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Linear Regulator Power Dissipation Calculator

Find how much heat a linear regulator or LDO produces, its junction temperature and efficiency.

Power dissipationJunction temperatureEfficiency7805, LDO
V
V
mA
°C/W
°C
Junction temperature exceeds 125 °C. Add a heatsink, reduce the input voltage or use a switching regulator.
Power dissipation3.56 W
Junction temperature203 °C
Efficiency41.3 %
Maximum current for Tj = 125 °C277.1 mA
How this result was calculated
  1. P = (Vin − Vout) × I + Vin × Iq = 3.56 W
  2. Tj = Ta + P × θJA = 25 + 3.56 × 50 = 203 °C
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Linear Regulator Heat

P = (Vin − Vout) × Iload + Vin × Iq
Tj = Ta + P × θJA
η ≈ Vout ÷ Vin

Take θJA from the regulator data sheet for your package and PCB copper area.

Worked example

12 V to 5 V at 0.5 A with 5 mA quiescent current: P = 7 × 0.5 + 0.06 = 3.56 W. With θJA = 50 °C/W, Tj = 25 + 178 = 203 °C, so a heatsink is essential.

Key insight: a linear regulator throws away the whole voltage difference as heat, so at more than a few volts of drop and a few hundred mA, a buck converter is the better choice.
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Frequently Asked Questions

How do I calculate linear regulator power dissipation?

Multiply the voltage drop (Vin − Vout) by the load current and add Vin × quiescent current.

What is the efficiency of a 7805 from 12 V?

About 5/12 = 42 %.

What is the maximum junction temperature?

Usually 125 °C or 150 °C; check the data sheet absolute maximum ratings.

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