Calculators / LDO power dissipation

LDO power dissipation and junction temperature

How much heat a linear regulator dissipates at each input voltage, how hot its junction gets, and what the package or heat sink must provide to keep it below the limit. The formulas are TI's, and they are shown under the results.

Related: power budget for a multi-rail circuit · buck converter power stage

Regulator

Input voltage V — min / nominal / max
The maximum sets the heat; the minimum sets the dropout check.
Both from the datasheet at this load current. Ground (quiescent) current flows from the input to ground and adds VIN × IQ of heat.
Temperature
Use the datasheet's recommended maximum operating junction temperature, not the absolute maximum or the thermal-shutdown point.
Thermal path
Presets are rounded, typical figures for a JEDEC test board; they vary by part and vendor. Use your part's RθJA where you have it, and read the note on θJA below.
The interface material. Essentially 0 for a part soldered to board copper.
The heat sink you are considering. The results give the largest θSA that still meets TJ max.
LDO or buck?
A fixed efficiency for the same rail from a switching regulator. First order only.
Worst-case dissipation
—W
Junction temperature
—°C
Max load current (thermal)
—A
Max ambient temperature
—°C
Required θJA (max)
—°C/W
Heat sink θSA (max)
—°C/W
Efficiency (nominal VIN)
—%
Min VIN for regulation
—V

Across the input range

Dissipation, junction temperature and efficiency at each input voltage you entered.

InputVIN (V)PD (W)TJ (°C)Margin (°C)EfficiencyInput current (A)Regulating?

LDO or buck? (nominal input)

The same rail from a switching regulator at the fixed efficiency you entered.

At nominal VINLDOBuck

Thermal derating: load current vs ambient

The largest load current the thermal path allows at each ambient, at the highest input voltage. ● marks your operating point.

Equations (TI SLVA118A)

PD= (VIN − VOUT) × IOUT + VIN × IQ TJ= TA + PD × θJA θJA(max)= (TJ(max) − TA) / PD θSA(max)= θJA(max) − θJC − θCS IOUT(max)= [(TJ(max) − TA)/θ − VIN × IQ] / (VIN − VOUT) η= VOUT × IOUT / [VIN × (IOUT + IQ)] VIN(min)= VOUT + VDO

How the calculation works

A linear regulator's pass element carries the whole load current and drops the difference between the input and the output voltage. The regulator's own ground current flows from the input to ground as well. Both end up as heat in the package, so the power dissipated is

  • PD = (VIN − VOUT) × IOUT + VIN × IQ
  • TJ = TA + PD × θJA

TI's Linear Regulator Design Guide for LDOs (SLVA118A) writes this as input power minus output power, with the quiescent term VIN × IQ added. Because PD rises with input voltage, the thermal check uses the highest input voltage. The dropout check uses the lowest: the regulator needs VIN ≥ VOUT + VDO to stay in regulation, and below that the output follows the input down. The maximum load current and the maximum ambient temperature are the same junction-temperature equation solved for IOUT and for TA.

The form opens on a worked example: a 3.3 V, 0.4 A rail from a 5 V ±10 % supply, in SOT-223 at 50 °C ambient.

Why the datasheet θJA is only a starting point

TI's Semiconductor and IC Package Thermal Metrics (SPRA953) explains that θJA (RθJA) is measured with the part on a standardized JEDEC test board, and that it is meant for comparing packages, not for predicting the temperature on your board. Copper area under and around the part, the number of layers, airflow and nearby hot components all change it. SLVA118A lists the same variables and notes that θJA for one package varies between parts and between vendors.

So treat the junction temperature here as a first estimate. If the margin is small, use the datasheet's thermal guidance for your copper area, or measure. The package presets are rounded, typical JEDEC-board figures for a quick start, not values for any particular part.

Sizing a heat sink

When no package meets θJA(max), a heat sink lowers the thermal resistance from junction to air. Junction to case, case to sink (the interface material) and sink to ambient add in series, so the sink must satisfy

  • θSA ≤ (TJ(max) − TA) / PD − θJC − θCS

θCS is essentially zero for a part soldered to board copper; for a bolted TO-220 it is set by the thermal interface material. Choose Heat sink as the thermal path to get θSA(max) and to check a particular sink. If θSA(max) comes out at or below zero, no heat sink is enough: the dissipation itself has to come down, with a lower input voltage, a series resistor or pre-regulator, or a switching regulator.

LDO or buck?

An LDO passes the whole load current at the input voltage, so its efficiency can never exceed VOUT / VIN. A switching regulator's loss is roughly POUT × (1/η − 1) instead. Setting the two losses equal gives the input voltage below which the LDO wastes less:

  • VIN(break-even) = VOUT × IOUT / [η × (IOUT + IQ)]

This comparison holds the buck's efficiency fixed at the number you enter, which is a first-order simplification: real switching-regulator efficiency falls at light load and moves with input voltage. Heat is also not the only criterion — an LDO has no switching ripple and needs fewer parts. For a buck loss model built from the switch, inductor and capacitor parameters, use the buck converter calculator.

References

  1. Texas Instruments, Linear Regulator Design Guide for LDOs, SLVA118A.
  2. Texas Instruments, Semiconductor and IC Package Thermal Metrics, SPRA953.

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