COM-power corporation APF-5060, Instruction Manual

Page: 35 / 40

P
P
a
a
g
g
e
e
3
3
5
5
o
o
f
f
4
4
0
0
I
I
N
N
S
S
T
T
R
R
U
U
C
C
T
T
I
I
O
O
N
N
M
M
A
A
N
N
U
U
A
A
L
L
A
A
P
P
F
F
-
-
5
5
0
0
6
6
0
0
M M A A G G N N E E T T I I C C F F I I E E L L D D G G E E N N E E R R A A T T O O R R
S
S
E
E
C
C
T
T
I
I
O
O
N
N
7
7
- - P P E E R R F F O O R R M M A A N N C C E E O O F F T T E E S S T T S S
1 1 9 9 1 1 2 2 1 1 E E L L T T O O R R O O R R O O A A D D ● ● S S I I L L V V E E R R A A D D O O , , C C A A L L I I F F O O R R N N I I A A 9 9 2 2 6 6 7 7 6 6 ● ● ( ( 9 9 4 4 9 9 ) ) 4 4 5 5 9 9 - - 9 9 6 6 0 0 0 0 ● ● c c o o m m - - p p o o w w e e r r . . c c o o m m
Rev091918
S
S
t
t
e
e
p
p
#
#
7
7
Using Equation 2, calculate the coil factor by dividing the
magnetic field strength measured during Step #3 by the current
measured during Step #4.
E
E
q
q
u
u
a
a
t
t
i
i
o
o
n
n
2
2
-
-
C
C
a
a
l
l
c
c
u
u
l
l
a
a
t
t
i
i
o
o
n
n
o
o
f
f
C
C
o
o
i
i
l
l
F
F
a
a
c
c
t
t
o
o
r
r
(
(
C
C
.
.
F
F
.
.
)
)
EXAMPLE: For a measured field strength of 25.5 A/m and measured coil
current of 30A , the coil factor is:
When testing is performed, the coil factor is entered into the APF-5060
USER test setup menu. The generator uses the entered coil factor to
determine the required output current to generate a given field strength
as shown in Equation 3.
E
E
q
q
u
u
a
a
t
t
i
i
o
o
n
n
3
3
-
-
C
C
a
a
l
l
c
c
u
u
l
l
a
a
t
t
i
i
o
o
n
n
o
o
f
f
G
G
e
e
n
n
e
e
r
r
a
a
t
t
o
o
r
r
O
O
u
u
t
t
p
p
u
u
t
t
/
/
C
C
o
o
i
i
l
l
C
C
u
u
r
r
r
r
e
e
n
n
t
t
EXAMPLE: To generate a 100
A/m magnetic field
using a loop with a
0.85 coil factor , the
required generator
output current, or loop coil current is calculated as follows: