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This lecture was delivered by Prof. Azhar Ali for Power Electronics course at Bengal Engineering and Science University. It includes: Inductor, Design, Constraints, Procedure, Winding, Magnetics, Filter, Ccm, Buck, Converter, Geometry, Magnetic
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Chapter 14: Inductor design
1
g^
Chapter 14: Inductor design
2
cu
rms
2
Objective:Design inductor having a given inductance
which carries worst-case current
max
without saturating,
and which has a given winding resistance
, or, equivalently,
exhibits a worst-case copper loss of
L R
i(
t) +–
L
i( t)^
i( t)
t
0
DT
s^
T^ s
I^
∆i
L
Example
: filter inductor in CCM buck converter
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Chapter 14: Inductor design
4
Given a peak winding current
max
, it is desired to operate the core flux
density at a peak value
max
The value of
max
is chosen to be less
than the worst-case saturation flux density
sat
of the core material.
From solution of magnetic circuit:Let
max
and
max
This is constraint #1. The turns ratio
n
and air gap length
l
g^
are
unknown.
ni
Rc
g
nI
max
=
B
max
A
c^
R
g^
=
B
max
lg μ
0
Chapter 14: Inductor design
5
Must obtain specified inductance
. We know that the inductance is
This is constraint #2. The turns ratio
n
, core area
, and air gap length c
l g
are unknown.
L
=
n
2 R
g
=
μ
0 A
c^
n
2
l g
Chapter 14: Inductor design
7
The window utilization factor
K
u
u^
is the fraction of the core window area that is filled by copper Mechanisms that cause
u^
to be less than 1:
Round wire does not pack perfectly, which reduces
u^
by a
factor of 0.7 to 0.55 depending on winding technique
-^
Insulation reduces
u^
by a factor of 0.95 to 0.65, depending on
wire size and type of insulation
-^
Bobbin uses some window area
-^
Additional insulation may be required between windings
Typical values of
u^
0.5 for simple low-voltage inductor0.25 to 0.3 for off-line transformer0.05 to 0.2 for high-voltage transformer (multiple kV)0.65 for low-voltage foil-winding inductor
Chapter 14: Inductor design
8
14.1.4 Winding resistance
The resistance of the winding iswhere
is the resistivity of the conductor material,
l
b^
is the length of
the wire, and
W
is the wire bare area. The resistivity of copper at
room temperature is
-cm
. The length of the wire comprising
an
n
-turn winding can be expressed as
where
is the mean-length-per-turn of the winding. The mean-
length-per-turn is a function of the core geometry. The aboveequations can be combined to obtain the fourth constraint:
R
=
ρ
n
(
MLT
)
A
W
R
=
ρ
l b A
W
l b^
=
n
(
MLT
)
Chapter 14: Inductor design
10
Core geometrical constant
K
g
A
2 c W
A
( MLT
)^
≥
ρ
L
2 I
2 max
B
2 max
RK
u
Elimination of
n
l g
, and
W
leads to
Right-hand side: specifications or other known quantities
-^
Left-hand side: function of only core geometry
So we must choose a core whose geometry satisfies the aboveequation.The core geometrical constant
g^
is defined as
K
g
=
A
2 c W
A
( MLT
)
Chapter 14: Inductor design
11
Discussion
K
g
=
A
2 c W
A
( MLT
)^
≥
ρ
L
2 I
2 max
B
2 max
RK
u
g^
is a figure-of-merit that describes the effective electrical size of magnetic cores, in applications where the following quantities are specified:
Copper loss
-^
Maximum flux density
How specifications affect the core size:
A smaller core can be used by increasing
max
use core material having higher
sat
allow more copper loss
How the core geometry affects electrical capabilities:
A larger
g^
can be obtained by increase of
c^
more iron core material, or
A^
larger window and more copper
Chapter 14: Inductor design
13
Determine core size
g
2 max
2 max
u
(^8)
(^5)
Choose a core which is large enough to satisfy this inequality
(see Appendix D for magnetics design tables). Note the values of
, c
and
for this core.
Chapter 14: Inductor design
14
with
c^
expressed in
cm
μ
0
H/m.
The air gap length is given in meters.The value expressed above is approximate, and neglects fringing flux
and other nonidealities.
l g^
=
μ
0 LI
2 max
B
2 max
A
c
10
(^4)
(m)
Chapter 14: Inductor design
16
max
max
c
(^4)
Chapter 14: Inductor design
17
A
W
≤
K
Wu
A n
(cm
2 )
Select wire with bare copper area
W
less than or equal to this value.
An American Wire Gauge table is included in Appendix D.As a check, the winding resistance can be computed:
R
=
ρ
n
(
MLT
)
A
w
(Ω
)
Chapter 14: Inductor design
19
n^1
: n
2 : n
k
rms current
I^1
rms current
I^2
rms current
I^ k
1
2
k
Core
Window area
W
A
Core mean lengthper turn
(MLT)
Wire resistivity
ρ
Fill factor
K
u
Given:
application with
k
windings
having known rms currents anddesired turns ratios
how should the window area
A^
be allocated among
the windings?
Chapter 14: Inductor design
20
Total windowarea
W
A
Winding
1
allocation
α
W 1
A
Winding
2
allocation
α
W 2
A etc.
0 <
α
j^
< 1
α
1
α
2
α
k^
= 1
