About this calculator
A linear regulator produces a clean output by burning off the difference between input and output as heat. That is fine when the difference is small and the current is low, and rapidly becomes a thermal problem when it is not.
This calculator has two modes. The thermal mode works out how much power your regulator dissipates, what junction temperature that produces, and whether you need a heatsink. The LM317 mode picks the resistor pair that sets a given output voltage.
How it works
Power dissipation is simply the voltage dropped across the regulator multiplied by the current through it: P = (Vin − Vout) × I. Note the current term — a linear regulator passes the same current it delivers, so dropping 7 V at 1 A means 7 watts of heat regardless of how efficient the load is.
That heat has to escape through a thermal path measured in degrees per watt. Junction temperature is ambient plus dissipation times total thermal resistance: Tj = Ta + P × θ. With no heatsink, θ is the package's junction-to-ambient figure, which for a bare TO-220 in still air is around 62 °C/W.
Fit a heatsink and the path changes: heat now flows junction to case, case to sink through the mounting interface, and sink to air. Those add: θ = θjc + θcs + θsa. Since θjc for a TO-220 is about 5 °C/W, the heatsink dominates and the improvement is dramatic.
Separately, a regulator needs a minimum dropout voltage across it to regulate at all. Classic 78xx parts need around 2 V; low-dropout designs manage a few hundred millivolts. Fall below it and the output simply follows the input down.
The LM317 works differently from fixed regulators: it holds 1.25 V between its output and adjust pins, so the current through the upper resistor is fixed, and the lower resistor converts that into any output voltage you want.
Worked example
A 7805 dropping 12 V to 5 V at 500 mA in a 25 °C room, in a TO-220 package with no heatsink.
- P = (12 − 5) × 0.5 = 3.5 W
- TO-220 in free air: θja ≈ 62 °C/W
- Tj = 25 + 3.5 × 62 = 242 °C
- That is far past the 125 °C limit — it will shut down or fail
- Required heatsink: θsa = (125 − 25)/3.5 − 5 − 0.5 = 23.1 °C/W
Without a heatsink the junction would reach 242 °C, so thermal shutdown kicks in almost immediately. A modest 23 °C/W heatsink fixes it. At only 42% efficiency, though, a buck converter would be the better answer.
Practical notes
- Efficiency is just
Vout / Vin. Dropping 12 V to 3.3 V is 27% efficient no matter what regulator you choose — the rest is heat, by definition. - Above roughly 1–2 watts of dissipation, seriously consider a switching regulator instead. A buck converter runs at 85–95% regardless of the voltage ratio.
- Thermal shutdown protects the regulator, not your circuit. A part that keeps cycling in and out of shutdown is not a working design.
- The θja figures assume still air and a modest copper pad. Real boards vary enormously — a large ground pour under a surface-mount part can improve things severalfold.
- Fit the recommended capacitors. A 78xx wants roughly 0.33 µF at the input and 0.1 µF at the output; many LDOs are picky about the output capacitor's ESR and will oscillate with the wrong type.
- For the LM317, keep R1 at the traditional 240 Ω. It sets the minimum load current the regulator needs to stay in regulation.
Frequently asked questions
How much heat does a linear regulator produce?
The voltage it drops multiplied by the current it passes. A 7805 taking 12 V to 5 V at 1 A dissipates 7 W — enough to require a substantial heatsink.
Do I need a heatsink for my 7805?
Multiply the voltage drop by the current. Below about 1 W a bare TO-220 usually copes in open air. Above that you need a heatsink, and above roughly 5 W you need a large one plus airflow.
What is dropout voltage?
The minimum input-to-output difference at which the regulator can still hold its output steady. Standard 78xx parts need about 2 V, so a 7805 requires at least 7 V in. Low-dropout parts manage a few hundred millivolts, which matters on batteries.
How do I set the LM317 output voltage?
Use Vout = 1.25 × (1 + R2/R1), with R1 the resistor between output and adjust, conventionally 240 Ω. For 5 V that needs R2 ≈ 713 Ω — so 715 Ω from the E96 series, or 750 Ω from E24 if you can live with 5.16 V.
Linear or switching regulator?
Linear when the drop is small, the current is modest, and you want low noise and simplicity — analogue and RF supplies especially. Switching when the drop is large, the current is high, or you are on battery power and the wasted heat matters.