## Schamplifier – Heat Sink Design

[Edit:  Whoa!  I was way off with my calculation of the heat sink surface area.  12″x3″ is 0.023 m2, not 0.21 m2.  This changes a lot, it makes $R_{\theta(hs-a)-conv} \approx 8.6 ^{\circ}C / W$, which is way too much.  If I up the heat sink size to 12″x6″, I get $R_{\theta(hs-a)-conv} \approx 4.3 ^{\circ}C / W$, which is not as awesome, but is probably doable.  I’ve updated the analysis to reflect all of this.]

## Introduction

Like the Squelette, I aim to design an enclosure that can also serve as the heat sink for the amplifier ICs (and any other power electronics that end up inside, like voltage regulators for control circuits, and so forth).  The heat-sink portion of the enclosure will be made of sheet and angle aluminum.

Since I’m don’t yet have a list of everything I want to include inside the enclosure, I’d like to provide a little more breathing room than the Squelette enclosure would offer.  Further, the two amplifier ICs will probably not be attached to the same panel of the enclosure.  Finally, I’d like to hook some kind of heat sink up to the ICs before I construct the enclosure so I can test out the circuit without burning everything up.  I could just use the dimensions specified for the Squelette enclosure, but that would be unwieldy for my test and wasteful of raw material for my enclosure, so I’d like to understand thermal design enough to figure out what kind of heat sink this device requires and determine with some level of confidence whether a particular enclosure design will function to dissipate heat adequately for the amplifier.

I had a helpful exchange with the author of the Squelette, Ross Herschberger, in the “Heatsink Design” thread on the Squelette main page.  This exchange led me, ultimately, to this analysis.

## Background

Knowing zilch about heat dissipation and transfer, I had to start learning from scratch in support of this project.  I found this EEVblog episode on thermal design the most helpful.  This short page and calculator on convective heat transfer helped me to understand how convection factors in.  This page on heat sink design provides a lot more general background information.  Here is a calculator for heat sink temperature rise that made it quick and easy to check my results as I was fiddling around with numbers.  The Wikipedia page on Thermal resistance in electronics helped me understand how the heat travels through heat sink materials.  Finally, I found this calculator that was somewhat helpful in finding the thermal resistance of an aluminum slab.

## Maximum Power Dissipation

Before we can determine what kind of heat sink capability we require, we must first determine how much power the amplifier will be dissipating.  From the LM1875T datasheet, given the power supply voltage and average load, we can compute the maximum power dissipation of the device.  The formula is

$P_{D(max)} \approx \frac{V_S^2}{2\pi^2 R_L} + P_Q \ \ \ (1)$

Where:

• $P_{D(max)}$ is the maximum power dissipation
• $V_s$ is the total power supply voltage (in this case, for the Squelette, ±18V = 36V)
• $R_L$ is the load resistance (in this case, 8Ω impedance speakers)
• $P_Q$ is the quiescent power dissipation of the device

The datasheet gives 100mA as the maximum quiescent current, and claims for example that a 60V system would have 6W of quiescent power, but it is not clear to me whether we can just use Joule’s law (P=IE) to determine the power here (since it’s not just a plain DC system).  If we did cheat and use Joule’s law, we’d get at most 3.6W for $P_Q$.  Plugging everything else in to solve for $P_{D(max)}$, we get roughly 11.8W for the power dissipation (not far off from the 11W given in the Squelette instructions).  This also looks pretty close to the value shown on the datasheet graph for power dissipation for a ±18V supply, which is further confirmation that we’re pretty close.(1)

The datasheet lists some example calculations for determining the maximum allowable heat sink to ambient thermal resistance.  The relevant formula is

$R_\theta = \frac{T_{J(max)} - T_{A(max)}} {P_{D(max)}} \ \ \ (2)$

Where:

