Heatsink Calculations and Efficiency Theory

Warning! A lot of math!

Hi Guys. Im doing research as usual.
This might be useful as well. Ill be calculating heats for diodes and drivers. Also, since metals have such high conductivity, I think I'll just omit it.

So the other day, I got Eudaimonium's heatsink in the mail. Before I usually get parts, I do heavy mathematical calculations, experiments, and the whole nine yards. The only thing that is out of my view is heatsinking calculations. I've been helping some people with heatsinking, but I need to find a CLEAR and CONCISE heatsinking definition for future reference and use. Lets get crackin' then.

As far as I know, its pretty usable.

To begin with, I use the specific heat equation.
Q = (specif. heat value) (mass) (delta temp.)

where Q is heat added in JOULES
delta temperature is change in temperature for all you terrible math noobs and people who are not in algebra. -.-

My heatsink is 22 grams as labeled in the mail, so Ill use that.
Aluminum has a C(m) value of .900 (arbitrarily), which is the specific heat value
Im going to use 330 K for my heat(f) value as a point where the laser should not be operating since... its kinda hot? yeah.
Also, 295 K shall be my init. value.

Q = (.9) (22) (330-295)
Q = 495 Joules.

495 Joules... doesnt that sound a bit absurd for a high power laser to heat it up?
According to a few formulas, wattage is equivalent to volts x amperes. Since a 445 diode produces only 1 watt from 4.5watts of imput, 3.5watts of heat is generated due to inefficiencies. A watt is #Joules per second. Arranging the formula gives

495 / 3.5 = 141.428 ... seconds

So it takes 141 seconds or approx 2mins ans 20 secs to heat up the heatsink from a 445 diode at ideal conditions, where a 445 is run under 'rated' conditions, pure aluminum, omitted heat conductivity, and no heat is escaping from the heatsink. This is excluding the host itself.

Does this sound reasonable for a 445 to heat up that much? What are your calculations?

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As for heatsinking a driver I ran a few more calculations...
Since Im using a boostdrive, Ill be using the worst possible conditions .. aka
75% efficiency

So I have an ideal setup with 4.2 Volts and 1 ampere. The boost will boost it up to 4.5 volts at the cost of current. Ohms law again...
s will stand for subscript

4.2V x 1A(s.1) = 4.5 x A(s.2)

A(s.2) will need to equal .9333 ... so when it converts it will only convert 93% of the current to extra volts needed to drive. Since Im using a boostdrive... I need to apply the 75% efficiency, which will be nasty. Voltage will be constant, so Ill apply it to current.

933mA (Current after the conversion) x 0.75 = ~.700mA

What does this mean? Well, under these conditions, Im feeding 1 ampere from the battery to the driver to obtain only a mere 700ma to the laser diode. Other factors include the efficiency bottleneck, so that means what happens is that at 4.5V, it will lose extra current as heat due to its 75% efficiency.

(4.5V)(.233A) = ~ 1 watt of heat will be generated.

For a device this small (9 x 12mm) a heatsink would be recommended. How much will it heat up without a heatsink? Thats gonna get quite complex, considering the conductivity of resins, the wires, metals on the board, resistors, inductors and the like...

SO LETS SAY THIS for simplicity. Remember that 3.5W of heat were generated during the operation of an ideal 445 diode? This is only 1 watt of heat, but its a LOT smaller and weighs even less the heatsink. Remember that resins have approx 1 C (heat cap.), so they will heat up similarly to the diode. According to really rough estimates under the worst conditions possible, the driver will heat up 20% faster than the heatsinked diode in comparison, but since its a very small device, once you put a decent connection to heatsinks or something on it, this is not bad at all. This is why DTR uses coins to heatsink. Why does this happen?

Remember that the larger things are, the slower they heat up, but they take in more heat to do so. The smaller things are, the faster they heat up with the same amount of energy. This is known as basic energy theory in both physics and chemistry. So using merely a coin is enough to heatsink drivers, since they are so small and generate roughly a fourth of the heat compared to a diode typically.

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If you are using a Flexdrive V5, the only difference is the efficiency, and not the boosting capability. Remember boosting from 4.2 to 4.5V is a static value and ratio i.e 7.6% converted current into extra volts. The key thing here to note is the efficiency of the Flex. The flex has a range of 86 - 95% effic. which is really good. Since we are in a laboratory-style setup, lets use 85% efficiency for simplicity and preparation if anything goes wrong... like if you LOST AN ELECTRONIC COMPONENT, or did damage to the driver. Substitute in your selected ampere range/value and the 85% efficiency, for all ranges, from 100ma to 1.5A so that you can be prepared for whatever happens.

Note that efficiency decreases as Amperage increases. Implied in the manuals. Since I cannot provide you with the proper answers for all you guys, I'll leave this as an exercise for those lazy brains. Also trying to exclude minors.
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Hope this helps.

