Discerning Thermal Limits in Hosts

[IsaacT]

So I think many of us have many builds going on in their head at any given time. I really want to make a 635nm laser to help fill in my wavelength collection, but as I look at my options, I realize that I want more power. So I am now leaning towards the 1W 638nm idea. But I worry about how much heat that would produce and how that would affect my host choices.

So, how can I discern the limits of a host in regards to how powerful a build I can put in it. This information will be largely useful to those considering builds and build options, but more specifically will help me reach a decision on my current dilemma.

Specifics:
1.4A Linear Driver
500mW 638nm Mitsu Diode
C6 Host w/ Copper or Aluminum Heatsink
1 x 18650 Battery


Do you think the above would work? I am hesitant to order any more custom hosts right now due to the 600 dollars I have spent on lasers in the past 2 weeks, so if I can use the C6 option that would be ideal....

Thanks,
Isaac

PS- hopefully information from this query will aid others who may have questions on the capabilities of other hosts.

 

[RA_pierce]

You can determine how much heat is produced at the diode by calculating:

I (current in Amps) x V (voltage) = W (power in watts)

Then take the power (W) and subtract the optical output of the diode.
This will tell you how much power is lost as heat.

Alternatively, you could do this calculation with the entire system to figure out how much power is lost as heat from the diode and driver together.

Red diodes have much lower Vf than blues and violets but require a little more current.
Overall red diodes are a little more efficient than their shorter wavelength counterparts so produce less heat for a given optical output.

 

[IsaacT]

So:

(1.4A)x(4.2V)=5.88W, 5.88W-1W=4.88W

So 4.88W of heat is being let into the host?


I don't really know how to interpret that....seems like a lot but who knows? Do you think a C6 with a Black Anondized Aluminum Heatsink would suffice? Copper Heatsink?

Also could I make it work with a Linear Driver such as the 17mm round boards that fit so nicely in the pill, or will a Buck Driver be necessary?

Thanks,
Isaac

I keep reading data sheets, and it helps some, but there is some that is just beyond my comprehension so far.

 

[RA_pierce]

That looks right.

Also, given the equation q = m x C x ΔT
where q is heat (in Joules), m is mass in grams, C is the specific heat of the heatsink material in J/g*C or Joules/gram degree Celsius, and ΔT is the change in temperature, we can figure out how heat will affect our system (the laser).

For example (using your value for heat above):
Given a Joule is a Watt second (J = Ws)
Given the C of Aluminum is about 0.9
and assuming a heatsink mass of 10g
and assuming the heat at the diode is 4.88W
and assuming operation for 10 seconds

We get:

4.88W x 10s = 48.8J = 10g x 0.9J/g*C x ΔT

Solving for ΔT we get:

ΔT = 48.8/9 *C
The units (grams and Joules) cancel so our units are left in *C

So our change in temperature, then, after ten seconds is:

ΔT = 5.4*C

Since this value is positive, we can say that a laser diode producing 4.88W of heat into an Aluminum heatsink with mass of 10g will raise the temperature 5.4*C after ten seconds.

Note: ΔT = Temperature final - Temperature initial ; knowing this and using a little more algebra, if we also know the initial temperature (room temp, for example), we can determine how long it will take for the heatsink to reach a given temperature (Temp final) if we can't measure it directly. For solid objects, like a heatsink, it will not be easy to measure the temperature with a thermometer.
If you use room temperature as your Tinitial, be sure that your heatsink has reached thermal equilibrium with the surroundings.

Another note: This does not take into account radiation of heat from the system to the surroundings.

Another note: I don't have much physics training so if anything above is erroneous, please correct me.
For now I'll just be embarrassed in advance...

 

[IsaacT]

So if we assume a linear rate of change, I would be seeing a duty cycle of like 30 seconds :/ ?

 

[RA_pierce]

So if we assume a linear rate of change, I would be seeing a duty cycle of like 30 seconds :/ ?

That seems about right. By that time, assuming my calculations are correct, the temp will have risen about 16*C.

BUT
Keep in mind that I used a 10g Aluminum heatsink for the example.

The 4.88W of heat figure you have above is about what I would expect from a 445nm diode at about 1.25A. That seems reasonable for such a small heatsink. 10g of Aluminum is not much; I have 4 Almuninum module "backs" and they add up to 5.8g. So 6.8 of them would equal about 10g. If you know what the backs of the modules look and feel like, 30 seconds is a reasonable amount of time for a high power laser dumping heat into such a tiny heatsink.

I have a heatsink for an SH-032 host sitting in front of me.
It is Cu and is 49.3g including a copper module.
The specific heat of copper is 0.385.
I'm not going to do the math this time because I am lazy but I would suggest you weigh the heatsink you intend to use and calculate it that way to get a better estimate of what to expect.

 

[IsaacT]

That's a good idea. I'll bring it to my physics lab on Thursday and weigh it then. Thanks