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Old 1st February 2007, 09:26 PM   (permalink)
Default Is this electrically safe?

Hi All,

I have a large LED array encased in resin, it might generate a lot of heat.. What if I had the negative go via a large block of exposed copper at the bottom of the board to which I attach a heatsink to. Is this electrically safe.. will have like 6V, 4A going through it...... Perhaps I have a non-electrically conductive, but heat conductive paste between heatsink and copper...

Just wondering how to dissipate heat if I have lots of LEDs in a matrix ( greater than 100 in one block with 5mm between each LED. The resin covers the board and a tiny bit of the LED casing.

Peter
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Old 1st February 2007, 09:30 PM   (permalink)
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don't know about heat sink paste, but you can get 'insulating' heat sink pads. they ensure that there's no electrical connection between the heatsink and whatever it's fastened to.

most of the heat created by an led is dissipated throught the legs, and onto the board. . . so if the resin covers this, it may cause problems with heat transfer.?

As for electrically safe? 6v isn't going to really do much if you touch it. But it's not good practice to leave any part of a circuit exposed.

BTW: if it's on the negative side, it won't be 6v. It will be 0v
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Last edited by fingaz; 1st February 2007 at 09:36 PM.
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Old 1st February 2007, 10:55 PM   (permalink)
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Does each LED have its own current-limiting resistor? If the LEDs are 1.8V red ones then the resistors will be much hotter than the LEDs.

If you don't have current-limiting resistors then the LEDs will all blow up. They are diodes, not light bulbs.
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Old 2nd February 2007, 08:16 AM   (permalink)
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Quote:
Originally Posted by audioguru
Does each LED have its own current-limiting resistor? If the LEDs are 1.8V red ones then the resistors will be much hotter than the LEDs.

If you don't have current-limiting resistors then the LEDs will all blow up. They are diodes, not light bulbs.
But of course audioguru, I have been listening you know They are super-bright white ones.. about 3.7V



Peter
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Old 2nd February 2007, 02:43 PM   (permalink)
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Quote:
Originally Posted by PeterDove
They are super-bright white ones.. about 3.7V
Then the series current-limiting resistor for each LED has 6V-3.7V= 2.3V across it and its value is 2.3V/23mA= 100 ohms.
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Old 2nd February 2007, 02:59 PM   (permalink)
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Quote:
Originally Posted by PeterDove
Hi All,

I have a large LED array encased in resin, it might generate a lot of heat.. What if I had the negative go via a large block of exposed copper at the bottom of the board to which I attach a heatsink to. Is this electrically safe.. will have like 6V, 4A going through it...... Perhaps I have a non-electrically conductive, but heat conductive paste between heatsink and copper...

Just wondering how to dissipate heat if I have lots of LEDs in a matrix ( greater than 100 in one block with 5mm between each LED. The resin covers the board and a tiny bit of the LED casing.

Peter
Hi Peter Dove,

If you are designing, why not try constant current sources for each bunch of series LEDs and such souces and leds can be more in parallel. Now the control devices can be outside resin(potting material-unless you deside to block access to circuitry)

this will help optimization of power. instead of series resistors for each LED.
you may perhaps consider this view point.
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Old 2nd February 2007, 04:40 PM   (permalink)
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If you're worried about short circuits then add a fuse or pollyfuse of the appropriate rating in series with your LEDs.
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Old 2nd February 2007, 04:49 PM   (permalink)
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I assume that he was asking if it is electrically safe to have the 'copper bar' on the negative side and the heatsink exposed.
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