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DFS06M 数据表(PDF) 36 Page - STMicroelectronics

部件名 DFS06M
功能描述  APPLICATION NOTE
PDF  42 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

DFS06M 数据表(HTML) 36 Page - STMicroelectronics

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AN1262 APPLICATION NOTE
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Flyback transformer design
Eqn. 4 gives the primary inductance (Lp = 1.37 mH, rounded up to 1.4 mH), while eqn. (5) gives the primary-to-
secondary turns ratio (n = 21.4). The design will be done considering Philip's E-cores in 3C85 ferrite and as-
suming a maximum peak flux of 0.25T, a temperature rise of 40 °C and 40% window utilization factor. Going
step-by-step:
1) Eqn. 6 provides a minimum AP of 0.042 cm
4. Table 10 shows that an E20/10/6 core could fit the design.
2) The primary turns number will be Npmin = 122.5.
3) The resulting secondary turn number will be 122.5/21.4=5.7 which will rounded up to 6. The primary turns
number will then become 6·21.4=128.4. Finally, the choice will be Np=128 turns and Ns=6 turns, which
yields an actual turns ratio of 128/6 = 21.33, very close to the target.
4) From eqn. 7, the air gap needed to get the desired value of Lp will be 0.63 mm.
5) Table 10 shows that the thermal resistance of the finished core is 46 °C/W, thus the maximum power dissi-
pation inside the transformer shall not exceed 40/46 = 0.87 W.
6) Equations 8, 9 and 10 will provide the actual flux swing (which will be lower than 0.25 T because Np>Npmin),
the actual core losses and the allowed copper losses respectively. The resulting flux swing is
∆B=180 mT:
the relevant core losses amount at 66 mW, thus it is possible to dissipate up to 0.8 W in the windings.
7) The required primary and secondary winding resistance will be 8.65
Ω and 30 mΩ respectively (resulting
from eqns. 11). The resulting primary resistance is quite high and the drop across it reduces significantly the
actual voltage applied at the primary inductance. The target primary resistance is then reduced at 4
Ω and
the secondary will be increased at 46m
Ω to maintain the same total copper losses.
The required primary and secondary copper area will be 2.87·10-4 cm2 and 1.2·10-3 cm2 respectively (eqns.
12, 13). Table 11 shows that this can be done with one AWG32 wire at the primary and four paralleled (twist-
ed) AWG32 wires at the secondary. This will both minimize high frequency effects and simplify the BOM. The
total occupied area will be 7 mm2 (eqn. 14), 20% of the total available area, thus the windings will fit.
On top of the primary and secondary winding, 14 turns of AWG32 wire will be wound to make the auxiliary
winding (eqn. 15).
8) The actual resistance of the primary and secondary windings will be 3.6
Ω and 42 mΩ respectively, for total
copper losses of 0.73 W. The total losses will be about 0.8 W and the resulting temperature rise 36.8 °C.
Zener clamp
To optimize losses at light load a zener clamp will be used. The clamp voltage should be around 200 V (eqn.
16), thus a BZW06-154 is first selected.
Assuming a leakage inductance of 30 µH (about 2% of the primary inductance), power dissipation will be about
0.6 W in normal operation and about 1.1 W in overcurrent limitation. The relevant clamping voltages would be
196 V and 209 V respectively. The initial choice will then be confirmed.
An STTA106 (1A / 600V turboswitch diode) will be used as the blocking diode.
Secondary rectifier
According to eqn. 17, and considering 25% margin, the blocking voltage of the diode should exceed 28 V, while
its current rating should be in excess of 4 A. Although table 14 suggests a bigger device, an 1N5822 (3A/40V)
Schottky diode is selected for this test board.



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