Matching your COB to a heat sink is one of the most difficult parts of designing your own COB grow light. Unless the manufacturer of the heat sink actually specifies the wattage you can get away with running, you’re left to do the calculations yourself. Luckily, there are tons of heat sinks available now that come pre-drilled for your COB or board, and are rated, in plain English, for their power handling ability. You might see smaller pin-fins (120mm or so) that are rated for 50 watts of power, larger pin fins at 140-150mm rated for 75 watts, or even bigger sinks that are 160mm+ and rated for 100W.

That being said, you might be a glutton for punishment and have a burning desire to source your own heat sink, or figure out if the one you’ve got kicking around will work. Fortunately, like most things in the DIY LED world, we can figure this out with a little research and perseverence. Fair warning though, this is pretty dry stuff.

The ultimate goal of a heat sink is to dissipate as much heat from the individual LED junctions (the actual point in each individual diode that electrical current flows through) as possible. Some COBs can have over 100 of these junctions, so quite often, manufacturers prefer to deal with the temperature of the “case”, which is the substrate that all these diodes are mounted to. Keeping each of these junctions under their maximum allowable temperature is critical so it’s very important that the heat sink you select is capable of shedding the heat your light produces.

Let’s start with an overview of how heat is created and how it moves through a light assembly, and then we’ll have a look at how to properly diagram this process. Finally, we’ll apply this knowledge to calculating our heat sink requirements with some examples.

## How Heat Flows in a COB LED

As power flows through the junctions of the LEDs in the COB, not all of it is converted to light. A good portion of this power is converted to heat, which is undesirable but a necessary evil. In a perfect world, 100% of the power would be converted into photons, but even LEDs are nowhere near that. Calculations done for many COBs by people far smarter than I have shown some COBs to be 50% efficient and higher (when driven at low currents), but manufacturers tend to err on the safe side. Taken from Cree’s Application note on Thermal Management of XLamp LEDs:

To be conservative, assume LEDs convert 25% of the input power to light and output 75% of the input power as heat. This estimate varies depending on current density, brightness and component, but is a good estimate for thermal design.

This waste heat that’s generated at each junction has to go somewhere, and that’s why we attach the COBs to heat sinks. Here is how heat travels through the light assembly, starting from the junction:

- Heat is conducted from the semiconductor junctions within the LED Modules to the case, which includes the inner components and substrate to which they’re mounted.
- Heat is then conducted from the case to the thermal interface material (TIM)
- Heat is then conducted through the TIM to the heat sink
- Heat is then conducted up through the heat sink, and finally is convected to the air around the heat sink by natural convection or forced convection. Some heat is also radiated to the ambient.

Below are 2 illustrations of COB LEDs mounted to a heat sink with a layer of thermal interface material (TIM) in between. The first image is from Bridgelux’s Application Note AN30 regarding Thermal Management – I’ll be drawing from this document for a lot of information in this article and I would suggest reading the entire application note if you’re interested in learning more. The second image is from the Thermal Management section of Citizen’s COB LED instruction manual.

Now, if we want to start plugging real numbers into this, we’ll need to define each of these points and the thermal resistances between them. We can simplify this by drawing out a thermal circuit.

## The Thermal Circuit

Below is a diagram and excerpt taken from Bridgelux’s Application Note AN30. This diagram represents the “Thermal Circuit” that heat must travel through on a single COB mounted to a heat sink. A thermal circuit can be drawn just like an electrical circuit: the flow of heat is equivalent to current, temperature itself can be likened to voltage, and thermal resistance is just like normal resistance. The specific definition of thermal resistance, according to Bridgelux is as follows:

Thermal resistance is a measure of the steady-state heat flow from a point of higher temperature to a point of lower temperature, calculated by dividing the temperature difference by the heat flow between the two points. The thermal resistance is usually denoted with an (Rx-y), where x and y are the points of different temperature.

