To this day, I feel like I don't have a good intuitive feel for how power dissipation turns into heating -- that is, if I waste 1 watt of power as heat into a device the size of a coffee mug, how hot does it get? How about 10 watts, 100, or 1,000?

I realize quite well that material selection, air flow, surface area, et cetra make huge differences. However, it would be nice to have some rules-of-thumb as a starting point to sanity check whether a device would be cool, warm, ridiculously hot, or an ignition hazard.

What are some of your approaches to estimate how hot your project will get without modeling or building the actual device?


Just to clarify, I am more interested in the steady-state temperature of the device (or at least the "touch surfaces") from continued operation; not the immediate heating effects of a momentarily on device.

  • \$\begingroup\$ There is detailed paper for electrolythic caps for example. au.newark.com/pdfs/techarticles/cornell/thermalapplet.pdf But it might be too detailed. What I remember that surprisingly, large can capacitors have thermal power about 100-200 milliwatt. \$\endgroup\$ – user924 Jun 20 '11 at 14:36

For a device you will often see a figured called \$\theta_{JA}\$. This is called thermal resistance.

This tells you that in a typical ambient environment for every watt dissipated, the device will heat up x°C above ambient. You must include ambient temperature into your calculation. In an open lab environment, it might be 25°C but in reality inside the casing of some electronics it can be much hotter.

If you add a heatsink you need to know \$\theta_{JC}\$ (junction-case resistance), \$\theta_{CI}\$ (case-insulator resistance, if any), \$\theta_{IH}\$ (insulation-heatsink resistance, if any), and finally \$\theta_{HA}\$ (heatsink-ambient resistance.) Like normal electrical resistance you can add these together to get a final figure for how much your device will heat up when it dissipates x watts.


When thinking about heating you have to go through a number of different units to get some sensible numbers.

The electrical heat dissipation is measured in Watts. The energy is measured in joules, and the heat itself is measured in calories.

Let's take a typical mug of water - say 300g of water (about 300cc, a typical coffee mug) Now let's say we have something that gives off 10W of heat dissipation. 10W is all very well, but how long do we count the 10W for? That's where the formula:

  • \$W=\frac{J}{t}\$

Where J is Joules, and t is time in seconds

comes in handy. One Watt is one Joule per second. So Joules = Watts × Seconds, ok? So if we heat at 10W for 10 seconds, we get 100 Joules.

Now, the calorie is the amount of heat required to heat 1g of water by 1°C, and is equivalent to 4.184 joules.

That means that our 100 Joules is equal to (EDIT: 23.9 calories [1 calorie = 4.184 J, so 100 J * 1 calorie/4.184 J = 23.9 calories, not 418.4 calories]). Over our 300g of water, that would be:

  • \$T=\frac{23.9}{300}\$

Which equals (EDIT: 0.08°C [not 1.395°C]) temperature rise.

So 10 watts of power for 10 seconds would raise the heat of the water in the coffee mug by a little under (EDIT: a tenth of a degree [not one and a half degrees]).

  • 2
    \$\begingroup\$ @whoever downvoted this - could you tell us why you did, so that Matt can improve his answer? \$\endgroup\$ – stevenvh Jun 18 '11 at 10:49
  • 1
    \$\begingroup\$ @whoever yes please - it would be nice to know what you didn't like about it. While it doesn't 100% answer all the question, it does demonstrate the mathematics behind the information that others have given. \$\endgroup\$ – Majenko Jun 18 '11 at 11:06
  • \$\begingroup\$ The calories part seems superfluous: surely you just need the specific heat of the water and the mug (in J/gK) and the mug's weight? Also, as edited, the question specifies a steady-state scenario, so you'd need to solve simultaneous equations such that the 300W heat input (microwave?) is balanced by total 300W heat output via radiance and convection from the mug's surface. For this you'd need the ambient temperature, the conductivity of the mug material, a surface made of a perfect insulator and a lid avoid losing water from the system... \$\endgroup\$ – Emyr Aug 10 '17 at 16:00

As an intuitive and very rough (but helpful) rule of thumb, I like to refer to resistors of different sizes. Pretty much everyone knows the "standard" 1/4 W resistors (a.k.a. 0207). Also, by looking at an electronics distributor's catalog (or with experience from continued hacking and repairing stuff), you get to know bigger and smaller resistors (SMD sizes for 1/4 W, 1/8 W, ... and larger power resistors for 2 W, 4 W, 5 W, 11 W, ...).

The way most resistors are designed is that you can run them at their rated power at an ambient temperature of 70 °C or 75 °C, and by doing so, you will cause them to reach their maximum allowable temperature of 125 °C or 155 °C (typical and common values, check data sheets for details).

Thus, you have a relationship between dissipated power and temperature rise (something in the order of 125 °C - 70 °C = 55 °C up to 155 °C - 70 °C = 85 °C), and, to get back to the core of your question, physical size (volume, surface area) of a part.

