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Lecture 18. Electric Motors

simple motor equations and their application

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I will deal with DC motors that have either permanent magnets or separately-excited field coils

This means that all I have to think about is the armature circuit.

We apply a current to the armature. The current cuts magnetic field lines creating a local force

This results in a global torque, making the armature accelerate

The motion of the armature means that a conductor is cutting magnetic field lineswhich generates a voltage that opposes the motion, the so-called back emf

All of this is governed by just three equations.

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€

τ =K1i, eb = K2ω

The torque is proportional to the armature current

The back emf is proportional to the rotation rate

These are connected by the voltage-current relation

€

e = Ldi

dt+ Ri

For most control applications the time scales are slow enough that we can neglect the inductance

Ohm’s law is good enough

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The voltage in Ohm’s law is the sum of the input voltage and the back emf

€

ei − eb = Ri

The two proportionality constants are generally more or less equal

€

K1 = K = K2

€

τ =Kei − eb

R= K

ei − Kω

R

Combine all this to get

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The maximum rotation rate is at zero torque: the no load speed

The maximum torque is at zero rotation: the starting torque

t

w

Power is torque times speed, so its maximum is at half the no load speed.

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You can find K and R from the starting torque and no load speedat whatever nominal voltage is given for the motor

€

τ s = Kei

R, ωNL =

ei

K

€

K =ei

ωNL

, R =Kei

τ S

=ei

2

ωNLτ S

There are other things you can do. This is discussed in the text.

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Take a look at some simple dynamics

One degree of freedom system — the red part moves

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You can easily verify that the ode is

€

C ˙ ̇ ψ = τ = Kei − Kω

R= K

ei − K ˙ ψ

R

which we can rearrange

€

C ˙ ̇ ψ +K 2

R˙ ψ = K

ei

R

We can solve this for

€

˙ ψ

€

˙ ψ = ω0 exp −K 2

CRt

⎛

⎝ ⎜

⎞

⎠ ⎟+

K

CRexp −

K 2

CRt − u( )

⎛

⎝ ⎜

⎞

⎠ ⎟ei u( )du

0

t

∫

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To visualize this let K, R, C = 1 and suppose the input voltage to be sinusoidal

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Let’s look at a simple control problem: find ei to move y from 0 to π

Define a state

€

x =ψ˙ ψ

⎧ ⎨ ⎩

⎫ ⎬ ⎭=

q

u

⎧ ⎨ ⎩

⎫ ⎬ ⎭

Write the state equations, which are linear

€

˙ x =0 1

0 −K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

0K

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ei

Define an error vector

€

e = x − xd( ) =ψ − π

˙ ψ

⎧ ⎨ ⎩

⎫ ⎬ ⎭

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The error equations are

€

˙ e =0 1

0 −K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪xd + e( ) +

0K

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ei

The xd term does not actually appear in the equation because the first column of A is empty

€

˙ e =0 1

0 −K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

π

0

⎧ ⎨ ⎩

⎫ ⎬ ⎭+

0 1

0 −K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪e +

0K

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ei

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So we have

€

˙ e =0 1

0 −K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪e +

0K

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪ei = Ae + Bei

Controllability

€

W = B AB{ } =0

K

CRK

CR−

K 3

C2R2

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

which is obviously of full rank (square with nonzero determinant)

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We can define a gain matrix, here a 1 x 2

€

G = g1 g2{ }

and write

€

ei = G.x = − g1 g2{ }e1

e2

⎧ ⎨ ⎩

⎫ ⎬ ⎭

The controlled (closed loop) equations

€

˙ e =0 1

−g1

K

CR−g2

K

CR−

K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪e

Characteristic polynomial

€

dets −1

g1

K

CRs + g2

K

CR+

K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= 0

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€

dets −1

g1

K

CRs + g2

K

CR+

K 2

CR

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= 0 = s s + g2

K

CR+

K 2

CR

⎛

⎝ ⎜

⎞

⎠ ⎟+ g1

K

CR

In this case we don’t actually need a g2 to stabilize the system —the motor can do that for us — but we can use it to place the eigenvalues

€

g2

K

CR+

K 2

CR= −2ζωn, g1

K

CR= −ωn

2

€

ei = −RC

Kωn

2 ψ − π( ) − 2ζωn

RC

K+ K

⎛

⎝ ⎜

⎞

⎠ ⎟˙ ψ

Substituting gives us a formula for the input voltage

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€

C ˙ ̇ ψ +K 2

R˙ ψ = K

ei

R

€

˙ ̇ ψ = −2ζωn˙ ψ −ωn

2 ψ − π( )

Combining these two equations leads us to the same input voltage

We can look at this from the nonlinear perspective, even though it it a linear problem

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What happens if we replace the torque as the controlling element for the robot with voltage?

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Each torque is given in terms of its motor constants, input voltage and shaft speed

€

τ 01 = K01

e01 − K01ω01

R01

€

τ12 = K12

e12 − K12ω12

R12

€

τ 23 = K23

e23 − K23ω23

R23

I’ll suppose the motors to be identical for convenience’s sake

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We need to figure out the ws and size the motor

€

ω01 = ˙ ψ 1

€

ω12 = ˙ θ 2

€

ω23 = ˙ θ 3 − ˙ θ 2

The motor has to have enough torque to hold the arms out straight

Speed is not a big issue

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The maximum torque required is that needed to hold the arms out straight

€

τMAX = m2gc2 + m3g 2c2 + c3( )

I can set the motor starting torque equal to twice this, from which

€

R =eMAX K

2g c2m2 + 2c2 + c3( )m3( )

Choose K = 1 and eMAX = 100 volts

The no load speed is 100 rad/sec = 955 rpm

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We can see how all this goes in Mathematica