NCP1215
http://onsemi.com
11
The EF16 core for transformer was selected. It has
cross−section area A
e
= 20.1 mm
2
. The N67 magnetic
allows to use maximum operating flux density
B
max
= 0.28 Tesla.
The number of turns of the primary winding is:
(eq. 20)
np
+
Lp · Ippk
B max · Ae
+
4.14 · 10−3 · 0.2047
0.28 · 20.1 · 10−6
+
150 turns
The A
L
factor of the transformer’s core can be calculated:
(eq. 21)
AL
+
Lp
(np)2
+
4.14 · 10−3 ·
(150)2
+
184 nH
For an adapter output voltage of 6.5 V, the number of turns
of the secondary winding can be calculated accounting
Schottky diode for output rectifier as follows:
(eq. 22)
ns
+
(Vs
)
Vfwd)(1
*
d
max)np
d
max · Vbulk− min
+
(6.5
)
0.7)(1
*
0.5)150
0.5 · 127
+
8.5
+
9 turns
The number of turns for auxiliary winding can be
calculated similarly:
(eq. 23)
ns
+
(Vs
)
Vfwd)(1
*
d
max)np
d
max · Vbulk− min
+
(12
)
1)(1
*
0.5)150
0.5 · 127
+
15.35
+
15 turns
The peak primary current is known from initial
calculations. The current sense method allows choosing the
voltage drop across the current sense resistor. Let’s use a
value of 0.5 V. The value of the current sense resistor can
then be evaluated as follows:
(eq. 24)
RCS
+
VCS
Ippk
+
0.5
0.2047
+
2.442
W
+
2.7
W
The voltage drop across the sense resistor needs to be
recalculated:
(eq. 25)
VCS
+
RCS · Ippk
+
2.7 · 0.2047
+
0.553 V
Using the above results the value of the shift resistor is:
(eq. 26)
Rshift
+
VCS
ICS
+
0.553
50 · 10−6
+
11.06 k
W
+
11 k
W
The value of timing capacitor for the off time control has
to be calculated for minimum bulk capacitor voltage since
at these conditions the converter should be able to deliver
specified maximum output power. The value of the timing
capacitor is then given by the following equation:
(eq. 27)
CT
+
1
fsw
*
Lp · Ippk
Vbulk− min
1.2 · 106
+
1
75 · 103
*
4.14 · 10
*
3 · 0.2047
127
0.12 · 106
+
55.5 pF
+
56 pF
The value of the startup resistor for startup time of 200 ms
and Vcc capacitor of 200 nF is following:
(eq. 28)
Rstartup
+
Vbulk− min
CVcc
Vstartup
tstartup
)
ICC−start MAX
+
127
200 · 10−9
12
0.2
)
10 · 10−6
+
5.77 M
W
+
5.6 M
W
The result of all the calculations is the application
schematic depicted in Figure 21.
