I do appreciate you taking the time to make this thread, and you do have some good information. However, there one point you make that is just plain wrong.
This is the biggest issue I have with your info. Titanium is in fact a lousy conductor. You acknowledged that copper is a superior conductor, but without numbers people might not understand the ENORMOUS difference. Taken from
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html. Copper has a thermal conductivity of 401
W/(m.K). Aluminum is 205
W/(m.K). Titanium is a mere 22
W/(m.K). In fact, aluminum is almost ten times as thermally conductive as titanium. And copper is close to twenty times more thermally conductive. Among metals, titanium is decidedly BELOW average in thermal conductivity.
The difference in electrical conductivity is similar. Aluminum is fifteen times as electrically conductive, and copper over 30 times as electrically conductive, compared to titanium.
There is no way you can say that titanium "conducts well" when you look at the numbers. This also happens to be among the most important properties in flashlights. Low electrical conductivity means you waste more power if you use a titanium body as part of the circuit, as many lights do. And arguably even more important, you have to substantially reduce drive levels if you build a light body out of titanium, or it will overheat. It just isn't capable of spreading the thermal load like aluminum or copper.
I am in fact a huge fan of titanium...ever since I learned about the SR-71 Blackbird. A design like that would not have been possible without titanium. But I don't think we should pretend it is a miracle metal...like all others, it has advantages and disadvantages. And frankly, in my opinion, for flashlights, the advantages of copper or aluminum are more suited.
thedoc007,
Titanium is a good conductor of electricity compared to a great many materials. It is a metal; it's kind of in the job description. Yes, relative to other metals, titanium is
relatively poor at conducting electricity. But total resistance is determined by resistivity and the geometry (i.e. the volume involved). You can achieve the same total circuit resistance by less of a very good conductor (such as copper) or more of a less good conductor (such as titanium). But in the case of a flashlight BODY, there is a LOT of material! If I tell you that the total resistance is, say, 0.01 ohms total for a titanium flashlight body, and .0001 ohms for copper, you could say that the resistance of titanium is a
hundred times higher! OMG! But I could say--with equal truth, and more relevance--that the resistance of both of them is very low relative to other resistances in the circuit. And that is in fact the case here. And more importantly, the contact resistances and resistances of the oxides are more significant, and here titanium has a decided advantage.
As for the thermal situation, you are mistaken. Here is a detailed analysis of the thermal situation in the LunaSol 20 from my
thread on the light. The situation is quite similar with the Haiku. Titanium has plenty of thermal conductivity to handle the heat dissipation loads involved in LED flashlights:
*****
TITANIUM
I have to confess that when I first saw the various custom made titanium lights on CPF I had exactly the same reaction that most people have. I was like "Titanium? Doesn't it have a higher resistance than aluminum? And doesn't aluminum conduct heat better than Titanium? Why would you pay so much more for a Titanium light? I don't get it!"
Now, some people like to "answer" these questions by saying that titanium is just "bling", just "man jewelry"; that it's a status symbol kind of thing, like a Rolex watch; that if you were just thinking practically and rationally, you'd use aluminum. However, this sort of psychological write-off really isn't true, as I've come to find out both from my experience owning titanium lights, and my scientific research into the question of titanium. So, let's look a bit more deeply and more thoughtfully into the question of titanium flashlights.
Electrical Resistance
Right off the bat, let's knock the electrical resistance thing on the head! Yes, titanium does indeed have 26 times higher resistivity than aluminum,
but that's still pretty good. Nickel has almost 3 times greater resistivity than aluminum, and yet it is used for 1/4 inch wide .005 inch thick ribbon strap connections between cells in a welded battery pack. And those straps are often asked to conduct four or five amps or even more. Metals—all metals—are good conductors of electricity. It's sort of in the job description. Further, the amount of material involved in a flashlight body's conduction pathway is enormous. There's a lot of titanium there—many orders of magnitude greater an amount than that involved in a battery packs' nickel ribbons; more even that that in a 12 gauge wire—and thus the total circuit path resistance involved in
both the aluminum and the titanium flashlight is so low as to be insignificant.
