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IR3558 Datasheet(PDF) 15 Page - International Rectifier

Part # IR3558
Description  The IR3558 integrated PowIRstage짰 is a synchronous buck gate driver co-packed with a control MOSFET and a synchronous MOSFET with integrated Schottky diode.
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Manufacturer  IRF [International Rectifier]
Direct Link  http://www.irf.com
Logo IRF - International Rectifier

IR3558 Datasheet(HTML) 15 Page - International Rectifier

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September 10, 2012 | FINAL DATASHEET
15
IR3558
45A Integrated PowIRstage®
If any of the application condition, i.e. input voltage,
output voltage, switching frequency, PVCC (HVCC/LVCC)
MOSFET driver voltage or inductance, is different from
those of Figure 9, a set of normalized power loss curves
should be used. Obtain the normalizing factors from
Figures 11-15 for the new application conditions; multiply
these factors by the power loss obtained from Figure 9 for
the required load current.
As an example, the power loss calculation procedures
under different conditions, VIN=10V, VOUT=1V, ƒSW = 300kHz,
L=210nH, PVCC (HVCC/LVCC) = 5V, IOUT=35A, TAMBIENT =
25°C, no heat sink, and no air flow, are as follows.
1) Determine the power loss at 35A under the default
test conditions of VIN=12V, VOUT=1.2V, ƒSW = 400kHz,
L=150nH, PVCC (HVCC/LVCC) = 7V, TAMBIENT = 25°C, no
heat sink, and no air flow. It is 5.2W from Figure 9.
2) Determine the input voltage normalizing factor with
VIN=10V, which is 0.97 based on the dashed lines in
Figure 11.
3) Determine the output voltage normalizing factor with
VOUT=1V, which is 0.90 based on the dashed lines in
Figure 12.
4) Determine the switching frequency normalizing factor
with ƒSW = 300kHz, which is 0.99 based on the dashed
lines in Figure 13.
5) Determine the MOSFET drive voltage normalizing
factor with PVCC (HVCC/LVCC) = 5V, which is 1.22
based on the dashed lines in Figure 14.
6) Determine the inductance normalizing factor with
L=210nH, which is 0.96 based on the dashed lines in
Figure 15.
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 5.2W x 0.97 x
0.90 x 0.99 x 1.22 x 0.96 = 5.3W.
SAFE OPERATING AREA
Figure 10 shows the IR3558 safe operating area 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), HVCC=LVCC=7V, TAMBIENT = 0°C to 90°C, no heat
sink, and Airflow = 0LFM / 100LFM / 200LFM / 400LFM.
If any of the application condition, i.e. input voltage,
output voltage, switching frequency, HVCC/LVCC MOSFET
driver voltage, or inductance is different from those of
Figure 10, a set of IR3558 case temperature adjustment
curves should be used. Obtain the temperature deltas
from Figures 11-15 for the new application conditions; sum
these deltas and then subtract from the IR3558 case
temperature obtained from Figure 10 for the required load
current.
The IR3558 safe operating area is obtained with the case
temperature controlled at or below 125°C. If a lower case
temperature is desired, reduce the highest ambient
temperature by the same delta.
As an example, the highest ambient temperature
calculation procedures for a different operating condition,
VIN=10V, VOUT=1V, ƒSW = 300kHz, L=210nH, PVCC
(HVCC/LVCC) = 5V, IOUT=35A, TAMBIENT = 25°C, no heat sink,
and no air flow, are as follows.
8) From Figure 10, determine the highest ambient
temperature at the required load current under the
default conditions, which is 65°C at 35A with 0LFM
airflow and the IR3558 case temperature of 125°C.
9) Determine the case temperature with VIN=10V, which
is -0.7° based on the dashed lines in Figure 11.
10) Determine the case temperature with VOUT=1V, which
is -2.2° based on the dashed lines in Figure 12.
11) Determine the case temperature with ƒSW = 300kHz,
which is -0.2° based on the dashed lines in Figure 13.
12) Determine
the
case
temperature
with
PVCC
(HVCC/LVCC) = 5V, which is +4.9° based on the dashed
lines in Figure 14.
13) Determine the case temperature with L=210nH, which
is -0.9° based on the dashed lines in Figure 15.
14) Sum the case temperature adjustment from 9) to 13),
-0.7° -2.2° -0.2° +4.9° -0.9° = +0.9°. Deduct the delta
from the highest ambient temperature in step 8), 65°C
- (+0.9°C) = 64.1°C.
15) If the desired IR3558 case temperature is 105°C
instead of 125°C, subtract 20°C ( =125°C - 105°C) from
the highest ambient temperature obtained from 14),
i.e. 64.1°C - 20°C = 44.1°C.


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