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Linear Regulator Power Dissipation Calculator

Estimate linear regulator or LDO power dissipation, efficiency, and thermal rise from input voltage, output voltage, and load current.

Inputs and tolerances

Estimate power loss, output power, input power, and ideal efficiency for a linear regulator or LDO stage.

Calculate
Input voltage (V)
Output voltage (V)
Output current (A)
Thermal resistance θJA (°C/W)
Ambient temperature (°C)

A linear regulator turns the input-output voltage difference into heat at the load current. Large Vin-Vout differential and high current quickly create package and PCB thermal problems.

Nominal results and guaranteed range

Regulator power loss1.4WRange: 1.008W to 1.848W
Output current200mARange: 160mA to 240mA
Input voltage12VRange: 11.4V to 12.6V
Output voltage5VRange: 4.9V to 5.1V
Ideal efficiency41.667%Range: 38.889% to 44.737%
Input power2.4WRange: 1.824W to 3.024W
Output power1WRange: 784mW to 1.224W
Voltage across regulator7VRange: 6.3V to 7.7V
Temperature rise112°CRange: 64.512°C to 177.41°C
Estimated junction temperature137°CRange: 89.512°C to 202.41°C
  • Warning: Regulator dissipation is above 1W. Package, copper area, airflow, and derating need attention.
  • Warning: Ideal linear efficiency is below 50%. A buck converter may be a better architecture.
  • Warning: Estimated junction temperature is at or above 125°C. Check the datasheet maximum and thermal shutdown margin.

This is a first-pass linear-regulator estimate. It does not model dropout, safe operating area, transient load, package protection behaviour, copper spreading, airflow, switching regulators, or datasheet-specific derating.
Why this matters

Linear regulators turn voltage headroom into heat

An LDO or linear regulator is simple, quiet, and useful, but the difference between input and output voltage is dissipated directly in the regulator at the load current. That makes thermal margin one of the first checks for simple power stages.

Dissipation check

Calculate how many watts the regulator package and PCB must remove.

Efficiency check

Ideal linear efficiency is Vout divided by Vin, before quiescent-current details.

Thermal screen

Use θJA and ambient temperature for a first-pass junction-temperature estimate.

Worked example

For a 12 V input, 5 V output, and 200 mA load, the regulator drops 7 V. The dissipation is 7 V x 0.2 A = 1.4 W. That is already large for many small packages unless the PCB provides a good thermal path.

Power balance

Pin = 12 V x 0.2 A = 2.4 W.

Pout = 5 V x 0.2 A = 1 W.

Ploss = 2.4 W - 1 W = 1.4 W.

Thermal implication

If θJA is 80 °C/W, 1.4 W gives about 112 °C temperature rise above ambient.

At 25 °C ambient, that estimates about 137 °C junction temperature before tolerance and layout uncertainty.

Common mistakes and limits

Confusing linear and switching efficiency

A linear regulator cannot trade voltage for current like a buck converter. The extra voltage becomes heat.

Ignoring dropout

This calculator requires Vin greater than Vout, but it does not verify dropout margin from the regulator datasheet.

Over-trusting θJA

Thermal resistance depends on package, copper area, vias, airflow, enclosure, and board stack-up. Treat it as a screening estimate.

Related calculators and next checks

Next check

Linear regulator thermal calculator

Turn regulator dissipation into junction temperature, thermal margin, and maximum output current.

Next check

Buck converter calculator

Compare against a switching regulator when linear dissipation is too high.

Next check

Battery life calculator

Fold regulator efficiency and current draw into a battery runtime estimate.

Next check

Regulator thermal guide

Review thermal resistance, ambient temperature, and junction-temperature margin.

Engineering reference

Equations, assumptions, and design guidance

Engineering approximation

Estimates linear regulator or LDO dissipation from input voltage, output voltage, and output current, with optional first-pass thermal rise.

Equations and variables
Power lossPloss = (Vin - Vout) * Iout
Output powerPout = Vout * Iout
Input powerPin = Vin * Iout
Ideal efficiencyeta = Vout / Vin
Temperature riseTrise = Ploss × θJA
Vin
Regulator input voltage (V)
Vout
Regulator output voltage (V)
Iout
Output current (A)
θJA
Junction-to-ambient thermal resistance (°C/W)
Assumptions and limitations

Assumptions

  • The device is a linear regulator or LDO, not a switching regulator.
  • Input current is approximated as output current, ignoring quiescent current unless it is folded into the load current.

Limitations

  • Dropout voltage, quiescent current, thermal shutdown, package copper spreading, airflow, safe operating area, transient loads, and datasheet-specific derating are not modelled.
Worked example and design use

12 V to 5 V at 200 mA

Inputs: Vin = 12 V, Vout = 5 V, Iout = 0.2 A

Outputs: Ploss = 1.4 W, Pout = 1 W, Pin = 2.4 W, ideal efficiency = 41.7%

Design guidance

  • Use this result early to decide whether a linear regulator is thermally realistic.
  • For high loss or low efficiency, compare against a buck converter before committing the architecture.