• $R_{\theta}$ is the total thermal resistance from the junction to ambient, the sum of
• $R_{\theta(j-c)}$ (junction to case) (listed in the datasheet as 2 oC/W)
• $R_{\theta(c-hs)}$ (case to heat-sink, e.g., the thermal resistance of your insulating pad or thermal compound layer) (in my case, the insulating pads are claimed 0.33 oC/W
• $R_{\theta(hs-a)}$ (heat sink to ambient) (this is what we’re trying to determine)
• $T_{J(max)}$ is the maximum junction temperature from the datasheet (listed as 150 oC)
• $T_{A(max)}$ is the maximum ambient temperature at which we are planning to operate the device (for example, 70 oC)
• $P_{D(max)}$ is the maximum power dissipation necessary (computed above as ~12W)

For our example, if we want to maintain a die temperature below the maximum rated 150 oC for ambient temperatures up to 70 oC, with our power dissipation at 12W, we would need a total thermal resistance from junction to ambient of less than (150 – 70)/12 = 6.67 oC/W.

Since the other components of $R_{\theta(hs-a)}$ are known,

$R_{\theta(hs-a)} \leq 6.67 - (2+0.33)$

or

$R_{\theta(hs-a)} \leq 4.33 ^{\circ}C/W \ \ \ (2b)$.

So, given a chunk of aluminum of arbitrary size, how can we determine whether its thermal resistance is less than 4.33 oC/W?  There are two factors at play here:

1. How quickly does the heat sink material transfer heat from the area where the heat source is connected (assuming it is smaller than the size of the heat sink itself) throughout the body of the heat sink?  We’ll call this value “heat sink thermal resistance due to conductance”, or $R_{\theta(hs-a)_{cond}}$
2. How quickly will the heat be dissipated from the heatsink due to convection and thermal emission?  We’ll call this value “heat sink thermal resistance due to convection” or $R_{\theta(hs-a)_{conv}}$

They both contribute to the total thermal resistance of the heat sink.  Imagine a heat sink made of a perfectly conductive material that instantly changed temperature across its entire mass — its thermal resistance would still be bound by the amount of heat that could be dissipated by convection.  Likewise, imagine a forced cooling system so effective that any heat on the surface of the heat sink would be instantly whisked away to the frigid depths of space — no matter how quickly the heat could be drawn off the surface of the heat sink material, it must still travel from the source through the material to the surface, and so presents some thermal resistance.  Such a cooling system would be of little use if the heat sink had a thermal conductance so poor (and therefore, that the heat took so long to travel from the source to the surface) that the IC had been fried by the time the heat could be dissipated.

Since I don’t know how these two factors affect one another, so will we take the naive, conservative approach and model thermal resistance of the heat sink as a whole as the sum of the two factors:

$R_{\theta(hs-a)} = R_{\theta(hs-a)_{cond}} + R_{\theta(hs-a)_{conv}} \ \ \ (3)$

## Thermal Resistance of the Heat Sink

The rate at which the body of the heat sink can conduct heat away from a warmer place (where the IC is attached) to a cooler place (presumably, the edge of the heat sink) is determined by the heat conductivity of the heat sink material and its shape.  For a simplified model, consider Fourier’s Law for heat conduction (from Wikipedia, Thermal Resistance in Electronics):

From Fourier’s Law for heat conduction, the following equation can be derived, and is valid as long as all of the parameters (x, A, and k) are constant throughout the sample.

where:

• Rθ is the thermal resistance (across the length of the material) (K/W)
• x is the length of the material (measured on a path parallel to the heat flow) (m)
• k is the thermal conductivity of the material ( W/(K·m) )
• A is the total cross sectional area of the material (measured perpendicular to the heat flow) (m2)

This gives us the general idea (that is, thermal resistance depends on how much material through which the heat must flow, and how fast the material carries heat).  Heat will be transferred faster through a shorter distance and through a wider cross section.  This equation might work for us if our heat sink were the shape of the IC package (TO-220, in this case) and extruded out for some length.  Since we want to incorporate the heat sink into the enclosure as a broad plate, rather than extruded solid mass, however, it’s not so simple.  The heat will be spreading along the (vary narrow) cross-section of the aluminum from some point (presumably, near the middle) to the edges.

