Transcript Document

Magnetics Design
Primary Constraints:
Peak Flux Density (B field) in the core : Bmax (T or Wb/m2)
Core losses
Saturation
Peal Current density in the windings : Jmax (A/m2)
Resistive losses
Heat
Wire with cross section Acond , carrying
Irms amperes of current, has current density :
Wire Cross Section:
Acond (m2)
If we have a limit on J < Jmax, then we
must choose a wire gauge with :
J
I rms
Acond
Acond
I rms

J max
Core Window Area: AW
(window through which all windings must pass)
Torroid
AW
E Cores
N Turns
of wire
with
cross
section
Acond
AW
Fill Factor kw, the fractional part of
the window actually occupied by
conductor cross sections.
kw Aw  NAcond
Applying our previous constraint to
Acond,
NI rms
Aw 
kw J max
For transformers with multiple windings, the different
windings must accommodate different currents, thus they
must have different gauges (different Acond):
kw Aw  N1 Acond ,1  N 2 Acond ,2 
  N y Acond , y
y
However, maximum current density must not be exceeded for
any winding, thus:
Aw 
N
I
y rms , y
y
k w J max
Inductor Core Cross Section: Acore
The core cross section must accommodate the
peak induced flux without exceeding the
maximum allowable flux density.
For an inductor, the peak flux is proportional
to peak current and inductance:
Acore
LIˆ

NBmax
Acore 
ˆ
Bmax
ˆ
LI
ˆ 
N
Transformer Core Cross Section: Acore
In a transformer, the flux density may not exceed +/- Bmax over one AC
operating cycle at the worst case operating condition.
Let T be the maximum excitation time for the primary. This will be at most, half the
period, or kc /fs , where kc is parameter reflecting the operating duty cycle of a power
converter, typically 0.5 for worst case flux density.
Faraday’s Law tells us that the
change in magnetizing flux density
will be :
Therefore, the change in flux
density is:
This places a lower limit on the
core cross section:
m 
V pri
N pri
V pri kc
T 
N pri f s
V pri kc
m

 Bmax
Acore N pri f s Acore
Acore 
kcV pri
f s N pri Bmax
**For sinusoidal AC operation, a value of kc = 0.225 is appropriate; Vpri is RMS.
Area Product
We now have expressions for the minimum core window and cross section,
as functions of maximum ratings and operating parameters, which tells us
how big the device must be. These expressions are used to express a useful
magnetic core design parameter we call the Area Product:
ˆ
 LIˆ   NI rms 
LII
rms
Ap  Acore,min A w,min  



NB
k
J
 max   w max  k w Bmax J max
 N y I rms , y  kc  Vy I rms , y
 kcV pri   

y
y
Ap  Acore,min A w,min  



 f s N pri Bmax   k w J max  k w f s Bmax J max


For Inductors
For
Transformers
Quiz
Draw the waveform, and compute the peak and RMS values for a 0.75 A
P-P triangular wave riding on a DC value of 5.0 A.
2
I RMS
2
0.75
2
0.75


 52  
  25  12  25.0469  5.005 5 Amp
3 

I peak  5  0.75 2  5.375 Amp
Core Selection
From vendor data, we select a core with Ap, Aw, and Acore which meet our
minimum requirements.
For a transformer, the number of turns
for each winding is computed using :
For an inductor, the number of turns is
computed using:
The inductor air gap length is computed to
provide the appropriate reluctance in the
magnetic path for the desired inductance
with the chosen number of turns:
Since Ag ~ Acore, and m0 << mcore, the gap
length can be approximated as:


kc
N y  Vy 

A
B
f
 core max s 
LIˆ
N
Acore Bmax
lg 
N 2 m0 Ag
L
 Ag m0 
 lcore 

 Acore mcore 
N 2 m0 Acore
lg 
L
Design Example: Inductor
Design Specs
L=100 mH
IDC = 5 A
I = 0.75 A p-p
Bmax = 0.25 T = 0.25 x 10-6 W/mm2
Jmax = 6.0A/mm2
fs = 100kHz
kw = 0.5
4
ˆ
10
 5.375 5


LII rms
4
Ap 


3587mm
kw Bmax J max  0.5   0.25x10-6   6.0 
Acore 
F2
4

H2
4


11.3

4
2
 5.52   97.44 mm 2
CF 
2
AW   2 D  
   5.5 18  11.3  36.85 mm
 2 
Ap  Acore AW   97.44  36.85   3950 mm 4
LIˆ
5.375 x 104
N

 22 Turns
Bmax Acore  0.25x 106   97.44 
Acond 
lg 
I rms 5.0

 0.83 mm 2
J max 6.0
N m0 Acore

L
2
This would be 18 gauge wire, which is
pretty stiff. The author suggests 5 strands
of 25 gauge (0.162 mm2)
 22 
2
 4 x 10   97.44   0.59 mm
10
104
Design Example: Transformer
Design Specs
v1 = v2 = v3 = 30 v
IRMS = 2.5 A
Bmax = 0.25 T = 0.25 x 10-6 W/mm2
Jmax = 5.0A/mm2
fs = 100kHz
kw = 0.5
kc = 0.5
Ap 
kc  Vy I rms , y
y
k w f s Bmax J max
 1800 mm 4
Acore 
F2
4

H2
 63.9 mm 2
4
CF 
2
AW   2 D  

29.2
mm

 2 




kc
0.5
  10 Turns
N1  N 2  N 3  V 
  30 
-6
5
  63.9   0.25 x 10 10  
 Acore Bmax f s 
Acond 
I rms 2.5

 0.5 mm 2
J max 6.0
This would be 20 gauge wire. The author
suggests 3 strands of 25 gauge (0.162 mm2)