EE 410/510: Electromechanical - The University of … 41… · · 2010-05-21EE 410/510:...

EE 410/510: Electromechanical SystemsChapter 4

• Chapter 4. Direct‐Current Electric Machines and Motion Devices

• Permanent‐Magnet DC Electric Machines• Radial Topology• Simulation and Experimental Studies • Generator Driven by a Motory• Electromechanical Systems with Power Electronics

• Axial Topology Permanent‐Magnet DC Electric Machines

• Device Fundamentals• Axial Topology Hard Drive Actuator

• Electromechanical Motion Devices: Synthesis and Classificationf

All figures taken from primary textbook unless otherwise cited.5/21/2010 1

Radial Topology Permanent magnetic DC Electric M hiMachines

• DC electric machines guarantee:

Hi h– High power

– High torque densities

– Efficiency

– Affordability

– Reliability

– RuggednessRuggedness

– Overloading capabilities

• Power range of modern DC electric machines W – 100 kWDi i f d d i 1 i di t d 5 l• Dimensions for modern devices 1mm in diameter and approx 5 mm long to 1 m in diameter

• Used widely in aerospace, automotive, marine, power, robotics, etc.

• Only permanent magnet synchronous machines which don’t have brushes surpass the use of DC machines in the field

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• Permanent magnet DC (PMDC) machines are rotating energy‐transforming electromechanical motion devices that convert energyelectromechanical motion devices that convert energy

• Motors convert electrical to mechanical

• Generators convert mechanical to electrical

• The same PMDC can serve as either a motor or a generator

• Electric machines will always have stationary and rotating components separated by an air gap

• The armature winding is placed on the rotor slots connected to a rotating commutator which rectifies the voltage

• One supplies the armature voltage, ua, to the rotor windingsa

• The rotor windings and permanent magnets on the stator are magnetically coupled

• The brushes ride on the commutator which is connected to the armatureThe brushes ride on the commutator which is connected to the armature windings

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• The armature winding consist of identical uniformly distributed coils

• Excitation of the magnetic field is produce by permanent magnets• Excitation of the magnetic field is produce by permanent magnets

• The commutator, armature windings, and permanent magnets produce stationary mmfs which are displaced by 90 electrical degrees

• The armat re magnetic force is along the rotor magnetic a is hile the• The armature magnetic force is along the rotor magnetic axis while the direct axis stands for a permanent magnet magnetic axis

• Torque is produced as a result of the interaction of these mmfs

5/21/2010 4http://www.displayresearch.com/education_DIYMotor.htm

Operation of a Radial PM DC MotorOperation of a Radial PM DC Motor

http://www.le.ac.uk/eg/research/groups/power/caecd/SRD1_pedrg.htm

Equations of Motion• The equation for the electronic circuit in the

motor described is

q

diddid

• Where the mechanical coupling term comes from the constant, ka, which depends on factors

dtdiLEir

dtdk

dtdiLir

dtdiru a

aaaaaa

aaaaaa

such as the number of turns in the armature winding, and the permeability of the magnet

Yi ldi h f ll i ODE f h l i• Yielding the following ODE for the electronic circuit

• Likewise the mechanical ODE is derived from the energy differential

N l ti th i f it th• Neglecting a the spring force one can write the following:

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Equations of Motionq

• Governing equations of motion

• These same equations can be written in Laplace form as

irsLu

• Equations set into a set of linear algebraic functions

a

rmLe

aaaa

Tk

BJsTTirsLu

1

Lm

amaa

ar T

BJsu

BJsrsL

5/21/2010 7

Steady State Torque Speed Characteristics

• Torque speed characteristics are mapped below for different applied voltages less than the

Steady State Torque Speed Characteristics

• DC operation stipulates that

• Thus one operates a DC motor under the

0dtdi

pp gmaximum rated voltage for the motor

Thus one operates a DC motor under the following steady‐state operation

• Where

• For the mechanical side, the electromagnetic torque must equal the applied torque load for steady state operation

• Under constant load, velocity is decreased by reducing the applied voltage. The angular l it t hi h th t t t i f d t

steady state ope at o

• Thus the torque speed of the system is described entirely by

Laae TikT

velocity at which the motor rotates is found at the intersection of the two curves

• Furthermore, if one neglects friction, then Newton’s second law states that

• Thus, angular velocity is

• Increased the applied armature voltage

• And in stead state, Te=TL providing a constant angular velocity with no load. ie

Increased the applied armature voltage

• Decreased with applied torque with a slope of –ra/ka2

Torque Speed ExampleTorque Speed Example

• Calculate and plot the torque‐speed characteristics for a 12V PM DC motor with the following parameters.

