Principles of Electromechanical Energy
Conversion
• Why do we study this?
– Electromechanical energy conversion theory is the
cornerstone for the analysis of electromechanical motion
devices.
– The theory allows us to express the electromagnetic force
or torque in terms of the device variables such as the
currents and the displacement of the mechanical system.
– Since numerous types of electromechanical devices are
used in motion systems, it is desirable to establish methods
of analysis which may be applied to a variety of
electromechanical devices rather than just electric
machines.
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Electromechanical Motion Fundamentals
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87
• Plan
– Establish analytically the relationships which can be used
to express the electromagnetic force or torque.
– Develop a general set of formulas which are applicable to
all electromechanical systems with a single mechanical
input.
– Detailed analysis of:
• Elementary electromagnet
• Elementary single-phase reluctance machine
• Windings in relative motion
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Electromechanical Motion Fundamentals
Kevin Craig
88
Lumped Parameters vs. Distributed Parameters
• If the physical size of a device is small compared to
the wavelength associated with the signal
propagation, the device may be considered lumped
and a lumped (network) model employed.
v
λ=
f
λ= wavelength (distance/cycle)
v = velocity of wave propagation (distance/second)
f = signal frequency (Hz)
• Consider the electrical portion of an audio system:
– 20 to 20,000 Hz is the audio range
186,000 miles/second
λ=
= 9.3 miles/cycle
20,000 cycles/second
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
89
Conservative Force Field
• A force field acting on an object is called
conservative if the work done in moving the object
from one point to another is independent of the path
joining the two points.
r
ˆ
ˆ
ˆ
F = F1i + F2 j + F3k
r
r uur
r
∫ F ⋅ dr is independent of path if and only if ∇× F = 0 or F = ∇φ
C
r uur
F ⋅ dr is an exact differential
Fdx
+ F2 dy + F3dz = dφ where φ (x, y,z)
1
( x2 ,y2 ,z2 ) r uur
( x2 ,y2 ,z2 )
F ⋅ dr = ∫
dφ = φ ( x 2 , y2 , z 2 ) − φ ( x1 , y1 , z1 )
∫
( x1 ,y1 ,z1 )
( x1 ,y1 ,z1 )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
90
Energy Balance Relationships
• Electromechanical System
– Comprises
• Electric system
• Mechanical system
• Means whereby the electric and mechanical systems can interact
– Interactions can take place through any and all
electromagnetic and electrostatic fields which are common
to both systems, and energy is transferred as a result of this
interaction.
– Both electrostatic and electromagnetic coupling fields may
exist simultaneously and the system may have any number
of electric and mechanical subsystems.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
91
• Electromechanical System in Simplified Form:
Electric
System
Coupling
Field
Mechanical
System
– Neglect electromagnetic radiation
– Assume that the electric system operates at a frequency
sufficiently low so that the electric system may be
considered as a lumped-parameter system
WE = We + WeL + WeS
• Energy Distribution
WM = Wm + WmL + WmS
– WE = total energy supplied by the electric source (+)
– WM = total energy supplied by the mechanical source (+)
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Kevin Craig
92
– WeS = energy stored in the electric or magnetic fields which
are not coupled with the mechanical system
– WeL = heat loss associated with the electric system,
excluding the coupling field losses, which occurs due to:
• the resistance of the current-carrying conductors
• the energy dissipated in the form of heat owing to hysteresis, eddy
currents, and dielectric losses external to the coupling field
– We = energy transferred to the coupling field by the electric
system
– WmS = energy stored in the moving member and the
compliances of the mechanical system
– WmL = energy loss of the mechanical system in the form of
heat due to friction
– Wm = energy transferred to the coupling field by the
mechanical system
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
93
• WF = Wf + WfL = total energy transferred to the
coupling field
– Wf = energy stored in the coupling field
– WfL = energy dissipated in the form of heat due to losses
within the coupling field (eddy current, hysteresis, or
dielectric losses)
Wf + WfL = ( WE − WeL − WeS ) +
• Conservation of Energy
( WM − WmL − WmS )
Wf + WfL = We + Wm
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Kevin Craig
94
• The actual process of converting electric energy to
mechanical energy (or vice versa) is independent of:
– The loss of energy in either the electric or the mechanical
systems (WeL and WmL)
– The energies stored in the electric or magnetic fields which
are not in common to both systems (WeS)
– The energies stored in the mechanical system (WmS)
• If the losses of the coupling field are neglected, then
the field is conservative and Wf = We + W m .
• Consider two examples of elementary
electromechanical systems
– Magnetic coupling field
– Electric field as a means of transferring energy
Actuators & Sensors in Mechatronics
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95
v = voltage of electric source
f = externally-applied mechanical
force
fe = electromagnetic or
electrostatic force
r = resistance of the currentcarrying conductor
l = inductance of a linear
(conservative)
electromagnetic system
which does not couple
the mechanical system
M = mass of moveable member
K = spring constant
D = damping coefficient
x0 = zero force or equilibrium
position of the mechanical
system (fe = 0, f = 0)
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Electromechanical Motion Fundamentals
Electromechanical System with Magnetic Field
Electromechanical System with Electric Field
Kevin Craig
96
di
v = ri + l + ef
dt
voltage equation that describes the
electric systems; ef is the voltage drop
due to the coupling field
d2 x
dx
f = M 2 + D + K ( x − x 0 ) − fe
dt
dt
WE = ∫ ( vi ) dt
 dx 
WM = ∫ ( f )dx = ∫  f
dt

