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I have lot of aluminum sheet 0.5mm thickness at home, normally heat sinks that I have seen had 1mm thickness.

Now I am wondering should I get that sheet welded as heatsink or do I absolutely have to invest in thicker sheet for efficient (at taking away and disposing heat) heat sinks ?

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  • \$\begingroup\$ What performance (thermal resistance Rθ degrC/W) do you expect from your heat sink and what restrictions in size you have? \$\endgroup\$
    – GR Tech
    Jun 1, 2014 at 8:02
  • \$\begingroup\$ I do not really know how to quantify performance here, I just want a good heat sink. I am going to have a 28cm(l) x 7 cm(b) with approx 2~3cm(h) and i am wondering should I get such a heat sink made out of 0.5mm thick aluminium sheet or not? \$\endgroup\$ Jun 1, 2014 at 8:07
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    \$\begingroup\$ If you want a good heat sink you must choose a good sized metal sheet. \$\endgroup\$ Jun 1, 2014 at 8:59
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    \$\begingroup\$ Theory is great, but by stacking many layers of aluminium, you are introducing some unknown factors. What is the thermal resistance form one (oxidized?) plate to another? I suggest you test the cooling capabilities of your creation. For example, Bolt a component to it (transistor / regulator) Make it dissipate a known amount off watts. and measure the temperature rize. If the Rth junction-heatink is given (datasheet component). You should be able to calculate the Rth heatsink-ambient. \$\endgroup\$
    – D.J.W.
    Jul 10, 2014 at 21:06
  • \$\begingroup\$ How many watts are you expecting to have to dissipate? \$\endgroup\$ Sep 8, 2014 at 22:15

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Rather than ask about how such a heat sink would work, why don't you try actually computing it? http://www.heatsinkcalculator.com/calculator.html will let you design your own.

Note that I entered "heat sink design software" into Google, and that was the third or fourth item on the list.

Let's take your example. You did not specify the number of fins, so I picked 10. The calculator says your 40 x 40 mm, 50 watt Peltiers will operate at about 160 C. Is that a good heat sink or not? Since do not provide a quantified performance goal, there is no way tell for sure, but most likely Not.

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  • \$\begingroup\$ I am just beginning to learn about heatsinks on my own, I am reading about these things on google but it's a little difficult because I am really really confused about how to differentiate a good heat sink from a bad one. \$\endgroup\$ Jun 1, 2014 at 10:09
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    \$\begingroup\$ Google can not replace school \$\endgroup\$
    – GR Tech
    Jun 1, 2014 at 11:10
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    \$\begingroup\$ Determining the degree of goodness of a heat sink is simple. Ask the question, "For a given amount of power being dissipated, how much does the temperature of the heat source rise?" The smaller the temperature rise, the better the heat sink. \$\endgroup\$ Jun 1, 2014 at 11:54
  • \$\begingroup\$ That was easy enough @WhatRoughBeast thank you for the clarification. GRTech : that's exactly the problem I am suffering from. \$\endgroup\$ Jun 1, 2014 at 14:02
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three ways for heat transfer: conduction- usually a good way of transferring heat. convection- fluid movement...caution:stagnant air is a good heat insulator. Radiation- often overlooked. Heat transfer to the air through use of a heat sink starts with conduction from the heat source through the source-sink interface. For best results, the interface should be bonded or as intimately made close to the source as possible. Some bonding materials, if used correctly, aid in heat transfer. It is best to have the sink enclose the device as much as possible. The selection of the sink should be based on its conductivity. Metals such as copper and brass are better than aluminum. Some porcelains are good under discrete devices to spread heat to the heat sink on top where the fins are... If possible position the device near a metal part of the container at a lower temperature so that the heat sink can be designed to contact it.

Others noted elsewhere have good ideas on the importance of fin design and relationship to total area exposed to air and fans to increase airflow.

To use radiation, the distance and field of view to a cooled surface plus the surface field of view are important. Radiation is a function of the surface absorptivity and emissivity. For most cases, a black surface with high a/e will create radiant flow depending on the temperature of the cooler surface.

I hope this will give a general perspective in planning to cool a hot spot.

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Thin metal will distribute the heat poorly; you will get a hotspot near the heat source and the outer regions of the metal will be relatively cool.

Nobody wants to spend much money on heatsinks : if the commercial ones could be made thinner and cheaper, they would be. (And 1mm already sounds thin to me, all the heatsinks I have seen approaching 28cm long have been considerably thicker)

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  • \$\begingroup\$ So, I should get a thick base probably 5mm with 2mm wide fins? would that be good enough? Should I get fins stretching from one side to another(along width) or even those should be slotted? \$\endgroup\$ Jun 1, 2014 at 8:57
  • \$\begingroup\$ @rijulgupta - did you use the calculator I linked to? It will answer your questions. \$\endgroup\$ Jun 1, 2014 at 17:03
  • \$\begingroup\$ @WhatRoughBeast : Please note the time difference b/w this comment and your answer! BTW I have already tried that link many times since yesterday I do not really understand it's working and implications! for ex source temperature, temp diff etc. \$\endgroup\$ Jun 1, 2014 at 17:26
  • \$\begingroup\$ I cannot say from the information given that it will be good enough : however it'll be better than the originally proposed solution. RoughBeast's calculator is worth learning, and may suggest a solution - perhaps including a fan for forced convection. \$\endgroup\$
    – user16324
    Jun 1, 2014 at 17:45
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Cooling electronics it is not an easy matter. First please look at here Designing a copper plate heat sink to have an idea about the relevant calculations.

I don’t recommend to invest in metal sheets as well, but -roughly- if you have a sheet of 0.5mm thick alum. and 20x20cm side, do not expect to dissipate more than 5W and to keep your heatsink at the temperature of 70 degrC.
About the same final temperature you can achieve with a piece of 12x12 cm alum. 2mm thick, or you can mount the semiconductor to one side of your (iron) metal enclousure, 2mm thick, if it is sized 18x18 cm

Finally the lower the thermal resistance needs your system, the most critical the calculations are, and the most difficult the selection process is.

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Since you seem pretty attached to using your thin sheet metal, there is a possibility you can try. Make a series of U channels with your .5 mm sheet, each channel the same length and about the same height, but with gradually decreasing width. Stack them together, and weld the areas in contact.

schematic

simulate this circuit – Schematic created using CircuitLab

This figure shows the idea in cross-section.

If you use, let's say, 10 pieces of aluminmum, the maximum base thickness will be 5 mm. If you can produce small differences in width you can produce a lot of fin area for a given overall width.

There are a couple of problems you need to think about.

1 - You must keep the bottom of each channel absolutely flat during welding, or the finished heatsink will not have a flat base, and you will not be able to make good thermal contact with your Peltiers. This probably means you will need a spot welder, rather than an arc or gas welder, and will require real attention to what you are doing.

2 - Because each channel is connected to the others at separate points, the thermal conductivity of your final base will be less than if you made it out of solid aluminum. It may not be too bad, though, since each sheet is pressed against its neighbors.

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Aluminium heat sinks are always made with the fins too thick. Because optimum-thickness aluminium heat sink fins are too soft and fragile.

Notice that any (old) steel heat sink you see will have thinner fins than any (new) aluminium heat sink: even though the thermal conduction is worse, the toughness of steel allows thinner fins.

Fortunately, the exact thickness of fins is not a critical parameter for a convective-flow heat sink. Making fins twice as thick only slightly reduces the fin spacing, which only slightly reduces the airflow, which only slightly changes the thermal resistance of the system.

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