My LDO circuit might be getting hotter than it should. Advise?

[guscrown]

Alright, so I'm making a small interface board to connect two USB devices to a single USB port on a host machine. I added the following:

2-port USB hub
FTDI Virtual COM port (the peripheral is a serial display).
ESD/EMI protection for the USB ports
1 connector for the other peripheral (USB Keypad)

The USB hub requires 3.3V supply, so I added an LDO MIC5209-3.3YS TR to go from 12V to 3.3v, this guy appears to be working fine.

For the VBUSes and the Serial display I added another LDO MIC39100-5.0WS TR to go from 12V to 5V. I measured the current consumption it is around 310mA.

Everything appears to be working fine, but I noticed that the board was getting warm, so I took out my Fluje 561 and measured a MAX temperature of 125F over the 5V LDO. I might have messed up the layout and I might be missing some copper that should be there to aid in temperature dissipation, although I am not drawing the maximum for the LDO, which according to the datasheet it is 500mA.

This is my schematic:



And this is a 3D View of that area:



This the top layer view (the top layer is copper flooded with GND)



This is the bottom side of that area:



This is a top view of the whole board:



And bottom:



12V comes in from the left side via J1, and it goes to a Ferrite Bead FBMH2012HM121-T (rated at 2A), I've highlighted 12V here:



And this is 5V0:




What did I miss?

[sleemanj]

a linear reg works by losing all the extra power as heat.

12v in, 5v out, at 300mA means ((12-5)*0.3) Watts to dissipate.

A surface mount reg is usually rated with a square inch of copper on the tab.  And it's "current rating" in he summary will always be some best case.

Look at the charts in the datasheet on page 13

Add a big heatsink

[helius]

The datasheet from Micrel has an in-depth discussion of heat dissipation budget and how to find it.
http://www.micrel.com/_PDF/mic39100.pdf look on page 12.

With a power dissipation of ~2.2W, the amount of copper needed to spread the heat will be very large.

[edavid]

Forget the LDO and buy a step-down module from eBay.

[wraper]

Forget the LDO and buy a step-down module from eBay.

Not for any serious job, as they they don't meet specs and very often made with fake ICs. But yeah, buck converter is adequate for this. Then 5V -> 3.3V LDO instead of 12V -> 3.3V to minimize loses.

[wraper]

BTW, adding 0.1uF ceramic capacitor on the output is really stupid idea. It is completely against what is suggested in the datasheets for MIC5209 and MIC39100 and can cause oscillation.

[T3sl4co1l]

As long as the lossy capacitor is dominant (a tantalum or electrolytic with ESR), a smattering of 0.1's is perfectly acceptable.

What matters is the impedance at the cutoff frequency, where the LDO is most sensitive.  If it isn't lossy (R dominant) at that frequency, it becomes unstable.

Regarding thermal: I see no vias around.  You're wasting half the power dissipation of your PCB there!

Tim

[Dago]

Your 5 V regulator (U7) is wasting 12-5*0.3 = 2.3 W of power which is a LOT without a heatsink. No wonder it is getting hot.

[sleemanj]

a tantalum or electrolytic with ESR

Or a ceramic with a series resistor.

[mariush]

Even if you get it right, a ldo is a bad choice. You will heat your whole board (and product overall) and waste power.

I'd suggest moving to a switching regulator, something like AP3211 or NCP3170  or AOZ1282 etc etc, something high frequency to keep inductors and capacitors tiny and not take much space.

[guscrown]

Thanks all for the help. I'm going to change over to a switching regulator, the thought was always in the back of my mind, but I thought I could manage it with the low current.