 dt 
di
v = ri + l + ef
dt
WE = ∫ ( vi ) dt
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Newton’s Law of Motion
Since power is the time rate of
energy transfer, this is the total
energy supplied by the electric
and mechanical sources
 di 
WE = r ∫ ( i )dt + l ∫  i dt + ∫ ( ef i )dt
 dt 
= WeL + WeS + We
2
Kevin Craig
97
d2 x
dx
f = M 2 + D + K ( x − x 0 ) − fe
dt
dt
 dx 
WM = ∫ ( f )dx = ∫  f
dt

 dt 
d x
 dx 
WM = M ∫  2 dx + D∫   dt + K ∫ ( x − x 0 )dx − ∫ ( f e )dx
 dt 
 dt 
2
2
Σ
WmS
Wf = We + Wm = ∫ ( ef i )dt − ∫ ( f e )dx
Actuators & Sensors in Mechatronics
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WmL
Wm
total energy transferred to
the coupling field
Kevin Craig
98
• These equations may be readily extended to include
an electromechanical system with any number of
electrical and mechanical inputs and any number of
coupling fields.
• We will consider devices with only one mechanical
input, but with possibly multiple electric inputs. In all
cases, however, the multiple electric inputs have a
common coupling field.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
99
J
K
j=1
k=1
Wf = ∑ Wej + ∑ Wmk
J
J
∑ W = ∫ ∑ e i dt
j=1
ej
j=1
K
K
∑W
k =1
fj j
mk
= − ∫ ∑ f ek dx k
k =1
J
Wf = ∫ ∑ efji jdt − ∫ f e dx
j=1
J
dWf = ∑ efji jdt − f edx
j=1
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Total energy supplied to the
coupling field
Total energy supplied to
the coupling field from the
electric inputs
Total energy supplied to
the coupling field from the
mechanical inputs
With one mechanical input
and multiple electric inputs,
the energy supplied to the
coupling field, in both
integral and differential form
Kevin Craig
100
Energy in Coupling Field
• We need to derive an expression for the energy stored
in the coupling field before we can solve for the
electromagnetic force fe.
• We will neglect all losses associated with the electric
or magnetic coupling field, whereupon the field is
assumed to be conservative and the energy stored
therein is a function of the state of the electrical and
mechanical variables and not the manner in which the
variables reached that state.
• This assumption is not as restrictive as it might first
appear.
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Electromechanical Motion Fundamentals
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– The ferromagnetic material is selected and arranged in
laminations so as to minimize the hysteresis and eddy
current losses.
– Nearly all of the energy stored in the coupling field is
stored in the air gap of the electromechanical device. Air is
a conservative medium and all of the energy stored therein
can be returned to the electric or mechanical systems.
• We will take advantage of the conservative field
assumption in developing a mathematical expression
for the field energy. We will fix mathematically the
position of the mechanical system associated with the
coupling field and then excite the electric system with
the displacement of the mechanical system held fixed.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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• During the excitation of the electric inputs, dx = 0,
hence, Wm is zero even though electromagnetic and
electrostatic forces occur.
• Therefore, with the displacement held fixed, the
energy stored in the coupling field during the
excitation of the electric inputs is equal to the energy
supplied to the coupling field by the electric inputs.
• With dx = 0, the energy supplied from the electric
0
J
system is:
Wf = ∫ ∑ efji jdt − ∫ f e dx