:huh:

Here, I calculate how much heat is generated when you use a heatsink.

The calculation of the time required to heat up the heat sink looks good to me. How about hooking up a 445 in that heat sink and making a plot of temperature vs. time to see how your approximation compares with reality? I'd like to see it!

Until I can get a decent temperature probe for my projects, which is not anytime soon, I probably cant produce a decent power/temp graph.

For my chemistry experiments, I will probably need a thermometer, but I can get can IR thermometer. Seeing as how I put this heatsink in the 'snow' ice-like crystals in my fridge, it becomes instantly cold. I think its reasonable to say that contacting the heatsink would undoubtedly skew data results. Plus, its great to use this type of device for chem too, since chemicals are pretty nasty when you get into it.

Maybe a DX cheap temperature device could work, but I need to see if its calibrated... Ill see next month when my workload decreases. Classes are pretty intense these first few weeks.

My concern is that your formula would seem to calculate the time it takes for the entirety of the heatsink to reach a certain temperature (in this case, 330 K, or 57 degrees C)

That ignores the reality that the temperature deep down inside that heatsink, at the heat source (the diode), will hit a much higher top temperature in the process of heating the entirety of that heatsink to a certain temperature. No?

You have a good point that I should probably address later. But here is what I can give you right now without me going through my papers. I am CERTAIN that once I get to thermodynamics and conductivity, I will solve this problem, but I gotta go finish my chem work.

Remember that we are not dealing with heat above 100W. The heat-transfer of aluminum, or heat conductivity to be more precise is only ~240W/m K. I need to do more research on this topic, but the reason why I did not factor this in because the wattage here is 3.5W in typical conditions. The gradient changes far too small for it to be considerable in this problem. I am aware that copper conducts heat at 400W/m K. Either way, you can perform simple experiments to see how much and how fast these metals conduct heat. Its truly interesting.

Another thing I can apply is the concentration rule of thumb. This is present in biochemistry, chemistry, physics, and some other sciences. When there is a lot of heat being generated in a specified area, in this case the substrate of the laser diode, the heat will move away from the heat source to places that have less heat, or energy. The more the the heatsink heats up, the slower the heatsink will be able to draw heat away. But heat conduction will continue until all components have reached thermal equilibrium, i.e all components are the same temperature.

This won't happen under normal, typical conditions because air will also draw heat away from the diode until it reaches equilibrium again, where in this case I mean that the heat is being drawn away from the diode at the rate of which it is generated, which is a rather high temperature. For now, I think its best to say that if the host gets hot to the point where it is significantly noticeable under touch, the laser should not be in operation, which is 130F or 57C. If the host can boil water on it, I have little to say besides questioning the life of the laser diode. It should be noted that this is for aluminum hosts only. Copper IMO can be heated a bit more... maybe 150F or 65C.

In my opinion, I'd say we can omit it for now because of the reasons above. 2-3 or even 4 if you want to push it minutes should be a fair operation time/duty cycle.

And lastly, the only thing that has 100% conductivity is Helium-II, where its conductivity is so great (approx. 100 times that of COPPER), it can be completely omitted form the equation. But you will never run into Helium in its special second state. Only maybe when you take an MRI, but thats a different story. Just for fun.

If I made strange mistakes, excuse me for being dead tired from another day.

@ rhd, Considering it that way, you would have to do a bit more calculation that would treat the diode+module as a separate reservoir from the heat sink because they are made of different materials that have different specific heats. You would then consider the module+diode to be in thermal equilibrium with the heatsink such that an infinitesimal change in the temp of one reservoir would result in an infinitesimal change in the temp of the other.

Perhaps though the ratio of the specific heats is close enough to 1 that you can treat the module+diode+heat sink as one entity so that heat is distributed uniformly throughout the mass?

I know the thermal conductivity of Al isn't really that great (used as handles for frying pans etc.) but would the small mass and volume of the heat sink make this negligible so that it heats and cools uniformly?

does it matter? lol :thinking:

Solidworks has a little suite of add-ons called solidworks simulation, you can set specific materials and points of heat generation and run simulations that will show you the temperature of each part over a period of time. A while back I modeled a 5.6mm diode all the way down to the die itself to see what I could come up with but I never really could figure the software out.

I think that having as much of the diode in contact with a material with a higher thermal conductivity is the best way to keep the thing cool. The brass the diode itself is made with is not the best material, but the point of the material being heated compared to the total surface area of the diode itself is very small. (Plus the temperature gradient between the die and base of the diode is very large at first.) The diode can should be able to evacuate heat from the die plenty quick enough, especially when running the diode within it's rated limits.

