BD9415FS
【setting example】
Output voltage = VOUT [V] = 40V
LED total current = IOUT [A] = 0.48A
DCDC input voltage of the power stage = VIN [V] = 24V
Efficiency of DCDC=η [%]=90%
Averaged input current IIN is calculated as follows.
VOUT [V] IOUT [A] 40[V]0.48[A]
IIN
0.89[A]ꢀ
VIN [V][%]
24[V]90[%]
If the switching frequency, fSW = 200kHz, and the inductor, L=100μH, the ripple current of the inductor L (ΔIL [A]) can be
calculated as follows.
(VOUT [V]VIN [V]) VIN [V]
(40[V] 24[V]) 24[A]
IL
0.48[A]ꢀ
L[H]VOUT [V] fSW [Hz] 100106[H]40[V]200103[Hz]
Therefore the inductor peak current, IPEAK is
IL[A]
0.48[A]
IPEAK IIN [A]
[A] 0.89[A]
1.13[A]ꢀ
…calculation result of the peak current
…RCS value confirmation
2
2
If RCS is assumed to be 0.3Ω
VCSPEAK RCS IPEAK 0.3[]1.13[A] 0.339[V] 0.45[V]
The above condition is met.
And IPEAK_DET, the current OCP works, is
0.45[V ]
IPEAK _ DET
1.35[A]
0.3[]
If the current rating of the used parts is 2A,
I
…current rating confirmation of DCDC
I
The current raying 1.33[A]1.35[A] 2.0[A]
peak
peak _det
This inequality meets the above relationship. The parts selection is proper.
And IMIN, the bottom of the IL ripple current, can be calculated as follows.
IL[A]
IMIN IIN [A]
[A] 1.13[A] 0.48[A] 0.65[A] 0ꢀ
2
This inequality implies that the operation is continuous current mode.
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23.May.2016 Rev.001
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