SC4612H
Applications Information (continued)
POWER FETS - The FETs are chosen based on several
criteria with probably the most important being
power dissipation and power handling capability.
TOP FET - The power dissipation in the top FET is a
combination of conduction losses, switching losses
and bottom FET body diode recovery losses.
a) Conduction losses are simply calculated as:
circuitry, either as extra output capacitance or, more
usually, additional input capacitors. Choosing low
ESR input capacitors will help maximize ripple rating
for a given size.
Low Side RDS(ON) Current Limit
PCOND = IO2 ⋅ RDS(ON) ⋅ D
where
D = Duty cycle ≈ VO
VIN
. . . . . . . . . (7)
b) Switching losses can be estimated if the switching
time is known or assumed:
( ) PSW
=
IO
⋅ VIN ⋅
tr + tf
2
⋅ fOSC
. . . . . . . . . (8)
c) Body diode recovery losses are more difficult to
estimate, but to a first approximation, it is reasonable
to assume that the stored charge on the bottom FET
body diode will be moved through the top FET as it
starts to turn on. The resulting power dissipation in
the top FET will be:
PRR = QRR ⋅ VIN ⋅ fOSC
. . . . . . . . . (9)
BOTTOM FET - Bottom FET losses are almost entirely
due to conduction. The body diode is forced into
conduction at the beginning and end of the bottom
switch conduction period, so when the FET turns on
and off, there is very little voltage across it resulting
in very low switching losses. Conduction losses for
the FET can be determined by:
( ) PCOND = I2O ⋅ RDS(ON) ⋅ 1− D
. . . . . . . . . (10)
INPUT CAPACITORS - Since the RMS ripple current in
the input capacitors may be as high as 50% of the
output current, suitable capacitors must be chosen
accordingly. Also, during fast load transients, there
may be restrictions on input di/dt. These restrictions
require useable energy storage within the converter
Fig3: Current Limit circuitry
1. Programming resistors Ra and Rb - Not installed:
2.75V − 100mV = 100mV − VPH . . . . . . (11)
R3
R2
solving for: VPH = -100mV, the circuit will trip
at RDS(ON) x ILOAD = 100mV
2. To increase trip voltage - install Ra.
Ra = − 772 − 20 ⋅ VPH
1+ 10 ⋅ VPH
. . . . . . . . . (12)
solving for double the current limit: VPH = -200mV.
Ra = 768kW.
3. To decrease trip voltage - install Rb
Rb = 8 − 20 ⋅ VPH
1+ 10 ⋅ VPH
. . . . . . . . . (13)
solving for half the current limit: VPH = -50mV.
Rb = 18kW.
NOTE: Allow for tempco and RDS(ON) variation of the
MOSFET- see the “Over Current Protection” section of
the datasheet.
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