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Transistor heat transfer capability
Transistor heat transfer capability
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Old 5th December 2020, 07:09 PM   #31
Duke58 is online now Duke58  United States
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From: eng-tips.com:
Millipede - a thousand footed bug

Millenium - a thousand years

Million (perhaps) - a thousand thousands?

Grazie mille - a thousand thanks

The possible incongruity of "million" aside, "milli" means thousand, or thousandth.

BTW, where I work, when I refer to a "mill" instead of "a thousandth (of an inch)", people look at me as if I have two heads. This might be a peculiar "linguistic island", but my point is that mil is not universal.
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Old 5th December 2020, 07:31 PM   #32
jneutron is offline jneutron  United States
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It gets interesting on a multi language forum.

Here are my daffynitions..googled

mil 1 (mĭl) n. 1. A unit of length equal to one thousandth (10-3) of an inch (0.0254 millimeter), used, for example, to specify the diameter of wire or the thickness of materials sold in sheets.


mill:: a machine that manufactures by the continuous repetition of some simple action

(course, by that definition, I could also be a mill, as I "by continuous repetition of a simple action")..... make a fool of myself..

jn
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Old 5th December 2020, 07:54 PM   #33
Duke58 is online now Duke58  United States
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Don't get me wrong- I think some of these amps are beautifully crafted, true works of art and craftsmanship. As a mechanical engineer/machinist, I'm wondering whether lapping the surface between the electronic device and the heatsink really makes a measurable difference in heat transferred?
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Old 5th December 2020, 08:30 PM   #34
jneutron is offline jneutron  United States
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No, it doesn't make any difference in the amount of heat transferred. That will be defined by the semiconductor's voltage and current, the total power dissipation. (note that for a diode, the vf will go down roughly 1 mV per degree C, so it bends the dissipation/derating curve a tad.)

What it does do is change the thermal resistance at the interface. If you dissipate 100 watts, and the thermal resistance is .5 degrees per watt, there will be a 50 degree difference across the interface. If you can get it down to .1 degree C per watt, there will be a 10 degree C difference.

Semiconductor lifetime halves every 10 degrees C.

Found my linear audio copy...

Dissipating 100 watts:

For a .5 inch by .5 inch interface, a .002 inch thick silicone grease interface will have .16 Deg C per watt, interface rise of 16 degrees C.

For a .5 inch by .5 inch mica .003 inches thick with .002 inch grease on both sides, the total is .96 degrees C per watt, interface rise of 96 degrees C.

The impact the thermal path has on semiconductor temperature and lifetime can be huge.


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Old 5th December 2020, 09:00 PM   #35
cbdb is offline cbdb  Canada
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Quote:
Originally Posted by Duke58 View Post
Has anyone measured a temperature difference between lapped and unlapped transistors and heat transfer? Does it make an audible difference?
Yes, burnt out transistors dont sound good.
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Old 5th December 2020, 09:20 PM   #36
Duke58 is online now Duke58  United States
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We're both saying the same thing, just differently.

The heat transfer mechanism can be via conduction, convection or radiation.

During conduction, the vibration amplitudes of the atoms and electrons of the metal (the heatsink and metal backing on the electronic device) become relatively large with the high temperature of the environment. Increased vibrational amplitudes (and thus the energy associated) pass from atom to atom during collisions between the atoms.

One of the laws of thermodynamics is that heat is always transferred from hot heat to cold. Call it a system, a reservoir. It's called a law because this is validated through experimentation and heat always goes from hot to cold. This happens through experimentation every time. No exceptions.

Experiments show the conduction rate P (amount of energy transferred per unit time) is:

P=Q/t = k/A[(Thot - Tcold)/(L)]

Where:
k is the thermal conductivity of the material
Q is the energy that is transferred as heat through material from it's hot area to cold area per unit of time
t is time
L is the thickness of the conducting material, in this case thermal paste

My entire point is that conduction rate P is inversely proportional to L. Thermal paste is added to aid heat conduction between the electronic device and the heatsink. Less heat equals more life for the electronic devices.

Is it worth the time and effort to lap the transistors? If so, what difference does it make? Do the transistors last many years longer?

Last edited by Duke58; 5th December 2020 at 09:24 PM.
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Old 5th December 2020, 09:22 PM   #37
Duke58 is online now Duke58  United States
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cbdb- yes, they don't work, so there's no sound. No sound from an audio device.

Y'all are saying that lapping transistors is going to double the life of the transistors?
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Old 5th December 2020, 09:44 PM   #38
jneutron is offline jneutron  United States
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Thermal paste is a dismally poor conductor of heat.

The only reason for thermal paste is to fill the gaps in the surface roughness, be it 200 microinch or 500.

The proper application of thermal paste is to simply put some on the surface and scrape it all off using the edge of a single edge razor. It should only fill the gaps.

Everybody goops it on, and uses the screw pressure to squeeze out the excess. Very bad technique, not how paste was designed to be used. For the .5 by .5 inch example I detailed, it's .08 degrees C per watt per .001 inch thickness.

Lapping the surfaces and then using just a film of paste will give the absolute best thermal transfer, the only thing better is a sheet of graphene used between a heat spreader and the case, using the thickness of the heat spreader to gain conductive area.

BTW, you included area (A) in the equation, but didn't mention it. If the package bottom is concave, the effective area no longer includes the middle of the bottom which is exactly where the die is located.

10 degrees C rise, half the life. Arrhenius equation. Used in MTBF calculations for large systems. And do NOT ask me about MTBF...That is a chapter in my life that I want to forget... (had to put together an MTBF presentation for very high level people for a 10e9 dollar machine, at the last minute they changed the presentation date but I was in Colorado on vacation. A person who used to be my friend gave the presentation. Now, he tries to run me over in the parking lot every chance he gets.)

But more importantly, many solid state circuits do not play well when the die gets hot. Bipolars change gain, bias points change, cooler is always better. The thermal time constant of a bipolar junction is 100 uSec to 1 mSec, but the thermal constant to the exterior of the case is the 1 to 5 second regime, and can be even longer if the thermal transfer is bad and there is any heatsink mass.. If your thermal tracking sensor for bias is on the heatsink and the die dissipates heavily, your sensor can be tens of degrees too low, that can cause runaway.

jn
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Last edited by jneutron; 5th December 2020 at 10:00 PM.
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Old 5th December 2020, 11:16 PM   #39
Duke58 is online now Duke58  United States
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Quote:
Originally Posted by Duke58 View Post
We're both saying the same thing, just differently.

The heat transfer mechanism can be via conduction, convection or radiation.

During conduction, the vibration amplitudes of the atoms and electrons of the metal (the heatsink and metal backing on the electronic device) become relatively large with the high temperature of the environment. Increased vibrational amplitudes (and thus the energy associated) pass from atom to atom during collisions between the atoms.

One of the laws of thermodynamics is that heat is always transferred from hot heat to cold. Call it a system, a reservoir. It's called a law because this is validated through experimentation and heat always goes from hot to cold. This happens through experimentation every time. No exceptions.

Experiments show the conduction rate P (amount of energy transferred per unit time) is:

P=Q/t = k/A[(Thot - Tcold)/(L)]

Where:
k is the thermal conductivity of the material
Q is the energy that is transferred as heat through material from it's hot area to cold area per unit of time
t is time
L is the thickness of the conducting material, in this case thermal paste
A is face area

My entire point is that conduction rate P is inversely proportional to L. Thermal paste is added to aid heat conduction between the electronic device and the heatsink. Less heat equals more life for the electronic devices.

Is it worth the time and effort to lap the transistors? If so, what difference does it make? Do the transistors last many years longer?
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