• The load is a nonlinear function of angular velocity:

• Solution: Torque speed characteristics are governed by:

2ar

mNT rL 2000002.002.0 rad

sVAmNka

05.005.0

Solution: Torque speed characteristics are governed by:

• One can use different values of the armature voltage to plot the steady state characterisitics

Practice ApplicationsPractice Applications

• Angular velocity of a PM DC motor is regulated by the applied armature voltage

• Note that one can use power converter electronics from the previous chapter to regulate the voltage of a PM DC motor

• To rotate the motor clockwise or counterclockwise, the bipolar voltage should be applied to the armature windingg

• Large motors require high end electronics capable of driving multiple amps through a circuit

• Small motors (1‐10W) can be driven using dual op‐amps as shown

PM DC Electric GeneratorsPM DC Electric Generators

• Assume a resistive electrical load, RL• The following equation is used for the electric circuit

• The induced emf is

• In steady state operation, the induced terminal voltage is proportional to the angular velocity.

• Voltage is therefore generated by applying torque, Tpm, by aerodynamic, thermal, or hydrodynamic forces.

• The resulting differential equation is:

• Note the sign change in the first and last terms of the angular acceleration equation This represents the flow of current back into the circuit instead from theequation. This represents the flow of current back into the circuit instead from the applied torque instead of driving the system “forward” as achieved in motor opreration

Simulation and Experimental StudiesSimulation and Experimental Studies• PM DC electric machines are among a very limited class of EM motion devices which can

theoretically be described by linear differential equations

Th j it f EM ti d i l d i li diff ti l ti hi h• The majority of EM motion devices are solved using nonlinear differential equations which can be solved using Matlab ODE solvers. (ex. ODE45 solver)

• Even though the equations of motion are described by linear ODE’s, linear theory may not always be applied to PM DC machines b/c of various applied voltage constraints

• For our immediate purposes, we will apply the state space model previously developed to analyze a simple system

Simulation and Experimental StudiesSimulation and Experimental Studies

St I tStep Input


Squarewave input

Applying a square wave using MatlabApplying a square wave using Matlab

• Homework: assign an applied armature voltage ua=10rect(0.5t) with a load torque TL=0

• Use the following Matlab code to aid in generating the plots presented in the previous slide

Loses and Efficiencies of EM DevicesLoses and Efficiencies of EM Devices

• Loses associated with these devices are the sum of the resistive and drag effects

• Efficiency can be determined by the ratio of input to output power

Example of a Physical SystemExample of a Physical System• Let us stop for a moment and examine the performance of a real world motor

• For this example we will examine the JDH2250 PM DC motor

• The following figure documents the acceleration of the unloaded motor at an applied• The following figure documents the acceleration of the unloaded motor at an applied armature voltage of 7.5 and 15 V.

• As the torque is applied, the angular velocity decreases as described by our steady state torque‐speed characteristics

Example of a Physical SystemExample of a Physical System

• For this example we will examine the JDH2250 PM DC motor

• The deceleration dynamics of a loaded vs. unloaded motor are presented

• Note that the experimental results presented did not match the numerical simulation derived for this system b/c of complex friction phenomenon that we will did not sufficiently describe.for this system b/c of complex friction phenomenon that we will did not sufficiently describe.

PM DC Generator Driven by a PM DC Motor

• Let us now analyze two PM DC electric machines integrated as a motor‐generator system.


• In this case, the prime mover (The PM DC motor) will drive the generator

pm = prime moverg = generator

• Assume that a resistive load, RL, is inserted in series with the generator armature winding.

