# pls reassure me about thermal values

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#### myleftear

Paid Member
Me again.

Thermal conductivity: More is better (more thermal energy is "tranported")
Thermal resistance: Less is better (thermal energy flows faster between the media)

Sorry for not being too accurate, it is just for getting the basic picture right...

Thank you
david #### Sangram

Paid Member
You're mostly correct. Note that thermal conductivity is specified with a dimensional parameter, which is inverse to the material thickness.

Thermal resistance has no such parameter and is an absolute value, hence it tends to be a bit of a hit and miss for interfaces as thickness is actually quite critical. Typically interfaces use the conductivity parameter, whereas devices will use the resistance parameter.

#### myleftear

Paid Member
Thank you, sangram!

I'd need a little clarification:

…thermal conductivity is specified with a dimensional parameter, which is inverse to the material thickness.

Does that mean, the thicker the material, the ... higher the conductivity? (Would make logical sense to me...)

…whereas devices will use the resistance parameter.

devices such as heat sinks etc., right?

Thanks again!
david

#### Sangram

Paid Member
No, thinner layers tend to exhibit higher conductivity.

W/m/K is three dimensional spec. The amount of heat that can be conducted for a 1K rise in temperature difference per unit thickness. For example, a material with 10W/m/K will be able to sustain a 1K temp rise with 10W heat input across a meter. If the thickness is reduced a hundred times, the amount of heat input *for the same temperature rise* increases a hundred fold.

To note that the equation is valid for a square meter of material. Smaller area tends to decrease conductivity. Also that this applies in only cases where the W/m/K expression is used, which is to say not always.

When I said devices I meant semiconductors but yes, heatsinks too use the resistance parameter rather than conductivity. However manufacturers will provide a graph for varying lengths.

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#### Mark Tillotson

Thermal and electrical resistance, resistivity, conductance and conductivity are direct parallels. Replace amps with watts, volts with kelvin, siemens with W/K and ohms with K/W
The formulas relating resistance/conductance and geometry to resistivity/conductivity are the same.

conductance = conductivity x area / distance
resistance = resistivity x distance / area

Thermal conductivity is in W/K/m

conductivities and resistivities are material properties
conductance / resistance are dependent on geometry as well, ie device properties.

#### myleftear

Paid Member
Thank you very much sangram and Mark.
I first mistook the conductivity with the "heat-storage".

For example, a material with 10W/m/K will be able to sustain a 1K temp rise with 10W heat input across a meter. If the thickness is reduced a hundred times, the amount of heat input *for the same temperature rise* increases a hundred fold.

But I really don't get your example…
say there's sample "A", 1mm thick has 10W/m/K
and sample "B", 0.01mm thick (the same material but 1% its thickness) ... has 1000W/m/K I try to let that sink in (my intellectual resistivity makes my head need bigger heatsinks ...) and read on in Wikipedia...

The formulas relating resistance/conductance and geometry to resistivity/conductivity are the same.

conductance = conductivity x area / distance
resistance = resistivity x distance / area

Mark, I'm struggling to understand the equations.
does the slash ("/") in area / distance mean [or], or does it mean [divided by] #### Mark Tillotson

+ - * / are arithmetic operators in every computer language I've ever seen, and that's a lot.

#### Sangram

Paid Member
But I really don't get your example…
say there's sample "A", 1mm thick has 10W/m/K
and sample "B", 0.01mm thick (the same material but 1% its thickness) ... has 1000W/m/K Thermal conductivity is the ability of an entity to transmit heat from one end to the other. This is not dissipation, but a measure of how well it can conduct thermal energy through itself. The W/m/K is a measure of self-heating. That's probably the simplest I can explain it.

A 1 meter cube of copper is a good example to start with.

Copper has a thermal conductivity about 380 W/m/K. This means that if you apply 380W at the bottom of the cube (evenly spread throughout the 1m square end), the cube will heat by only 1 degree assuming a black body in contact with the top, ignoring dissipation along the sides.

Now if you take a shim from this cube of 1mm thickness, logically it will heat itself less for the same heat input - i.e it becomes more efficient at transmitting heat. So yes, to increase the temperature of this shim by a degree, we have to apply 380kw to it.

