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MSU ECE 480 - st_mot_3

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Ampere’s Law & Biot-Savart LawMagnetic Fields for a Long Thin WireThe SolenoidThe ElectromagnetBasic Model for a Stepper MotorControlling Stepper-Motor State TransitionsStepper-Motor Interface Circuit ModelDiode ProtectionStepper-Motor Operation and Interfacing FundamentalsPrepared by: P. David Fisher and Diane T. RoverAmpere’s Law & Biot-Savart LawAn electrical current I in a ware causes (induces) a magneticfield B. The direction of B is given by the “right-hand rule”.Magnetic Fields for a Long Thin WireFor a long thin wire, the strength of the magnetic field B a distance R from the wire isB = (U0I)/(2R) (1) where 0 = permeability of vacuum, 0 = 4x10-7 henry/meterThe SolenoidA coil of wire with N turns creates a magnetic field B in the direction illustrated, whereB = kNI, (2) with k being a constant. Hence, B is proportional to N and I.1 st_mot_3.docIBRMagneticFieldLinesIBThe ElectromagnetIf you place a compass and in the vicinity of the iron core, you would discover thatone end (say A) would be similar to the “South Magnetic Pole” of the earth, while the other end (say B) would be similar to the Earth’s “North Magnetic Pole”.Two important properties of electromagnets are the following:1. All Electromagnets are dipoles; i.e., they have a North Pole (N) and a South Pole (S).2. The position of the Poles (at A or B) is determined by the direction of the current I and the direction of the winding.2 st_mot_3.docIron Core(South) SN (North)A BBasic Model for a Stepper MotorConsider the four electromagnets physically arranged as illustrated.3 st_mot_3.docP1(L11,L12)P4(L41,L42)P3(L31,L32)P2(L21,L22)ABAAA BBBR11L11i11L31R31i31vaR12L12i12L32R32i32vbR21L211i21L41R41i41vcR22L22i22L42R42i42vdWindingsforP1 & P3WindingsforP2 & P4whereControlling Magnetic Polarities with Winding Voltages (va, vb, vc and vd)The magnetic polarities of the electromagnets can be controlled by varying the winding voltages va, vb, vc and vd. Consider the following two cases.Case I – P1 and P3Note: Winding voltages va and vb control the polarities for electromagnets P1 and P3.4 st_mot_3.docABABNSP1(L11,L12)ill > 0 va = 5V il2 = 0 vb = 0V ABABSNP(L31,L32)i3l > 0 va = 5V i32 = 0 vb = 0V ABABSNP1(L11,L12)ill = 0 va = 0V il2 > 0 vb = 5V ABABNSP3(L31,L32)i3l = 0 va = 0V i32 > 0 vb = 5VCase II – P2 & P4Note: Winding voltages vc and vd control the polarities of electromagnets P2 and P4.5 st_mot_3.docABABNSP2(L21,L22)i2l > 0 vc = 5V i22 = 0 vd = 0V ABABSNP4(L41,L42)i41 > 0 vc = 5V i42 = 0 vd = 0V ABABSNP2(L21,L22)i2l = 0 vc = 0V i22 > 0 vd = 5V ABABNSP4(L41,L42)i4l = 0 vc = 0V i42 > 0 vd = 5VControlling Stepper-Motor State TransitionsThe “state” of a stepper motor can be controlled by controlling the winding voltages of the electromagnets. Consider the following example.Note: The polarities of electromagnets P1 and P3 can be reversed by simultaneously changing va from 0V to 5V and vb from 5V to 0V.6 st_mot_3.docP4P3P2BBBNP1BNSSCP1 & P3va = 0Vvb = 5VP2 & P4vc = 5Vvd = 0VP4P3P2BBBNP1BSNSCP1 & P3va = 5Vvb = 0VP2 & P4vc = 5Vvd = 0VNext StatePresent StateImportant Questions and ConclusionsWith respect to the previous example, answer the following questions.1. Assume that the stepper motor is in its “initial state.” If a compass is positioned with the pivot point of its needle at point C, in what direction would the needle point?2. Assume that the stepper motor is in its “next state.” If a compass is positioned with the pivot point of its needle at point C, in what direction would the needle point?3. Did the compass needle move clockwise or counter clockwise?4. What voltages do we need to change to have the compass needle rotate in the opposite direction?5. How many “steps” does it take to make a 360 rotation?6. How might you add the number of steps for a 360 rotation? Identify two distinct approaches.7. Why might you want to add steps?8. What are the engineering design considerations that must be addressed as a newstepper-motor assembly is designed for a new commercial application?7 st_mot_3.docStepper-Motor Interface Circuit ModelThere are a number of significant challenges facing the computer engineer who must interface a stepper motor to a microcontroller. For example, consider the following transient circuit response problem.Case 1: Switch closes at t = 0i11(0-) = i11(0+) = VS/RS  0A (3)i11(t) = (VS/R11)e-t/, where  = R11/L11(4)VL11(t) = Vse-t/(5)Case 2: Switch Opens at t = 0Because currents through an inductor cannot change discontinuously, ill(0+) = i11(0-) = VS/R11(6)Applying Kirchhoff’s Voltage Law (KVL) around the loop at time t = 0+ yields:-VS + VS1 + R11i11 + VL11 = 0 (7)-VS + RS1i11 + R11i11 + VL11 = 0 (8)VL11 = VS – (RS1 + R11)i11  VS – (RS1/R11)VS, when RS1 >> R11(9)VL11  -(RS1/R11)VS  -(1M/0.1k)VS  -104VS (10)These large voltages will destroy the solid-state switch.8 st_mot_3.docVSRS1+ VS1 -R11L11 +VL11 -i11S1RS1 = Shunt Resistance of SwitchRS1 >> R11Diode ProtectionThere exists a very standard solution to the problems which arise due to the desire to rapidly switch electrical currents in circuits containing inductive loads. The following example illustrates the solution.The winding of an electromagnet can be modeled as a resistance in series with an inductance, as illustrated in the figure. Under computer control, current i11 is to be controlled by controlling voltage v1. As we saw with the stepper-motor example, i11 will assume one of two steady-state values—i.e., i11 = 0A and i11 = VS/R11. In the circuit illustrated, diode D1 protects the interface logic from large transient voltages.Case 1: v1 = 0VThe voltage drop across the diode is v1 = 0V, and the diode is turned off (an open circuit). Also, i11 = 0A.Case 2: v1 = VS, where VS > 0VThe voltage drop across the diode is v1 = VS, and the diode is turned off (an open circuit). Also, i11 = VS/R11.Case 3: At t = 0-, v1 = VS, where VS > 0V. Then at time t = 0, the interface logic switches and presents a high impedance to the rest of the circuit.At time At t = 0+, the current i11 = VS/R11 and passes through the diode. The diode isforward biased with v1 = -0.7V. With time, i11 drops to 0A, v1 returns to 0V and the diode is turned off (an open circuit).This is the solution to only one interfacing problem. Another common problem is the fact that actuators, such as the stepper motor, do not operate at standard “logic voltages.”


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MSU ECE 480 - st_mot_3

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