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IP1202 Datasheet(PDF) 7 Page - International Rectifier |
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IP1202 Datasheet(HTML) 7 Page - International Rectifier |
7 / 29 page www.irf.com 7 iP1202 Adjusting the Power Loss and SOA curves for different operating conditions To make adjustments to the power loss curves in Fig. 2, multiply the normalized value obtained from the curves in Figs. 4, 5, 6 or 7 by the value indicated on the power loss curve in Fig. 2. Remember that the power loss in Fig 2. is the power loss for 2 outputs operating with the same output voltage. If differing output voltages are used the initial power loss for each channel needs to be divided by 2. Then 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. 2. 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. 3, determine your maximum PCB Temp & Case Temp at the maximum operating current of each iP1202. Then, add the correction temperature from the normalized curves in Figs. 4, 5, 6 or 7 to the TX axis intercept (see procedure no. 2 above) in Fig. 3. When multiple adjustments are required, add all of the temperatures together, then add the sum to the TX axis intercept in Fig. 3. See example no. 2. Operating Conditions for the following examples: Output1 Output Current = 12A Input Voltage = 8V Inductor = 1.75µH Output Voltage = 1.2V Sw Freq= 300kHz Output2 Output Current = 10A Input Voltage =8V Inductor = 1.75µH Output Voltage = 3.3V Sw Freq= 300kHz Example 1) Adjusting for Maximum Power Loss: Output1 (Fig. 2) Maximum power loss = W 8.25/2 = 4.125W (Fig. 4) Normalized power loss for input voltage ≈ 0.93 (Fig. 5) Normalized power loss for output voltage ≈ 0.98 (Fig. 6) Normalized power loss for frequency ≈ 1.0 (Fig. 7) Normalized power loss for inductor value ≈ 0.98 Adjusted Power Loss = 4.125 x 0.93 x 0.98 x 1.0 x 0.98 ≈ 3.7W 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. (see AN-1047 for further explanation of T X ) 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. 4) If no top sided heatsinking is available, assume T CASE temperature of 125°C for worst case performance. 0 10 203040 5060 7080 90 100 120 110 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 0 10 203040 5060 7080 90 100 110 120 PCB Temperature (ºC) Case Tem pe rature ( 8Ã T X V IN = 12V V OUT 1 = V OUT 2 = 1.5V I OUT = 15A f SW = 300kHz L = 1.8uH Safe Operating Area 1 3 2 |
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