In the diagram above:

**Pth**is heat dissipated at the junction and conducting through the LED Modules**Tj**is the temperature at the junction of the device**Tc**is the case temperature**Th**is the temperature at the point where the heat sink is attached to the substrate**Ta**is the ambient air temperature**Rj-c**is the thermal resistance from junction to case**RTIM**is the thermal resistance between the case and the heat sink (TIM resistance)**Rh-a**is the thermal resistance of the heat sink to ambient.**This is the number we’re after!****Rc-a**is the sum of thermal resistance from case to ambient**Rj-a**represents the total thermal resistance from junction to ambient (system thermal resistance)

Now that the basics are covered, let’s apply them to find out which heat sinks will work for our COBs.

## How to Calculate the Required Thermal Resistance of Your Heat Sink

To find a heat sink with proper thermal resistance, you’ll need to figure out a few variables first, including:

- The power you’re running through the COB (more specifically, the
*thermal power,*which is “heat watts” not “light watts”). - Ambient air temperature (Ta).
- Your desired maximum case temperature (Tc).
- The thermal resistance of your thermal interface (RTIM).

### 1. Calculate Your Thermal Power (Pth)

To find your thermal power, the first step is to multiply your drive current by your forward voltage. Remember, forward voltage changes as you change your current. The easiest way to check your forward voltage is to use the manufacturer’s product simulation tools. Check out this thread to learn how. Let’s say you’re going to be driving a COB at 700mA and determine that its forward voltage at that current is 34V. Multiply these together (34 * .7) to find your total power, which equals 23.8 Watts.

Now, figure out how much of this power is going to be wasted as “heat watts”. Cree and Bridgelux recommend using a factor of .75 so multiplying 23.8 by .75 gives us **17.85** heat watts. You may choose to use different factors based on which COB you’re using (also, based on what color temperature you’re using, as cooler whites are more efficient – Bridgelux actually specifies a factor of .65 for cooler whites like 5000K). If you trust the efficiency calculations you see coming from other hobbyists around the web, you could use these numbers they’ve calculated instead (if a COB is 55% efficient at a certain current, you’d use .45 as your factor for heat watts).

### 2. Measure Your Ambient Air Temperature (Ta)

You’ll need to know what the air temperature is in the room you’ll be running your lights. This one’s pretty easy.

### 3. Determine Your Maximum Case Temperature (Tc)

How hot do you want to run your COB? Check the data sheet and see how the chip performs at different temperatures – most will list the maximum allowable case temperature. Here’s an example of a Cree CXB3590:

Viewing the graphs above, I would choose 85 degrees as my max case temperature. At 85 degrees, the case is cool enough that the COB can still receive its max current, even though I’d never run that much current anyway.

Some data sheets will explicitly state a case temperature maximum like this Citizen CLU048-1212:

### 4. Find the Resistance of Your Thermal Interface Material (RTIM)

Thermal resistance (lower is better) is not to be conflated with thermal conductance (higher is better), which is the spec that many thermal compound manufacturers provide. If you’re using regular old thermal grease like most people and want to just throw a number at it without wasting time researching this, an estimate of **.05 °C/W** should do the trick. This is almost negligible and will only add a few degrees (if that) in temperature rise, but is worth factoring in nonetheless. If you’re hell-bent on getting the exact number, check the data sheet for your compound.

### 5. Apply Your Parameters to the Thermal Resistance Formula

Now that we’ve got those parameters defined, we can start plugging them into our formula. The actual formula we’ll use to make this calculation is:

##### Rθxy = (Tx – Ty) / Pth

Essentially, this equation states that the thermal resistance between two points is equal to the temperature of the first minus the temperature of the second, divided by the thermal power.

## Examples

1. Let’s say we’ve figured out the variables listed above and are ready to apply the formula to find the required thermal resistance of our heat sink. Here are our example figures:

- Thermal Power (Pth)= 24W
- Ambient Air (Ta)= 25°C
- Maximum Case Temperature (Tc)= 85°C
- Thermal Interface Material Resistance (RTIM) = .05°C/W

First thing we need to do is find out what the thermal resistance is from the case all the way to the ambient air. Once this is done, we can isolate the resistance of the heat sink to ambient.

To solve for Rc-a, we apply the formula **Rθxy = (Tx – Ty) / Pth**, Since we are solving for Tc to Ta, the “x” in our formula will become “c” (our maximum case temperature in degrees Celsius) and “y” in the formula will become “a” (our ambient air temperature in degrees Celsius).

