BD9111NV_09 Datasheet, PDF (9/14 Pages) Rohm – Output 2A or More High Efficiency Step-down Switching Regulator with Built-in Power MOSFET

BD9111NV
Technical Note
●Selection of components externally connected
1. Selection of inductor (L)
IL
VCC
IL
L
ΔIL
V_OUT
Co
Fig.27 Output ripple current
The inductance significantly depends on output ripple current.
As seen in the equation (1), the ripple current decreases as the
inductor and/or switching frequency increases.
ΔIL=
(VCC-VOUT)×VOUT
L×VCC×f
[A]・・・(1)
Appropriate ripple current at output should be 20% more or less of the
maximum output current.
ΔIL=0.3×IOUTmax. [A]・・・(2)
(VCC-VOUT)×VOUT
L=
ΔIL×VCC×f
[H]・・・(3)
(ΔIL: Output ripple current, and f: Switching frequency)
* Current exceeding the current rating of the inductor results in magnetic saturation of the inductor, which decreases efficiency.
The inductor must be selected allowing sufficient margin with which the peak current may not exceed its current rating.
If VCC=5V, VOUT=3.3V, f=1MHz, ΔIL=0.3×2A=0.6A, for example,(BD9111NV)
(5.0-3.3)×3.3
L= _{0.6×5.0×1M} =1.87μ → 2.2[μH]
* Select the inductor of low resistance component (such as DCR and ACR) to minimize dissipation in the inductor for better efficiency.
2. Selection of output capacitor (CO)
VCC
Output capacitor should be selected with the consideration on the stability region
and the equivalent series resistance required to smooth ripple voltage.
VOUT
L
ESR
Co
Fig.28 Output capacitor
Output ripple voltage is determined by the equation (4)：
ΔVOUT=ΔIL×ESR [V]・・・(4)
(ΔIL: Output ripple current, ESR: Equivalent series resistance of output capacitor)
*Rating of the capacitor should be determined allowing sufficient margin
against output voltage. Less ESR allows reduction in output ripple voltage.
22μF to 100μF ceramic capacitor is recommended.
3. Selection of input capacitor (Cin)
VCC
Cin
VOUT
L
Co
Fig.29 Input capacitor
Input capacitor to select must be a low ESR capacitor of the capacitance
sufficient to cope with high ripple current to prevent high transient voltage.
The ripple current IRMS is given by the equation (5):
√VOUT(VCC-VOUT)
IRMS=IOUT×
VCC
[A]・・・(5)
< Worst case > IRMS(max.)
When Vcc is twice the V_OUT, IRMS=
IOUT
2
If VCC=5.0V, VOUT=3.3V, and IOUTmax.=2A, (BD9111NV)
IRMS=2×
√3.3(5.0-3.3)
5.0
=0.947[ARMS]
A low ESR 22μF/10V ceramic capacitor is recommended to reduce ESR dissipation of input capacitor for better efficiency.
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9/13
2009.05 - Rev.A