Overview of motors and motion control

Transcription

1 Overview of motors and motion control. Elements of a motion-control system Power upply High-level controller ow-level controller Driver Motor. Types of motors discussed here; Brushed, PM DC Motors Cheap, rugged, high-reliability ow Torque ripple tepper Motors Cheap, rugged, high-reliability o brushes, suitable for any environment Do not require feedback At low speeds, provide up to 5 torque of brushed motor, torque of brushless motor uffers from resonance and long settling times Consume current regardless of load or motion, run hot osses at speed are high Undetected position loss as a result of open loop operation Brushless DC Motors o brushes, suitable for any environment Brushless AC Motors o brushes, suitable for any environment

2 tepper Motors Half tepping Phase A Phase B

3 Full tepping: Phase A Phase B

4 tepper (cont.) Wiring arrangements for various teppers 6Ω 4-lead 5-lead 6-lead 8-lead Behavior of stepping-motor ) A typical step motor has full step positions. Thus, the typical stepper will move.8 degrees in one step. ) When the stator portion of the stepper makes a step (but prior to the rotor actually moving), the stepper sees its maimum torque. As the rotor moves under the influence of this torque, the torque drops to a zero value when the rotor reaches the end of its step (.8 degrees). ) This results in a static torque curve that may look something like: Torque Ma T. tart step table degrees.8.6 4) Considering the static torque curve above, the rotor on a step motor could lag commanded motion by as much as.8 degrees during an acceleration phase, or lead by as much as.8 degrees during deceleration. ote that the static torque curve of a stepper behaves similar to a mass-spring system with a non-linear spring. From this model, the response to a single step move would look something like;

5 angle time 5) Due to this behavior of the step motor, driving the motor at a step speed near its natural frequency can greatly increase the oscillations during response. The natural frequency of the motor depends on the stiffness (electro-magnetic field) and rotational inertia. Higher inertia will decrease the natural frequency of the system and provide more separation between typical driving frequencies and ω n. Typical ω n for an unloaded stepper may be - Hz. 6) An additional consideration in driving steppers is the acceleration profile. Remember that the maimum torque of the system cannot be eceeded. Therefore, to operate at a desired rotation or slew rate, the motor must be started (and stopped) in a profiled manner, approaching the desired speed such that the motor is not driven past maimum acceleration.

6 Brushed DC motor: The iron-core brushed DC motor has been one of the most common motors used in servomotor systems. A basic schematic of the motor is shown in the following figure; Permanent magnet i F=iB B Armature/ rotor brushes +V commutator -V The armature contains many more windings than the one shown in this figure, and therefore generates an output with fairly low torque ripple. The commutators provide a mechanical means to sense the position of the motor and send current to a new set of windings. Brushed DC motors have two primary difficulties; ) Contact between the brushes and commutator creates carbon dust, sparks, wears out ) The armature generates heat as a function of the power loss equation i R, and this heat must escape through the magnets or motor shaft. Motor Dynamics: The behavior of this motor can be described through a system of equations that relates the input to the motor, voltage, to the motor output, motion (position, velocity, acceleration) under a given load. This model will require; ) a description of the current flow in the motor, ) equations of motion defining rotation of the motor, ) electro/mechanical relation in the motor. Motor Electronics: + R V in - + umming voltages throughout the circuit gives, V = k e ω emf

7 V in = di + Ri + k e ω, dt a first order ODE in current, i, with, R the motor inductance and resistance, k e the electric constant, V emf the back emf and ω the rotational speed of the motor (rad/s). Motor Mechanics: θ, ω, α J motor g motor J load g T The rotational inertia, J as seen by the motor is, d g J = J + motor J load, d g accounting for motor inertia (rotor) as well as gear train and load inertia. Motor Dynamics: J && θ + C & θ = T T friction This second order ODE in rotation, θ, with C viscous damping in the system, T the motor torque and T friction the friction torque describes the motor dynamics. Finally, the electro-mechanical relation is approimately given as an equation that describes motor torque as a linear function of current in the motor, T = kti where k t the motor torque constant. These equations can be combined to result in two equations, a st and nd order ODE with two unknowns, i and q; J && θ + C & θ kti = T friction i + Ri + k & e θ = V in These will be cast in state-space form to result in a system of three st order DE s = θ & = = & k θ & C t = T J + J friction = i k R V & = b + in or;,

8 = + = y y y V R R k J k J C b t & & & Matlab provides a variety of tools to easily model this system. A simple first step could be to observe a step response of this system (step input of the motor voltage) using the command; >step(a,v*b,c,d) with A,B,C the matrices shown above (in order). teady-tate Motor Behavior: At steady state, the dynamic motor model above can be greatly simplified (i_dot, θ_dot = ) to yield the equations; k e V in Ri = + ω + i k T t = ω = R k k V k R T t e in t, This equation represents a linear torque/speed relation for a PM dc motor. The response (assuming a constant input applied voltage) looks like; Torque Power T,P T stall ω no load ω The maimum generated torque occurs at rest (stall), and decreases to zero at the maimum motor speed under no load. At this speed, the back-emf voltage in the motor is equal to the input voltage (minus a small amount to overcome inefficiencies in the motor).

