Page 653 - Mechatronics with Experiments

ELECTRIC ACTUATORS: MOTOR AND DRIVE TECHNOLOGY 639
Then,
N ⋅ i
x
H = ( ) (8.153)
x + A x y
g g
2 A y
A x N ⋅ i
H = ( ) (8.154)
y
2A y
x + A x y
g g
2 A y
Another way to look at this is in terms of MMFs and reluctances. The flux due to MMF
and effective reluctance in the circuit,
Φ=Φ = 2 ⋅ Φ (8.155)
x y
MMF N ⋅ i
= = (8.156)
R eqv R + R y,eqv
x
Ni A x
o
= ( ) (8.157)
x + A x y g
g
2 A y
where the effective reluctances
x g
R = (8.158)
x
⋅ A
o x
y g
R = (8.159)
y
⋅ A
o y
and R from the parallel connection of two reluctances,
y,eqv
( ) −1
1 1
R y,eqv = + = R ∕2 (8.160)
y
R y R y
Flux linkage is a function of variable quantities i and x (y is constant)
g
g
(x , i) =Φ ⋅ N = L(x ) ⋅ i (8.161)
g
x
g
2
N ⋅ i
= ( ) A (8.162)
o x
x + A x y g
g
2 A y
N 2
g
o
L(x , i) = ( ) A x (8.163)
x + A x y
g g
2 A y
The co-energy expression as a function of flux linkage and current is
1 1 2
W co = (x , i) ⋅ i = L(x ) ⋅ i (8.164)
g
g
2 2
W (x , i)
g
co
F(x , i) = (8.165)
g
x
g
2
1 N A x 2
o
=− ⋅ i (8.166)
2 ( ) 2
x + A x y
g g
2 A y
The generated force is a function of actuator geometery, A , A , x , y , permeability of gap,
y
x
g
g
, coil turn, N, and current, i.
0