MB39A104
PC = ID 2 × RDS (ON) × Duty
=
3
2 × 0.05 × 0.174
=: 0.078 W
VD (Max) × ID × tr × fOSC
PS (ON)
=
=
6
19 × 3 × 100 × 10−9 × 500 × 103
6
=: 0.475 W
VD (Max) × ID (Max) × tf × fOSC
PS (OFF)
=
=
6
19 × 3.18 × 100 × 10−9 × 500 × 103
6
=: 0.504 W
PT = PC + PS (ON) + PS (OFF)
=: 0.078 + 0.475 + 0.504
=: 1.057 W
The above power dissipation figures for the TPC8102 are satisfied with ample margin at 2.4 W (Ta = +25 °C) .
• Inductors
In selecting inductors, it is of course essential not to apply more current than the rated capacity of the inductor,
but also to note that the lower limit for ripple current is a critical point that if reached will cause discontinuous
operation and a considerable drop in efficiency. This can be prevented by choosing a higher inductance value,
which will enable continuous operation under light loads. Note that if the inductance value is too high, however,
direct current resistance (DCR) is increased and this will also reduce efficiency. The inductance must be set at
the point where efficiency is greatest.
Note also that the DC superimposition characteristics become worse as the load current value approaches the
rated current value of the inductor, so that the inductance value is reduced and ripple current increases, causing
loss of efficiency. The selection of rated current value and inductance value will vary depending on where the
point of peak efficiency lies with respect to load current.
Inductance values are determined by the following formulas.
The L value for all load current conditions is set so that the peak to peak value of the ripple current is 1/2 the
load current or less.
Inductance value : L
2 (VIN − VO)
L ≥
ton
IO
26
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