Most of the time, that is . . . because the story of electrical resistance doesn't end there. The thing is that both aluminum and titanium (and most metals) oxidize in air, and a surface layer forms, which may or
may not be conductive. Aluminum's surface oxide layer is most emphatically NOT very conductive. This is why aluminum lights are Chemkoted so that they have a different surface layer that
is electrically conductive, and which is environmentally stable and corrosion resistant. Otherwise, any bare aluminum to aluminum joints in the conduction pathway
will eventually develop significant resistance, totally negating the low resistivity of pure aluminum. Titanium's surface oxide layer, in sharp contrast, conducts electricity just fine and bare titanium to bare titanium is a totally viable and environmentally stable corrosion resistant joint just as it is with no special treatment. So, if we were to reckon up titanium's advantages, ironically "electrical conductivity"
might actually be one of them, considering it's surface oxide, but you could argue against this view if you have complete trust in the Chemkote surface layer on the aluminum light.
Heat Conductivity – The Long Version
Heat conductivity is a much more complicated consideration, unfortunately. I will show below that regardless of any theoretical calculations and discussions that may follow, the LunaSol 20 manages heat very well even in worst case scenarios (as does the Ti-PD-S), and honestly, I don't know why that should surprise anyone as many
plastic LED lights exist that also perform just fine, if it comes to that, and I can guarantee you that titanium has better heat conductivity than any plastic. However, let's take a look at the heat transfer problem from the theoretical perspective, just for fun. And don't blame me if it starts to get a little tedious! Feel free to skip the rest of this section at any point and jump to "Heat Conductivity – The Short Version" below.
Heat transfer can happen via conduction, convection, or radiation, and a general heat transfer scenario is painfully difficult to calculate theoretically. Conductive heat transfer, or transfer scenarios dominated by conduction, or which can be reduced to an equivalent conductive (and thus simpler) problem, are much easier to deal with, and can be thought of just like an electrical resistance problem with wires and current and voltage and a number of resistors. The current is the heat flow and the resistors are the
junctions or conductive elements. An LED, like a transistor or MOSFET, has a degrees C/W rating, which predicts what the temperature drop across the junction will be when a certain number of watts are dissipated through the device. This is analogous to a resistor's voltage drop when a known current flows through it. The equation is very simple:
dQ/dT = deltaT / R
q = deltaT / R
where dQ/dt is the rate of heat flow in watts, deltaT is the temperature difference, and R is the resistance of the junction (or in general, of the material conducting the heat flow). For simplicity, I will let small q = dQ/dt. Some may recognize this as Newton's Law of Cooling, by the way. So, for example, a 10 C/W junction dissipating 2 watts, will have a 20 degrees C temperature differential across it. This, however, doesn't tell you the die temperature unless you know the temperature of the other side of the junction—usually a heat sink. The heat sink, in turn, has a temperature drop across
it. This can be calculated by expanding the "R" term above. For a simple rectangular block of material:
R = L / (k * A)
where L is the length over which the heat must flow, k is the thermal conductivity, and A is the area through which the heat must flow. Notice that R increases with L, but decreases with both k and A. Notice also the k must have units of watts / ( meters * degrees C), although usually it appears as degrees Kelvin, actually, but 1 degree C = 1 degree K for differences, as Kelvin is just Celsius with a different zero. This makes R have units of (degrees C / watts), which you can work out for yourself if you wish.
