The power dissipation calculator above does two jobs that belong together. First it works out how many watts a component is turning into heat, from whichever pair of electrical quantities you happen to know. Then it converts that figure into a thermal requirement: the total thermal resistance the design is allowed, the heatsink rating needed to meet it, and what happens if no heatsink is fitted at all.
Arb Digital builds free tools that finish the thought rather than stopping at the easy half. Plenty of pages will tell you that a resistor is dropping eleven watts. Far fewer will tell you that eleven watts leaving a bare TO-220 package in still air implies a junction temperature several hundred degrees above ambient, which is the number that actually decides whether the design works. This page publishes no heatsink catalogue, no allowable temperature table and no component ratings of its own. Every temperature and thermal resistance figure must come from the manufacturer's data sheet for your exact part.
What This Power Dissipation Calculator Does
The electrical half is straightforward. As OpenStax University Physics Volume 2, section 9.5 on electrical energy and power, puts it, the power dissipated by a resistor takes the form P = I2R = V2/R. All three of those forms, plus the direct product P = VI, describe the same physical quantity, and the tool lets you enter whichever pair you measured.
The thermal half is where the tool earns its keep. Heat flow obeys a relationship deliberately built to mirror Ohm's law, described in the open engineering text at Engineering LibreTexts, section 7.7 on heat transfer, as heat transfer rate equals temperature difference divided by thermal resistance. Temperature difference plays the part of voltage, watts play the part of current, and thermal resistance in degrees Celsius per watt plays the part of ohms.
That analogy makes the whole problem a series circuit. The junction sits at the top, ambient air at the bottom, and between them sit the resistance from junction to case, from case to heatsink through whatever interface material is used, and from heatsink to air. Those three add up, and the sum multiplied by the watts gives the temperature rise from air to junction.
How to Use It
- Choose what you actually know. Current and resistance is the most reliable pair for a resistor; voltage and current suits a regulator or a transistor where the resistance is not a fixed number.
- Use the voltage across the part. For a linear regulator that is the input minus the output, not the input rail. Entering the rail voltage inflates the answer badly.
- Take the temperatures from the data sheet. The rated junction maximum belongs to your exact part number, and the ambient figure should be the air inside the enclosure on the hottest day it will see.
- Enter the two package thermal resistances. Junction to case comes from the data sheet; case to heatsink depends on the mounting method and the thermal interface material.
- Read the required heatsink rating. A heatsink is specified by its own degrees per watt, and you need one whose figure is at or below the number shown.
The Formula: Watts In, Degrees Out
The electrical step gives P in watts. The thermal step starts from the total budget: RθJA = (TJ,max − TA)/P. That is the largest total thermal resistance the whole path from die to air is allowed to have. Subtract the fixed parts and what remains is the heatsink requirement: RθSA = RθJA − RθJC − RθCS.
Work the defaults. A part carrying 1.5 A through 4.7 Ω dissipates 1.52 × 4.7 = 10.58 W. With a 150 °C rating, a 20 per cent design margin brings the working ceiling down to 125 °C. From 25 °C ambient that leaves a 100 degree rise for 10.58 W, so the total budget is 9.46 °C/W. Take away 1.5 for junction to case and 0.5 for the interface and the heatsink must be rated at 7.46 °C/W or better.
The no-heatsink column is the same arithmetic run backwards. A bare package quoted at 62 °C/W would reach 25 + 10.58 × 62 = 681 °C, which is not a temperature so much as a statement that the package cannot do this alone. Its honest limit is (125 − 25)/62 = 1.61 W.
Why the Free-Air Number Is Usually Optimistic
Data sheets quote a junction-to-ambient figure that looks convenient and is frequently misleading. It is measured under standardised laboratory conditions, typically a single device on a defined test board of a defined copper area, in still air, alone. Almost none of that describes a real product.
Three things make the installed figure worse. Copper area is the main one for surface-mount parts, because for many packages the board itself is the heatsink, and halving the copper pour can add tens of degrees per watt. Neighbours are the second: a part sitting between two other warm components is not in ambient air, it is in air already heated by them. Enclosure temperature is the third, and it is the one people forget entirely, because a sealed box at 25 °C outside can easily sit at 50 °C inside once everything is running.
The practical response is to treat the quoted figure as a best case and to design against the worst internal air temperature you can justify. That is what the ambient field on this page is for, and it is why the default is not room temperature dressed up as a guarantee.
The Interface Nobody Budgets For
The case-to-heatsink resistance is small, boring and routinely responsible for the failure. It covers the joint between the component's mounting face and the heatsink, and it depends on the flatness of both surfaces, the mounting pressure, and what sits between them. A dry metal-to-metal joint might manage 0.5 °C/W. The same joint with a mica washer and no compound could be three or four times worse.
The counterintuitive part is that thermal compound is not there because it conducts heat well. Compared with metal it conducts heat badly. It is there because it fills the microscopic air gaps between two nominally flat surfaces, and air is far worse still. Too much compound is a real fault, not just waste: an excess layer holds the surfaces apart and adds resistance instead of removing it.
