How do I find a fitting passive cooler/heatsink for a TO-220 Mosfet? What parameters do I have to search for?

  • 1
    \$\begingroup\$ Are you looking for a heatsink? \$\endgroup\$
    – The Photon
    Commented Aug 30, 2012 at 23:12

2 Answers 2


First you need to determine how much power is dissipated in the transistor. The second factor is the maximum ambient temperature that you expect the transistor to experience. Finally, decide what maximum temperature you want at the transistor chip itself...maybe 75°C is a good target. Subtract the ambient temperature from the maximum transistor die temperature and you have the maximum temperature increase from the transistor chip to the outer surface of the heatsink. Divide the temperature difference by the power dissipated and you have a value for thermal resistance in °C/W (degrees C per watt). This is the maximum thermal resistance you can allow from the transistor die to the ambient air. Find out the thermal resistance of the transistor package itself. For the transistor you list above, this is given in the data sheet as "Thermal Resistance, Junction to Case" and it has a value of 6.25°C/W. Now subtract the thermal resistance of the transistor package from the total allowable thermal resistance you calculated earlier. Now go to the catalogs and look for a heatsink that fits the TO-220 package and has a thermal resistance, in still air, equal to or less than this value.

  • 1
    \$\begingroup\$ Just wanted to add that, in addition to subtracting out the Rth junction-case, you also need to subtract the thermal resistance between the case and the sink. Many data sheets, including this one, don't list that value, but it's approximately consistent within a package style. For a TO-220, .5 K/W is a good value. \$\endgroup\$ Commented Oct 10, 2012 at 14:31

As a component dissipates energy (power during some time) it will heat up. How much will depend on the materials used and the mass of them. So temperature will keep rising, but at the same time the part will start losing heat to the environment, and it will lose more heat when the temperature difference with the environment becomes larger. At a certain temperature an equilibrium is reached when the added heat is equal to the drained heat.

You can compare the temperature difference with a voltage difference, and the energy flow with current. In electricity the ratio between the two is resistance, and so is the corresponding parameter in thermodynamics: thermal resistance. The lower thermal resistance the more heat can be drained, and the lower the temperature difference will become.

The datasheet says in Absolute Maximum Ratings that the junction temperature shouldn't exceed 150 °C. Let's say we're safe with 125 °C. The datasheet also gives two thermal resistances:

Junction to case: 6.25 K/W (*)
Junction to ambient: 83.3 K/W

The latter is much higher because it's convection heat exchange, while the former is conducted. Convection will also allow lower values, but then a much larger contact surface with the surrounding air is needed, and that's exactly what a heatsink gives us.

Before we can go on we need some information: how much power do you want the FET to dissipate? Let's say 10 W (you can always do the calculation again for a different power). Also let's assume the environment temperature is 30 °C. The temperature difference between junction and environment is then 95 °C for 10 W, so we can only afford a total thermal resistance of 9.5 K/W. That's not much when you see that we already used 2/3 of that for the junction to case (6.25 K/W is a rather high value). Now just like electrical resistances in series you add thermal resistances to get a total value. If we estimate 1.5 K/W for the resistance between case and heatsink (will require enough thermal paste and a good mechanical contact), then we have 9.5 K/W - 6.25 K/W - 1.5 K/W = 1.75 K/W.

Shopping time! Digikey has a few heatsinks that fit the bill, most of them rather expensive (heatsinks are expensive!), but this one seems OK: suitable for TO-220 and 0.5 K/W, natural air flow. It will be lower with forced air flow.

You can also calculate the reverse: how much power dissipation can I afford with this heatsink?

\$ P = \dfrac{125 °C - 30 °C}{6.25 K/W + 1.5 K/W + 0.5 K/W} = 11.5 W \$

(*) Some people prefer K/W, others °C/W, but since it's about a temperature difference, both are equivalent.

  • \$\begingroup\$ Most TO-220 packages I've seen that actually list the Rth case-sink give .5 K/W. Any particular reason you said 1.5 K/W? \$\endgroup\$ Commented Oct 10, 2012 at 14:32
  • \$\begingroup\$ @Remiel - a conservative guesstimate. 0.5 K/W is really a good value, especially since you have to rely on heat transfer paste, instead of a solid thermal contact like between die and case, and that has a much higher thermal resistance. \$\endgroup\$
    – stevenvh
    Commented Oct 10, 2012 at 15:05

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