An IC with maximum junction temperature Tj(max) = 125 °C and θJA = 50 °C/W is used at an ambient temperature Ta = 85 °C. What is the upper limit on allowable power dissipation?

Answer: A. 0.8 W

Junction temperature is given by Tj = Ta + P × θJA. Solving for P gives P = (Tj(max) − Ta)/θJA = (125 − 85)/50 = 0.8 W. The higher the ambient temperature, the smaller the allowable temperature rise, and so the smaller the allowable power. 2.5 W is the mistake of forgetting to subtract the ambient temperature (125/50), and 1.7 W is the mistake of computing 85/50. In practice, considering that θJA varies with the board and allowing derating margin, design to keep power below the calculated value.