Reducing the temperature of a device stored inside a waterproof case - Page 1
Forum section: electronics / projects

author: mcinque
author_rank: Unknown
timestamp: April 17, 2014, 12:00:53 pm
content: I'm building a device that will be housed in a plastic, 50x50x10cm waterproof briefcase and will be placed outside my house, in the countryside. Basically it's a radio datalogger that will monitor the temperature and the moisture level and send me the information thru 433MHz packets.The climate sensors will be external, connected thru a waterproof connector.I'm using a linear regulator for the 5V rail, and the total power drain by the ciruitry is 1,4W.With a so low efficency device (about 35%), there is 0,9W to heat to dissipate (a lot of wasted energy), so I must have an external heatsink, otherwise with moderate or even warm climates, the heat will grow inside the briefcase and possibly will blow up the circuitry . The waterproof sealing don't allow any air flow inside.I will do some tests in the next days using a DC-DC converter with an efficency of 78% instead the linear regulator, having so "only" 0,3W of heat to dissipate, but I'd like to have a couple of advices from you:are 0,3W of heat, something that I can ignore since the poor thermal insulation of the briefcase will equalize any difference?I mean, in summer time the direct sunlight would raise quickly the internal temperature up to 60°C, and I guess that 0,3W of internal heat could be ininfluent for the circuitry (while 0,9W is too much).What do you think about it?

author: turbo!
author_rank: Regular Contributor
timestamp: April 17, 2014, 12:26:40 pm
content: With the level of power you're talking about, there's no way this is dry cell powered. Just use an external 5.0v regulated AC adapter.

author: SL4P
author_rank: Super Contributor
timestamp: April 17, 2014, 12:32:57 pm
content: Direct sunlight is probably going to be your biggest headache.Possible options area heatpipe to an externalheatsink,a heatsink sealed to the case with a gasket,and active components inside,a peltier setup.All ways - make sure to accommodate condensation

author: turbo!
author_rank: Regular Contributor
timestamp: April 17, 2014, 01:01:45 pm
content: Quote from: SL4P on April 17, 2014, 12:32:57 pmDirect sunlight needs to be avoided anyways if you expect meaningful temperature reading. You're reading the air temperature. Not how hot the sunlight can get the thermistor tip.

author: max666
author_rank: Frequent Contributor
timestamp: April 17, 2014, 01:11:25 pm
content: To deal with direct sunlight, I would mount a plate in front of your briefcase, offset by a few millimetres so that there's air in between. This way the plate will catch most of the heating due to sunlight.

author: Codemonkey
author_rank: Regular Contributor
timestamp: April 17, 2014, 01:17:54 pm
content: Paint it white, that should help prevent it warming up due to sunlight a bit or as others have suggested, shade it somehow Other than that I would just try it. You could always drill a few small holes under the bottom and place it such that rainwater cannot flow uphill into it.

author: bookaboo
author_rank: Frequent Contributor
timestamp: April 17, 2014, 01:42:03 pm
content: I've never heard of low power electronics needing any sort of cooling, even with high ambient temperatures. Think of some of the industrial applications electronics are used in, ambients of 60 Celsius plus.You could do an experiment with a power resistor and a datalogger before you build your actual circuit.

author: KedasProbe
author_rank: Frequent Contributor
timestamp: April 17, 2014, 01:47:04 pm
content: How big is it? 0.3W is probably not enough to worry about.If you want to know how much energy is transferred lock up some warm object (water?)and record start time/temp and end time/temp and try to keep surrounding temp constant.This will tell you how fast energy gets out (watts) for a certain temp diff.

author: macboy
author_rank: Super Contributor
timestamp: April 17, 2014, 04:00:58 pm
content: Quote from: mcinque on April 17, 2014, 12:00:53 pm50x50x10 cm gives 4500 cm2of surface area. The plastic does not conduct heat like a metal will, but it does conduct some heat, and with that impressive surface area, you probably won't have a problem. You can look up the specifics for the type of plastic you are using if you are interested in doing the engineering on this.I would think of this as a two part system. One part is the electronics dissipating their heat into theinternalambient temperature, which will be elevated above theoutsideambient. The second part is the dissipation of heat from the internal (heated) air to the external air, through those 4500 cm2of plastic (having thickness of...? and thermal resistance of ...? per ?). It is not terribly difficult to do the math to come up with the thermal resistance of your enclosure system (air to air) in degC/W, and once you have that you can determine the elevation of temperature from inside to outside. Make an assumption for the maximum outside temperature, then you have the maximum internal temperature. Then determine if your electronics will be happy in that. They may simply require adequate heatsinking to be fine.Or just build it and see. If it's a one-off personal use thing, I'd use a big heatsink on the regulator (maybe a large aluminum plate in contact with one of the enclosure walls), and just try it and see.

