Introduction / Context:
Thermal design ensures semiconductor devices operate within safe junction temperatures. The key parameters are junction temperature limit, ambient temperature, and the device's effective thermal resistance to ambient. This question applies the basic thermal equation to find allowable power dissipation.
Given Data / Assumptions:
- Maximum junction temperature Tj,max = 120 °C.
- Ambient temperature Ta = 40 °C.
- Thermal resistance Rθ = 1.6 °C/W (junction-to-ambient).
- Steady-state operation; heat-sinking performance encapsulated in Rθ value.
Concept / Approach:
Use the thermal relation analogous to Ohm's law: ΔT = P * Rθ, where ΔT = Tj,max − Ta. Solve for P (power dissipation) allowed by the temperature rise limit.
Step-by-Step Solution:
Compute allowable temperature rise: ΔT = 120 − 40 = 80 °C.Thermal equation: P = ΔT / Rθ.Substitute: P = 80 / 1.6 = 50 W.Hence, the device may dissipate up to 50 W without exceeding 120 °C junction temperature.
Verification / Alternative check:
If a heat sink reduces Rθ, allowable P would increase. Conversely, higher ambient would reduce allowable P. The computed value aligns with typical datasheet thermal math.
Why Other Options Are Wrong:
20 W: Underestimates; corresponds to ΔT = 32 °C at Rθ = 1.6.92 W or 128 W: Would require much lower Rθ or lower ambient to be feasible.62.5 W: Would imply ΔT = 100 °C, exceeding the available 80 °C margin.
Common Pitfalls:
Mixing up junction and case temperatures; using the wrong temperature difference; forgetting units °C/W.
Final Answer:
50 W
Discussion & Comments