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AN1889 датащи(PDF) 12 Page - STMicroelectronics

номер детали AN1889
подробное описание детали  The need to choose a high value of the fly-back voltage
PDF  33 Pages
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производитель  STMICROELECTRONICS [STMicroelectronics]
домашняя страница  http://www.st.com
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AN1889 датащи(HTML) 12 Page - STMicroelectronics

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Application description
AN1889
12/33
Equation 13
Before carrying on with the wire dimension calculation, we can preliminary verify if the
selected core is acceptable from a thermal point of view during its normal operation.
Through the Ampere's law we can calculate the B field in normal operation:
Equation 14
Being
µfe the magnetic permeability in the iron much higher than mo (the magnetic
permeability in the air) we can neglect the last item in the Equation 14:
Equation 15
From the ferrite datasheet the power dissipation is 150 mW/cm3, knowing that the total
volume of the core is about 5 cm3 the total power dissipation in the core is about
Pfe = 750 mW.
The thermal resistance of the selected core is 24oC/W, knowing that
Equation 16
This value allows the designers to operate in safe condition since we have to add also the
losses due to the winding with a further consequent increase in temperature.
The next step is the wire dimension calculation, from both primary and secondary side. First
we must assume the maximum losses on the copper. Normal practice is to choose a
dissipation level comparable with the one achieved in the core, practically we can chose 1W.
From the Joule's law we can calculate the resistance of both primary and secondary
winding.
Equation 17
Equation 18
From that, knowing the copper resistivity at 100 oC (=2.303 10-6
Ω cm), and the average
wind length Lt (Lt = 4.1cm), we can easily calculate the wire sections (in cm
2).
Equation 19
N
P
N
Aux
-------------
V
fl
V
Aux
V
F diode
,
+
-------------------------------------------
500
15
1
+
----------------
31
N
Aux
5
=
⇒
=
=
=
N
pIp
Hdl
H
ili
H
glg
H
felfe
B
µ
0
-----l
g
B
µ
fe
-------l
fe
+
=
+
=
∑
=
∫°
=
N
PIP
B
µ
0
-----l
g
B
170mT
=
⇒
=
P
fe
∆T
R
th
--------
∆T18° C
=
⇒
=
R
P
P
CU
2I
2
PMRS
----------------------
R
P
3.46
Ω
=
⇒
=
R
s
P
CU
2I
2
SMRS
----------------------
R
s
0.0174
Ω
=
⇒
=
A
PCU
ρ
100
N
P
•
L
t
•
R
P
------------------------------------
4.08
10
4
–
d
p
0.022 cm
[]
=
⇒
•
=
=



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