lecture 18. electric motors
DESCRIPTION
Lecture 18. Electric Motors. simple motor equations and their application. 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. - PowerPoint PPT PresentationTRANSCRIPT
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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 = L didt
+ 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 =K ei − 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
τ
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 = K ei
R, ωNL = ei
K
€
K = ei
ωNL
, R = Kei
τ S
= ei2
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 = K ei − KωR
= 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 ⎧ ⎨ ⎩
⎫ ⎬ ⎭=
qu ⎧ ⎨ ⎩
⎫ ⎬ ⎭
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
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪π0 ⎧ ⎨ ⎩
⎫ ⎬ ⎭+
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
C 2R2
⎧
⎨ ⎪
⎩ ⎪
⎫
⎬ ⎪
⎭ ⎪
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
−g1K
CR−g2
KCR
− K 2
CR
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪e
Characteristic polynomial
€
dets −1
g1K
CRs + g2
KCR
+ K 2
CR
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪= 0
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€
dets −1
g1K
CRs + g2
KCR
+ K 2
CR
⎧ ⎨ ⎪
⎩ ⎪
⎫ ⎬ ⎪
⎭ ⎪= 0 = s s + g2
KCR
+ K 2
CR
⎛ ⎝ ⎜
⎞ ⎠ ⎟+ g1
KCR
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
€
g2K
CR+ K 2
CR= −2ζωn, g1
KCR
= −ωn2
€
ei = − RCK
ωn2 ψ − π( ) − 2ζωn
RCK
+ 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 = K01e01 − K01ω01
R01
€
τ12 = K12e12 − K12ω12
R12
€
τ 23 = K23e23 − 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
€
w01 = ˙ ψ 1
€
w12 = ˙ θ 2
€
w23 = ˙ 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 K2g 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