• Kirchhoff’s voltage law yields

• The torsional dynamics of the generator prime mover system is• The torsional dynamics of the generator‐prime mover system is


• The applied torque on the PM DC motor is


– Where iapm is the armature current in the prime mover

– Kapm is the torque constant of the prime mover

• While the load torque on the prime mover is that created by the generator

• Thus one obtains the torsional‐mechanical dynamics by the following differential equation

• The dynamics of the electric circuit in the prime mover is given by



• The resulting three differential equations must therefore govern the system



Example: PM DC MotorExample: PM DC Motor

• The following parameters are used to model an electric machine:

Prime Mover GeneratorPrime Mover

rapm = 0.4 OhmLapm = 0.05 Hkapm = 0.3 V-sec/rad

Generator

rag = 0.3 OhmLag = 0.06 Hkag = 0.25 V-sec/radapm

Bapm = 0.0007 N-m-s/radJpm = 0.04 kg-m2

ag Bag = 0.0008 N-m-s/radJg = 0.05 kg-m2

Example: PM DC Motor

• The following parameters are used to model an electric machine:


Prime Mover

rapm = 0.4 OhmL = 0 05 H

Generator

rag = 0.3 OhmL = 0 06 HLapm = 0.05 H

kapm = 0.3 V-sec/radBapm = 0.0007 N-m-s/radJpm = 0.04 kg-m2

Lag = 0.06 Hkag = 0.25 V-sec/radBag = 0.0008 N-m-s/radJg = 0.05 kg-m2

• Simulate and examine the state and dynamic operation of a PM DC generator driven by 100V PM DC motor. Study the transient dynamics and the voltage generation, uapm, for different resistive loads, RL, and angular velocities, rpm

Using the state equation:

Assuming steady state operation and an infinite resistive load, RL=

One then finds:


• A Simulink model can be created for the system using system using the Matlab inputs


provided as:

http://www.mathworks.com/access/helpdesk/help/toolbox/simulink/ug/bq3qcn_.html#bq31ya1Simulink help file for creating custom blocks

Example: PM DC Motor for constant RL

• A Simulink model can be created for the system using system using the Matlab inputs


provided as:

Example: PM DC Motorfor constant RL

• Where:

Example: PM DC Motorfor constant RL

• Where:


• Where:


• Where:

Example: PM DC Motor with RL = 5 Ohmsp L


INCORRECT FIGUREINCORRECT FIGURE IN TEXTBOOK ?

MAKE THIS SIMULATION ANDSIMULATION AND PROVIDE THE

CORRECT FIGURE FOR HOMEWORK

Example: PM DC Motor driven at constant uapmp apm

Example: PM DC Motor driven at uapm = 50 Vp apm

PM DC Motor Driven by a Buck Converter

• Let us now examine the application of a high frequency step down switch (buck) converter to


control a PM DC motor

• The duty ratio of the converter is

Th ti f th b k t• The equations for the buck converter

developed in Chapter 3 are

• Yielding the following 4 differential equations for the system


• Recall that duty ratio is regulated by the signal–level control voltage, uc, which is bound


cbetween utmax and utmin. Assume in these systems that utmin =0.

• utmax = max voltage rating for the motor

• uc = input contol voltage for the motor = drive voltage

With the nonlinear term

PM DC Motor Driven by a Buck ConverterPM DC Motor Driven by a Buck Converter

PM DC Motor Driven by a Boost Converter

• The resulting four differential equations govern the system

PM DC Motor Driven by a Boost Converter

PM DC Motor Driven by a Cuk Converter

• The resulting six differential equations govern the system

PM DC Motor Driven by a Cuk Converter

Axial Topology PM DC Electric MachinesAxial Topology PM DC Electric Machines

• Motors using planar segmented permanent RotorStator

magnet arrays that are driven by windings above or below the magnet

• We know that a planer current loop of any size and shape generates Torque, T, in a uniform magnetic field

Rotor

North and magnetic field

• Where i is the current, s is the area of the loop, B is the magnetic field, and m is the magnetic

o a dSouth PM poles

dipole moment generated

• Using the relation one can show that the torque generated causes motion II to the plane of the coil

• One can also write the force applied to the rotor as

• It is important to note that this type of motor design works for both linear translation and

Stator

Rotor

grotary motors alike

• Axial motors are used in hard disk heads, cooling fans, linear axis drive systems, etc. etc. etc.

http://etd.lsu.edu/docs/available/etd-07062006-185252/unrestricted/Challa_thesis.pdf

Axial Topology Example:Axial Topology Example:

• Assume a current of 10 Amps is applied around a square loop with dimensions 10 x 20 cm

• Parallel to the loop and located slightly below, is a permanent magnet generating the following magnetic flux density:ag et c u de s ty:

• Using the equation for torque,

Tesla

where the vector s is normal to the surface

aaa zyx

ˆˆˆˆ

mNamNT x

ˆ12.0

8.06.00100*2.0*1.0*10

1D Axial Topology Linear Motor1D Axial Topology Linear Motor

• Using the previous equations. Consider a series line filament, l, each carrying current in or out out of the page.

