Power dissipation comes down to three equivalent formulas: P = V × I, P = I²R, and P = V²/R, all measured in watts. Pick whichever matches the values you already have, and use P = V²/R when voltage and resistance are known to limit rounding error. To turn that wattage into a temperature, apply ΔT = P × θJA, where θJA comes from the component's datasheet.
TL;DR:
- Series circuits dissipate most power in the largest resistor; parallel circuits dissipate most in the smallest branch resistance because voltage is shared.
- For AC, use RMS rather than peak voltage or current, and convert prefixes correctly; entering 10 instead of 0.01 for 10 mA corrupts thermal estimates.
- MOSFET loss includes conduction and switching; because RDS(on) rises with junction temperature, recalculate both loss and temperature iteratively using the hot resistance.
- A datasheet θJA value assumes its reference board; larger copper areas or thermal vias can change thermal resistance by a factor of two or more.
- For pulsed loads, use the datasheet’s transient Zth curve and safe operating area, not steady state θJA; model interacting heat sources with simulation.
Table of Contents
- Power dissipation formulas: derivation and when to use each
- Calculating power dissipation in series and parallel resistor circuits
- Power dissipation in diodes, LEDs, and MOSFETs
- Turning watts into temperature: θJA, transient Zth, and thermal modeling
- Practical takeaways from a circuit designer
- Check wattage and heat before you commit to a layout
- FAQ
- Sources
Power dissipation formulas: derivation and when to use each
All three formulas describe the same physical quantity because they all trace back to Ohm's law, V = IR, combined with Joule's law for heat generated by current flowing through resistance. Substitute IR for V in P = VI and you get P = I²R. Substitute V/R for I instead and you get P = V²/R. They are not three different rules, just three algebraic rearrangements of one relationship, and the one you reach for depends on which two quantities you already measured.
If you know the current and resistance precisely, P = I²R avoids compounding error from a derived voltage. If voltage and resistance are your direct readings, P = V²/R is the more numerically stable choice, since dividing a measured value by a measured value introduces less rounding drift than multiplying two derived figures.
A few quick numeric checks:
- A 100 ohm resistor carrying 50 mA: P = I²R = (0.05)² × 100 = 0.25 W.
- A 12 V supply across a 240 ohm load: P = V²/R = 144 / 240 = 0.6 W.
- A 5 V rail drawing 2 A: P = V × I = 10 W.
The most common mistakes are using peak voltage instead of RMS in AC circuits, which inflates the result, and dropping or mishandling unit prefixes, like entering 10 for 10 mA instead of 0.01. Both errors throw off every downstream thermal calculation.
Calculating power dissipation in series and parallel resistor circuits
Series and parallel networks follow the same general sequence, but the result lands on different resistors depending on topology.
For a series circuit, follow these steps:
- Add the resistances to get the equivalent resistance, Req.
- Find total current with I = V / Req.
- Find each resistor's dissipation with Pi = I² × Ri.
Say a 10 V source drives three series resistors: 100 ohm, 200 ohm, and 300 ohm. Req = 600 ohm, so I = 10 / 600 ≈ 0.0167 A. The 300 ohm resistor dissipates P = (0.0167)² × 300 ≈ 0.083 W, more than either of the others, since in series the same current flows through every resistor and power scales with resistance.
The largest resistor in a series string always dissipates the most power, because current is shared equally but P = I²R rewards the higher R.
Parallel circuits flip that logic. Compute Req as the reciprocal of the sum of reciprocals, then find each branch's power directly with P = V²/R, since every branch sees the same voltage. For a 10 V source across 100 ohm and 200 ohm branches, the 100 ohm branch dissipates 1 W while the 200 ohm branch dissipates 0.5 W. Here the smallest resistance dissipates the most power, the opposite of the series case, because a lower resistance draws more current at a fixed voltage.
Power dissipation in diodes, LEDs, and MOSFETs
Resistor formulas assume a linear V to I relationship, which breaks down for semiconductors. For diodes and LEDs, dissipation is simply forward voltage times forward current: an LED dropping 2.1 V at 20 mA dissipates P = 2.1 × 0.02 = 42 mW, with no resistance term involved because the device is not ohmic.
MOSFETs add a second loss mechanism. Total dissipation is the sum of resistive and switching losses: PD = PD_resistive + PD_switching, where the resistive term is I² × RDS(on) and the switching term is commonly estimated for early design passes from switching frequency, drain voltage, and transition times.
The complication is that RDS(on) itself rises with junction temperature, so a single pass through the formula understates real losses at operating temperature. The fix is an iterative loop:
- Assume a starting junction temperature, Tj.
- Look up or interpolate RDS(on) at that temperature.