j=1
J
Wf = ∫ ∑ efji jdt
j=1
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Kevin Craig
103
• For a singly excited electromagnetic system:
dλ
ef =
dt
Wf = ∫ ( i )dλ with dx = 0
Wf = ∫ ( i )dλ
Area represents energy stored
in the field at the instant
when λ = λa and i = ia.
Graph
Stored energy and coenergy in
a magnetic field of a singly
excited electromagnetic
device
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
For a linear magnetic system:
Curve is a straight line and
1
Wf = Wc = λi
2
Wc = ∫ ( λ )di
Area is called
coenergy
λi = Wc + Wf
Kevin Craig
104
• The λi relationship need not be linear, it need only be
single-valued, a property which is characteristic to a
conservative or lossless field.
• Also, since the coupling field is conservative, the
energy stored in the field with λ = λa and i = ia is
independent of the excursion of the electrical and
mechanical variables before reaching this state.
• The displacement x defines completely the influence
of the mechanical system upon the coupling field;
however, since λ and i are related, only one is needed
in addition to x in order to describe the state of the
electromechanical system.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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105
• If i and x are selected as the independent variables, it
is convenient to express the field energy and the flux
linkages as Wf = Wf ( i,x )
λ = λ ( i, x )
∂λ ( i, x )
∂λ (i,x )
dλ =
di +
dx
∂i
∂x
∂λ ( i, x )
dλ =
di with dx = 0
∂i
i ∂λ ( ξ, x )
∂λ ( i, x )
Wf = ∫ ( i )dλ = ∫ i
di = ∫ ξ
dξ
0
∂i
∂ξ
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
Energy stored
in the field of a
singly excited
system
Kevin Craig
106
• The coenergy in terms of i and x may be evaluated as
Wc ( i, x ) = ∫ λ ( i, x )di = ∫ λ ( ξ, x )dξ
i
0
• For a linear electromagnetic system, the λi plots are
straightline relationships. Thus, for the singly excited
magnetically linear system, λ ( i, x ) = L ( x ) i , where
L(x) is the inductance.
• Let’s evaluate Wf(i,x). dλ = ∂λ ( i, x ) di with dx = 0
∂i
dλ =L ( x ) di
1
Wf ( i,x ) = ∫ ξL ( x )dξ = L ( x ) i 2
0
2
i
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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107
• The field energy is a state function and the expression
describing the field energy in terms of the state
variables is valid regardless of the variations in the
system variables.
• Wf expresses the field energy regardless of the
variations in L(x) and i. The fixing of the mechanical
system so as to obtain an expression for the field
energy is a mathematical convenience and not a
restriction upon the result.
1
Wf ( i,x ) = ∫ ξL ( x )dξ = L ( x ) i 2
0
2
i
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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108
• In the case of a multiexcited electromagnetic system,
an expression for the field energy may be obtained by
evaluating the following relation with dx = 0:
J
Wf = ∫ ∑ i jdλ j
j=1
• Since the coupling field is considered conservative,
this expression may be evaluated independent of the
order in which the flux linkages or currents are
brought to their final values.
• Let’s consider a doubly excited electric system with
one mechanical input.
Wf ( i1 ,i 2 , x ) = ∫ i1dλ1 ( i1 ,i 2 , x ) + i 2 dλ 2 ( i1 ,i 2 , x ) 
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
with dx = 0
Kevin Craig
109
• The result is:
Wf ( i1 ,i 2 , x ) = ∫
i1
0
∫
i2
0
∂λ1 ( ξ, 0, x )
ξ
dξ +
∂ξ
 ∂λ1 ( i1 , ξ, x )
∂λ 2 ( i1 , ξ, x ) 
+ξ
i1
dξ
∂ξ
∂ξ


