hyperboost driver question

Axkiker · Aug 23, 2010

Has anyone ever used a hyberboost driver from Task led where they needed to mount the driver seperatly from the leds in an aluminum box of some sort.

I know that the hyperboost will need to be heat sinked so I am thinking abotu using one of the aluminum project boxes you can buy. My thoughts are to mount the driver to the top of the box and hope it will conduct enough heat to keep the driver inside cool.

I also thought about using a small aluminum box and potting the boost driver.

Any thoughts?

bshanahan14rulz · Aug 23, 2010

Test before implementing. Boost drivers and long output leads don't seem to combine so well...

Axkiker · Aug 23, 2010

bshanahan14rulz said:
Test before implementing. Boost drivers and long output leads don't seem to combine so well...

Well that brings up a good point

What would you consider long.. I need to take that into account.

I did email Task LED and they said as long as I provide a thermal path for the heat to dissipate out that potting was fine....

Now that brings up a whole other question. I need some sort of aluminum box which i could place the driver in then pour potting material on top of. I really need the box to have some good fins to help dissipate the heat.

Anyone seen something like what I would need. There are all sorts of project boxes on ebay that I could use.. However im looking for more fins.

georges80 · Aug 23, 2010

Before you try to choose a box, you need to at least work through some numbers. Without more information no one here (or me) will be able to answer most of your questions.

You haven't specified in your posts:

a) how many LEDs (total Vf)?
b) What drive current?
c) Input voltage range?

Without those answers you can't determine approximate efficiency and therefore dissipation losses in the driver. Once you know how many watts are being dissipated as heat you have some chance of deciding on a box etc.

Why do you need to pot the driver? Is there some environmental reasons like moisture etc?

What length do you expect the wires from the driver to the LEDs to be. A few feet will be fine with hyperboost, it is very stable on the output side.

cheers,
george.

Axkiker · Aug 23, 2010

georges80 said:
Before you try to choose a box, you need to at least work through some numbers. Without more information no one here (or me) will be able to answer most of your questions.

You haven't specified in your posts:

a) how many LEDs (total Vf)?
b) What drive current?
c) Input voltage range?

Without those answers you can't determine approximate efficiency and therefore dissipation losses in the driver. Once you know how many watts are being dissipated as heat you have some chance of deciding on a box etc.

Why do you need to pot the driver? Is there some environmental reasons like moisture etc?

What length do you expect the wires from the driver to the LEDs to be. A few feet will be fine with hyperboost, it is very stable on the output side.

cheers,
george.

Okay here we go....

4 led's ran in series
12.8V a piece
1.05A
input voltage to hyperboost 12v

I may be able to get away without potting however moisture is a concern.

wires will be no more than a few feet from LED

Axkiker · Aug 23, 2010

I ran the numbers and if I did everything correctly I should have almost 5A on the input... So not ideal however within specs..

I believe that if I run it off 24V I will have 1/2 the current which would help me out.

Any advice on heat issues

Justin Case · Aug 23, 2010

24V input probably won't work too well when using a boost driver to run a 12.8V load.

Not sure how you get 5A in if Vbatt is 12V while Vload is 12.8V. Your Vbatt to Vload ratio is nearly unity, which is as good as it gets. Assuming 90% driver efficiency

0.9*12V*Ibatt = 12.8V*1.05A

giving a calculated Ibatt of 1.24A.

If the driver efficiency is 90%, then the driver waste heat will be only about 1.5W. That should be easily manageable.

Axkiker · Aug 23, 2010

Justin Case said:
24V input probably won't work too well when using a boost driver to run a 12.8V load.

Not sure how you get 5A in if Vbatt is 12V while Vload is 12.8V. Your Vbatt to Vload ratio is nearly unity, which is as good as it gets. Assuming 90% driver efficiency

0.9*12V*Ibatt = 12.8V*1.05A

giving a calculated Ibatt of 1.24A.

If the driver efficiency is 90%, then the driver waste heat will be only about 1.5W. That should be easily manageable.

Sorry I should have been more clear

its 12.8V a piece... so with 4 in series im over 50V

Justin Case · Aug 23, 2010

The more favorable you can get your Vbatt to Vload ratio, the better off you are.

If Vload is 4*12.8V = 51.2V and Vbatt is 12V, then

Ibatt = 51.2V*1.05A/0.9/12V ~5A (assuming 90% driver efficiency)

Ibatt is right at the driver's limit (generally not a good thing). The driver waste heat is about 6W, which is a lot of heat to dissipate.

If you increase Vbatt to 24V, then Ibatt is halved to about 2.5A. Driver efficiency should also improve by 1%-2%, which could cut the heat that you need to dissipate down to about 4.7W to 5.3W (call it 5W). You can save about a watt.

georges80 · Aug 23, 2010

24V input would certainly help reduce input current and hence increase driver efficiency and thus drop waste heat losses.

Hyperboost does have an EXCELLENT thermal path from the components that generate heat losses to the back side of the PCB. Mounting the PCB via the supplied thermal pad or high quality double side adhesive thermal tape will provide a very good path for the components to transfer the heat to the heatsink.

