4-2. Calculating the power dissipation and junction temperature of an LDO regulator

๐‘ƒ = (๐‘‰๐ผ๐‘ โˆ’ ๐‘‰๐‘‚๐‘ˆ๐‘‡) ร— ๐ผ๐‘‚๐‘ˆ๐‘‡ + ๐‘‰๐ผ๐‘ ร— ๐ผ๐ต ๏ฝฅ๏ฝฅ๏ฝฅ(1)

The power dissipation (P) of an LDO regulator can be calculated as follows:

๐‘ƒ = (๐‘‰๐ผ๐‘ โˆ’ ๐‘‰๐‘‚๐‘ˆ๐‘‡) ร— ๐ผ๐‘‚๐‘ˆ๐‘‡ ๏ฝฅ๏ฝฅ๏ฝฅ(2)

When the term VIN ร— IB is negligibly small, Equation 1 can be approximated as follows:

๐‘‡๐‘— = ๐‘ƒ ร— ๐‘…๐‘กโ„Ž (๐‘— โˆ’ ๐‘Ž) + ๐‘‡๐‘Ž ๏ฝฅ๏ฝฅ๏ฝฅ(3)

The junction temperature Tj of the LDO regulator can be calculated as follows:

๐‘…(๐‘กโ„Ž(๐‘—โˆ’๐‘Ž)) = (๐‘‡๐‘—(๐‘š๐‘Ž๐‘ฅ) โˆ’ 25) / ๐‘ƒ๐ท ๏ฝฅ๏ฝฅ๏ฝฅ(4)

where Rth(j-a) is the thermal resistance of the LDO regulator when it is mounted on a board. If Rth(j-a) is not shown in the datasheet, you can calculate it from the power dissipation (PD) value:

๐‘‡๐‘— = ๐‘ƒ ร— (๐‘‡(๐‘—(๐‘š๐‘Ž๐‘ฅ)) โˆ’ 25) / ๐‘ƒ๐ท + ๐‘‡๐‘Ž ๏ฝฅ๏ฝฅ๏ฝฅ(5)

Substituting Equation 4 into Equation 3, we obtain:

๐‘‡๐‘— = (3.6 โˆ’ 1.8) ร— 0.1 ร— (150 โˆ’ 25) / 0.6 + 40 = 77.5ยฐC

Suppose that Tj(max) = 150ยฐC, PD = 600 mW, VIN = 3.6 V, VOUT = 1.8 V, IOUT = 100 mA, Ta = 40ยฐC. Then, Tj is calculated as follows:

Note that the power dissipation and thermal resistance values in the datasheet are the values under the specified board conditions. It is therefore necessary to evaluate LDO regulators carefully to ensure that they work properly under actual usage conditions.

The following application notes also provide a description of the power dissipation and junction temperature of an LDO regulator.

Application note: Basics of Low-Dropout (LDO) Regulator ICs

Application note: Thermal Design to Maximize the Performance of LDO Regulators

Application note: Power Efficiency Optimization and Application Circuits Using Dual-Supply LDO Regulators

Chapter4 Efficiency and power loss calculation for LDO

4-1. Efficiency of LDO regulators
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Related information

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