MB39A102 Datasheet, PDF (21/24 Pages) Fujitsu Component Limited. – Evaluation Board

MB39A102
(2) Schottky Barrier Diode (SBS004 (SANYO product) )
VF (forward voltage) = 0.35 V (Max) : at IF = 1 A, VRRM (repeated peak reverse voltage) = 15 V
IFSM (surge forward current) = 10 A, IF (mean output current) = 1 A
• Diode current: Peak value
The peak current of this diode must be within its rated current.
If the diode’s peak current is IFSM, it is obtained by the following formula.
IFSM ≥
VO+VIN (Min)
VIN (Min)
× IO +
1
2
( ) 1
L3
+
1
L4
× VO × tOFF
( ) ^≥
3.3+2.5
2.5
×
0.5
+
1
2
1
10×10^−6
+
1
15×10^−6
× 3.3
×
1
500×10³
×
(1−0.569)
≥ 1.397 A
• Diode current: Average value
The mean value of diode current must be within its rated current.
If the mean value of diode current is IF, it is obtained by the following formula.
IF ≥ IO
≥ 0.5 A
• Repeated peak reverse voltage
The repeated peak reverse voltage of this diode must be within its rated voltage.
If the diode’s repeated peak reverse voltage is VRRM, it is obtained by the following formula.
VRRM ≥ VIN (Max) +VO
≥ 6+3.3
≥ 9.3 V
(3) Inductor (L3 : RLF5018T-100MR94, TDK product)
10 µH (tolerance ± 20%) , rated current = 0.94 A
The condition for L to be a continuous current within the operating voltge range is obtained by the following
formula.
L≥
VIN (Max) ²
2IOVO
× tON
≥
6²
2×0.5×3.3
×
1
500×10³
× 0.355
≥ 7.7µH
The load current satisfying the continuous current condition is obtained by the following formula.
IO
≥
VIN (Max) ²
2LVO
× tON
≥
6²
2×10×10^−6×3.3
×
1
500×10³
× 0.355
≥ 0.387 A
Note : The continuous current condition becomes a large current value compared with the
current value obtained by L4.
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