DPAK Regulator (MC78M05CDTG) Overheating at 24V Input?

[Censtry]

Hi everyone,I've been lurking on this forum for a while, and I've finally decided to make my first post because physics is beating me up.I am currently working on a small DIY board and running into a classic thermal issue with a linear regulator. I chose the ON Semi MC78M05CDTG (5V, 500mA output in a TO-252/DPAK package) to step down a 24V rail to 5V for a local microcontroller.The MCU and a few status LEDs draw around 150mA in total. Based on my understanding:$$P = (24\text{V} - 5\text{V}) \times 0.15\text{A} = 2.85\text{W}$$Even though 150mA is way below the 500mA maximum rating of the chip, the DPAK package is getting blazing hot to the touch within 2 minutes (hitting around 85°C on a standard 2-layer PCB). I poured a decent amount of copper plane on the tab pin (connected to GND), but clearly, it's not enough to dissipate nearly 3W of heat.A few quick questions for the seasoned pros here:Is 2.85W simply pushing a DPAK package too far without a dedicated clip-on heatsink?Should I bite the bullet and switch to a buck converter, or is there a clever PCB layout trick I am missing to keep a linear regulator cool?Would love to hear your thoughts!

[Ian.M]

Assuming the load current doesn't spike above 150 mA, feed Vin through a 100 ohm 3W resistor!

[dobsonr741]

Yep, your math is the whole story here.

Back-of-napkin:

P = (24V - 5V) * 0.15A
P = 19V * 0.15A
P = 2.85W

Now take a rough DPAK thermal resistance. On a real 2-layer board, even with some copper, call it maybe 40–60°C/W unless the layout is very generous.

Temperature rise = 2.85W * 50°C/W ≈ 143°C

So even at 25°C ambient, the junction wants to be around:

25°C + 143°C = 168°C

That is already in thermal-shutdown / bad-day territory. The fact that the package is hitting ~85°C externally makes perfect sense; the silicon junction is hotter than the case.

The 500mA rating is not the useful number here. That rating assumes you can get the heat out. With 24V in and 5V out, even 150mA is a lot for a linear regulator.

A buck converter is the right answer. Even a small cheap buck dropping 24V to 5V at 150mA will waste maybe a few hundred mW instead of 2.85W.

PCB tricks can help around the edges: more copper, vias to the other side, big ground pours, airflow, heatsink tab, etc. But they do not change the basic problem: you are burning almost 3W in a little plastic package.

So: not a current-limit problem, a dissipation problem. Use a buck, or at least pre-drop the voltage before the 7805 if you insist on keeping the linear.

[Smokey]

https://www.onsemi.com/pdf/datasheet/mc78m00-d.pdf

Page2.

DT Package:
Thermal Resistance, Junction-to-Air = 92 degC/W
Thermal Resistance, Junction-to-Case = 5 degC/W

Page 7 plot:


Figure out how much actual copper you have around the regulator and then you can see what the plot says about how hot it should get at 2.85W.

[Whales]

EDIT: 3 posts before I finished writing!  I'm slow today

Datasheet max limit numbers are all independent requirements; ie the 500mA should be treated only as a max current limit, not a thermal limit.  For example the chip also claims a max voltage of 35V, but it has zero chance of doing that at the same time as 500mA (*= 18W) continuously.

3W from an IC into a 2-layer PCB, assuming you have no forced air or other heatsinking, does seem like a bit much.

If the outside of the D2PAK is 85degC and the thermal resistance between the junction (silicon) and case (metal tab) is 5degC/W then the silicon die is probably about 100degC.  That's well within the chip's 150degC limit.  Alas, you might be measuring the temperature of the plastic bit of the case, not the metal tab?  And you only did a short term test? 

Sidenote:Thermal cameras are awful at reading small reflective metal surfaces like IC metal tabs, you will probably want to use something like a small thermocouple taped on.

