My LDO circuit might be getting hotter than it should. Advise? - Page 1
Forum section: electronics / projects

author: guscrown
author_rank: Regular Contributor
timestamp: July 17, 2015, 09:56:58 pm
content: 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 hubFTDI Virtual COM port (the peripheral is a serial display).ESD/EMI protection for the USB ports1 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?

author: sleemanj
author_rank: Super Contributor
timestamp: July 17, 2015, 10:17:45 pm
content: 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 13Add a big heatsink

author: helius
author_rank: Super Contributor
timestamp: July 17, 2015, 10:22:46 pm
content: The datasheet from Micrel has an in-depth discussion of heat dissipation budget and how to find it.http://www.micrel.com/_PDF/mic39100.pdflook on page 12.With a power dissipation of ~2.2W, the amount of copper needed to spread the heat will be very large.

author: edavid
author_rank: Super Contributor
timestamp: July 17, 2015, 10:24:45 pm
content: Forget the LDO and buy a step-down module from eBay.

author: wraper
author_rank: Unknown
timestamp: July 17, 2015, 10:41:35 pm
content: Quote from: edavid on July 17, 2015, 10:24:45 pmNot 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.

author: wraper
author_rank: Unknown
timestamp: July 17, 2015, 10:51:07 pm
content: 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.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: July 18, 2015, 12:05:55 am
content: 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

author: Dago
author_rank: Frequent Contributor
timestamp: July 18, 2015, 12:11:28 am
content: 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.

author: sleemanj
author_rank: Super Contributor
timestamp: July 18, 2015, 12:32:00 am
content: Quote from: T3sl4co1l on July 18, 2015, 12:05:55 amOr a ceramic with a series resistor.

author: mariush
author_rank: Super Contributor
timestamp: July 18, 2015, 01:14:46 am
content: 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 likeAP3211orNCP3170orAOZ1282etc etc, something high frequency to keep inductors and capacitors tiny and not take much space.

author: guscrown
author_rank: Regular Contributor
timestamp: July 18, 2015, 03:01:17 am
content: 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.