BD8966FVM Datasheet, PDF (9/15 Pages) Rohm – Low Noise High Efficiency Step-down Switching Regulator with Built-in Power MOSFET

BD8966FVM
Technical Note
●Consideration on permissible dissipation and heat generation
As this IC functions with high efficiency without significant heat generation in most applications, no special consideration is
needed on permissible dissipation or heat generation. In case of extreme conditions, however, including lower input
voltage, higher output voltage, heavier load, and/or higher temperature, the permissible dissipation and/or heat generation
must be carefully considered.
For dissipation, only conduction losses due to DC resistance of inductor and ON resistance of FET are considered.
Because the conduction losses are considered to play the leading role among other dissipation mentioned above including
gate charge/discharge dissipation and switching dissipation.
P=IOUT²Ã—(RCOIL+RON)
RON=D×RONP+(1-D)RONN
D：ON duty (=VOUT/VCC)
RCOIL：DC resistance of coil
RONP：ON resistance of P-channel MOS FET
RONN：ON resistance of N-channel MOS FET
IOUT：Output current
1000
800
①587.4mW
600
①using an IC alone
θj-a=322.6℃/W
②mounted on glass epoxy PCB
θj-a=212.8℃/W
If VCC=5V, VOUT=1.5V, RCOIL=0.15Ω, RONP=0.35Ω, RONN=0.25Ω
IOUT=0.8A, for example,
D=VOUT/VCC=1.5/5=0.3
RON=0.3×0.35+(1-0.3)×0.25
=0.105+0.175
=0.28[Ω]
P =0.8²Ã—(0.15+0.28)
≒275.2[mW]
②387.5mW
400
200
0
0 25 50 75 85 100 125 150
Fig. 24
As RONP is greater than RONN in this IC, the dissipation increases as the ON duty becomes greater. With the consideration
on the dissipation as above, thermal design must be carried out with sufficient margin allowed.
●Selection of components externally connected
1. Selection of inductor (L)
IL
VCC
IL
L
ΔIL
VOUT
Co
Fig.25 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 30% 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=1.5V, f=1MHz, ΔIL=0.3×0.8A=0.24A, for example,
(5-1.5)×1.5
*Select the inLd=uctor0o.2f l4o×w5r×e1sMistance=c4o.3m7p5o^μne^→nt4(s.7u[c^μhHa]s DCR and ACR) to minimize dissipation in the inductor for better efficiency.
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9/14
2009.05 - Rev.A