Calculator KB-16
Linear Regulator Calculator
The entire voltage difference turns into heat. See exactly how much.
Result
In brief: Calculate power dissipation, efficiency, junction temperature, and the required heatsink thermal resistance for 7805, LM317, AMS1117, and LDO linear regulators. Example: Dissipated as heat: 3.5 W.
Calculation example
| Input voltage | 12 V |
|---|---|
| Output voltage | 5 V |
| Load current | 0.5 A |
| Minimum dropout voltage | 2 V |
| Junction → ambient | 65 K/W |
| Ambient temperature | 25 °C |
| Dissipated as heat | 3.5 W |
|---|---|
| Efficiency | 41.7 % |
| Delivered to the load | 2.5 W |
| Junction temperature | 253 °C |
| Required thermal resistance to ambient | 28.6 K/W |
How it is calculated
η = Uo / Ui
Tj = Ta + P · Rth
A linear regulator works like a controlled resistor: its input current equals the load current, and the excess voltage is dropped across the chip. So its efficiency is simply the ratio of the voltages.
What to look for
- The junction temperature limit is usually 125 °C, but for reliability, keep it below 100 °C.
- If it dissipates more than 1–2 W, it’s usually easier and cheaper to use a switching module.
Frequently asked questions
How much heat does a 7805 dissipate?
P = (Uin − Uout) · I. With 12 V at the input and a current of 0.5 A, that’s 3.5 W. Without a heatsink, a TO-220 package rated at 65 K/W will overheat: the calculated temperature is 253 °C.
What is the efficiency of a linear regulator?
Approximately Uout / Uin: about 42% for 12→5 V. The rest is dissipated as heat, so a switching converter is a better choice when the voltage difference is large.
How is an LDO different from a regular regulator?
An LDO works with a small difference between input and output: 0.2–1 V instead of 2 V for a 7805. It dissipates the same amount of heat for the same voltage difference, and the capacitors for stability should be selected according to the datasheet.