author: georges80
author_rank: Frequent Contributor
timestamp: April 17, 2014, 04:50:44 pm
content: Quote from: mcinque on April 17, 2014, 12:00:53 pm5V ?? Seems like most everything these days can be had as 3.3V and highly efficient. What's your design? Seems like you are drawing close to 300mA (from 5V 1.4W). I can't imagine something that just measures temperature and moisture drawing that much current. Even a Digi 900MHz multi-mile range radio draws well under 200mA WHILE it is transmitting.78% efficient DC-DC converter sounds horrific, you should easily find something in the 90%+ range.Since you are after suggestions, I'd suggest you look over your power management for your design. Seems you could easily be down in the 10's of mA or less average consumption, with peaks when you transmit a data packet and that would only take fractions of a second.Take care of your average current in the design and most of your thermal worries are resolved. For the case, I'd get a sheet of aluminium and create a 'roof' that goes over the box to create an air gap and to keep the sun off the case and only if I was in a HOT area of the world. Otherwise just spray your case with one of the chrome/metalic spray paints that you can buy over the counter.cheers,george.

author: dannyf
author_rank: Super Contributor
timestamp: April 17, 2014, 04:55:51 pm
content: QuoteIt probably will blow up more from the sunlight than from internally generated heat.Thus, how you insulate it will depend on where the heat comes from: if the heat comes from inside, you want to spread the heat out; if the heat comes from the outside, you want to insulate your device from it - the exact opposite approach.So try to define your problem before you attempt to find a solution for it.

author: ion
author_rank: Regular Contributor
timestamp: April 17, 2014, 05:41:27 pm
content: Probably the best way to see what effect that 0.9W has is to dissipate it in the case and monitor the temperature rise. At these power levels, something like a LM317 is a good source of heat.The calculator at the bottom of this page is a quick way of figuring out the voltage and current you need for the heat dissipation you want:http://mikeyancey.com/hamcalc/LM317resistors.phpProbably the simplest setup would be to configure the LM317 as a current regulator (just add a resistor) and short it out. Don't forget to add the power dissipated through the resistor in this case. This calculator could be useful:http://diyaudioprojects.com/Technical/Current-Regulator/A small heatsink on the regualtor would probably be a good idea.And as others have said, the outside environment will effect the result.

author: Dago
author_rank: Frequent Contributor
timestamp: April 17, 2014, 06:08:03 pm
content: Just use some modern switching regulator and the efficiency should be close to 95% and then you have no heat issues to worry about.

author: mcinque
author_rank: Unknown
timestamp: April 17, 2014, 08:11:10 pm
content: First of all thank you for all your replies, I appreciate them and I took ALL of them in consideration.I really like the macboy's mathematical approach of the problem. Unfortunately I'm illiterate on this@Georges80: I'm not been much descriptive about the device but there is a lot of other components that needs to stay there and that drain that current.@Dannyf: the heat is coming from inside, I can avoid direct sunlight: I want to prevent excessive inner heat buildup, I fear that without air ventilation the heat could be trapped inside and that can grow exponentially in few hours.After reading a couple of replies about efficency I'm pretty confident that a switching regulator should do the job, I'm gonna investigate in the next daysThank you again for sharing your ideas and providing advices.

author: dannyf
author_rank: Super Contributor
timestamp: April 17, 2014, 08:17:39 pm
content: QuoteI thought there are a few other forms of heat transfer, aside from convection.

author: hagster
author_rank: Frequent Contributor
timestamp: April 17, 2014, 08:51:16 pm
content: The maths is pretty simple if you ignore convection and radiation and focus on conduction. It's really just like calculating voltage drops a cross resistors. I.e temperature drop a across thermal resistance. Don't worry about specific heat capacity as this just effects the lag and not the end temperature rise in the box.Or try this neat free simulator. It can even do solar loading and wind.http://energy.concord.org/energy2d/

author: mcinque
author_rank: Unknown
timestamp: April 17, 2014, 08:54:01 pm
content: Thank you for the link!!