• Now consider a series of magnetic poles on the rotor facing up or down from top to bottom of the page

• The force generated by current in the line element and the magnetic field generated by the magnet generate motion horizontally along the length of the page.g y g g p g

Rotor

Stator

htt // td l d /d / il bl / td 07062006 185252/ t i t d/Ch ll th i dfhttp://etd.lsu.edu/docs/available/etd-07062006-185252/unrestricted/Challa_thesis.pdf

1D Axial Topology Rotational Motor1D Axial Topology Rotational Motor

• By wrapping the linear motor into a circular shape, the linear motion becomes that of a rotation about a central axis.

• Motion is then described by angular velocity with an effective flux density that depends on angular displacement

RotorStator

angular displacement, r.• The magnetic flux density, B(r) applied

depends on the magnet magnetization, geometry, and shape of the rotor/stator system.

North and South PM polesy

• For permanent magnets, the flux density is viewed from the windings as a periodic function of angular displacement.

• If the rotor design is produced such that there

poles

are no gaps between magnet segments, then one may use the following relation to accurately describe the magnetic flux density relation

• Where Bmax is the maximum effective flux density produced by the magnets from the winding

• N is the number of magnets (segments)• Nm is the number of magnets (segments)

• n is the integer function of the magnet magnetization, geometry, shape, thickness, separation, and so on.

1D Axial Topology Rotational Motor lExample

• Consider the three different magnetic flux density values given Where Bmax = 0.9 T and Nm = 4. max m

We can plot B(r) using the following statements

Rotary PM DC Motor

• The electric circuit equation for torque can be derived as

Rotary PM DC Motor

where leq is the effective length, which includes the winding filament length and the lever arm, and N is the number of turns in the coil

O l d i d th i f ti• One can also derived the expression for magnetic energy,

where Aeq is the effective area that takes into account the number of turns, magnetic field no uniformity, etc.

• Applying

b• One obtains:

Hard Drive ActuatorHard Drive Actuator• Consider an axial topology PM hard drive actuator

assembled with two permanent magnet segments in an array

• Rotation of the motor is achieved by applying voltage across the current loop.

• The polarity voltage applied sets the current and p y g pptherefore the direction of the motor

• The relative change in magnetization of the two motor segments also contributes to the direction of the rotationrotation

• In hard drive actuators, a mechanical limiter restricts the angular displacement to

• Typical hard drives operate with displacement limiters of

• Applying Kirchhoff’s Voltage law to the problem

Rotational Hard Disk Motor ExampleRotational Hard Disk Motor Example

• The equation of the circuit must be further limited by use of two (left and right) filamentsof two (left and right) filaments

• Likewise Newton’s 2nd law of motion results in• Likewise, Newton’s 2nd law of motion results in

• Consider two practical cases when two magnetic strips are magnetized to ensure

• For these cases, we will let k = 1 and a =10 and 100 for a maximum magnetic flux density of 0.7 Tesla

Rotational Hard Disk Motor Example

• For the case where


• The electromagnetic torque can then be described as where L and R are the left and right angular displacements respectively.

• We will solve for the system using the following limiting factorsWe will solve for the system using the following limiting factors

• Torque is then expressed as

• One can further match the system by applying a nonideal Hook’s law (for spring forces) to the systemsystem


• For this example, let us use the following parameters


Rotational Hard Disk Motor Example Case N b 1Number 1:

Rotational Hard Disk Motor Example Case Number 2:Number 2:

Rotational Hard Disk Motor Example Case Number 3: Simplified Linear ModelNumber 3: Simplified Linear Model

Geometrical Variations of l h i l i iElectromechanical Motion Devices

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