A Pro Mini needs to be placed in a 100mmx50mmx25mm, no holes but not air tight ABS plastic enclosure, with a 12v power supply connected to the raw input. It will be consuming max 40mA, ambient temperature 30-32C, and needs to be run 24/7. Upon prototyping in open air, I tried to touch the regulator MIC5205, after 30 seconds upon powering on with my bare fingers and it was too hot to touch. The rest of the components were cool to touch. I would use a buck converter to step down to 5v, however there isn't space for it in the current setup.

I used to run a Mega 24/7 on 12v, consuming much more current about 700mA, the onboard regulator malfunctioned after 1 year of use, which I believe was due to the excessive heat, as the PCB around the regulator turned a little brownish.

Despite being hot to touch, I understand that electronics run in heat all the time like computers, routers, etc for 24/7. Therefore, I would like to know if 12vx0.04A=0.48w will be too much heat/feasible to run in the above mentioned setup, without damaging the regulator in the long run?

  • Why is it running hot? 40mA shouldn't even warm up a 150mA regulator? Somethings wrong... Can you post the circuit (after checking it for shorts!)? – Mark Williams May 9 '15 at 10:55
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    @MarkWilliams - heat depends on power, which is voltage as well as current. With a very high voltage drop, even a small current can result in a hot regulator. – Chris Stratton May 9 '15 at 15:58
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    There almost certainly exist buck converters which will fit, but you can also see if you can sleep the processor a larger percentage of the time or reduce its clock speed, and so reduce the average current draw to a lower level. – Chris Stratton May 9 '15 at 16:00
  • Datasheet says 220 deg/Watt. 0.04A * 7v (drop) = 0.28 = 61 degrees. Not that hot, but yeah, uncomfortable. It has a thermal shutdown, so you could get brownouts if it gets hot - but much hotter than this. – Mark Williams May 9 '15 at 17:49
  • @MarkWilliams That's Junction to Ambient, so you need to add another 20C for ambient temperature (actually more, it's in a box) - so really 80C range and a burned finger. – Chris Stratton May 10 '15 at 5:36


Simple, effective, low cost solution - choose any 3:
A series resistor in the regulator input, of properly designed resistance and wattage will greatly reduce regulator temperature and increase overall reliability. Reason and design details given below.


Assuming a 5V processor dissiaption in the regulator is
(Vsupply - Vregulator_out) x Current
= (12-5)x 0.040 = 280 mW - usefully less than the 480 mW if you are dissipating all the power in the regulator.
For a 3V3 processor dissipation is (12-3.3)* 0.040 ~= 350 mW

"Too hot to touch" is a minimum of 55C and probably hotter - say 70C for now.
At 30C ambient delta T = (70-30) = 40 C.
Thermal resistanc = dT/Watts =40/0.280 = 140 C/W or 114 C/W at 3V3.

Both of those thermal resistances are very high. You do not describe what heatsinking is on the regulator but that suggests PCB copper only and not much of it. Even adding a very rudimentary heat sink should help immensely. If there is no obvious way to add a formal heat sink then adding a 'flag' of copper or brass by soldering should be doable.

Using Mark William's 220C/Watt figure (see datasheet for conditions).
Say 300 mW (see above).
Delta t = 220 C/W x 0.3 W = 66 C rise.
Ambient is 30-32 so allow 35 for very minor safety factor.
66 C rise + 35 C ~= 101C - water boils.
And temperature inbox will rise mildly from energy dissipated so hotter again.
BUIT a series resistor inside the box or external will make an immense difference.

An easy and effective solution:

The regulator needs some "headroom" - some voltage above the Vout value to allow some dropacross the internal switch. Check the data sheet, but a modern LDO probably allows as little as 1V headroom. So, for 5V out you need 6V in. The extra 6v from 12V to 6V is not "needed" by the regulator. So, it can be dissipated in a low cost high reliability resistor.

  • Vsupply_minimum - the absolute lowest value supply will assume.

  • Vreg_in_minimum I'll use 6V to start.

  • I_load_maximum - you said 40 ma. This must be abs_max or if of very short duration must be able to be supported by capacitors.