You could take some of your 3W away from the regulator IC and share it with some large series (24V-side) resistors.  They can handle high temperatures well.  The system will still dissipate the same 3W total, but a smaller fraction of it will be in the regulator IC.    Make sure to calculate their resistances assuming a highest-worst-case current draw so that the regulator will still see its minimum acceptable voltage (usually the input voltage needs to be a few volts above the output voltage for regulation to work reliably, check the datasheet).

I would always encourage trying some buck converters.  On this note, please don't use a 34063, it is a very old design that requires bigger parts and has far fewer protections than is considered normal these days (it blows up violently if you make a mistake).  There are lots of 6-pin buck regulators (usually SOT23-6) out there that need nothing more than some resistors, capacitors and one inductor to work.  Eg the DIODES AP63/64 series parts like this one, or a TI TPS part.  I found these parts by sorting by highest stock levels, but there are also dedicated online tools like TI WEBBENCH.

[Psi]

When dealing with 24V my go to solution is a 7805 7803 style drop in DCDC module.
Unless I can be bothered to design a dedicate one myself to lower costs.
The drop in 78xx style DCDC is the quickest option.

You usually spend more trying to add thicker copper to the PCB than trying to replace with a DCDC.

[Smokey]

I did a design with a AP64060QWU-7 recently.  Tiny, inexpensive, integrated fets.  2.2MHz switching so tiny inductor.  Low capacitance requirements.  Super straightforward layout. 

https://www.diodes.com/datasheet/download/AP64060Q_AP64060TQ_AP64060ZQ.pdf

Just checked. Total layout footprint was only slightly bigger than a DPAK (and just the package, not the huge surrounding LDO pour).

[xecor]

1. Why is it getting so hot within two minutes?A 2.85W power dissipation is way too high for a DPAK package. Even though your 150mA current draw is well within specs, the voltage drop is massive ($24\text{V} - 5\text{V} = 19\text{V}$). A linear regulator acts just like a giant heating resistor, calculating as:$$(24\text{V} - 5\text{V}) \times 0.15\text{A} = 2.85\text{W}$$Without a large external heatsink, a standard DPAK package can usually only handle around 1W max safely.2. How do you fix it? (Two Options)The most hassle-free fix: Switch directly to a DC-DC Buck Converter. With over 85% efficiency, the heat dissipation will drop to around 0.1W, and the chip will stay completely cool. You can search the Xecor platform for pin-to-pin compatible 3-terminal switching regulator modules that can directly drop into your existing 7805 footprint without a board redesign.If you cannot modify the board layout (PCB Tweak): Drill a tight matrix of thermal vias directly inside the large DPAK exposed ground pad to dump the heat into the bottom layer or inner ground (GND) planes.Bottom line: Don't waste your sanity fighting thermal limits on a linear regulator. Head over to the MOD REMOVED LINK (SAME COMPANY NAME AS USER) Component Marketplace and grab a drop-in DC-DC buck module to fix it once and for all!

[squadchannel]

Head over to the https://www.xecor.com/ Component Marketplace and grab a drop-in DC-DC buck module to fix it once and for all!

What was that all about? Just a spammer? 
weird marketing.

Hi everyone,

I’ve been reading the EE Vblog forum and watching Dave's videos for years, and I finally decided to stop lurking and register this account (xecor).

I'm just a regular electronics hobbyist who loves spending weekends with a soldering iron, troubleshooting weird circuit behaviors, and trying not to let the magic smoke out of my components. My bench is currently a mess of microcontrollers and random analog parts.

I created this account so I can finally join the discussions, ask some dumb questions when I get stuck, and learn from the massive amount of experience here.

Glad to finally be a registered part of the community!

Cheers,
xecor

[Smokey]

Long con... I like it..

[stj]

incase it hasnt been mentioned yet, your generating heat and transfering it to the copper track - until it's saturated.
then the whole mess will just get hotter and hotter until your regulator goes into thermal shutdown.

you have to DISIPATE the heat with airflow.
or preferably not try to convert 19v into heat to begin with.
use a buck convertor and if noise is an issue you buck to say 7v and then use an LDO like the 78m05 or a 1117 afterwards.