BD9486F
Datasheet
[setting example]
Output voltage = VOUT [V] = 40V
LED total current = IOUT [A] = 0.48V
DCDC input voltage of the power stage = VIN [V] = 24V
Efficiency of DCDC =η[%] = 90%
Averaged input current IIN is calculated as follows.
IIN[A]
VOUT[V] IOUT[A]
VIN[V] [%]
40[V] 0.48[A]
24[V] 90[%]
0.89
[A]
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.
ΔIL
(VOUT [V] VIN[V]) VIN[V]
L[H] VOUT [V] fSW [Hz]
100
(40[V] 24[V]) 24[V]
106[H] 40[V] 200 103[Hz]
0.48
[A]
Therefore the inductor peak current, Ipeak is
Ipeak
IIN[A]
IL[A] [A]
2
0.89[A]
0.48[ A ]
2
1.13
[A ] …calculation result of the peak current
If Rcs is assumed to be 0.3Ω
VCS peak Rcs Ipeak 0.3[] 1.13[A] 0.339 [V] 0.4V
…Rcs value confirmation
The above condition is met.
And Ipeak_det, the current OCP works, is
Ipeak _ det
0.4[ V ]
0.3[]
1.33
[A]
If the current rating of the used parts is 2A,
Ipeak Ipeak _ det The current rating 1.13[A ] 1.33[A ] 2.0[A ] …current rating confirmation
of DCDC parts
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.
IMIN
IIN[A]
IL[A] [A]
2
1.13[A]
0.48[ A ]
0.65[ A ]
0
This inequality implies that the operation is continuous current mode.
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13.Feb.2014 Rev.004