- Calculate PD using the hot RDS(on) value.
- Calculate the new Tj using ΔT = PD × θJA, then repeat until Tj stops changing meaningfully between iterations.
This iterative approach is what separates a rough estimate from a design you can trust at full load.
Pro Tip: For pulsed or switching loads, check the manufacturer's transient Zth curve and safe operating area plot rather than relying on steady-state θJA alone, since a short pulse can exceed limits that a DC calculation would miss.
Turning watts into temperature: θJA, transient Zth, and thermal modeling
Steady-state temperature rise follows ΔT = P × θJA, where θJA is the junction-to-ambient thermal resistance listed on the datasheet. That number is not fixed to the device alone: it is measured on a specific test board, and changing copper area or adding thermal vias can shift θJA by a factor of two or more, so a θJA figure only applies cleanly when your board resembles the datasheet's reference layout.
Steady-state math works for continuous loads, but a pulsed or transient load needs a different tool: transient thermal impedance, or Zth. Datasheets publish Zth curves showing how temperature responds to a single pulse of a given width, which is the basis for predicting junction temperature under switching or burst conditions rather than constant DC.
For more detail than a single θJA number but less setup than full simulation, you can build a simple RC network fitted to the Zth curve, often called a Foster network. Each RC stage represents a layer in the thermal path, letting you estimate temperature versus time in a spreadsheet. Reach for SPICE or dedicated thermal software when you need multiple heat sources interacting on the same board or a layout too complex for a hand-fitted network.
A worked steady-state example: a MOSFET dissipating 2 W with a θJA of 50°C/W from its datasheet gives ΔT = 2 × 50 = 100°C above ambient. At 25°C ambient, that puts the junction near 125°C, close to many parts' limits, which is exactly the kind of number that should push you toward adding copper area or vias before committing to a layout.
| Thermal factor | Typical effect on θJA or temperature |
|---|---|
| Larger copper pour under the package | Can lower θJA, reducing temperature rise for the same wattage |
| Added thermal vias to inner/bottom copper | Can lower θJA by helping heat reach additional copper layers |
| Pulsed load vs continuous load | Requires transient Zth data instead of steady-state θJA |
| Neighboring heat sources on the board | Raises effective ambient temperature seen by the part |
- Steady-state θJA calculations assume isolated, continuous dissipation on the datasheet's reference board.
- Transient Zth and RC/Foster models are the right tool for pulsed loads or when you need temperature versus time.
- Simulation tools earn their setup time once multiple components share a thermal path.
Practical takeaways from a circuit designer
Assume worst-case junction temperature, pull θJA and transient Zth straight from the datasheet, and iterate the RDS(on) loop before trusting a single pass. A pencil-and-paper thermal estimate catches most surprises faster than jumping straight to simulation, and a thermal map or BOM import in a tool like Circuit Center's viewer speeds the same checks once the hand math lines up.
— Matthew
Check wattage and heat before you commit to a layout
Once you have the watts per component, the slow part is usually mapping that number onto an actual board and a real bill of materials. Our BOM Tool reads a KiCad, Altium or Eagle export in your browser, matches every line to real parts, and prices the build at your quantity, so you can compare a cheaper part against one with a lower RDS(on) before you order either.
From there, our Circuit Designer opens the same KiCad project as a schematic, board, and 3D view, with a thermal map that estimates how hot each part and the board will run from the watts you enter, heat sinks and chassis mounting included. If a part is missing locally, parts for pickup maps nearby suppliers with stock on hand. Start with the Integrated Circuits Directory to search parts, price the BOM, and preview the thermal picture before you commit to a board spin.
FAQ
How do I calculate energy dissipation?
Energy dissipation is power multiplied by time, E = P × t, with power in watts and time in seconds to get energy in joules. If a component dissipates power for some time, it releases heat energy equal to power multiplied by time over that interval.
How do I calculate power dissipated in an AC circuit?
Use the same P = V × I, P = I²R, or P = V²/R formulas, but with RMS voltage and current rather than peak values, since using peak values will overstate the real dissipated power. For purely resistive AC loads the math is otherwise identical to a DC calculation.
What is the 3 formula of power?
The three equivalent formulas for power dissipated by a resistive element are P = V × I, P = I²R, and P = V²/R, all derived from Ohm's law and all giving the same result in watts. Which one to use depends on which two values, voltage, current, or resistance, you already know.
How do I calculate power loss?
Power loss in a resistive component follows the same three formulas as power dissipation, since the energy is lost as heat; for semiconductors like MOSFETs, total loss adds a switching term to the resistive term. For a quick estimate of the resulting temperature rise, multiply the loss in watts by the component's θJA from its datasheet.