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A7985ATR bảng dữ liệu(PDF) 29 Page - STMicroelectronics

tên linh kiện A7985ATR
Giải thích chi tiết về linh kiện  2 A step-down switching regulator for automotive applications
PDF  44 Pages
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nhà sản xuất  STMICROELECTRONICS [STMicroelectronics]
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A7985ATR bảng dữ liệu(HTML) 29 Page - STMicroelectronics

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A7985A
Application information
44
6.5
Thermal considerations
The thermal design is important to prevent the thermal shutdown of the device if the junction
temperature goes above 150 °C. The three different sources of losses within the device are:
a)
conduction losses due to the not-negligible RDS(on) of the power switch; these are
equal to:
Equation 36
where D is the duty cycle of the application and the maximum RDS(on) overtemperature is
220 m
. Note that the duty cycle is theoretically given by the ratio between VOUT and VIN,
but actually it is quite higher to compensate the losses of the regulator. So the conduction
losses increase compared with the ideal case.
b)
switching losses due to Power MOSFET turn-on and turn-off; these can be
calculated as:
Equation 37
where TRISE and TFALL are the overlap times of the voltage across the power switch (VDS)
and the current flowing into it during turn-on and turn-off phases, as shown in Figure 16.
TSW is the equivalent switching time. For this device the typical value for the equivalent
switching time is 40 ns.
c)
Quiescent current losses, calculated as:
Equation 38
where IQ is the quiescent current (IQ = 2.4 mA).
The junction temperature TJ can be calculated as:
Equation 39
where TA is the ambient temperature and PTOT is the sum of the power losses just seen.
Rth(JA) is the equivalent thermal resistance junction to ambient of the device; it can be
calculated as the parallel of many paths of heat conduction from the junction to the ambient.
For this device the path through the exposed pad is the one conducting the largest amount
of heat. The Rth(JA) measured on the demonstration board described in the following
paragraph is about 40 °C/W for the HSOP8 package.
PON
RDS on

IOUT

2 D

=
PSW
VIN IOUT
TRISE TFALL
+

2
------------------------------------------- Fsw

VIN IOUT TSW FSW

==
PQ
VIN IQ
=
TJ
TA RthJA PTOT
+
=



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