IP2002PBF データシートの表示（PDF） - International Rectifier
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IP2002PBF Datasheet PDF : 12 Pages
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iP2002PbF
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.
Case Temperature (8Ã
Procedure
1) Draw a line from Case Temp axis at TCASE to the PCB
Temp axis at TPCB.
2) Draw a vertical line from the TX 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.
0
32
10
20
30
28
26
24
22
20
18
16
14
12
10 VIN = 12V
8 VOUT = 1.3V
6
4 fSW = 1MHz
2 L = 0.30uH
0
0
10
20
30
40
50
60
70
80
3
Safe
Operating
Area
30
40
50
60
70
80
PCB Temperature (ºC)
90
100
110
120
1
2
TX
90
100
110
120
Adjusting the Power Loss and SOA curves for different operating conditions
To make adjustments to the power loss curves in Fig. 1, multiply the normalized value obtained from the curves in Figs. 3,
4, 5 or 6 by the value indicated on the power loss curve in Fig. 1. If multiple adjustments are required, multiply all of the
normalized values together, then multiply that product by the value indicated on the power loss curve in Fig. 1. The resulting
product is the final power loss based on all factors. See example no. 1.
To make adjustments to the SOA curve in Fig. 2, determine your maximum PCB Temp & Case Temp at the maximum operating
current of each iP2002PbF. Then, add the correction temperature from the normalized curves in Figs. 3, 4, 5 or 6 to the TX axis
intercept (see procedure no. 2 above) in Fig. 2. When multiple adjustments are required, add all of the temperatures
together, then add the sum to the TX axis intercept in Fig. 2. See example no. 2.
Operating Conditions for the following examples:
Output Current = 30A
Sw Freq= 900kHz
Input Voltage = 10V
Inductor = 0.2uH
Output Voltage = 3.3V
Example 1) Adjusting for Maximum Power Loss:
(Fig. 1)
(Fig. 3)
(Fig. 4)
(Fig. 5)
(Fig. 6)
Maximum power loss = 11W
Normalized power loss for input voltage ≈ 0.98
Normalized power loss for output voltage ≈ 1.28
Normalized power loss for frequency ≈ 0.955
Normalized power loss for inductor value ≈ 1.02
Adjusted Power Loss = 11W x 0.98 x 1.28 x 0.955 x 1.02 ≈ 13.44W
6
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