• The first integral results from the first step of the
evaluation with i1 as the variable of integration and
with i2 = 0 and di2 = 0. The second integral comes
from the second step of the evaluation with i1 equal to
its final value (di1 = 0) and i2 as the variable of
integration. The order of allowing the currents to
reach their final state is irrelevant.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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110
• Let’s now evaluate the energy stored in a
magnetically linear electromechanical system with
two electrical inputs and one mechanical input.
λ1 ( i1 ,i 2 , x ) = L11 ( x ) i1 + L12 ( x ) i 2
λ 2 ( i1 ,i 2 , x ) = L 21 ( x ) i1 + L 22 ( x ) i 2
• The self-inductances L11(x) and L22(x) include the
leakage inductances.
• With the mechanical displacement held constant (dx
= 0):
dλ1 ( i1 ,i 2 , x ) = L11 ( x ) di1 + L12 ( x ) di 2
dλ 2 ( i1 ,i 2 , x ) = L21 ( x ) di1 + L22 ( x ) di 2
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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• Substitution into:
Wf ( i1 ,i 2 , x ) = ∫
i1
0
∫
i2
0
∂λ1 ( ξ, 0, x )
ξ
dξ +
∂ξ
 ∂λ1 ( i1 , ξ, x )
∂λ 2 ( i1 , ξ, x ) 
+ξ
i1
dξ
∂ξ
∂ξ


• Yields:
Wf ( i1 ,i 2 , x ) = ∫ ξL11 ( x ) d ξ + ∫ i1L12 ( x ) + ξL 22 ( x )dξ
0
0
1
1
2
= L11 ( x ) i1 + L12 ( x ) i1i 2 + L22 ( x ) i 22
2
2
i1
Actuators & Sensors in Mechatronics
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i2
Kevin Craig
112
• It follows that the total field energy of a linear
electromagnetic system with J electric inputs may be
expressed as:
1 J J
Wf ( i1 ,K ,i j , x ) = ∑∑L pqi p iq
2 p=1 q=1
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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113
Electromagnetic and Electrostatic Forces
• Energy Balance Equation:
J
Wf = ∫ ∑ efji jdt − ∫ f e dx
j=1
J
dWf = ∑ efji jdt − f edx
j=1
J
f e dx = ∑ efji jdt − dWf
j=1
• To obtain an expression for fe, it is first necessary to
express Wf and then take its total derivative. The total
differential of the field energy is required here.
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Electromechanical Motion Fundamentals
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114
• The force or torque in any electromechanical system
may be evaluated by employing: dWf = dWe + dWm
• We will derive the force equations for electromechanical systems with one mechanical input and J
electrical inputs.
J
• For an electromagnetic system: f e dx = ∑ i jdλ j − dWf
j =1
• Select ij and x as independent variables: W = W ( ri , x )
r
r
 ∂Wf i , x
J  ∂W
f i,x
dWf = ∑ 
di j  +
dx
∂i j
∂x

