I need to select a LDO with output voltage of 3.3V for my design.

My load requirement is 300 mA maximum.

The LDO is powered by either USB 5 V (VBUS) or by an external power source ranging from 6 V to 8.5 V (VBEC.)

I have used 2 diodes to switch between the two input sources. I assembled and tested the circuit using a MaxLinear SPX3819M5-L-3-3/TR as show in the below schematic.

The circuit is working fine and the load is around 200 mA now but I am noticing that the LDO is getting much too hot.

Did I choose the wrong LDO?

Can somebody suggest a better alternative, knowing that I don't want anything bigger than SOT-89 because the space on my PCB is very limited?

enter image description here

  • 3
    \$\begingroup\$ "getting much too hot" Meaning what? It's normal for them to sit somewhere between 40-70°C during operation, depending on currents and heat sinks. There's usually a thermal shutdown around 85°C or so to prevent damage to the part, but you don't want that to happen anyway - because when a LDO goes into thermal shutdown, it starts to cool off, wakes up again, heats, thermal shutdown... all of it causing wild oscillations on the output voltage. \$\endgroup\$
    – Lundin
    Commented Apr 25, 2023 at 6:49
  • 1
    \$\begingroup\$ You're trying to drop a lot of heat for such a small space so any linear circuit that little will get hot, whatever its components are. Which means you need a switching regulator. Selecting one that fits whatever space you can find on your board is a job best done by yourself on a decent distributor's website with a lot of examination of parts and possibilities. Only you understand the size and space concessions you will inevitably have to make. \$\endgroup\$
    – TonyM
    Commented Apr 25, 2023 at 7:08
  • 2
    \$\begingroup\$ Any LDO converting 5V to 3.3V at 300mA is going to waste about half a watt of heat. That's inherent to their design. If this isn't okay with you, use a buck converter instead. If this is okay with you but you think the temperature is too much, add a heatsink to spread out the heat over a bigger area. \$\endgroup\$ Commented Apr 25, 2023 at 15:04
  • \$\begingroup\$ Thanks guys for all the information and explanation, I will proceed and use a buck converter \$\endgroup\$ Commented Apr 26, 2023 at 5:41

3 Answers 3


Since you didn't specify what input voltage you used during your test, I will be using the maximum values you design your circuit for in this answer: an input of 8.5 V with a load current of 300 mA.

  1. Your diode has a forward voltage of ~0.6 V -> your LDO input voltage will be 8.5 V - 0.6 V = 7.9 V
  2. At an output of 3.3 V the LDO has to drop 7.9 V - 3.3 V = 4.6 V
  3. With a load current of up to 300 mA you will have a maximum power dissipation of 4.6 V * 0.3 A = 1.38 W

Let's check the LDO's datasheet to see what the maximum allowed power dissipation is (image taken from Maxims' SPX3819 datasheet): enter image description here

The wording is quite confusing, so I might be wrong on the following part:

  • The maximum allowed power dissipation is a function of (Tjmax - Ta)/ THETAja.
  • Tjmax would be the maximum junction temperature which is +125 °C
  • Ta is the ambient temperature, lets say 25 °C
  • THETAja depends on your package, I believe you are using the SOT23-5 in which case its 191 °C/W; you might want to recalculate it if that is wrong.

Using the formula that's (125 °C - 25 °C)/191 °C/W = 0.52 W

That means your LDO can dissipate at maximum 0.52 W of power, but you designed your circuit in a way that it might have to drop up to 1.38 W. I would definitely not recommend using that LDO/that package.

You'll either need to use a switching regulator (less losses) or another package (with the DFN-8 package you have a max. power dissipation of 1.69 W which would barely be in your range).

  • 3
    \$\begingroup\$ A big resistor in series with the LDO comes to mind too. \$\endgroup\$
    – winny
    Commented Apr 25, 2023 at 6:37

Linear regulators work by converting the voltage drop (multiplied by current) to heat. No LDO will do better. The only way to drop voltage without generating so much heat is to use a switching regulator.

There are now switching regulator modules designed as drop-in replacements for some LDOs, for example: Digikey. I can't address if any are small enough for your need.


You should provide a datasheet link for the LDO.

You can reduce the input voltage somewhat by using ordinary silicon diodes for DS1 and DS2. Your problem probably occurs mostly when powering from VBEC when it approaches the maximum of 8.5 VDC. The LDO will work down to a dropout voltage of 500 mV, So you could use a 2V zener so that the input will be a maximum of 6.5 V and a minimum of 4 V. A 2 V zener may be hard to find, but you could use three silicon diodes which will give you 1.8 to 2.1 V drop.

  • 1
    \$\begingroup\$ Or replace DS1 with a large-area RED LED that can take 300mA without overheating...that's roughly 2V. \$\endgroup\$
    – glen_geek
    Commented Apr 25, 2023 at 11:37
  • 1
    \$\begingroup\$ @glen_geek Heh, just a decade ago that would have been quite ineffective, but with best LEDs now radiating roughly 75% of their energy input as light, what you propose might actually reduce heating significantly (at least if it is not inside an enclosure). \$\endgroup\$
    – jpa
    Commented Apr 25, 2023 at 16:54

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