heatsink size, watts, current capacity?

mercrazy

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Jul 28, 2014
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making a light with pwm driver.
1. is total watts all that matters when sizing heatsink?
as supply voltage goes up, current goes down and vice versa, so watts stay the same.
if the number of LEDs change, and watts stays the same, is the same amount of heat generated?

2. when supply voltage is in normal range, total current on my reverse polarity protection diode is .9amp. if voltage drops, current goes to 1.2amp before decreasing when voltage drops too low. should i use a 2amp diode?
thanks
 

DIWdiver

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Yes, total watts is all you care about for the heatsink. But that's total watts of HEAT. A pretty significant amount of the input power leaves as light.

Oh, well that and max die temp, thermal path, environment, etc.

Yes, I would use a 2A diode. They are cheap, and there's no sense stressing your parts to save a few cents. I don't like to operate anything near, at, or especially above it's rating as a matter of normal operation.
 

mercrazy

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thanks as usual, you always come through.
my heatsink is getting to 140F running 6 3535 LEDs at .7amp.
i have good heat transfer with this proven heatsink running at .7amp.
i want/need to take it up to .85amp with same 6 LEDs.
temp goes up to 180F which is too high. need to keep it safe to touch.
this is same circuit you looked at before with 2 strings of 2 LEDs in parallel running at 1amp so each side gets .5amp.
also have a separate string of 2 LEDs on different driver.
don't like running it this way but overhead can drop too low and cause a bunch of trouble.
when i had strings of 3 LEDs and bad connections made overhead too low, customers would blame the light.
don't want this any more.
any ideas how to run higher current without overheating the heatsink?
do the site owners know how valuable you are?
thanks again.
 

DIWdiver

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do the site owners know how valuable you are?
thanks again.

I doubt it, but it doesn't matter. I don't do it for them.

To stay cool with more power, add a fan or use a bigger heatsink. There aren't really many other options. Physics is so unyielding.
 

KENN MOSSMAN

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Sep 11, 2018
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making a light with pwm driver.
1. is total watts all that matters when sizing heatsink?
as supply voltage goes up, current goes down and vice versa, so watts stay the same.
if the number of LEDs change, and watts stays the same, is the same amount of heat generated?

Total watts is a major part of sizing a heat-sink but you also have the thermal coefficients from
the junction to case, and case to heat-sink. As well as ambient temperature,and the thermal coefficient of the heat-sink.
If you are using PWM then I don't know what you mean by the supply voltage going up and down. It doesn't with PWM.

Otherwise as V goes up, C goes up......................... V = I * R
 

MeMeMe

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Aug 27, 2018
Messages
125
thanks as usual, you always come through.
my heatsink is getting to 140F running 6 3535 LEDs at .7amp.
i have good heat transfer with this proven heatsink running at .7amp.
i want/need to take it up to .85amp with same 6 LEDs.
temp goes up to 180F which is too high. need to keep it safe to touch.
this is same circuit you looked at before with 2 strings of 2 LEDs in parallel running at 1amp so each side gets .5amp.
also have a separate string of 2 LEDs on different driver.
don't like running it this way but overhead can drop too low and cause a bunch of trouble.
when i had strings of 3 LEDs and bad connections made overhead too low, customers would blame the light.
don't want this any more.
any ideas how to run higher current without overheating the heatsink?
do the site owners know how valuable you are?
thanks again.

I looked at your other thread and I think I understand, but perhaps I am missing something. Are all parts mounted on the heat-sink or just the LEDs? Something seems not right here. Assuming 70F ambient, at 0.7A, you are getting a 70C rise. At 0.85A, you are getting a 110F rise. The forward voltage of the LEDs could be a bit higher, but I doubt more than a few percent. The 110F versus 70F rise suggests 57% more dissipation for a <25% power increase to the LEDs. Something is not right.

Either the heat sink really is hotter, or it is measurement error. Are you using a mechanical measurement method (thermocouple) or optical temp sensor? I would go with the former, and make sure you glue it to the heatsink with a thermal glue.

I think the first thing to do is figure out why the increase is so much larger than the estimated power increase. It's either measurement error or an unexpected power draw.

I assume you are doing this test "free air", i.e. with no movement of air? What color is your heat-sink? Is it aluminum? Bare aluminum is a poor radiator of heat, oxidized better, anodized can be very good or so-so depending on the treatment. Can always test with a light layer of paint sprayed onto the heat sink. Color really is not overly important.
 

KENN MOSSMAN

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Sep 11, 2018
Messages
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making a light with pwm driver.

2. when supply voltage is in normal range, total current on my reverse polarity protection diode is .9amp. if voltage drops, current goes to 1.2amp before decreasing when voltage drops too low. should i use a 2amp diode?
thanks

WHY do you think you need reverse protection????

1) waste of power
2) unnecessary; LEDS do have a reverse break-down threshold
3) extra cost & trouble
4) you can use a MOSFET or Schottky (spelling) diode
 

MeMeMe

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Aug 27, 2018
Messages
125
WHY do you think you need reverse protection????

1) waste of power
2) unnecessary; LEDS do have a reverse break-down threshold
3) extra cost & trouble
4) you can use a MOSFET or Schottky (spelling) diode

Virtually every automotive electrical product, and almost all other products powered by high current batteries without a keyed polarity connection are going to have reverse polarity protection.

The reverse breakdown threshold on an LED can be very low and many LEDs explicitly state they are not to be reverse biased.

A FET based reverse bias protection will waste minimal power. The trouble is at best minimal, and while there is a cost, it is better than frying your product, starting wires on fire, etc.
 
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