LDO Power Loss and Thermal Calculator
LDO Power Loss and Thermal Calculator
Work out how much heat a linear regulator dissipates, its efficiency and junction temperature, and the most current it can deliver before it reaches its temperature limit — after checking that the input clears the dropout voltage.
Linear regulator dissipation
5 V in, 3.3 V out at 0.5 A, 0.5 mA ground current, 0.3 V dropout, θJA 50 °C/W, 40 °C ambient, 125 °C maximum
Linear regulator dissipation
- IGND
- ground (quiescent) current, which flows from input to ground
- θ
- θJA, or θJC + θCS + θSA on a heatsink
- Vin − Vout
- must be at least the dropout voltage
Worked example
5 V in, 3.3 V out at 0.5 A, 0.5 mA ground current, 0.3 V dropout, θJA 50 °C/W, 40 °C ambient, 125 °C maximum
Headroom 5 − 3.3 = 1.7 V, above the 0.3 V dropout, so it regulates
PD = 1.7 × 0.5 + 5 × 0.0005 = 852.5 mW
Efficiency = 1.65 ÷ 2.5025 = 65.9%
Tj = 40 + 0.8525 × 50 = 82.6 °C
Most current before 125 °C: ((125 − 40)/50 − 0.0025) ÷ 1.7 = 998.5 mA
Heat in a linear regulator
A linear regulator — an LDO or an older three-terminal part — works by dropping the difference between input and output across a pass transistor. The output current flows through that transistor, so the power it turns into heat is simply (Vin − Vout) × Iout, plus the small ground current the control circuit draws from the input. The efficiency can never be better than Vout/Vin: 3.3 V from 5 V is at best 66%, and in the example 65.9%.
Dropout first. Below its dropout voltage an LDO cannot hold its output; it becomes a switch passing Vin minus a small drop. This page refuses a result when Vin − Vout is below the dropout you enter. Dropout rises with current and temperature, so read it at your load and allow for ripple on the input, which must stay above the dropout at its lowest point.
Temperature. The dissipation times the thermal resistance gives the rise above ambient. A datasheet θJA comes from a standard JEDEC test board (JESD51) with a set copper area; a small board or a hot enclosure can make it much worse, generous copper much better. With a heatsink, use θJC + θCS + θSA instead. The chart shows the junction temperature climbing in a straight line with load current until it meets the limit; at 5 V in the example that is at 998.5 mA.
When it runs hot. Lowering the input voltage is the most effective fix: the dissipation is proportional to the drop. A resistor in series with the input can take part of the heat off the regulator (keep the input above dropout at full load). If the drop and current are both large, a switching regulator such as a buck converter wastes far less — see the buck converter designer. For the heat path itself use the heatsink thermal resistance calculator, and for a battery-powered design the battery life calculator.
Frequently asked questions
How do you calculate LDO power dissipation?
PD = (Vin − Vout) × Iout + Vin × Iground. 5 V to 3.3 V at 0.5 A with 0.5 mA ground current is 852.5 mW.
What is the efficiency of a linear regulator?
Vout × Iout ÷ (Vin × (Iout + Iground)), which can never exceed Vout ÷ Vin.
How hot will my LDO get?
Tj = Ta + PD × θJA. 0.8525 W on 50 °C/W at 40 °C ambient is 82.6 °C.
What happens if the input is below the dropout voltage?
The regulator can no longer regulate: the output follows the input minus the pass transistor’s drop, and noise and ripple pass straight through.
Related calculators
References
- MaxLinear (Exar). Thermal Considerations for Linear Regulators. Application note ANP-2, 2006. PD = (VIN − VOUT)·IOUT + VIN·IGND; TJ = TA + PD·θJA.
- Fwu W. LDO Basics: Thermals — How Hot Is Your Application? Texas Instruments technical article SSZT950, September 2017.
- JEDEC JESD51-2A. Integrated Circuits Thermal Test Method Environmental Conditions — Natural Convection (Still Air). JEDEC Solid State Technology Association, 2008. The conditions under which a datasheet θJA is measured.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Ch. 1 (RC circuits, the time constant, RC low-pass and high-pass filters).