# How to calculate thermal heat from a switching voltage regulator?

I found this answer here which explains how to calculate the watts that are dissipated in a switching voltage regulator, but once you calculate the watts dissipated how do you know how much temperature your device will dissipate?

I am using an LM2576HVT-15. There will be an input voltage of 42 volts and the output voltage will be 15 volts. The load will draw somewhere between .2 amps and 1 amp so I calculated the watts dissipated being 15V * 1A = 15W. In the datasheet it shows an efficiency of about 87% so we get 15W / .87 = 17.24W which means about 2.24 watts will be dissipated.

Now that I know how many watts will be dissipated what do I do from here? The datasheet shows all of these temperature values, but I don't which ones I need to take into account for the heat dissipated? I'm just hooking it up to a breadboard with TO220 packaging.

• I think your calculation of dissipation is right, but it is not correct to assume that all of the power loss occurs in the controller. Quite a bit of the power loss will be in the inductor and/or external diode. Sep 21, 2022 at 0:25
• "once you calculate the watts dissipated how do you know how much temperature your device will dissipate?" - do you mean, what temperature will the device reach when dissipating the calculated watts? Sep 21, 2022 at 4:42
• Yes Bruce that is what I mean. Sep 21, 2022 at 6:36

Your goal is keeping the die temperature below the maximum rated on the datasheet.

You calculate how hot the die ("junction") will get by taking the ambient temp (TA it's usually called), and add to it the heating caused by power dissipation.

In your case, the TO-220... without a heatsink you're primarily looking at RJA -- Junction to ambient thermal resistance. It's 32.4 C/W.

If you dissipate 2W in the device with 25C ambient, this means your junction will reach 32.4 * 2 + 25 = 89.8C This is below the max junction temp allowed (150C per datasheet). A wise designer doesn't come anywhere near limits however ... 90C is tolerable but still pretty hot. Especially if more power is sunk than you anticipate, or ambient climbs higher than you expect.

Even a relatively small heatsink attached to the TO-220 would go a long way towards keeping the junction temperature down. Your IC would appreciate it! Heat is the killer of all things electrical... keep the temp down, it'll last longer.

• If the junction to ambient is the only measurement I have to worry about, then what do all the other temperature readings mean? When do you have to take those into account? Sep 21, 2022 at 6:35
• Depends how you mount it. TO-220 is typically a through-hole part that sits above the PCB. If you mount an SMT device to the PCB, you can build a heatsink out of copper on the PCB. Junction-To-Case-Top would be if you have a heatsink mounted to the plastic part of the TO220. Junction to case bottom is if the heatsink is attached to the metal part on the back of the device (with some extra thermal resistance due to it not being a 'perfect' mating surface... Heatsink compound helps. Basically, where are you dumping the heat? Even a small heatsink has a pretty good thermal resistance, Sep 21, 2022 at 6:59
• and you're (by calculation) not quite red-lining, is why I say a small heatsink probably will do. My own personal (non scientific) method of determining if something is too hot is to touch it. If you can hold your finger on the IC indefinitely, it's probably OK. If you pull away immediately, well... too hot. Thermocouples can be used to determine some data also.. If you know the case temperature, and you can measure the current/voltage (i.e. power dissipated) you can back-calculate to the junction temperature. Sep 21, 2022 at 7:01
• That makes sense. Thank you Kyle B! Sep 21, 2022 at 19:29
• @Trev347 Those other numbers are useful for finite element thermal models. Sep 21, 2022 at 21:59