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 15 15 ME375 Handouts Block Diagram Reduction • Solve for the output signal in terms of the input signals • Substitute the transfer functions’ label with the actual formula and simplify Ω (s) = θ (s) = KT LA s + RA ⋅Τ L ( s ) ⋅ Ei ( s ) − 2 LA J A s + ( BLA + RA J A )s + ( RA B + Kb KT ) LA J A s + ( BLA + RA J A )s + ( RA B + Kb KT ) 2 KT LA s + RA ⋅ Ei ( s ) − ⋅Τ L ( s ) 2 s( LA J A s + ( BLA + RA J A )s + ( RA B + K b K T )) s( LA J A s + ( BLA + RA J A )s + ( RA B + K b K T )) 2 School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 16 16 ME375 Handouts Transfer Function from Block Diagram Disturbance D(s) Reference Input R(s) + Error − E(s) GC (s) Control Input + + U(s) Output GP(s) Y(s) Plant Controller H(s) Sensor In general, for a negative feedback loop, the transfer function from an input to the output is: School of Mechanical Engineering Purdue University ME375 Feedback Control - 17 17 ME375 Handouts Example – TF from Block Diagram Disturbance D(s) Reference Input R(s) + Error GC (s) − E(s) Control Input + + U(s) Output GP(s) Y(s) Plant Controller H(s) Sensor Y (s) = ⋅ R( s ) + GYR ( s ) ⋅ D( s ) GYD ( s ) School of Mechanical Engineering Purdue University ME375 Feedback Control - 18 18 ME375 Handouts Example – TF from Block Diagram Disturbance D(s) Reference Input R(s) + Error GC (s) − E(s) Control Input + + U(s) Output GP(s) Y(s) Plant Controller H(s) Sensor E (s) = ⋅ R(s) + G ER ( s ) ⋅ D (s) G ED ( s ) School of Mechanical Engineering Purdue University ME375 Feedback Control - 19 19 ME375 Handouts Example – DC motor with load Schematic DC Motor + eRa − + ei(t) _ RA + eLa − iA LA + Eemf _ τm θ, ω θL, ωL Assumptions: JL N2 • Gears and shafts are rigid and massless. JA N1 BL B Block diagram of the load inertia: Block diagram of the gear train: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 20 20 ME375 Handouts Example – DC motor with load Block diagram of the DC motor subsystem: + − 1 LA s + RA − KT + 1 JA s + B Kb Reduce the mechanical portion of the block diagram: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 21 21 ME375 Handouts Example – DC motor with load Simplified block diagram: + − 1 LA s + RA KT Kb Transfer Function from input voltage Ei(s) to the angular velocity or position of the load: School of Mechanical Engineering Purdue University ME375 ElectroMechanical - 22 22 ...
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue University-West Lafayette.

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