j=1 


r
r
 ∂λ j i , x
J  ∂λ
j i,x
dλ j = ∑ 
di n  +
dx
∂i n
∂x

n =1 


( )
( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
f
f
r
λj = λj i,x
( )
( )
Kevin Craig
115
• The summation index n is used so as to avoid
confusion with the subscript j since each dλj must be
evaluated for changes in all currents to account for
mutual coupling between electric systems.
• Substitution:
r
r
 ∂Wf i , x
J  ∂W
f i,x
dWf = ∑ 
di j  +
dx
∂i j
∂x

j=1 


r
r
 ∂λ j i , x
J  ∂λ
j i,x
dλ j = ∑ 
di n  +
dx
∂i n
∂x

n =1 


( )
( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
( )
into
J
f e dx = ∑ i jdλ j − dWf
j =1
Kevin Craig
116
r
r
 J  ∂λj i , x

 ∂λ j i , x
J
r
 


f e i , x dx = ∑ i j ∑
di n +
dx 
∂x

j=1
 n =1  ∂i n


r
r
 ∂Wf i , x
J  ∂W
f i,x
−∑ 
di j  +
dx
∂i j
∂x

j=1 


r
r

 ∂Wf i , x 
J  ∂λ
i
,
x
r
j


−
f e i , x dx = ∑ i j
 dx
∂x
∂x

 j=1 


r
r


 ∂Wf i , x
J 
 J  ∂λ j i , x


+ ∑ i j ∑
di n −
di j 
∂i n
∂i j

j =1  n =1 




• Result:
( )
( )
( )
( )
( )
( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
( )
( )
( )
Kevin Craig
117
• This equation is satisfied provided that:
r
r
 ∂Wf i , x
J  ∂λ
r
j i,x
−
f e i , x = ∑ i j
∂x
∂x

j=1 


r
r


 ∂Wf i , x
J 
 J  ∂λ j i , x


0 = ∑ i j ∑
di n −
di j 
∂i n
∂i j

j=1  n=1 




( )
( )
( )
( )
( )
• The first equation can be used to evaluate the force on
the mechanical system with i and x selected as
independent variables.
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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118
• We can incorporate an expression for coenergy and
J
obtain a second force equation:
Wc = ∑ i jλ j − Wf
j =1
• Since i and x are independent variables, the partial
derivative with respect to x is:
r
∂Wc i , x
( )
∂x
• Substitution:
r
r
 ∂Wf i , x
J  ∂λ
j i,x
−
= ∑ i j
∂x
∂x

j=1 


( )
( )
r
r
r
 ∂Wf i , x
J  ∂λ
∂Wc i , x
r
j i,x
−
f e i , x = ∑ i j
=
∂x 
∂x
∂x
j =1 


( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
( )
( )
Kevin Craig
119
• Note:
– Positive fe and positive dx are in the same direction
– If the magnetic system is linear, Wc = Wf.
• Summary:
r
r
 ∂Wf i , x
J  ∂λ
r
j i,x
−
f e i , x = ∑ i j
∂x
∂x

j=1 


r
∂Wc i , x
r
r
J  ∂λ
fe i , x =
r
j i, θ
∂x
Te i , θ = ∑ i j
∂θ
j=1 

r
fe
Te
∂Wc i , θ
r
Te i , θ =
x
θ
∂θ
( )
( )
( )
( )
( )
( )
( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
( )
r
 ∂Wf i, θ
−
∂θ


( )
Kevin Craig
120
• By a similar procedure, force equations may be derived
with flux linkages λ1, …, λj of the J windings and x as
independent variables. The relations, given without
proof, are:
r
r
( )
( )
J 
∂i j λ , x  ∂Wc λ, x
r
+
f e λ , x = − ∑ λ j
∂x 
∂x
j=1 


r
∂Wf λ, x
r
fe λ, x = −
r
∂x
J 
∂i j λ, θ
r
Te λ , θ = − ∑ λ j
∂θ
j=1 

r
∂Wf λ, θ
r
Te λ , θ = −
∂θ
( )
( )
( )
( )
( )
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
( )
( )
r
 ∂Wc λ, θ
+
∂θ


