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© 2022 ROHM Co., Ltd.
65UG028E Rev.001
2022.7
User's Guide
BD7F105EFJ-EVK-001
Transformer design-continued
The maximum peak current on the primary side of the IC is determined by I
LIMIT
of electrical characteristics.
I
LIMIT
minimum-value determines the secondary min-peak current I
SPK1(Min)
.
𝐼
𝑆𝑃𝐾1(𝑀𝑖𝑛)
= 𝐼
𝐿𝐼𝑀𝐼𝑇(𝑀𝑖𝑛)
×
𝑁
𝑃
𝑁
𝑆
The secondary peak current I
SPK2(Max)
is calculated from the maximum output current I
OUT(Max)
by the following
equation.
𝐼
𝑆𝑃𝐾2(𝑀𝑎𝑥)
=
2 × 𝐼
𝑂𝑈𝑇(𝑀𝑎𝑥)
(1 − 𝐷
𝑀𝐴𝑋
) × (2 − 𝑘)
×
1
𝜂
𝜂
: Use a power supply efficiency of 70 % as a guideline.
𝐼
𝑂𝑈𝑇(𝑀𝑎𝑥)
: Max. secondary output current (Determined by maximum power of 3 outputs ÷ output voltage 1)
I
SPK2(Max)
< I
SPK1(Min)
must be met in order for I
OUT(Max)
to be printed.
If the conditions cannot be satisfied, change k to redesign. With higher k values in discontinuous mode
The operating load area becomes wider. When k = 1, discontinuous mode operation is performed in all areas.
This IC is continuous
A low k-value is recommended to achieve high-speed response and low EMI characteristics by mode operation.
Even if the k value is high, there is no problem with power supply operation.
The secondary-side index L
S(Max)
is calculated by the following equation.
𝐿
𝑆(𝑀𝑎𝑥)
=
(2 − 𝑘) × (𝑉
𝑂𝑈𝑇
+ 𝑉
𝐹
) × (1 − 𝐷
𝑀𝐴𝑋
)
2
2 × 𝐼
𝑂𝑈𝑇(𝑀𝑎𝑥)
× 𝑓
𝑠𝑤(𝑀𝑎𝑥)
× 𝑘
𝐿
𝑆(𝑀𝑎𝑥)
=
(2 − 0.25) × (6.2𝑉 + 0.6𝑉) × (1 − 0.44)
2
2 × 0.85 × 430𝑘𝐻𝑧 × 0.25
= 21𝜇𝐻
𝑓
𝑆𝑊(𝑀𝑎𝑥)
: Switching frequency This switching frequency should be calculated at 430 kHz.
𝐼
𝑂𝑈𝑇(𝑀𝑎𝑥)
: Max. secondary output current (Determined by maximum power of 3 outputs ÷ output voltage 1)
At this time, the primary inductance Lp is obtained by the following equation.
𝐿
𝑃
= 𝐿
𝑆
× (
𝑁
𝑃
𝑁
𝑆
)
2
𝐿
𝑃
= 21𝜇𝐻 × (0.92)
2
= 18𝜇𝐻
From the above, we will proceed with the design as Lp:18μH, Ls:21μH in this design.
[A]
[A]
[µH]
[µH]
