# Trouble with overheating LDOs

#### kramzar

Joined Mar 7, 2022
28
I am designing a custom pcb with a 3.3V and 5V LDO power regulator that converts 24V DC input to 3.3 and 5V to power a microcontroler , ethernet modbus transcievers, and some sensors. The whole system drawns maximum 200mA of current during operation

5V LDO = L78M05ABDT-TR
3.3V LDO= LM317AG-TN3-R

The board works well when I power it directly on the 3.3V rail via usb. But when I power it through the LDOs, they work for about 2 mins before they get very hot and the current starts droping (presumebly because the internal thermal shutdown kicks in).

My question is why is this happening? The LDOs are both rated for much higher currents and voltages, yet already start failing at 24V input.
The system consumes about 200mA of current during normal operation, which is also far below the rated limit.
I am using an external adjustable power supply to power it, which can supply up to 5A.

I also tried powering the system with 8V DC and it worked fine for an extended period, and while the LDOs did get hot, they didnt get too hot to touch or started droping current.

Do i need better cooling or is there an input/load resistor i am missing that i should add?

I am attaching a picture of the schematic and PCB layout of the power supply section.

#### ericgibbs

Joined Jan 29, 2010
17,410
hi k,
Have you calculated the power dissipation in the LDO's.
E

#### MrSalts

Joined Apr 2, 2020
2,767
0.2A * (24v - 3.3v) = 0.2A * 21.7v = 4.2 W. Oh, that poor LDO! Scalded to death!

#### Sensacell

Joined Jun 19, 2012
3,180
Better to use a switching regulator for the 5V, then use an LDO for the 3.3V, powered from the 5V.

Linear regulators suffer terrible power efficiency when dropping large voltages.

#### crutschow

Joined Mar 14, 2008
32,026
As noted, as well as a current rating, the devices have a power dissipation rating (as determined their heat -sink, if any).
That power equals (Vin-Vout) * Iout.
That power needs to be dissipated by an appropriate heat-sink.
Alternately, you can reduce the input voltage by putting a power resistor in series with the LDO input.

#### MrSalts

Joined Apr 2, 2020
2,767
Linear regulators suffer terrible power efficiency when dropping large voltages.
They work by turning the excess energy to heat. I'd call that inefficient unless the heat is needed somewhere.

#### dl324

Joined Mar 30, 2015
15,535
5V LDO = L78M05ABDT-TR
3.3V LDO= LM317AG-TN3-R
Why are you calling these regulators low drop out? They look like typical step down regulators to me. Both have a drop out voltage of about 2V.

At 200mA, the LM317 would be dissipating over 4W. Thermal resistance (junction to ambient) is 62 degrees C per W, so it wouldn't take long for thermal protection to kick in at half that current.

Do as recommended in post #4. Use a switching regulator for 5V and use an LDO regulator on the 5V output to generate 3.3V.