( )
Kevin Craig
121
• One may prefer to determine the electromagnetic
force or torque by starting with the relationship
dWf = dWe + dWm
rather than by selecting a formula.
• Example:
– Given:
λ = 1 + a ( x )  i 2
– Find fe(i,x)
Actuators & Sensors in Mechatronics
Electromechanical Motion Fundamentals
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122
Elementary Electromagnet
• The system consists of:
– stationary core with a winding of N turns
– block of magnetic material is free to slide relative to the
stationary member
x = x(t)
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dλ
v = ri +
voltage equation that describes the electric system
dt
λ = Nφ
flux linkages
φ = φl + φm
(the magnetizing flux is common to
φl = leakage flux
both stationary and rotating members)
φm = magnetizing flux
Ni
φl =
ℜl
Ni
φm =
ℜm
If the magnetic system is considered to be
linear (saturation neglected), then, as in the
case of stationary coupled circuits, we can
express the fluxes in terms of reluctances.
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124
 N2 N2 
λ =
+
i

 ℜl ℜm 
= ( Ll + Lm ) i
ℜm = ℜi + 2ℜg
ℜi
ℜi =
li
µ riµ 0 A i
x
ℜg =
µ0 A g
ℜg
flux linkages
Ll = leakage inductance
L m = magnetizing inductance
reluctance of the magnetizing path
total reluctance of the magnetic material
of the stationary and movable members
reluctance of one of the air gaps
Assume that the cross-sectional areas of
the stationary and movable members are
equal and of the same material
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Ag = Ai
This may be somewhat of an oversimplification,
but it is sufficient for our purposes.
ℜm = ℜi + 2ℜg

1  li
=
+ 2x 

µ 0 Ai  µ ri

2
N
Lm =

1  li
+ 2x 

µ 0 Ai  µ ri

Assume that the leakage inductance
is constant.
The magnetizing inductance is
clearly a function of displacement.
x = x(t) and Lm = Lm(x)
When dealing with linear magnetic circuits wherein mechanical
motion is not present, as in the case of a transformer, the change
of flux linkages with respect to time was simply L(di/dt). This is
not the case here.
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126
λ(i,x) = L(x)i = [ L l + L m (x) ] i
The inductance is a
function of x(t).
dλ (i,x) ∂λ di ∂λ dx
=
+
∂i dt ∂x dt
dt
di dLm (x) dx
v = ri + [ Ll + Lm (x)] + i
dt
dx dt
N2
Lm ( x ) =

1  li
+ 2x 

µ0 Ai  µ ri

k
L m (x) =
k0 + x
N 2µ0 A i
k=
2
li
k0 =
2µ ri
Actuators & Sensors in Mechatronics
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The voltage equation is
a nonlinear differential
equation.
Let’s look at the magnetizing
inductance again.
k N 2µ 0µ ri Ai
L m (0) =
=
k0
li
k
L m (x) ≅
x
for x > 0
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Detailed diagram of electromagnet
for further analysis
Electromagnet
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128
k
L m (x) ≅
x
for x > 0
Use this approximation
k
L ( x ) ≅ Ll + Lm ( x ) = L l +
x
for x > 0
λ(i,x) = L(x)i = [ L l + L m (x) ] i
The system is magnetically linear:
r
r
 ∂Wf i , x
J  ∂λ
r
j i,x
−
f e i , x = ∑ i j
∂x
∂x

