Transcript Lab 7
Motors Discussion D10.2 Chapter 15 Hans Christian Oersted (1777 – 1851) X 1822 In 1820 he showed that a current produces a magnetic field. Ref: http://chem.ch.huji.ac.il/~eugeniik/history/oersted.htm André-Marie Ampère (1775 – 1836) French mathematics professor who only a week after learning of Oersted’s discoveries in Sept. 1820 demonstrated that parallel wires carrying currents attract and repel each other. attract A moving charge of 1 coulomb per second is a current of 1 ampere (amp). repel What do Maxwell’s Eqs. Predict? B = 0 B = magnetic flux density (magnetic induction) B = mH m = magnetic permeability Magnetic field lines must be closed loops Force on moving charge q Lorentz force F = q ( v B) B Vector Multiplication Scalar (Dot) Product Vector (Cross) Product A A q q B B C=A B C = AB C = A B cosq C = A B sin q Direction of C given by "Righthand rule" (into the screen) Force on current in a magnetic field Force on moving charge q -- Lorentz force F = q ( v B) Current density, j, is the amount of charge passing per unit area per unit time. N = number of charges, q, per unit volume moving with mean velocity, v. L S j F = N V q(v B) F = ( j B)V F = (i B)L vt V = S L dQ NqSvt i= = = j S dt t j = Nqv Force per unit length on a wire is iB Rotating Machine B Force out i + i X Force in Rotating Machine B Force in X i + i Force out Rotating Machine B Force out commutator i + i brushes X Force in What do Maxwell’s Eqs. Predict? E = B t Corresponds to Faraday’s law of electromagnetic induction A changing magnetic flux B density induces a curl of E B nda S t E ds = E nda = S E ds = t S B nda = t flux through S Back emf B B dq = dt Force out +- i a b r q l i X X Force in area = A = lw = l 2r cos q w = 2r cos q flux = BA = Β2rl cos q emf = E ds = flux = B2rl cos q t t q emf = eab = B2rl sin q = 2Brl sin q t B Back emf eab = 2Brl sin q Force out + - i a b i X eab Force in q B Back emf Force out commutator i + - eab = 2Brl sin q i brushes X Force in eab q Armature with four coil loops eab S X X X X N q Motor Circuit Ia Ra Vt Ea Vt = Ea I a Ra Ia = Vt Ea / Ra Ia = Vt Ka / Ra Ea = Ka Power and Torque Pd = Ea I a = d d = Ka I a Vt K a K a d = Ra Ra 2 Armature Current, Speed, Power and Torque 1 Ia = d K a d = Ka I a Vt K a K a d = Ra Ra Vt Ra = 2 d Ka Ka Pd = Ea I a = d Vt Ra 2 Pd = d 2 d Ka Ka 2 FF-130 Series Motors Vt Ra 2 Pd = d 2 d Ka Ka 1 Ia = d K a Vt Ra = 2 d Ka Ka Vt Ra = 2 d Ka Ka Vt Ra = =0 2 d Ka Ka dω=0 Vt Ka = = 6.4 mN-m Ra Vt Ra = 2 d Ka Ka d = 0 d =0 Vt = = 7100 rpm K a Ia = 1 d K a I aω=0 Vt = Ra Vt 4.8 Ra = = = 4.0 I a 1.21 At = 0 Vt K a d = = 6.4 mN-m Ra Pd max Vt 2 = 4 Ra Vt Ra 2 Pd = d 2 d Ka Ka Maximum Power dPd V 2Ra = t =0 2 d d d Ka Ka d max Vt K a = 2 Ra d =0 Vt K a = Ra d =0 Vt = = 7100 rpm K a K a = I aω=0 Vt 4.8 = = 0.00646 volt-sec 7100 rpm 7100(2 / 60) Vt = Ra Vt 4.8 Ra = = = 4.0 I a 1.21 Pd max Vt 2 4.82 = = = 1.4 W 4Ra 4 4 dω=0 Vt K a 4.8 0.00646 = = = 7.6 mN-m Ra 4 1.18 W 6.4 mN-m Generator Circuit Ia Ra Vt Ea Vt = Ea I a Ra Ea = Ka Ea = 0.00646 7100 2 / 60 = 4.8 V Ka = 0.00646 volt-sec Vt = 4.8 0.15 4 = 4.2 V +5v V Motor Driver Circuit 470 1 SSR V PWM p31 6 2 14 13 12 11 10 9 4 8 MOTOR 7406 1 2 3 4 5 6 7 GND MOS FET Relays G3VM-61B1 Pulse-Width Modulation count[3] set = ~| count pwm duty period reset = (count == duty) Vpwm duty = Vref period pwm.v module pwm(clk,clr,duty,pwm); input clk; count[3] input clr; input [3:0] duty; set = &count output pwm; pwm reg [3:0] count; reg pwm; wire set, reset; // duty period reset = (count == duty) 4-bit counter always @(posedge clk or posedge clr) if(clr == 1) count <= 0; else count <= count + 1; pwm.v assign set = ~| count; assign reset = (count == duty); always @(posedge clk) begin if(set == 1) pwm <= 1; if(reset == 1) pwm <= 0; end endmodule count[3] set = &count pwm duty period reset = (count == duty) Lab 11 Lab 11 +5V +5V +5V 470 Ω 14 6 7406 1 2 7 1 4 G3VM-61B1 Solid-State 2 Relay pwm 1N4004 Motor Generator VOUT