Electronic Product Design Heat Sink

8/3/2019 Electronic Product Design Heat Sink

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1.Basic Theory

A common rule of thumb (although not exactly accurate - see rule 9 here) is that every 10Creduction in the junction temperature of a semiconductor will double the life expectancy of thatsemiconductor clearly then it is in our interest to keep the junction as cool as is practical and in

order to achieve this we must consider three factors:

The sum of the thermal resistances between the junction and the ambient air (see diagrambelow)

The amount of heat to be dissipated

The ambient air temperature

Equivalent schematic

Where

Tj = Maximum semiconductor junction temperature

jc = Thermal resistance, junction to case.

Tc = Case temperature

cs = Thermal resistance, case to heatsink (see Hot Tips)

Ts = Heatsink temperature

sa = Thermal resistance, heatsink to ambient.

Ta = Ambient temperature

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Notes:

jc is a function of the semiconductor design and is a fixed number (see manufacturersdata). It cannot be influenced by the addition of a heatsink or any external agent.

cs can be minimised by the application of silicon grease on the semiconductor mountingsurface area.

sa is the most important and most controllable parameter. The aim is to reduce this to thesmallest possible level that is both economical and practical.

sa is a function of the convection coefficient (hc) and the heatsink surface area (A), and can beexpressed by the formula

Clearly the greater the heatsink surface area or convection coefficient then the smaller will be sa.Surface area (A) can be increased either by improved heatsink design or moving to a largerversion of the same heatsink.

The convection coefficient (hc) can be improved by moving from natural convection to forcedconvection (air or liquid).

2. Operating conditions

Under operating conditions (natural and forced convection) the permissible power dissipation (Q)

for a semiconductor/heatsink assembly is defined by the formula

In most applications the only unknown will besa. Therefore re-arranging equation 2 in favour ofsgives

Equation 3 now enables us to consider a practical application.

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Example

Assume a semiconductor (TO3 case) has a maximum operating junction temperature of 125C at10W dissipation, in an ambient air temperature of 50C. jc is given as 1.5C/W (manufacturersdata) and cs is estimated at 0.09C/W using thermal heatsink compound. Find the maximum saunder these conditions:

Solution: Using equation 3,

sa = 5.9C/W

This is the largest value ofsa that can be tolerated. Clearly smaller values would be acceptable athese would result in a lower junction temperature.

3. Extruded heatsink selection

3.1. Natural convection mode use of volume

We have seen from the previous section that, given certain information, we can calculate a figurefor the heatsink performance (sa) under given operating conditions. But having established thefigure, how can it be used? How can we relate it to the size of the heatsink required to achieve theparticularsa? Experience here can obviously help but failing that, we can check published datalooking for a suitable product.

However, a shortcut would be to use Graph 1 below. It illustrates the volume of heatsink requiredover a range of thermal resistances for natural convection. It is not exact, as it represents theaverage data from many profiles, but can be relied upon to a first approximation. The volume of aheatsink is the outline envelope times the length of the heatsink:



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Graph 1

Typically, to reduce thermal resistance by 50% the heatsink volume must be quadrupled. Thisassumes all other parameters remain constant.

Having established a figure for the volume and presumably knowing the maximum available widthand height, we can calculate the length. Alternatively, fixing any two parameters allows us todetermine the third. Armed with this information, and providing it is acceptable, we can now look foa suitable heatsink profile. Clearly, under natural convection conditions and assuming no othervariables, the heatsink volume must be increased to reduce sa.

3.2. Forced convection mode use of surface area

Again we are faced with the problem of establishing heatsink performance under a particular set oconditions and again we have adopted a generalised approach. The following curves combinetheory and practice plus some basic assumptions and we have considered the many questionsconcerning the relationship between thermal performance and extrusion length.



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Graph 2

The engineering data supporting the curves is based on air movements in the laminar region(frontal velocities of 120 to 240 metres per minute) and will provide good approximations of thermaperformance. As with natural convection, it is assumed that device quantity and location add nounusual heat distribution effects. Note that the outlet air temperature from the extrusion is going tobe higher than at the inlet, and it is imperative that this outlet temperature never exceeds thedesired maximum surface temperature of the heatsink. In fact it should remain as far below as ispractical.

The rise in temperature of the outlet air is given by the equation

where

T = increase in air temperature

Q = heat dissipation, in Watts

V = volumetric flow rate, in litres per second

The 0.83 constant is based on a 25C ambient air temperature. For ambient air temperatures above 25C,multiply T by the following correction factors:

Air temperature C Correction factor

30 1.02

40 1.06

50 1.09



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Multiply by 45C correction factor:

Tcorrected = 3.2 x 1.08 = 3.45C

Now calculate the heatsink temperature rise:

Tsink = sa x input power = 0.65 x 100 = 65C

The heatsink temperature at the air outlet is

Tair+ Tair+ Tsink = 45C + 3.2C + 65C = 114C

This suggests that the performance will stay within acceptable limits.

