Regarding using the LD39050PU33R voltage regulator which is 5V to 3.3V fixed voltage regulator without any heatsink, its maximum current is given as 500mA. I was trying to figure out how they calculated this 500mA. So I tried to calculate the maximum current by myself by using the thermal resistances and maximum junction temperature as follows:

From the thermal data below: enter image description here The junction to case thermal resistance R_jc is sum of those above; so R_jc = 65 °C/W.

For 5V input the voltage drop across the regulator will be Vd = 5V-3.3V = 1.7V.

Max junction temperature is 125°C. And in the worse case if we have 55°C ambient temperature the ΔT = 70°C.

That means maximum power dissipated Pd can be found from:

ΔT = Pd * R_jc

Pd = 70 / 65 = 1.077W

This makes the max current I = Pd / Vd then I = 1.077 / 1.7 = 633mA.

In my calculation I took max ambient temperature 55°C. To obtain 500mA, the ambient temperature would be taken as around 70°C. Is my conclusion correct?


1 Answer 1


The reason for the 500mA rated current is actually much simpler than that: The chip has an integrated current limit that may be as low as 600mA depending on tolerances (typically 800mA). Even if the chip is sufficiently cooled, it won't output any more than that. The manufacturer therefore rates it for 500mA, with some safety margin to the 600mA current limit.

Your calculation also breaks down for the other variants of the regulator (i.e. the fixed 1.0V version, or the adjustable version), which all have the same 500mA rated current - assuming sufficient cooling.

You are right in your conclusion that you can use the 3.3V variant of the regulator with 5V input voltage without any additional cooling at 55°C ambient temperature.


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