Obviously real life will be a little different with losses, contact area, etc.

Now assume you are dealing with a transistor that has a shim of 1mm to couple the die to the case. This is normally a small piece of copper, say 10mm x 10mm. Because of the much smaller area this will not be as efficient as our large flat sheet. By my inexpert guessing, a 100 sq. mm x 1mm thickness of our original cube will heat by a degree at 38W power input, giving it a thermal resistance of 0.02C/w. The transistor itself will not be as efficient, as there are inefficiencies in bonding the die to the shim etc.

Now you see how we took a material spec of thermal conductivity and used it to compute thermal resistance of a finished product.

The problem with your example in the quote is that the material spec applies only to a 1m thickness. By specifying other thickness, you have to then remove it from the units. When you specify material thickness as 1mm for a 10 W/m/K material, you are already making redundant the 'm' unit. This is basic math, your resultant measure will be W/K and no longer W/m/K.

Therefore a 1mm piece of a such a material will have thermal conductivity of 10000 W/K. For an even thinner layer of 0.01mm, the conductivity will improve to 1 MW/K (megawatt, not meter).

#### Mark Tillotson

Thermal conductivity is the ability of an entity to transmit heat from one end to the other. This is not dissipation, but a measure of how well it can conduct thermal energy through itself.
Correct
The W/m/K is a measure of self-heating. That's probably the simplest I can explain it.
No, W/m/K is the unit of thermal conductivity
A 1 meter cube of copper is a good example to start with.

Copper has a thermal conductivity about 380 W/m/K. This means that if you apply 380W at the bottom of the cube (evenly spread throughout the 1m square end), the cube will heat by only 1 degree assuming a black body in contact with the top, ignoring dissipation along the sides.
Nothing to do with black bodies. If 380 watts of heat is flowing uniformly through the cube, the temperature difference between the ends will be 1 kelvin.
Now if you take a shim from this cube of 1mm thickness, logically it will heat itself less for the same heat input - i.e it becomes more efficient at transmitting heat. So yes, to increase the temperature of this shim by a degree, we have to apply 380kw to it.
Nothing to do with increasing the temperature of the material, everything to do with the temperature difference across the material. So a sheet of copper 1mm thick with ice on one side and boiling water the other sees a temperature difference of 100degC (same as 100K), the temperature gradient is thus 100,000 K/m, and the heat flux is thus 380 x 100,000 = 38MW/square metre.

A more realistic example, that 1mm sheet of copper has 100W per square cm flowing through it from a big transistor. The heat flux is 100 x 10,000 W/square metre = 1MW/m^2, so the temperature gradient = 1M/380 = 2600K/m, which is 2.6K per mm

All of this is straightforward application of the equations for conductivity and conductance and for heat flow and temperature gradient

conductance = conductivity x area / distance (units are W/K, W/m/K, m^2, m respectively)

heat flow = temperature difference x conductance (units are W, K, W/K)

heat flow density = temperature gradient x conductivity (units are W/m^2, K/m, W/m/K)

You can check equations like this by substituting in the units, for instance in the first equation:

W/m/K x m^2 / m = W/K

[ and remember this is directly equivalent to the electrical units:

A/m/V x m^2 / m = A/V (or S/m x m^2 / m = S, where S == siemen, the unit of conductance (used to be mho, its the reciprocal of ohm). ]

You do need simple algebra, you do need to know all the relevant SI unit names and symbols (case always matters for SI, note).

#### myleftear

Paid Member
The W/m/K is a measure of self-heating.

No, W/m/K is the unit of thermal conductivity

You do need simple algebra, you do need to know all the relevant SI unit names and symbols (case always matters for SI, note).

Thank you again, both Sangram and Mark.
I have to admit that my algebraic resistance is rather high, as my conductivity of knowledge of physics is low.

But my mind tends to be rather adhesive to un-understood equations.
I'm reading over and over both this thread and neurochrome's "Taming the LM3886".
Already think I'm not that lost at last, although I still don't get it. Once I may...
david Status
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