I could not find a good example set of equations with which to model this.  However, I did come across this online calculator for “Slab Thermal Resistance With Constriction”, which models thermal resistance from a smaller source (like our IC) to a larger slab of material (like our heat sink).  It does not take into account convection, but that’s okay, we’re handling that elsewhere anyway.

If we plug in the following values:

• heat source width (TO-220 package dimensions, roughly 10mm)
• heat source length (TO-220 package width, roughly 15mm)
• slab width (heat sink width, 6″ or 0.1524m)
• slab length (heat sink length, 12″ or 0.3048m)
• slab height (heat sink thickness, 1/8″ or 3.175mm)
• thermal conductivity (of aluminum, 200 W/moC)
• film coefficient (of air, 5 W/m2oC), though this value doesn’t seem to matter much

We get a result of $R_{\theta(hs-a)_{cond}} \approx 0.52 ^{\circ}C/W$  (see note (2))

## Thermal Resistance Due to Convection

Ultimately, all heat dissipated by the heat sink must be dissipated by free convection (since I don’t want to use a fan or other active cooling system).  Since we know the amount of power we want to dissipate (12 W) and the size of the heat sink we are trying to use, we can use Newton’s Law of Cooling to determine the temperature at which the heat sink will dissipate that power.  According to this handy explanation of convective heat transfer,

The equation for convection can be expressed as:

$q = k A dT\ \ \ (4)$

where

$q$ is the heat transferred per unit time
$A$
is the heat transfer area of the surface (m2)
$k$ is the convective heat transfer coefficient of the process (W/m2K or W/m2oC)
$dT$
is the temperature difference between the surface and the bulk fluid (K or oC)

In our case, we know:

• $q$ (the power to dissipate, 12 W)
• $A$ (the area of our heat sink, for a 12″x6″ plate, roughly 0.046 m2)
• $k$ (for air, 5 W/m2 oC.  Various places I looked had widely differing values for the convective heat transfer coefficient of air, but 5 W/m2 oC seemed generally to be the worst value, so we’ll be conservative and start with that)

We want to solve for dT (the temperature rise above ambient), which works out to

$dT = \frac{q}{kA} \ \ \ (5)$

or, substituting our values, ~48.2 oC.  This means that, with a 12″x6″ heat sink attached to a device dissipating 12W of power, we can expect the heat sink to rise ~48.2 oC above the ambient temperature.  To compare, by the same formula, if we had a 3″x3″ heat sink, we would expect a rise of ~386 oC above ambient at the same power dissipation.

Now that we know the temperature our heat sink will be when dissipating a given amount of power, it is easy to compute the effective thermal resistance.  In this case, it will be:

$R_{\theta(hs-a)conv} = \frac{dT}{q} \ \ \ (6)$

Substituting Eq. 4 into Eq. 5 yields:

$R_{\theta(hs-a)conv} = \frac{\frac{q}{kA}}{q} \ \ \ (7)$

which simplifies to:

$R_{\theta(hs-a)conv} = \frac{1}{kA} \ \ \ (8)$

So, for our 12″x6″ chunk of 1/16″ aluminum plate, substituting 5 W/m2 oC for k and 0.046 m2 for A, we get an $R_{\theta(HS-A)conv}$ of 4.3 oC/W.

## Conclusion

Equation 8 seems to be a severe simplification, but perhaps we can determine experimentally whether it is an oversimplification.  The question remains whether it is useful as a rule-of-thumb in the absence of better information.