The key is you have 5 to 6W of heat you need to radiate/convect etc - so you need to determine whether your enclosure and heatsink can dissipate that heat while maintaining a stable temperature to the PCB.

From my own measurements, the supplied thermal pad material easily will keep the 'hot' components within about 10C of the heatsink. So, if your box/heatsink runs at say 50C stable, then the hottest a component on hyperboost will reach is 60C. Given all the 'hot' components have excellent thermal paths from the junctions to the PCB (all have thermal pads) you will have a very conservatively running driver configuration.

So, YOUR job is to see what temp your case (that hyperboost is mounted to) stabilizes at when running long term. 60C is what I call the ouch test - you can touch 60C for a while, but after a bit you'll go ouch and want to remove your finger. So, if you can touch the outside of the box where hyperboost is mounted on the other side and keep your finger there, then I'd say you have a safe operating temperature - ASSUMING you have hyperboost firmly mounted with the thermal pad/tape material to the box and the box is made of aluminium etc...

Oh, also, the 5A max input current is a conservative figure, so running it at 5A long term is not a problem IF the heat dissipation issues are addressed.

cheers,
george.

Axkiker · Aug 24, 2010

georges80 said:
24V input would certainly help reduce input current and hence increase driver efficiency and thus drop waste heat losses.

Hyperboost does have an EXCELLENT thermal path from the components that generate heat losses to the back side of the PCB. Mounting the PCB via the supplied thermal pad or high quality double side adhesive thermal tape will provide a very good path for the components to transfer the heat to the heatsink.

The key is you have 5 to 6W of heat you need to radiate/convect etc - so you need to determine whether your enclosure and heatsink can dissipate that heat while maintaining a stable temperature to the PCB.

From my own measurements, the supplied thermal pad material easily will keep the 'hot' components within about 10C of the heatsink. So, if your box/heatsink runs at say 50C stable, then the hottest a component on hyperboost will reach is 60C. Given all the 'hot' components have excellent thermal paths from the junctions to the PCB (all have thermal pads) you will have a very conservatively running driver configuration.

So, YOUR job is to see what temp your case (that hyperboost is mounted to) stabilizes at when running long term. 60C is what I call the ouch test - you can touch 60C for a while, but after a bit you'll go ouch and want to remove your finger. So, if you can touch the outside of the box where hyperboost is mounted on the other side and keep your finger there, then I'd say you have a safe operating temperature - ASSUMING you have hyperboost firmly mounted with the thermal pad/tape material to the box and the box is made of aluminium etc...

Oh, also, the 5A max input current is a conservative figure, so running it at 5A long term is not a problem IF the heat dissipation issues are addressed.

cheers,
george.

I like what I am hearing....

I am going to put together a package and see wht happens. ill measure the outside case temp and go from there....

What I really need is a heat sink type case... Something that is small, hollow on one side but has some serious fins on the top side to dissipate heat..

I have yet to find something like that

Axkiker · Aug 24, 2010

by the way here is the case I thought about using

!Bo)gkI!BGk~$(KGrHqMH-CUEuN4MRrlZBLo)fSSl9g~~_12.JPG

I thought about mounting the hyperboost to the underside of the unit.. That way heat would be dissipated outward..... I would mount with supplied thermal pad material and screws... or however you recommend.

Justin Case · Aug 24, 2010

Thermal resistance is given by bond line thickness/(thermal conductivity * effective contact area). The TGF120K thermal pad material has a thermal conductivity of 1.2 W/m-K and an uncompressed thickness of 1mm. Let's say that the pad thickness reduces by 1/10 to 0.1mm (1x10-4 m) when clamped.

The effective contact area is harder to measure. I'm going to approximate the total contact area by measuring the LxW of the three rectangular thermal via areas. I get about 0.15"x0.16" (1.5x10-5 m^2), 0.25"x0.15" (2.4x10-5m^2), and 0.07"x0.22" (9.9x10-6 m^2), or a total area of about 4.9x10-5 m^2.

Thus, thermal resistance calculates to (1x10-4)/(1.2 * 4.9x10-5) = 1.7 K/W.

At 6W, that translates to about 10K temp differential between the case and the board.

George has measured a 10K differential.

Can I cook the numbers any better than that?

Axkiker · Aug 24, 2010

Justin Case said:
Thermal resistance is given by bond line thickness/(thermal conductivity * effective contact area). The TGF120K thermal pad material has a thermal conductivity of 1.2 W/m-K and an uncompressed thickness of 1mm. Let's say that the pad thickness reduces by 1/10 to 0.1mm (1x10-4 m) when clamped.

The effective contact area is harder to measure. I'm going to approximate the total contact area by measuring the LxW of the three rectangular thermal via areas. I get about 0.15"x0.16" (1.5x10-5 m^2), 0.25"x0.15" (2.4x10-5m^2), and 0.07"x0.22" (9.9x10-6 m^2), or a total area of about 4.9x10-5 m^2.

Thus, thermal resistance calculates to (1x10-4)/(1.2 * 4.9x10-5) = 1.7 K/W.

At 6W, that translates to about 10K temp differential between the case and the board.

George has measured a 10K differential.

Can I cook the numbers any better than that?

WEll not sure I understand any of that but I think it means its all good LOL

hyperboost driver question

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