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Old 05-10-2009, 04:02 PM #17
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Originally Posted by heruursciences View Post
that way the heat generated- READ: 10's to 100's of KILOWATTS PER SQUARE CENTIMETER power density- does not have to go through the bulk of the chip to get dissipated.
As Leibnitz said (and I quoted recently): let's calculate.

An LPC diode has an operating current of around 400mA the way it's being done here, the voltage drop over the diode is ~2.5V, so the power is just about 1W. Around 250mW is generated as laser light; that means 750mW are generated as heat.

From a picture in the internet, you can measure the dimensionso the LPC long die diode as approximately 0.3 x 0.015 x 0.01 cm (for the sake of familiarity, I'm not using SI units). The contact area with the brass case is therefor 0.0045 cm2 (sigh... still no sub/superscripts).

This translates to 0.75/0.0045 = 167W/cm2. Fully 3 orders of magnitude lower than heruursciences assertion.

The thickness of the die is ~0.1mm, like a thin piece of paper. How significant is a thermal gradient over such a small thickness? Again... you guessed it... math comes to the rescue.

Here you can find information on the thermal conductivity of Al(x)Ga(1-x)As (*sigh* again): depending of the value of x (i.e. the exact composition), it varies between 0.11..0.55 W/(cmK), which is pretty low - Silicium is 3 to 15 times better. All the waste heat has to be transferred through the contact area with the brass case (see above). The formula for the necessary temperature gradient is:

delta T [K] = P [W] x d [cm] / A [cm2] / k [W/cmK]

Plugging in the worst case for k (0.11 W/cmK) and the other numbers, we get (this also assumes that the heat gets generated on only the "far side" of the die so it has to be transported the max distance):

delta T = 15 K (or deg C)

This is for x=0.5; for pure GaAs (x=0) the thermal conductivity is 5 times better and delta T would be 3 K.

It seems unlikely that anyone would introduce major changes in the manufacturing process for such a meager difference - especially as the operating conditions (current) I cited are way above the rated spec of the diode; for them, the temp difference would be correspondingly less. So it's safe to conclude that we can exclude thermal considerations from the case positive/negative debate.


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Old 05-10-2009, 11:13 PM #18
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It depends a bit on what end of the range that ends up - 15K seems worthwhile to do something about, 3K doesnt.

Ordinarily, in semiconductors, the substrate is n-type material, so the case-negative result seems easiest to achieve. I would assume both IR and 660 diodes are constructed starting with a n-type substrate, so the only way a diode would end up case-positive if the completed semiconductor was mounted upside down to the casing.

Doing so would bring the active region closer to the contact point on the case, but would require growing a massive metal contact layer on top of the chip first.

I suppose we won't figure it out by just thinking about it.. any chance of anyone contacting a manufacturer about this issue? The answer will probably be surprisingly simple and logical once we know
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Old 05-13-2009, 03:33 AM #19
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Glad to see the discussion is growing (yes I read all the answers, keep 'em coming!)
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Old 05-13-2009, 11:22 PM #20
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167W/cm2 is the total dissapation of the whole chip- The current flow and heat is generated only in the lasing channel, If you take the dimensions of a typical lasing channel say 3uM by 1500um you'll finds that the power density is well in the MEGAWATT range. The math is: 4500um sq= 0.0045mm sq, but were not done yet... 0.0000000045 cm square Now take your heat of 0.75W and devide by...0.0000000045 cm square you get 16,666,666W/cm2 or about 16.7 MEGAWATTS per square centimeter power density generated in the lasing channel! So yes after doing the math, I was a little off on actual powert density... by a few orders of magnitude...
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Old 05-14-2009, 12:50 AM #21
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But it still wont explain why its worthwhile to mount IR diodes upside down, but visible diodes in normal orientation at similar power output... unless the thermal conductivity of the materials in IR's is much lower.
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Old 05-14-2009, 10:22 AM #22
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Sorry... now you're off by 4 orders of magnitude (factor 10,000) in the area calculation and partially compensate that by another factor 10 error in the final division - check your figures.

I'm not convinced that heat is only generated in the lasing channel (as a side note, if you calculate the energy density of the laser beam at the point of emission - several 100mW in an area of about 10 um^2 - you get a figure of about 10MW/cm^2). What's more, it wouldn't significantly affect the temperature gradient across the die.
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Old 05-14-2009, 05:58 PM #23
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Ummm, why are we talking about electrical energy dissipation across an area? That doesn't even make sense. Current can flow through an area, a la a flux, but to dissipate electrical energy, it has to go through a voltage drop, so it really only makes sense to have energy dissipation per VOLUME, since the voltage drop happens across a specific volume of the diode. The volume you would need to use is the volume of the active region, not the whole diode, and the active region is only nanometers thick, so the power dissipation density in that volume is very high.

And as far as whether heat is only generated in the lasing channel, that's correct, because the active region is where the voltage drop is. But the active region, or the "lasing channel" needs to be defined in edge direction. In the longitudinal/axial direction (direction of output, along the ridge), the active region is typically the length of the diode. In the transverse direction (down through layers, perpendicular to the top of the diode), the active region is the thickness of the layers, typically quantum wells, so that length is only nanometers. The more complicated one as far as the size of the active region is the one that is in question, which is the "width" of the "lasing channel", which is the lateral direction. dr-ebert, you took this value to be the full width of the diode, 150 microns if I'm reading correctly, and heruur said that the width of the lasing channel would be smaller. Basically, that one is impossible to know fully without knowing how the diode is made. If the diode is a simple broad area design with a contact that is the full width of the diode, then using the full width of the diode is correct. However, it is almost certainly not a simple broad area laser with a full-width contact, therefore the width through which carriers flow in the active region in that direction will almost certainly not be the full width of the diode. If it is a ridge design, or a buried heterostructure, and even a simple gain-guided design where the well is the full width but the contact is not, then carriers will not flow through the full cross-section of the diode.

So, the volume in which energy is dissipated would be: (thickness of the active region on the order of nanometers) * (length of the diode in the long direction) * (something less than the width of the diode, but yet unknown unless you know how the diode was made). The cross-sectional area through with current flows in the active region would then be: (length of the diode in the long direction) * (something less than the full width of the diode, since current likely will not spread to the edges of the diode).

If anyone wants a possilby slightly better treatment of how energy is dissipated from the active region, there's a treatment of it in a textbook by Coldren, "Diode Lasers and Photonic Integrated Circuits". He teaches a laser diode course here, and really knows his stuff. He did "write the book" on the subject, after all.
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Old 05-14-2009, 11:30 PM #24
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Well, if he teaches at your place of education, why not ask about the case positive/negative issue - i'm sure anyone with good insights into the innards and production of laser diodes would be able to answer it straight away...
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Old 05-15-2009, 07:36 AM #25
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Hi pullbangdead, you're talking about the volume where the heat is generated. I'm talking about the cross-sectional area through which it has to be transported to the heatsink. In my original post I made the simplifying assumption that it is generated at one full side of the die and thus needs to be transported to the other size, that's a simple linear case. If instead you assume that the heat is essentially generated along a line on one side of the die, it's a bit more complicated because the T drop is no longer linear across the thickness, you'd have a two-dimensional T field. This would take far more computation than I'm willing to do, but I guess the end result wouldn't be much different.

I too think that we don't really have a chance to decide the question by mere thinking and discussion. The real reason might be something really deep, it might be something where we say "well of course...!" afterwards, or it might be something really trivial, like "the first customer to order a commercial quantity wanted a case positive diode because that's how he designed his device".
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