To visualise this more clearly it is useful to show these temperature profiles as follows:

Example 2: Determine the required length of extrusion for a particularsa

Again, consider the use of Wakefield 1371 extrusion, and assuming a design goal of sa =0.3C/W.

Because of variables associated with the length of heatsinks, we will average the information fromthe curve of each of the four lengths.

Locate 0.3C/W on the vertical axis of the graph and move horizontally right to intersect the curvesand read vertically down, recording the HDS for each length:

7.5cm length, HDS = 1500cm2



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15cm length, HDS = 2000cm2

22.5cm length, HDS = 2350cm2

30cm length, HDS = 2750cm2

Thus the average HDS is 2150cm2.

The HDS of Wakefield 1371 extrusion is 58.17cm2/cm and the required length can be determinedby simple division:

= 2150/58.17 = 40cm.

Clearly any greater length would be acceptable. For intermediate lengths it would be reasonable toextrapolate the required values.

4. Hot tips

4.1. Thermal resistance from case to heatsink (cs)

Typical values are as follows, depending on the mounting medium between the semiconductordevice and the heatsink:

Mounting medium Typical cs range, C/W

Wakefield thermalcompound 0.1 0.2

Beryllia washer0.2

Wakefield "delta

pads" 0.25 0.5

Mica washer0.5



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An excellent article about different thermal compounds that can be used to improve cs can befound at http://www.electronics-cooling.com/Resources/EC_Articles/MAY97/article3.htm.

A more extensive list of thermal interface materials can be found at http://www.peltier-info.com/tims.html

4.2. Distributed load extruded heatsinks

In order to achieve a balanced temperature rise along the heatsink (assuming there is an equalload sharing among transistors) the following mounting arrangements are recommended:

4.3. Extrusion fin design forced air application .v. natural convection

When extruded heatsinks are used in forced air applications, the fin spacing can be considerablycloser than for natural convection, due to the reduction in the boundary (or blanket) layer of airsurrounding the fins.

4.4. Thermal resistance .v. length (for a given profile)

For lengths in the range 4cm to 30cm, the thermal resistance is inversely proportional to the squarroot of the length (and therefore volume):

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4.5. Conduction paths

The thermal resistance through a solid is given by the equation

where K is constant for the material. Therefore to keep thermal resistance low, conduction pathsshould be short and have a large cross-sectional area.

Any finned shape will have a lower thermal resistance in forced convection than in naturalconvection, on account of the higher heat transfer rate. Therefore under these conditions choose aheatsink with thicker sections.

4.6. Thermal performance .v. mounting attitude

Thermal data produced by manufacturers generally relates to the most efficient mountingarrangement, with the fins vertical. The following is a guide to the reduction in performance thatcan be expected for different mounting attitudes.

However, it is stressed that the performance of a heatsink is dependant on many variables such aslocation in the equipment, effective airflow, fin spacing, fin height, fin thickness, base thickness,shape, and overall length. Consequently the impact on the performance of a heatsink mountedother than vertically is not a fixed number, and may depend on the inter-relationship between twoor more of these variables.

Vertical 100% effective



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Horizontal 85% effective

Horizontal Up 70% effective

Horizontal Down 60% effective

4.7. Specifications

Thermal efficiency improves (and therefore thermal resistance, sa, reduces) with increaseddissipation. Beware of data providing only one figure forsa. It is probably the best figure atmaximum dissipation. At low dissipation, sa would typically increase by 50%, or worst case 100%



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4.8. Black surfaces

Under natural convection conditions, the performance of a heatsink with a black surface will be 6%to 8% better than that with a plain or bright surface. However, this differential disappears underforced air conditions.

5. Peltier effect devices

These devices are solid state devices that function as heat pumps. They are electrically powered,and pump heat from one side of their body to the other. The effect is that one side gets hotter andthe other side gets cooler. They can be used to improve cooling of semiconductor devices by fixingthe semiconductor to the cool side of the Peltier effect pump, and mounting the heatsink on the hoside of the pump. They can be stacked to reduce the temperature further. They are quite expensivhowever!

In general, they are not very efficient, taking large amounts of power. For more information on thePeltier effect, devices, and manufacturers, see http://www.peltier-info.com/

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Electronic Product Design Heat Sink