🌡️ Thermal Design for Power Semiconductors

Thermal design is the bridge from watts to degrees. First you estimate how much heat the device makes. Then you estimate how hard it is for that heat to travel from the junction to ambient air. Multiply the two and you get temperature rise.

By the end, you can

1. Separate MOSFET conduction loss from switching loss
2. Estimate junction temperature from $T_J = T_A + P_\text{loss}\theta_{JA}$
3. Explain why switching frequency and transition time trade against heat
4. Choose practical thermal fixes: lower resistance, lower frequency, better copper, airflow or a different package

Intuition first

A MOSFET wastes power in two main ways.

When it is on, it is not a perfect short. It has an on-resistance, $R_{DS(on)}$. Current through that resistance makes heat:

$$P_\text{cond} = I^2 R_{DS(on)} D$$

where $D$ is the fraction of time the device is on.

When it is switching, it briefly has voltage across it and current through it at the same time. That overlap happens twice per cycle and costs energy. A useful first estimate is

$$P_\text{sw} \approx \frac{1}{2} V I t_\text{sw} f_s$$

where $t_\text{sw}$ is the combined transition time and $f_s$ is switching frequency. Raise frequency and you pay that transition loss more often. Slow the transition and you spend longer in the lossy middle. Fast edges reduce switching heat but can raise EMI. That is the trade.

Once you have total heat, the thermal path turns watts into degrees:

$$T_J = T_A + P_\text{total}\theta_{JA}$$

$\theta_{JA}$ is thermal resistance from junction to ambient in °C/W. A device dissipating 2 W through 40 °C/W rises about 80 °C above ambient.

Drive it

Move current, voltage, duty cycle, on-resistance, transition time, switching frequency, thermal resistance and ambient. The widget is deliberately simplified, but it shows the right shape: current punishes conduction loss quadratically, frequency punishes switching loss linearly, and poor thermal resistance turns a few watts into a hot junction.

Try making three fixes to a hot design:

1. Lower $R_{DS(on)}$ or current to attack conduction loss.
2. Lower switching frequency or transition time to attack switching loss.
3. Lower $\theta_{JA}$ with more copper, thermal vias, airflow, a heat sink or a better package.

?A MOSFET carries 10 A with Rds(on) = 20 mΩ and duty cycle 50%. What is the conduction loss?

• A 1 W
• B 2 W
• C 5 W
• D 10 W

?A device dissipates 3 W, has θJA = 30 °C/W, and ambient is 40 °C. What junction temperature does the simple model predict?

• A 70 °C
• B 90 °C
• C 130 °C
• D 160 °C

↘ Why lower Rds(on) can make switching loss worse

A larger MOSFET die often has lower $R_{DS(on)}$, so conduction loss falls. But a larger die also has more gate charge and capacitance. Driving that gate takes more energy per transition, and the device may switch more slowly unless the driver is stronger. That can raise switching loss and EMI stress. Power design is full of these coupled trades: the part that wins at 20 kHz may lose at 500 kHz.

Key takeaways

• ▸Conduction loss is $I^2R$ times duty cycle, so current is squared.
• ▸Switching loss rises with bus voltage, current, transition time and switching frequency.
• ▸Junction temperature is ambient plus heat multiplied by thermal resistance.
• ▸Thermal fixes can reduce loss or improve the heat path. Good designs do both.
• ▸Datasheet thermal numbers are starting points, not proof that the real board is safe.

Practice 1 core

A design runs at 140 °C junction in an 85 °C enclosure. Name two electrical fixes and two mechanical or layout fixes.

↘ Show worked solution

Electrical fixes: choose a lower $R_{DS(on)}$ MOSFET if conduction loss dominates; lower current; lower switching frequency if switching loss dominates; improve gate drive so transition time shrinks; reduce bus voltage if the system can tolerate it.

Mechanical or layout fixes: add copper area connected to the drain or thermal pad; add thermal vias to an internal or backside copper pour; choose a package with a better thermal pad; add airflow; attach a heat sink; move other hot parts away. The right fix depends on which term is making the heat and which thermal resistance is blocking the heat from leaving.