Sadly, the MCPCB heat sink in the LunaSol 20 isn't rectangular, with the area constant across the length of heat flow. The Golden Dragon LED is at the center of a disc of aluminum, and thus heat flows out from the center to the edges, and also out of the top and bottom. Since I want to calculate a worst case scenario, and because the top and bottom are much more complicated heat transfer situations, I will pretend that
no heat escapes from the top or bottom of the .75 inch diameter, 2 mm thick heat sink. Now, if we think of a very thin cylinder where heat flows from the inside to the outside, it should be clear that the rectangular equation above easily applies because the cylinder can be unfolded, without any significant distortion, to a rectangle of 2*pi*radius width, h height, and t thickness, where t is small. If t were large, then we couldn't unfold it without distortion. But, with that in mind, we can think of the MCPCB heat sink as a big collection of nesting cylinders, all which have a small thickness, and then we can just add them up. Such is the power of integral calculus! LOL! So, thickness becomes dr, width is 2*pi*r, and height is still h. Rearranging terms above we get:
deltaT = q / k * integral (dr / (2*pi*r*h)) from r=inside to r=outside
We can pull 2, pi, and h from inside the integral as they are constants:
deltaT = q / (2*pi*k*h) * integral (dr/r) from r=inside to r=outside
But the integral of dr/r is just ln(r), so,
deltaT = q / (2*3.14159*237W*m^-1*K^-1 * .002m) * (ln(.375inch) – ln(.125inch))
deltaT = q / (2*3.14159*237W*m^-1*K^-1 * .002m) * (ln(.375/.125))
To determine q, we can put an upper limit on the heat dissipated by just making it equal to the power of the light. (The actual heat dissipated would be less than this, as some of the power goes into the production of
light energy.) Back of the envelope calculation of that is simple: take the Wh of a CR123A cell and divide by the runtime in hours. The Wh of a good CR123A cell at a .5 amp draw rate (the current to the Golden Dragon is 425 mA) is right around 3.7Wh. This makes for 1.85 Watts due to the 2 hour runtime of the LunaSol 20 on high. So q = 1.85W, and to clean up some more, .375/.125 = 3, and ln3 = 1.0986. Here's what we have now:
deltaT = 1.85 / (6.28318*.002*237)*1.0986 K
where I canceled some units and moved K^-1 at the bottom to K at the top. Work this out and we get the result:
deltaT(heatsink) = 0.68 degrees C
This is the drop
across the heat sink. And we can also add to that the drop across the LED to MCPCB solder joint. From the spec sheet the Golden Dragon has an 11 C/W thermal resistance. So, q*R will give us a result of
deltaT(junction) = 20.35 degrees C
Having fun yet? Let's move on to the temperature drop across the titanium body. Using the entire body for the calculations would be too optimistic, because the heat has to flow from the heat sink to head mating surface, and then all the way down the body before it gets to the tail, and this doesn't happen without a temperature drop. The tail section does indeed draw away a significant amount of heat, but since that is hard to calculate, and since I'm interested in a worst case scenario, I will pretend that the titanium body only consists of the head (with maybe a bit extra length to partially compensate). So, I will use a .125 inch thick, 1.5 inch high, 1 inch outer diameter cylinder. It is interesting to compare the simple rectangular equation's results with our more accurate one in this case, since the thickness (.125 inches) is not exactly large with respect to the diameter (1 inch), but I will leave that as homework. (Yes, I'm having fun with this. *cough* Sorry. *sheepish grin*) OK. So, let's use the same formula, but this time we use titanium's heat conductivity of 21.9W*m^-1*K^-1, an h of .0381m, an inner radius of .375inches, an outer radius of .5inches. The result is,
deltaT(titanium case) = .10 degrees C.
MY GOD! That's crazy! Clearly we needed aluminum there, didn't we? After all, if we had used aluminum, the result would have been,
deltaT(aluminum case) = .009 degrees C,
and isn't that a whole order of magnitude better, after all?