9 If we consider power in the motor, write a linear equation for T as a function of motor speed; T ( ω ) = T ω s ω ma and then epress power as; P ( ω ) = T ( ω ) = ωt ω s. ω ma The maimum power occurs at; dp ( ω ) = T ω = dω ω s ma or ω = ω ma T,P T stall Torque ma Power Power ω no load / ω no load ω Eample Problem: Motor election This eample will demonstrate the process of analyzing a particular motor in the mechanical system demonstrated below. This system requires a DC brushed motor to move payloads along an inclined conveyor. The motor drives the conveyor through a gear reducer simply shown with pulleys g and g. θ, ω, α J motor g motor M J load g T θ

10 Problem givens; M=kg; r w =8cm; g:g=gr=5; J load =. kgm ; θ = deg., consider friction in the system as a percentage of the static load. Analysis Approach; First we need to get our state equations. These will consist of equations of motion, our KV equations and our electro-mechanical relations as derived above. The KV relation will be identical to that above. Create a simple FBD of the motor system to get the equations of motion; FBD: J equiv θ, ω, α T Mgsin(θ)/g Equations: J & equivθ + C & θ kti = Mg sin( θ ) T friction GR GR i + Ri + k & e θ = V in ote that the variable q is the motor rotation. Jequiv. is the equivalent inertia of the system as seen by the motor and determined using the relation described above. Remember that inertia s are magnified by the square of intermedi gear reductions. In this case, Jequiv. is given as; J equiv = J armature + J conveyor + J payload GR GR J = Mr payload w Equations in tate-pace Form: = θ = & θ = i

11 & ( ) = C k + Mg sin T friction & t θ J J J GR J GR & k b R RV rw y GR r y = w GR y ote that in describing that observers or outputs, y a conversion is made to output displacement and velocity in meters and m/s respectively, y still represents current. olution approach: Implement using the step function in Matlab. In this case, the system will be described in state space form with the matrices A, B, C (D=[]) as above. Eample: Consider the MA motor from Parker automation (page 88 in 99 computmotor catalog). The specifications on this motor are as follows; J armature =.e-4 kg m ; k t =.45.A; k b =5.68*6/(*pi*) m s; = 4.8e- H, R = 9.65O, V=4 Matlab code: M=; %Payload mass g=9.8; %gravity Jconv=.; %Conveyor inertia rw=.8; %Wheel radius dg=;dg=5;gr=dg/dg; J_motor=.e-4; %Motor inertia J=J_motor+Jconv*(/GR)^+M*rw^*(/GR)^; %Total inertia at motor C=.79e-5; %Viscous damping theta=*pi/8; kt=.45; % torque constant kb=5.68*6/(*pi*); %Back emf constant =4.8*e-; %Inductance R=9.65; %Resistance V=; %Input voltage Tf=.5; %frictional torque as a percent of dead load A=[,,,-C/J,kt/J,-kb/,-R/]; B=[;(-M*g*rw*sin(theta)/GR)*(+Tf)/J;V/]; C=[rw/GR,,;,rw/GR,;,,]; D=[;;]; step(a,b,c,d)

12 tep Response 6 - From: U() To: Y() 4. Am plitude To: Y().5 To: Y() Tim e (sec) ote that this eample shows the motor provides a reasonable response time, and low current at steady state. On the other hand, our final speed may be a little low (.65 m/s), so try decreasing the size of the gear ratio to 5: This yields the following results;

13 .5 tep Response From: U() To: Y() Am plitude To: Y()..5 To: Y() Tim e (sec) ow we have a maimum speed of.5 m/s, a maimum current draw of appro..5 A and a fast response. Perhaps this motor is larger than needed (note also that this motor is intended to run at -7V). o, an M6A motor is tried with the 4V input voltage; J armature = 5e-6 kg m ; k t =.8.A; k b =4.*6/(*pi*) m s; =.5e- H, R=.4 O

14 .6 tep Response From : U() To: Y().4..6 Am plitude To: Y().4. 4 To: Y() Tim e (sec) otice that the response time is still acceptable, but the final speed is low and the current draw is high. If a higher motor volatage is tried (4V):.4 tep Response From: U(). To: Y().. Amplitude To: Y().5 To: Y() Tim e (sec)

15 Final velocity is now acceptable, but the current may be to high, (near to slightly above the motor maimum ratings). Therefore, this motor is too small for this application.