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AN701 Datasheet(PDF) 18 Page - Vishay Siliconix

Part No. AN701
Description  reduce the size of energy storage components
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Maker  VISHAY [Vishay Siliconix]
Homepage  http://www.vishay.com
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AN701 Datasheet(HTML) 18 Page - Vishay Siliconix

 
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AN701
Vishay Siliconix
www.vishay.com
18
Document Number: 70575
16-Jan-01
Appendix B
FLYBACK CONVERTER TRANSFORMER DE
SIGN
The flyback magnetic component is often referred to as a
transformer, but should in fact be viewed as an inductor with
energy storage capability.
Two modes of operation are possible. In discontinuous mode,
all the energy is transferred to the output(s). In continuous
mode, some energy remains in the inductor. Discontinuous
mode provides the advantages of a smaller inductor and the
absence of unexpected or difficult transfer modes in the control
mechanism. As energy transfer is complete, furthermore,
rectifier currents always fall to zero before reverse voltages are
applied. The result is lower rectifier switching losses.
For the sake of simplicity, it is recommended that only highly
experienced
designers
use
continuous
mode, further
information on which can be reviewed in some of the
publications listed in the reference section of this application
note. In this appendix, only the discontinuous mode is
explained.
Due to the discontinuous nature of the energy transfer
mechanism, flyback inductors are usually larger than the
transformers encountered in forward and other Buck
topologies, where only voltage transformation—and no
energy storage—is performed.
To design a Flyback Inductor, the following information must be
known:
Fsw
operating switching frequency (usually 20 to
500 kHz)
Dmax
maximum duty cycle (usually 50% in
Vishay Siliconix products)
n
target efficiency (usually 0.75 to 0.85)
VMIN
minimum input voltage used
POUT
total output power
VOUT
output voltage(s) required.
In this case, a design for a 24-V (16- to 30-V), 5-V, 2-A (10 W)
inductor will be demonstrated.
The first step is to determine the minimum primary inductance
from:
L
pmin +
V
in min xton max
2
xfsw x h
2x Pout
L
pmin +
(18 IV x 2
ms)
2
x250 kHzx0.75
2x 10 W
L
pmin + 12.2 mH [ 12 mH
The next step is to determine the peak primary current. For a
given inductor, the applied voltage will be:
V
+ L
di
dt
Re-arranging yields:
i
pk +
V
in min xton max
Lpmax
i
pk +
18 V x 2
ms
12
mH
+ 3A
The R.M.S. value for a triangular waveform can be used to
calculate the power that will be dissipated in the MOSFET:
ims
+ i
pk x
dmax
3
ims + 3A x
0.5
3 + 1.22 A [ 1.2 A
To calculate the number of turns required for the 5-V output the
following quotation can be used:
N
S + Np
V
O ) VF
(1–Dmax)
V
inmin Dmax
in this case, for Dmax = 0.5 and VF = 0.6 V for a 5 V output, the
secondary turns will be:
N
S + Np
(5V
) 0.6 V)(1–0.5)
16 V x 0.5


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