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IR3553 Datasheet(PDF) 16 Page - International Rectifier

Part No. IR3553
Description  40A Integrated PowIRstage
Download  22 Pages
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Manufacturer  IRF [International Rectifier]
Direct Link  http://www.irf.com
Logo IRF - International Rectifier

IR3553 Datasheet(HTML) 16 Page - International Rectifier

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July 16, 2014 | DATASHEET V3.3
16
IR3553
40A Integrated PowIRstage®
SW
VIN
PGND
VCC
BOOST
PWM
CSIN+
CSIN-
LGND
IOUT
BBRK#
REFIN
IR3553
PHSFLT#
VIN
VOUT
C2
47uF x4
C5
0.22uF
C4
0.22uF
R2
2.49k
L1
150nH
C6
470uF x3
C1
0.22uF
VCC
R1
10k
IVCC
IIN
IOUT
C3
1uF
C7
1nF
VSW
Figure 32: IR3553 Power Loss Measurement
Figure 7 shows the measured single-phase IR3553
efficiency under the default test conditions, VIN=12V,
VOUT=1.2V, ƒSW = 400kHz, L=150nH (0.29mΩ), VCC=7V,
TAMBIENT = 25°C, no heat sink, and no air flow.
The efficiency of an interleaved multiphase IR3553
converter is always higher than that of a single-phase
under the same conditions due to the reduced input RMS
current and more input/output capacitors.
The measured single-phase IR3553 power loss under the
same conditions is provided in Figure 8.
If any of the application condition, i.e. input voltage,
output voltage, switching frequency, VCC MOSFET driver
voltage or inductance, is different from those of Figure 8, a
set of normalized power loss curves should be used.
Obtain the normalizing factors from Figure 10 to Figure 14
for the new application conditions; multiply these factors
by the power loss obtained from Figure 8 for the required
load current.
As an example, the power loss calculation procedures
under different conditions, VIN=10V, VOUT=1V, ƒSW = 300kHz,
VCC=5V, L=210nH, VCC=5V, IOUT=30A, TAMBIENT = 25°C, no
heat sink, and no air flow, are as follows.
1)
Determine the power loss at 30A under the default
test conditions of VIN=12V, VOUT=1.2V, ƒSW = 400kHz,
L=150nH, VCC=7V, TAMBIENT = 25°C, no heat sink, and no
air flow. It is 4.7W from Figure 8.
2)
Determine the input voltage normalizing factor with
VIN=10V, which is 1.02 based on the dashed lines in
Figure 10.
3)
Determine the output voltage normalizing factor with
VOUT=1V, which is 0.90 based on the dashed lines in
Figure 11.
4)
Determine the switching frequency normalizing factor
with ƒSW = 300kHz, which is 0.99 based on the dashed
lines in Figure 12.
5)
Determine the VCC MOSFET drive voltage normalizing
factor with VCC=5V, which is 1.18 based on the dashed
lines in Figure 13.
6)
Determine the inductance normalizing factor with
L=210nH, which is 0.94 based on the dashed lines in
Figure 14.
7)
Multiply the power loss under the default conditions
by the five normalizing factors to obtain the power
loss under the new conditions, which is 4.7W x 1.02 x
0.90 x 0.99 x 1.18 x 0.94 = 4.74W.
THERMAL DERATING
Figure 9 shows the IR3553 thermal derating curve with the
case temperature controlled at or below 125°C. The test
conditions are VIN=12V, VOUT=1.2V, ƒSW=400kHz, L=150nH
(0.29mΩ), VCC=7V, TAMBIENT = 0°C to 90°C, with and without
heat sink, and Airflow = 0LFM /100LFM /200LFM /400LFM.
If any of the application condition, i.e. input voltage,
output voltage, switching frequency, VCC MOSFET driver
voltage, or inductance is different from those of Figure 9, a
set of IR3553 case temperature adjustment curves should
be used. Obtain the temperature deltas from Figure 10 to
Figure 14 for the new application conditions; sum these
deltas and
then
subtract
from the IR3553 case
temperature obtained from Figure 9 for the required load
current.
8)
From Figure 9, determine the maximum current at the
required ambient temperature under the default
conditions, which is 34.5A at 45°C with 0LFM airflow
and the IR3550 case temperature of 125°C.
9)
Determine the case temperature with VIN=10V, which
is +0.6° based on the dashed lines in Figure 10.
10) Determine the case temperature with VOUT=1V, which
is -3.0° based on the dashed lines in Figure 11.
11) Determine the case temperature with ƒSW = 300kHz,
which is -0.4° based on the dashed lines in Figure 12.


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