The module's mass should be large enough to hold the heat generated by the diode, as that module heats up it's ability to remove heat from the diode will decrease. (A large temperature gradient is what moves heat around a material fast.) Adding a heatsink for the module itself will increase the temperature gradient which will pull heat from the module (and the diode as well.) If the surface area of the heatsink is large enough and the material has a high heat transfer coefficient the heatsink will evacuate heat to the air. That will maintain the thermal gradient and the heatsink/module/diode combo will continue to evacuate heat.

Then thermal capacity comes into play, that is the amount of energy required to raise a specific mass of material by a certain unit of temperature. Thermal capacity directly affects thermal continuity, so I guess I should have put that first (energy making the temperature gradient which is what moves the heat through the material.)

I guess what I am trying to say is the diode itself is not a bottleneck in heatsinking, the module the diode is pressed into is the largest bottleneck. Also, the crazy numbers of variables involved in all of this stuff makes calculating the temperature of a diode over a specific time so difficult I don't really think it's possible. Someone should prove me wrong (and reword my post so that it says what I wanted it to say, it looks like a lot of gibberish to me!)

Sounded good to me! :beer:

You would use conductivity to calculate how much heat gets by from the material transition. Pans are insulative at the handles BTW

poor conductor of heat = an insulator.

Also, Pontiacg5... this is for people who do not like to have a whole load of program downloads on their computer. I wanted to show people a formula for stuffs.

Isoleucine said:

I can get can IR thermometer.

Click to expand...

Be careful with this method. Shiny metals do not emit the same way dull/painted/anodized metals emit.

rhd said:

the temperature deep down inside that heatsink, at the heat source (the diode), will hit a much higher top temperature in the process of heating the entirety of that heatsink to a certain temperature. No?

Click to expand...

The biggest difference will be at junctions. The temp throughout one material will be virtually the same.

Isoleucine said:

Copper IMO can be heated a bit more... maybe 150F or 65C.

Click to expand...

Weird how it was all objective before this. Where did opinion come in?

Ibuprofen200mg said:

I know the thermal conductivity of Al isn't really that great (used as handles for frying pans etc.)

Click to expand...

Um... what? Most heat sinks are made out of Al. Explain that one.

The thermal conductivity of Al is 237 W/mK, which is about of half that of copper, around 400 W/mk. Yes, copper is a better conductor of heat. Aluminum and copper both have similar molar heat capacities, that is a given molar mass of Al would experience roughly the same change in temp as an identical molar mass of copper if the same amount of energy was put into both. Aluminum has a molar mass of less than half that of copper, so kg for kg, aluminum has a lesser ability to absorb and store heat. Aluminum is used in heat sinks because it is per unit volume much less massive than copper, ie it dissipates heat much quicker. It is also relatively cheap and lightweight, things that copper and steel are not.
I tried.

Copper has a thermal conductivity almost 4 times that of aluminum alloys (There is a big difference between aluminum and aluminum alloys. I can guarantee that every single heatsink sold on this site is an aluminum alloy, not pure aluminum. The thermal conductivity of most aluminum alloys is around 100-130W/mK) That means copper will distribute the heat evenly through the heatsink 4 times faster than an aluminum heatsink. The heatsink will heat up faster at the point where the temperature is greatest though. I don't really know if the time that difference in temperature is sustained for a very long time but it does happen.

Copper also has a higher heat capacity than aluminum, meaning it takes more energy to raise the temperature of one kg of copper 1 degree than it does for aluminum. Copper is also three times as dense as aluminum so a copper heatsink the same size as an aluminum heatsink will be able to sustain runtimes more than three times as long as aluminum.

Copper also has a higher heat transfer coefficient than aluminum, meaning it can dissipate heat from itself to air/water around the sink much faster than aluminum. Copper will take longer to disperse the heat though if both copper and aluminum are raised to the same temperature but only because the copper would be "holding" much more energy. (since most of our heatsinks can't disperse heat to the air around them easily copper is still the best choice for a laser heatsink.) The heat transfer coefficient of a material will determine the temperature at a junction. If the materials have a similar coefficient than the temperature should be almost equal between both parts, if one has a lower coefficient it will remain colder than the other material as it can't "suck" the heat out of the other material as fast as the other material can push it out.

Copper cannot be heated more than aluminum though, it will take longer to heat up and dissipate that heat faster but it is not safe to run a diode to 150 degrees in a copper heatsink assuming it is safe to run the same diode to 100 degrees in aluminum.

As far as the pan handles thing, I have personally never seen a pan with an aluminum handle. If the handle was aluminum it was covered in a insulator of some sort. If you put the prongs of a steel fork and aluminum fork in a pan of boiling water the aluminum fork would heat up a good deal faster than the steel fork. The "goodness" of a materials thermal conductivity is subjective, it really depends on your application of that material. Aluminum happens to have an excellent thermal conductivity for heatsinks, second only to copper (and silver or diamonds, those are much more cost prohibitive though!)