j=1 


r
∂Wc i , x
r
fe i , x =
∂x
( )
( )
( )
( )
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1
Wf ( i,x ) = Wc ( i, x ) = L ( x ) i2
2
( )
1 2 ∂L ( x )
f e ( i, x ) = i
2
∂x
ki 2
=− 2
2x
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• The force fe is always negative; it pulls the moving
member to the stationary member. In other words, an
electromagnetic force is set up so as to minimize the
reluctance (maximize the inductance) of the magnetic
system.
• Equations of motion:
di
v = ri + l + ef
dt
d2 x
dx
f = M 2 + D + K ( x − x 0 ) − fe
dt
dt
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Steady-State Operation
(if v and f are constant)
v = ri
f = K ( x − x 0 ) − fe
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Steady-State Operation
of an
Electromagnet
f = K ( x − x0 ) − fe
−f e = f − K ( x − x 0 )
 ki 
−  − 2  = f − K (x − x0 )
 2x 
Parameters:
r = 10 Ω
K = 2667 N/m
x0 = 3 mm
k = 6.283E-5 H m
v=5V
i = 0.5 A
2
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Stable Operation: points 1 and 2
Unstable Operation: points 1´ and 2´
(f = 4 N)
(f = 0)
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Single-Phase Reluctance Machine
• The machine consists of:
– stationary core with a
winding of N turns
– moveable member which
rotates
θr = angular displacement
ωr = angular velocity
θr = ∫ ωr ( ξ ) dξ + θr ( 0 )
t
0
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dλ
v = ri +
dt
voltage equation
φ = φl + φm
φl = leakage flux
φm = magnetizing flux
λ = ( Ll + Lm ) i
It is convenient to express the flux
linkages as the product of the sum of the
leakage inductance and the magnetizing
inductance and the current in the winding.
L l = constant (independent of θr )
L m = periodic function of θ r
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Lm = Lm ( θr )
N2
L m (0) =
ℜm ( 0 )
 π
Lm   =
2
N2
 π
ℜm  
2
ℜm is maximum
L m is minimum
ℜm is minimum
L m is maximum
The magnetizing inductance varies between maximum and
minimum positive values twice per revolution of the rotating
member.
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Assume that this variation may
be adequately approximated
by a sinusoidal function.
L m ( θr ) = L A − L B cos ( 2θr )
L m ( 0 ) = LA − LB
L ( θr ) = L l + L m ( θ r )
= Ll + L A − L B cos ( 2θr )
di dLm (θr ) d θr
v = ri + [ Ll + L m (θ r ) ] + i
dt
dθr
dt
Actuators & Sensors in Mechatronics
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 π
Lm   = LA + LB
2
LA > LB
L A = average value
voltage equation
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• This elementary two-pole single-phase reluctance
machine is shown in a slightly different form.
Winding 1 is now winding as and the stator has been
changed to depict more accurately the configuration
common for this device.
dλ as
v as = rsi as +
dt
λ as = L asasi as
L asas = L ls + L A − L B cos ( 2θr )
θ r = ∫ ω r ( ξ )dξ + θ r ( 0 )
t
rs = resistance of as winding
Lasas = self-inductance of as winding
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0
Lls = leakage inductance
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• Electromagnetic torque:
– Magnetic system is linear, hence Wf = Wc.
1
Wc ( i as , θ r ) = ( Lls + L A − L B cos ( 2θ r ) ) i as2
2
r
r
 ∂Wf i, θ
J  ∂λ
r
j i, θ
−
Te i , θ = ∑ i j
∂θ 
∂θ
j=1 