Working with our conservative, naive approach, we can go back to equation (3) to get what should be an upper bound for the thermal resistance of the heat sink to ambient, using our example values:

$R_{\theta(hs-a)} = R_{\theta(hs-a)_{cond}} + R_{\theta(hs-a)_{conv}}$

$R_{\theta(hs-a)} = 0.52^{\circ}C/W + 4.3^{\circ}C/W$

$R_{\theta(hs-a)} = 4.8^{\circ}C/W$

Since we have a max $R_{\theta(hs-a)}$ limit of 4.33 oC/W (from (2b)), it looks like we won’t be able to run quite at 70oC ambient temperature.  If our total thermal resistance with this heatsink is 2 + 0.33 + 4.8 = 7.1 oC/W, we can see that at an ambient temperature of 20 oC, dissipating 12 W of power, our junction temperature should be:

$T_J = 20 ^{\circ}C + 12 W * 7.1 ^{\circ}C/W$

$T_J = 20 ^{\circ}C + 85 ^{\circ}C$

$T_J = 105 ^{\circ}$

This is within the max operating temperature, but is quite hot.  If we recompute our maximum allowable ambient temperature, we get 64oC.  At the very least, this means we must be careful to make sure the ambient temperature stays within reason (for example, we shouldn’t put the amplifier, enclosure and all, into some other sealed container).

An oft-repeated quotation, commonly attributed to Albert Einstein, goes something like

Everything should be made as simple as possible, but not simpler.

Now that we have made the heat sink design for this amp simple, we need to determine whether we have made it too simple.  There was certainly a lot of assumptions and fudging going on in these calculations, and there’s nothing better than to confirm them (or prove them wrong) than by some experimentation.

• Our computation of $P_Q$ using the power rule and the eye-balling of the graphs in the datasheet is pretty close to the right value
• The value of $R_{\theta(c-hs)}$ from the insulating pad datasheet is accurate
• My model of isolating the convection and conduction factors and summing them for an approximation of the total thermal resistance is not too far off
• The value of the conduction thermal resistance factor, which came from an online calculator (the correctness of which I cannot examine and verify), is pretty close to the right value
• It is sufficient to use in that calculator the total package dimensions instead of the dimensions of the die itself (see note 2)
• The value of 5 W/m^2K for the convective heat transfer coefficient of air is a good lower bound, and is not made even worse by size, shape, and orientation factors
• I didn’t screw up the algebra and arithmetic anywhere

To put these assumptions to the test, I will actually construct a heat sink, re-do this analysis (based on the final size and shape of it), and then run some tests to see if the measured case and heat sink temperatures line up with the expectations.  That at least should give me some sense of how close this approach will get me, or if it’s so far off as to burn up my amplifier chip.

## Notes:

1. The datasheet also provides a graph for supply current vs. supply voltage.  If this is to mean quiescent current, then it looks like the chip draws ~60mA (instead of ~100mA) at ±18V, which (again, cheating with Joule’s law) would give us 2.16W instead of 3.6W for $P_Q$ (quiescent power dissipation) in our total equation), which would bring $P_{D(max)}$ to just under 11W instead of just under 12W.  Working on the upper side of these values will provide us with a bit more headroom.  We can always go back and recompute with more precision if the ballpark answers are too difficult to work with.
2. In email discussions, a technical associate of Novel Concepts, Inc. (the organization which produced the calculator) advised me to use not the dimensions of the TO-220 package as the source but rather the dimensions of the die inside the chip (which he gave as ~4.27 mm).  I am not so sure about this approach, mainly because we already have from the datasheets the $R_{\theta(j-c)}$, but also because most datasheets I have seen do not provide the dimensions of the die itself.  Using the reduced dimensions results in a value of ~0.8 oC / W, rather than ~0.5 oC / W.  That seems relatively insignificant to me, but I’m certainly open to other feedback on the matter.
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### 2 Responses to Schamplifier – Heat Sink Design

1. schazamp says:

I n00bed the heat sink surface area conversion to meters, so the original thermal resistance due to convection was about an order of magnitude too low. I upped the heat sink size, and corrected the figures, and arrived at a doable, though not awesome, new result.

2. jaycollett says:

Funny, I’ve been obsesed with dealing with thermal issues correctly, just yesterday I started the layout of a new power distribution board and while researching ran accross this article talking about the use of thermal vias for bottom side pcb cooling. It was particularly intresting as I happen to use d2pak package voltage regulators alot.

http://powerelectronics.com/mag/602PET22.pdf

Thanks for the post!