Design guide IDP2303(A)
LLC design
Application Note
24
Revision 2.0
2017-05-03
4.2.2
Main transformer and resonant network
4.2.2.1
Transformer turns ratio
In this design, an integrated transformer is considered, where the leakage inductance is used as a series
inductor, while the magnetizing inductor is used a shunt inductor. The all-primary-referred model of the
transformer is shown in Figure 17, where
𝑛
𝑒
is the equivalent turns ratio. All the elements related to leakage
flux are located on the primary side.
Figure 17
Transformer primary referred model
When a separate resonant inductor is used, the equivalent turns ratio
𝑛
𝑒
is equal to the physical turns ratio.
With an integrated resonant inductor, the equivalent turns ratio
𝑛
𝑒
is less than the physical turns ratio, due
to the leakage inductance in the transformer secondary.
The equivalent turns ratio can be estimated as:
𝑛
𝑒
=
𝑛
𝑀
𝑣
where
𝑀
𝑣
= √
𝑚
𝑚 − 1
𝑚
is the ratio between the primary inductance
𝐿
𝑝
(= 𝐿
𝑚
+ 𝐿
𝑟
)
and resonant inductance
𝐿
𝑟
:
𝑚 = 𝐿
𝑝
/𝐿
𝑟
In this design example,
𝑚 = 4.35
is pre-selected as a starting point, based on the transformer bobbin
available and the rule of thumb. Verification and optimization is required for a given implementation,
switching frequency and design power level. Guidance on selection will be provided in a later section.
The required equivalent turns ratio can be calculated as:
𝑛
𝑒
=
𝑉
𝑏𝑢𝑠_𝑛𝑜𝑚
2(𝑉
𝑜
+ 𝑉
𝑓
)
=
390
2(24 + 0.5)
= 8
Assuming the forward voltage drop of secondary-side rectifier diode
𝑉
𝑓
= 0.5 𝑉
.
Correspondingly the physical turns ratio of the transformer
𝑛
will be -
𝑛 = 𝑛
𝑒
𝑀
𝑣
= 𝑛
𝑒
√
𝑚
𝑚−1
= 8 ∗ 1.14 = 9
.1
Consider that, in an actual design,
𝑛
is adjusted to be 8.5. Then the corresponding
𝑛
𝑒
is 7.5. The
corresponding gain at nominal input voltage is
𝑀
𝑛𝑜𝑚
=
2(𝑉
𝑜
+ 𝑉
𝑓
)𝑛
𝑒
𝑉
𝑏𝑢𝑠_𝑛𝑜𝑚
= 0.94
