Thermal design and heatsinks

The idea

Every watt a component loses becomes heat inside its silicon junction. That heat has to flow out through the case, into a heatsink, then into the air. Each step resists the flow, so the junction ends up hotter than the air. Treat it exactly like Ohm’s law: heat flow behaves like current, temperature like voltage, and thermal resistance like resistance.

Electrical	Thermal
Current $I$ (A)	Heat flow $P$ (W)
Voltage difference (V)	Temperature difference $\Delta T$ (°C)
Resistance $R$ (Ω)	Thermal resistance $R_{th}$ (°C/W)
$V=IR$	$\Delta T=P{R}_{th}$
Series resistors add	$R_{th1}+R_{th2}+R_{th3}$

Resistance	From	To	Where you get it
$R_{th1}$ ($R_{thJC}$)	junction	case	Datasheet
$R_{th2}$ ($R_{thCS}$)	case	heatsink	Datasheet (greased), more with an insulating pad
$R_{th3}$	heatsink	air	You choose it by picking a heatsink
$R_{thJA}$	junction	air, with no heatsink	Datasheet

Junction temperature

The same heat $P$ flows through every step, like current through series resistors. Start from the air, which you know, and climb:

$T_{hs}=T_{amb}+P{R}_{th3}\to T_C=T_{hs}+P{R}_{th2}\to T_J=T_C+P{R}_{th1}$

$\boxed{T_J=T_{amb}+P(R_{th1}+R_{th2}+R_{th3})}$

With no heatsink, use the single datasheet value instead: $T_J=T_{amb}+P{R}_{thJA}$.

The limits:

• Silicon’s absolute maximum is 175 °C. Design for 100–120 °C, and never above 120 °C.
• Safety margin: $\text{margin}=100-\frac{T_J}{175}\times 100$, aiming for at least 30%. 120 °C gives 31.4%.

Sizing a heatsink

Rearrange for the one resistance you choose:

$\boxed{R_{th3}=\frac{T_J-T_{amb}}{P}-(R_{th1}+R_{th2})}$

Put in your target $T_J$, and the answer is the largest $R_{th3}$ you can get away with. Pick a heatsink at or below it. Lower °C/W means a bigger heatsink.

A negative answer

If $R_{th3}$ comes out negative, even a perfect heatsink ($R_{th3}=0$) can’t keep the part cool, because $R_{th1}+R_{th2}$ alone already make it too hot. The fixes:

• a better interface (thermal grease, about 0.24 °C/W, instead of an insulating pad)
• less loss: lower switching frequency, a part with lower $R_{DS(on)}$, or MOSFETs in parallel
• a lower ambient temperature, or forced-air or liquid cooling

Worked example

The IRFP140 from MOSFET power loss dissipates 5.36 W in a robot whose electronics bay reaches 35 °C. Datasheet: $R_{thJC}=0.83$, $R_{thCS}=0.24$ (greased), $R_{thJA}=40$ °C/W.

1. Can it run without a heatsink?
2. What heatsink keeps $T_J$ at 100 °C?
3. You fit an 8 °C/W heatsink. What are $T_J$ and the margin?

Answer

1. $T_J=35+5.36\times 40=249$ °C. No. It’s far over 175 °C.
2. $R_{th3}=\frac{100-35}{5.36}-(0.83+0.24)=12.13-1.07=11.1$ °C/W. Any heatsink at or below 11 °C/W works.
3. $T_J=35+5.36\times (0.83+0.24+8)=35+48.6=83.6$ °C. Margin $=100-\frac{83.6}{175}\times 100=52\%$ ✓

Common mistakes

• Adding the thermal resistances but forgetting to add $T_{amb}$ at the end.
• Reading “lower $R_{th3}$” as “smaller heatsink”. It’s the opposite.
• Mixing $R_{thJA}$ (no heatsink) with the three-step chain. Use one or the other.

Check yourself

1. 40 W, $R_{th1}=0.7$, $R_{th2}=2.0$ (insulating pad), 50 °C ambient, target 120 °C. What heatsink?

$R_{th3}=70/40-2.7=1.75-2.7=-0.95$ °C/W. Negative, so it’s impossible. Swap the pad for grease: $1.75-0.94=0.81$ °C/W, which is a big but possible heatsink.

2. P = 10 W, $T_{amb}$ = 60 °C, $R_{th}$ = 0.5, 1, 2 °C/W. Find $T_{hs}$, $T_C$ and $T_J$.

$T_{hs}=60+20=80$ °C, $T_C=80+10=90$ °C, $T_J=90+5=95$ °C.