National Semiconductor LM3424, Руководство пользователя

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10. INPUT CAPACITANCE
Solve for the minimum C
IN
:
To minimize power supply interaction a 200% larger capaci-
tance of approximately 20 µF is used, therefore the actual
Δ
v
IN-PP
is much lower. Since high voltage ceramic capacitor
selection is limited, four 4.7 µF X7R capacitors are chosen.
Determine minimum allowable RMS current rating:
The chosen components from step 10 are:
11. NFET
Determine minimum Q1 voltage rating and current rating:
A 100V NFET is chosen with a current rating of 32A due to
the low R
DS-ON
= 50 m
Ω
. Determine I
T-RMS
and P
T
:
The chosen component from step 11 is:
12. DIODE
Determine minimum D1 voltage rating and current rating:
A 100V diode is chosen with a current rating of 12A and V
D
=
600 mV. Determine P
D
:
The chosen component from step 12 is:
13. INPUT UVLO
Solve for R
UV2
:
The closest standard resistor is 150 k
Ω
therefore V
HYS
is:
Solve for R
UV1
:
The closest standard resistor is 21 k
Ω
making V
TURN-ON
:
The chosen components from step 13 are:
14. OUTPUT OVLO
Solve for R
OV2
:
The closest standard resistor is 499 k
Ω
therefore V
HYSO
is:
Solve for R
OV1
:
The closest standard resistor is 15.8 k
Ω
making V
TURN-OFF
:
The chosen components from step 14 are:
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LM3424