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IP2003ATR Datasheet(PDF) 5 Page - International Rectifier

Part # IP2003ATR
Description  Full function muliphase builiding block
Download  10 Pages
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Manufacturer  IRF [International Rectifier]
Direct Link  http://www.irf.com
Logo IRF - International Rectifier

IP2003ATR Datasheet(HTML) 5 Page - International Rectifier

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5
iP2003A
Applying the Safe Operating Area (SOA) Curve
The SOA graph incorporates power loss and thermal resistance information in a way that allows one to solve for maximum
current capability in a simplified graphical manner. It incorporates the ability to solve thermal problems where heat is drawn
out through the printed circuit board and the top of the case.
Procedure
1) Draw a line from Case Temp axis at T
CASE to the PCB
Temp axis at T
PCB.
2) Draw a vertical line from the T
X axis intercept to the SOA
curve.
3) Draw a horizontal line from the intersection of the vertical
line with the SOA curve to the Y-axis. The point at which
the horizontal line meets the Y-axis is the SOA current.
Calculating Power Loss and SOA for Different Operating Conditions
To calculate power loss for a given set of operating conditions, the following procedure should be followed:
Determine the maximum current for each iP2003A and obtain the maximum power loss from Fig 1. Use the curves in
Figs. 3, 4, 5 and 6 to obtain normalized power loss values that match the operating conditions in the application. The
maximum power loss under the operating conditions is then the product of the power loss from Fig. 1 and the normal-
ized values.
To calculate the SOA for a given set of operating conditions, the following procedure should be followed:
Determine the maximum PCB temperature and Case temperature at the maximum operating current of each iP2003A.
Obtain the SOA temperature adjustments that match the operating conditions in the application from Figs. 3, 4, 5 and
6. Then, add the sum of the SOA temperature adjustments to the Tx axis intercept in Fig 2.
The example below explains how to calculate maximum power loss and SOA.
Example:
Operating Conditions
Output Current = 40A
Input Voltage = 10V
Output Voltage = 3.3V
Sw Freq= 900kHz
Inductor = 0.2µH
TPCB = 100°C, TCASE = 110°C
Calculating Maximum Power Loss:
(Fig. 1)
Maximum power loss
= 15W
(Fig. 3)
Normalized power loss for input voltage 0.98
(Fig. 4)
Normalized power loss for output voltage 1.14
(Fig. 5)
Normalized power loss for frequency 0.94
(Fig. 6)
Normalized power loss for inductor value 1.013
Calculated Maximum Power Loss for given conditions = 15W x 0.98 x 1.14 x 0.94 x 1.013 15.96W
120
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102030405060708090
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PCB Temperature (ºC)
Safe
Operating
Area
VIN = 12V
VOUT = 1.3V
fSW = 1MHz
L=0.3uH
Case Temperature (ºC
)
TX


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