Where electrical isolation is needed the penalty grows again, because an insulating pad has to carry heat through a material chosen for its dielectric properties. This is the point at which a specification stops being a calculation and becomes a component selection made against real data sheets, which is why this tool asks you for the number rather than inventing one.
Steady State, Pulses and Thermal Mass
Everything above assumes a steady state: constant power, long enough for temperatures to settle. Many real loads are nothing like that. A part switching on for two seconds every minute dissipates the same peak watts but a small fraction of the average, and its junction never reaches the steady-state temperature because the thermal mass of the package and heatsink absorbs the pulse.
Handling that properly needs transient thermal impedance curves, which data sheets publish as a family of lines for different pulse widths and duty cycles. Using a steady-state resistance for a short pulse gives an answer that is far too pessimistic; using it for a duty cycle above roughly half is usually close enough to be safe. The dangerous middle ground is a long pulse, where the junction has time to climb but the heatsink has not yet warmed up, and only the transient curve tells you the truth.
The same caution applies in the other direction. A heatsink with a large thermal mass takes minutes to reach equilibrium, so an assembly that seems comfortable after thirty seconds of testing may be well over its limit after an hour. If you are measuring rather than calculating, wait for the temperature to stop climbing before you write the number down.
How This Differs From the Adjacent Electrical Tools
The boundary in one sentence: this page turns watts into a thermal resistance requirement, while the Ohm's law calculator and the electrical power calculator stop once the watts are known. The Joule heating calculator takes the same watts in a different direction, towards the energy delivered over time and the temperature rise of a mass being heated deliberately.
For a resistor specifically, the resistor power rating calculator takes the dissipation figure and picks a standard wattage rating with a derating margin, which is a component selection question rather than a heatsink one. The LED resistor calculator handles the same arithmetic for the specific case of a current-limiting resistor. Where mains wiring rather than a component is doing the heating, the voltage drop calculator and the breaker size calculator apply, and both defer to a licensed electrician for the same reason this page defers to a data sheet.
Arb Digital builds free tools like this one because useful pages earn attention. If you want tools, calculators or content built for your own audience, we can help.
Browse All Free Tools Talk to Arb DigitalCommon Mistakes to Avoid
- Using the supply rail as the voltage — dissipation depends on the drop across the part itself, which for a regulator is input minus output.
- Taking room temperature as ambient — the air inside a working enclosure is often twenty or thirty degrees hotter, and that comes straight off the budget.
- Forgetting the interface resistance — the joint between case and heatsink is easy to ignore and can be a large share of a tight budget.
- Designing to the absolute maximum rating — the rated junction temperature is a limit, not a target, and a design margin is what keeps normal variation from crossing it.
- Applying steady-state figures to short pulses — thermal mass changes the answer completely, and transient impedance curves exist for exactly that case.
Related Free Tools From Arb Digital
For the underlying electrical relationships, use the Ohm's law calculator and the electrical power calculator. For heating as an intended outcome rather than a nuisance, see the Joule heating calculator and the heat transfer calculator. Component selection follows through the resistor power rating calculator and the LED resistor calculator. For mains circuits rather than components, see the breaker size calculator, and browse the full free online tools hub for everything else.
Frequently Asked Questions
It is how many degrees hotter something gets for every watt of heat passing through it. A path rated at 5 °C/W carrying 4 W sits 20 degrees above whatever is on the other side of it, in exactly the way a resistor develops a voltage across it.
In series, yes. Junction to case, case to heatsink and heatsink to air add to give the total from junction to ambient. Parallel paths, such as heat leaving through both a heatsink and the circuit board, combine the same way resistors in parallel do.
The difference between the input and output voltages, because that is the drop appearing across the regulator itself. A part converting 12 V to 5 V at 1 A dissipates 7 W, not 12 W.
It means no heatsink can rescue the design, because the junction-to-case and interface resistances alone already exceed the whole budget. The fixes are less power, a lower ambient temperature, a better package or forced airflow.
Because the rated junction temperature is an absolute limit rather than an operating point. Sizing to it leaves nothing for component tolerance, a hot day, a blocked vent or the ageing of thermal compound.
Not compared with metal. Its job is to displace the air trapped between two imperfectly flat surfaces, and air is a far worse conductor than the compound is. That is also why a thick layer is harmful rather than helpful.
Only as a worst case. Steady-state thermal resistance ignores the thermal mass that absorbs short pulses, so it will overstate the junction temperature. Transient thermal impedance curves in the data sheet cover pulsed operation properly.
No. It is a first-order estimate from figures you supply, and it takes no account of airflow, board layout, altitude, enclosure design, safety standards or the transient behaviour of the real load. Verify with measurement on the finished assembly.
This tool is provided for educational and estimating use only. It is not electrical or thermal design advice, and it publishes no component ratings, temperature limits or heatsink data of its own. Any equipment connected to mains electricity must be designed and installed by a suitably qualified person in accordance with the standards that apply in your jurisdiction.
