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IR3553 Datasheet(PDF) 16 Page  International Rectifier 

IR3553 Datasheet(HTML) 16 Page  International Rectifier 
16 / 22 page July 16, 2014  DATASHEET V3.3 16 IR3553 40A Integrated PowIRstage® SW VIN PGND VCC BOOST PWM CSIN+ CSIN LGND IOUT BBRK# REFIN IR3553 PHSFLT# VIN VOUT C2 47uF x4 C5 0.22uF C4 0.22uF R2 2.49k L1 150nH C6 470uF x3 C1 0.22uF VCC R1 10k IVCC IIN IOUT C3 1uF C7 1nF VSW Figure 32: IR3553 Power Loss Measurement Figure 7 shows the measured singlephase IR3553 efficiency under the default test conditions, VIN=12V, VOUT=1.2V, ƒSW = 400kHz, L=150nH (0.29mΩ), VCC=7V, TAMBIENT = 25°C, no heat sink, and no air flow. The efficiency of an interleaved multiphase IR3553 converter is always higher than that of a singlephase under the same conditions due to the reduced input RMS current and more input/output capacitors. The measured singlephase IR3553 power loss under the same conditions is provided in Figure 8. If any of the application condition, i.e. input voltage, output voltage, switching frequency, VCC MOSFET driver voltage or inductance, is different from those of Figure 8, a set of normalized power loss curves should be used. Obtain the normalizing factors from Figure 10 to Figure 14 for the new application conditions; multiply these factors by the power loss obtained from Figure 8 for the required load current. As an example, the power loss calculation procedures under different conditions, VIN=10V, VOUT=1V, ƒSW = 300kHz, VCC=5V, L=210nH, VCC=5V, IOUT=30A, TAMBIENT = 25°C, no heat sink, and no air flow, are as follows. 1) Determine the power loss at 30A under the default test conditions of VIN=12V, VOUT=1.2V, ƒSW = 400kHz, L=150nH, VCC=7V, TAMBIENT = 25°C, no heat sink, and no air flow. It is 4.7W from Figure 8. 2) Determine the input voltage normalizing factor with VIN=10V, which is 1.02 based on the dashed lines in Figure 10. 3) Determine the output voltage normalizing factor with VOUT=1V, which is 0.90 based on the dashed lines in Figure 11. 4) Determine the switching frequency normalizing factor with ƒSW = 300kHz, which is 0.99 based on the dashed lines in Figure 12. 5) Determine the VCC MOSFET drive voltage normalizing factor with VCC=5V, which is 1.18 based on the dashed lines in Figure 13. 6) Determine the inductance normalizing factor with L=210nH, which is 0.94 based on the dashed lines in Figure 14. 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 4.7W x 1.02 x 0.90 x 0.99 x 1.18 x 0.94 = 4.74W. THERMAL DERATING Figure 9 shows the IR3553 thermal derating curve 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), VCC=7V, TAMBIENT = 0°C to 90°C, with and without heat sink, and Airflow = 0LFM /100LFM /200LFM /400LFM. If any of the application condition, i.e. input voltage, output voltage, switching frequency, VCC MOSFET driver voltage, or inductance is different from those of Figure 9, a set of IR3553 case temperature adjustment curves should be used. Obtain the temperature deltas from Figure 10 to Figure 14 for the new application conditions; sum these deltas and then subtract from the IR3553 case temperature obtained from Figure 9 for the required load current. 8) From Figure 9, determine the maximum current at the required ambient temperature under the default conditions, which is 34.5A at 45°C with 0LFM airflow and the IR3550 case temperature of 125°C. 9) Determine the case temperature with VIN=10V, which is +0.6° based on the dashed lines in Figure 10. 10) Determine the case temperature with VOUT=1V, which is 3.0° based on the dashed lines in Figure 11. 11) Determine the case temperature with ƒSW = 300kHz, which is 0.4° based on the dashed lines in Figure 12. 
