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Power dissipation

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Figures 6 and 7 of the data sheet show the efficiency of the regulator versus input voltage and load current. Efficiency is the ratio of output power over the total power. Thus, for example, if the output power were 5V x 5A = 25W and you were operating at the 85% efficiency point, then the input power would be 25W / 85% = 29.4W. The power dissipated for that particular condition is then 29.4W - 25W = 4.4W. That power includes losses in the IC, the flyback diode, and the inductor.
 
Very good explanation. Thanks a lot, i understand it now
additional question: continuing your example, if a had a 25W output and 29 (lets say 30) W was the input, we could then find the amperage from the input given a fixed voltage input? so if i had a 15V input for that 5V 5A output, the current at Vin of the IC would be 30/15 = 2A?
 
Very good explanation. Thanks a lot, i understand it now
additional question: continuing your example, if a had a 25W output and 29 (lets say 30) W was the input, we could then find the amperage from the input given a fixed voltage input? so if i had a 15V input for that 5V 5A output, the current at Vin of the IC would be 30/15 = 2A?
That is correct.
 
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