Rmax = V/I = (V_supply_min-Vregin_nax) / Imax
= (12 - 6 )/.040 = 150 Ohm.
This is the absolute maximum resistor that meets the specification. A slightly smaller value will be safer. Say 120 R in this case.
Pmax in R = I^2R = .04^2 x 120 = 192 mW.
A 500 mW resistor would be "safe enough" and a 1 watt one very safe.
Pregulator max = (6-5) x 0.040 = 40 mW, 14% of the 280 mW initially. A W resistor with adequate cooling will be very safe. Safest is probably a purpose built air cooled wire wound.

Note that the dissipated energy will raise internal ambient. A resistor is still an effective solution.

Place a suitable decoupling capacitor at the regulator input. Even if not normally used here it is needed due to the series input resistor raising the supply resistance.


Temperature rise inside box due to heat dissipation.

Summary: Using pessimistic assumptions for box thickness, available area and power dissipation an internal temperature rise of under 15 degrees C can be expected. As the regulator is now dissipating under 15% of the original dissipation due to the series resistor it will have a far longer operating lifetime.


Temperature rise (assuming no air transfer occurs) is limited by heat transfer through the plastic. This is controlled by the thermal conductivity of th e plastic, its thickness and its area. Obviously:
- Thicker plastic leads to higher temperature rise.

  • More surface area leads to lower temperature rise.

  • Higher thermal conductivity of the plastic leads to lower temperature rise

So given Thermal conductivity = Ktc

  • Delta T = Watts x thickness / (Ktc x area)

So Ktc = Watts x thickness /(Delta_Tx area)
So units of Ktc must be Watts/metre/degree_C
Which is indeed how Ktc is usually expressed.

This table of Thermal properties of plastic materials lists Ktc for ABS as 0.17 W/m/K
This means that 170 mW will cause a 1 degree K (or C) rise when plastic ratio of thickness to area is 1.

Assume thickness of box = 5mm = much higher than is likely.
Box area for a box L x W x D = 2 x (LW + WD + DL)
Here = L D W = 100mm x 50mm 25 mm or 0.1 x 0.05 x 0.025m Area = 0.0175 m^2.
Use a smaller more pessimistic area of 0.01 m^2
Use a pessimistic heat dissipation of 0.5 Watts

Using the formula - Delta T = Watts x thickness / (Ktc x area) from above

Temperature rise = 0.5 x 0.05 / (0.17 x 0.01) ~= 15 degrees C (or K)
This is based on a thicker than likely box, smaller than actual area, and higher than expected energy dissipation. A 15C rise is bearable and actual is lower, and as the regulator now dissipates under 20% of original it will survive very well.

| improve this answer | |
  • The series resistor is a great idea. But you need to know that a series resistor with a linear regulator will not reduce total heat generated: it will only take some of the heat away from the regulator and heat up the resistor. This is a benefit because it increases the surface are you are radiating from. BUT if it's all going inside the box it's not that much of a difference. If you can inline this resistor into the cable leading into the box this might give you the best cooling solution – portforwardpodcast May 11 '15 at 13:32
  • @portforwardpodcast Your points are covered in my existing answer. If you allow 1V regulator headroom, a 5V regulator with 6V into regulator and 12V supply will reduce dissipation in the regulator by a factor of 7 times. ie it dropped 7V in the regulator before and now drops 1V at full current. A 1 watt resistor operating at 240 mW inside the box with some internal temperature rise will be reliable and a 2 Watt resistor could be used with still very compact sizing. Removing 80%+ of the heat energy from the regulator will make a vast difference. | ... – Russell McMahon May 11 '15 at 14:02
  • ... The statement "... if it's all going inside the box it's not that much of a difference ..." is essentially not correct. You can calculate delta temperature from the total dissipation for a typical box and its not vast - the local temperature rise of the regulator is what kills things. I'll add a calculation for box temperature rise to show that it's minimal. But, as I noted in the answer* - the resistor can be external if desired. | [ *"BUT a series resistor inside the box or external will make an immense difference."] – Russell McMahon May 11 '15 at 14:22

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