Typical value = 100KΩ
C1 = 4.7µF
R1 = 100KΩ
C2 = 0.1µF
R2 = 499Ω
To limit the voltage drop on the input supply caused by transient
in-rush currents when the switch turns-on into a discharged load
capacitor or a short-circuit, a capacitor needs to be placed
between VIN and GND. A 4.7µF ceramic capacitor, CIN, must be
placed close to the VIN pin. A higher value of CIN can be used to
further reduce the voltage drop experienced as the switch is
turned on into a large capacitive load.
A 0.1µF capacitor COUT, should be placed between VOUT and
GND. This capacitor will prevent parasitic board inductances
from forcing VOUT below GND when the switch turns-off. For the
FPF2148, the total output capacitance needs to be kept below a
maximum value, COUT(max), to prevent the part from
registering an over-current condition and turning-off the switch.
The maximum output capacitance can be determined from the
ILIM(max) x tR(max)
For best performance, all traces should be as short as possible.
To be most effective, the input and output capacitors should be
placed close to the device to minimize the effects that parasitic
trace inductances may have on normal and short-circuit
operation. Using wide traces for VIN, VOUT and GND will help
minimize parasitic electrical effects along with minimizing the
case to ambient thermal impedance.
The middle pad (pin 7) should be connected to the GND plate
of PCB for improving thermal performance of the load switch.
An improper layout could result higher junction temperature and
triggering the thermal shutdown protection feature. This concern
applies when the switch is in an overcurrent condition or the
worst case when output is shorted to ground.
During normal operation as a switch, the power dissipation is
small and has little effect on the operating temperature of the
part. The parts with the higher current limits will dissipate the
most power and that will only be,
P = (ILIM)2 x RDS = (0.4)2 x 0.12 = 19.2mW
If the part goes into current limit the maximum power dissipation
will occur when the output is shorted to ground. For the
FPF2148, a short on the output will cause the part to operate in
a constant current state dissipating a worst case power as
calculated in (3) until the thermal shutdown activates. It will then
cycle in and out of thermal shutdown so long as the ONB pin is
active and the short is present.
P(max) = VIN(max) x ILIM(max)
= 5.5 x 0.4 = 275mW
FPF2148 Rev. H