r
∂Wc i , θ
r
Te i , θ =
∂θ
( )
( )
( )
( )
( )
Te ( i as , θ r ) = L Bi as2 sin ( 2θr )
Valid for both transient and steady-state operation
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• Consider steady-state operation: ias is constant
Te = K sin ( 2θr )
K = L Bi as2
stable operating point
unstable operating point
Electromagnetic torque versus angular displacement of a
single-phase reluctance machine with constant stator current
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138
• Although the operation of a single-phase reluctance
machine with a constant current is impracticable, it
provides a basic understanding of reluctance torque,
which is the operating principle of variable-reluctance
stepper motors.
• In its simplest form, a variable-reluctance stepper
motor consists of three cascaded, single-phase
reluctance motors with rotors on a common shaft and
arranged so that their minimum reluctance paths are
displaced from each other.
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Windings in Relative Motion
• The rotational device shown will be used to illustrate
windings in relative motion.
Winding 1: N1 turns on stator
Winding 2: N2 turns on rotor
Assume that the turns are
concentrated in one position.
Air-gap size is
exaggerated.
end view
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Electromechanical Motion Fundamentals
cross-sectional view
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140
dλ1
v1 = r1i1 +
dt
dλ 2
v 2 = r2i 2 +
dt
λ1 = L11i i + L12i 2
λ 2 = L 21i i + L22 i 2
L11 = L l1 + L m1
N12 N12
=
+
ℜl1 ℜm
L 22 = L l 2 + L m2
=
2
2
2
2
N
N
+
ℜl 2 ℜ m
voltage equations
The magnetic system is assumed linear.
The self-inductances L11 and L22 are
constants and may be expressed in
terms of leakage and magnetizing
inductances.
ℜm is the reluctance of the complete
magnetic path of ϕm1 and ϕm2 , which
is through the rotor and stator iron and
twice across the air gap.
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Let’s now consider L12.
θr = angular displacement
ωr = angular velocity
θr = ∫ ωr ( ξ ) dξ + θr ( 0 )
t
0
When θr is zero, then the coupling between
windings 1 and 2 is maximum. The magnetic
system of winding 1 aids that of winding 2
with positive currents assumed. Hence the
mutual inductance is positive.
N1N 2
L12 ( 0 ) =
ℜm
When θr is π/2, the windings are orthogonal.
The mutual coupling is zero.
π
L12   = 0
2
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Assume that the mutual
inductance may be adequately
predicted by:
L12 ( θr ) = L sr cos ( θr )
N1 N2
Lsr =
ℜm
Lsr is the amplitude of the
dλ1
v1 = r1i1 +
sinusoidal mutual inductance
dt
between the stator and rotor
dλ 2
v 2 = r2i 2 +
windings.
dt
In writing the voltage equations, the
λ1 = L11i1 + ( L sr cos θr ) i 2
total derivative of the flux linkages is
λ 2 = L 22i 2 + ( Lsr cos θr ) i1
required.
di1
di 2
v1 = r1i1 + L11
+ Lsr cos θr
− i 2ωr Lsr sin θ r
dt
dt
di 2
di1
v 2 = r2i 2 + L22
+ Lsr cos θr
− i1ωr Lsr sin θr
dt
dt
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 v1  r1 0   i1  d  λ1 
 v  =  0 r  i  + dt λ 
 2 
2 2
 2
 λ1   Ll1 + L m1 Lsr cos θr   ias 
λ  =  L cos θ L + L  i 
 2   sr
r
l2
m2   bs 
Since the magnetic system is assumed to be linear:
1
1
2
Wf ( i1 ,i 2 , θr ) = L11i1 + L12i1i 2 + L22i 22 = Wc ( i1 ,i 2 , θr )
2
2
r
r
 ∂Wf i, θ
J  ∂λ
r
j i, θ
−
Te i , θ = ∑ i j
∂θ 
∂θ
j=1 
Te ( i1 ,i 2 , θ r ) = −i1 i2 Lsr sin θ r


r
∂Wc i , θ
r
Te i , θ =
∂θ
( )
( )
( )
( )
( )
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• Consider the case where i1 and i2 are both positive
Te = −K sin θ r
and constant:
K = i1 i2 Lsr
stable operation
unstable operation
Electromagnetic torque versus angular displacement with constant winding currents
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• Although operation with constant winding currents is
somewhat impracticable, it does illustrate the
principle of positioning of stepper motors with a
permanent-magnet rotor which, in many respects, is
analogous to holding i2 constant on the elementary
device considered here.
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