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IR3558 Datasheet(PDF) 15 Page - International Rectifier |
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IR3558 Datasheet(HTML) 15 Page - International Rectifier |
15 / 21 page 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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