Also, you can use light bulbs (old-school filament style) and other things of which you know the size and power (a.k.a. wattage). Think for example of a 40 W bulb: At room (ambient) temperature, the surface gets just so hot that you can still barely touch it (which translates to maybe 60 °C). A water boiler (for tea water) takes something in the order of 2 kW and with 1 l of water, it rises from 20 °C to 100 °C in about one or two minutes (and would self-destruct if not shut off by its thermostat. Extend this concept to other everyday devices of which you know: dissipated power, size, temperature rise.

Works very well in many cases if you just need to get a feeling for something you are considering to build.


You rightly mention material as a factor. Every material has a specific heat, which tells you how much energy in the form of heat you have to add for a 1K temperature rise on a 1g sample. For instance, to warm 1g of water from 14.5°C to 15.5°C you need 4.186 J. (This is the definition of the old unit of 1 calorie).
When talking about the flow of this heat you're interested in thermal resistance (just like you want to know the electrical resistance to find out electrical current). Thermal resistance is expressed in K/W (Kelvin per Watt), and tells you how much difference in temperature you get between two points when heat flows at a certain rate (energy per unit of time = power). When you read the datasheet of a power component you'll see thermal resistance between die and housing, and from housing to ambient.

edit (regarding your edit)
For an equilibrium state the same factors play: specific heat determines the die's temperature and the series of thermal resistances how much heat can be drained to the environment. Equilibrium means that the latter is equal to the energy you dissipate.


Perhaps a list of what real-world devices dissipate would be a good reference. Smartphone 1-2W, laptop 10-30W, 50" LCD TV 100W, desktop computer 200-500W, space heater 1500W.

Surface area and air movement (fans) can allow for several orders of magnitude more heat dissipation at the same temperature, so mechanical design is a big deal for anything that runs hot. A hair dryer is about the size of a coffee mug, runs over 1000W, and is only warm in front of the blower, but if you took it apart, the heating coil might ignite paper. Even 1W is plenty to start a fire if concentrated on a small enough area, say by a laser. A desktop CPU putting 100W into 1cm^2 can blow a hole in the motherboard if left running with no heatsink, but properly cooled will only make the heatsink hot and the case warm.

If your project runs below 0.1W, you probably don't need to worry about heat. At 1W the metal in the circuit board might spread the heat enough to allow ambient cooling. At 10W you'll probably need a decent sized heatsink (which could be the case) &/or a fan. At 100W you will probably need a fan. Above 1000W you've effectively built a space heater, and whether or not it burns things will depend on how fast you can move heat into the surrounding air. Above 5000W you may need to vent the heat outdoors to keep the room from getting too hot.

Most people don't have anything in their house that draws over a few thousand watts, the highest single load probably being the clothes dryer. Keep in mind that 1W costs about $1/year to run all the time, so anything over a few hundred watts is going to be expensive to own unless it's only used intermittently.


In response to "it would be nice to have some rules-of-thumb" ..

  • if you can't hold your thumb on it, it's too hot. It will need a heat sink on it.
  • I've found that more than 2W dissipated in a 40-pin DIP makes for a surface too hot to touch.
  • even just 1W is a lot in a TO-220 w/o a heat sink

You probably won't encounter too many 40 pin DIP packages these days, and if you do, it seems doubtful they will be dissipating as much as 2W. I mention it though as it provides a handy sense of scale.

The TO-220 package is still going strong though, and is basically designed to be used with heat sinks. That metal tab is there for a reason, so there's little point in running one of these hot when an aluminum sink and a dab of contact thermal grease are so cheap and easy.

  • \$\begingroup\$ .. and before someone decides to nuke me for not applying any mathematics, the OP did say "What are some of your approaches to estimate how hot your project will get without modeling or building the actual device?" ergo, no mathematical models here. \$\endgroup\$ – JustJeff Jun 20 '11 at 23:15
  • \$\begingroup\$ Instead of nuking you, I'm 1-up'ing you. It's good to know the math, but a lot of Engineering won't go well without some rules of thumb and gut (or fingertip) feeling. \$\endgroup\$ – zebonaut Jun 21 '11 at 5:45
  • \$\begingroup\$ 1W is a lot in a TO-220 if there's no heat-sinking. If there is good heat-sinking it's not very much at all. \$\endgroup\$ – Jason S Jun 21 '11 at 11:03
  • \$\begingroup\$ @Jason S - didn't phrase it clearly enough, i guess. i was trying to indicate i'd consider a 1W TO-220 as overheated as a 2W 40pin DIP. (and now I think about it more, even 1/2W in a TO-220 might be a bit much) \$\endgroup\$ – JustJeff Jun 21 '11 at 23:01
  • \$\begingroup\$ @JustJeff: are we talking about the same package? As in, the IRF640N in a TO-220? (irf.com/product-info/datasheets/data/irf640npbf.pdf) The Rjc + Rcs is 1.5 C / W. Even for 10W, that's only a 15 C rise above the heatsink. You can't get that kind of heat out of a DIP, but you can with a TO-220 since it's got a metal tab. I agree that a TO-220 without a heatsink can't dissipate much heat (IRF640N datasheet estimates 62 C/W), but TO-220s get used all the time for power electronics. \$\endgroup\$ – Jason S Jun 22 '11 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.