Now, I've left two important pieces out so far, and those are the case to air junction and the heat sink to case, and MCPCB solder joint to heat sink contact resistances. If you take any two blocks of materials, A and B, and push them together, and cause heat to flow, there will be a discontinuity of the temperature from the left end of A, to the right end of B, where they meet. This is called the thermal contact resistance, and it is notoriously difficult to calculate. I don't know much about it, but I did do enough research to know two things: a value of .5 C*in^2*W^-1 for our situation is conservative, and that
contact pressure is the
most important parameter for smooth mating surfaces with little to no air gap (hence the need for heat sink grease in such cases, by the way). Why do I mention the pressure thing? Well, when aluminum heats up, it expands at a greater rate than titanium which greatly increases the contact pressure on that mating surface, ensuring a nice thermal joint there. It's a nice side benefit to using a titanium body in your flashlight. Anyway, moving on, the delta T across that junction will just be q * (Rc / A) where A is the area, which we calculate from a .75inch diameter and a 2mm height. Thus,
deltaT(heatsink to case junction) = 1.85 W * .5 C * inch^2* W^-1 / (pi*d*h)
deltaT(heatsink to case junction) = 1.85 * (.5 / (3.14159*.750*.07874) C
deltaT(heatsink to case junction) = 4.99 degrees C
For the solder joint to heat sink junction, we need to use an area of .45 in by .4 in (from the golden dragon spec sheet), which yields
deltaT(solder pad to heatsink junction) = 5.14 degrees C
Now, all that's left is the case to air junction. Oh, that's all, is it? Good grief! Calculating that from theoretical considerations would
not be trivial. What I can tell you about it is that the greater conductivity of aluminum wouldn't play much of a role. Here, emissivity is more important, and titanium has a pretty good emissivity, although hard anodized aluminum is also good. I'm not going to get into any numbers or make any arguments. This part of the thread is already way too long. All I am going to do is to steal some of Don's
measurements. And thank God for those. Nothing like predicting reality from reality. OK. So, Don set up an FLIR measurement and measured 47.7 C at the hottest part of the head, and this was a worst case scenario, being constant on just sitting in open air. See for yourself, as the setup is kind of neat, and I already uploaded those pics to my server space:
Note that the ambient temperature was over 28 C! Which means that the temperature delta from head to air was 19.2 C, yielding an R value of 10.4—very close to the R of the Golden Dragon LED itself, interestingly enough. More importantly, though, consider that if the air had been at 20 C, then the head would have been just below 40 C instead of 47.7 C. But, again, since we are interested in the worse case scenario, let's use the 47.7 C value.
So, let's add it all up
47.7C + 0.10C + 4.99C + 5.14C + 0.68C + 20.35C
die temperature of LED = 78.96 C,
and the maximum junction temperature of the Golden Dragon LED is 125 C.
So, note a number of things here. First, that even in a worst case scenario and making an upper bound sort of estimate, the die temp is well within limits. Second, notice that
the largest contributions to the die temperature do not come from the heat conductances of the aluminum or titanium parts, but rather come from the body to air junction and the LED junction. They dominate the calculation. And the next largest component comes from the contact resistances. So . . . maybe we could all please agree that titanium's thermal and electrical "disadvantages" are non-issues for most flashlight use situations?
Heat Conductivity – The Short Version
There was actually a much easier way to demonstrate that the LunaSol 20's die temperature is well below the limit, and that is to look at the runtime / output plot which appears below.
Note that the output
does not diminish with runtime. If the thermal situation was pushing the edge, then the output would start at a maximum then ramp down to a lower level due to rising die temperature, which causes loss of efficiency and output over a low temperature die running the same current. This is something I have seen before in multi-level flashlights. It happened with the Arc4 and appeared in this_is_nascar's output graphs of the Arc4 at various levels. The highest level graph showed this ramp down, and TIN complained about it, and Peter Gransee told him it was thermal and that if he just held onto the light the whole time it wouldn't happen. TIN did that, and lo and behold, the output stayed at the high starting level for the whole run due to the lower surface temperature of the body of the light (and thus also of the die) when held by TIN's hand.