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**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.

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