When we substitute our values, we get Rc-a = (Tc – Ta)/Pth, which, with our actual numbers, is Rc-a = (85-25)/24

The result is Rc-a = 2.5°C/W

Now that we have Rc-a, we can solve for Rh-a because we know RTIM. Our equation for this is **Rh-a = Rc-a – RTIM**

When we substitute our values, we get Rh-a = 2.5-.05

And the final result is **Rh-a = 2.45°C/W**. In other words, we will need a heat sink that rises no more than 2.45 °C above ambient for every watt it’s trying to dissipate.

2. As another example, what if we tripled our thermal power but left all other factors the same?

- Thermal Power (Pth)= 72W
- Ambient Air (Ta)= 25°C
- Maximum Case Temperature (Tc)= 85°C
- Thermal Interface Material Resistance (RTIM) = .05°C/W

Rc-a = (Tc – Ta)/Pth, which, with our actual values, is Rc-a = (85-25)/72

The result is Rc-a = .83°C/W

Now we use **Rh-a = Rc-a – RTIM**

When we substitute our values, we get Rh-a = .83-.05

And the final result is **Rh-a = .78°C/W**. In other words, we will need a heat sink that rises no more than .78 °C above ambient for every watt it’s trying to dissipate. This heat sink will need to be considerably less resistant to heat flow than the first one and will need to be much larger in size.

## Using Rh-a Values to Find Suitable Heat Sinks

When you’re looking at heat sinks, many manufacturers will provide thermal resistance specifications. Now that you can calculate your requirement, all you need to do is find a model that can meet or exceed it. Have a look at these data sheets for examples:

We can prove our formula works by applying it against the Cooliance data sheet above. In their notes, they state that they are assuming a case temperature of 85°C and only 20% efficiency of the LED (they word it a little funny by calling it “input power to output power conversion efficiency of 80%”, but this 80% actually refers to the percentage of heat watts they’ve assumed).

Under the power dissipation column, we’ll use the ambient 25°C values. For the CPLI4050-XXX, they say dissipation is 83W.

- Thermal power will equal 83 x .80 = 66.4W
- Ambient Air (Ta)= 25°C
- Maximum Case Temperature (Tc)= 85°C
- Thermal Interface Material Resistance (RTIM) = N/A (unaccounted for here)

Then, Rc-a = (85-25)/66.4

Rc-a = .90°C/W, which is what they list as the thermal resistance of that particular heat sink. Nice!

Below are a couple MechaTronix heat sinks that also list their thermal resistance in the specs:

## Rules of Thumb When Thermal Resistance Figures are Unavailable

You won’t always be lucky enough to find thermal resistance specs for heat sinks you’re looking at. In this case, you can follow these “rule of thumb” recommendations. Every rule I found also requires you to calculate the total surface area of the heat sink in question – this isn’t too hard to do though. Just figure out the surface area of each fin, multiply by the number of fins, and then figure out the surface area of the base.

- The Bridgelux AN30 Thermal Management paper states that as a general rule of thumb, for a well-ventilated heat sink, there should be 64.5 cm2 (10 in2 ) of the heat sink in contact with cooling air for every 1 Watt of thermal power dissipated.
- As per Cree’s XLamp Thermal Management paper, a rough estimate of approximately 5-10 in2 of heat sink surface area per watt of heat can be used for a first-order estimate of heat sink size.
- GrowMau5’s video on choosing a heat sink recommends 17 square inches per heat watt for a passive heat sink and 6 square inches per heat watt for an active heat sink.

March 3, 2017 at 12:30 pm

Just ordered 6 of these for my next build, whatever it may be. 996 square inches for 9 bucks, couldn’t pass them up. http://www.ebay.com/itm/162352212802?_trksid=p2055119.m1438.l2649&ssPageName=STRK%3AMEBIDX%3AIT

March 4, 2017 at 9:45 am

Cool. Do you have any other parts picked yet?

March 4, 2017 at 11:12 am

Not sure yet, might stay with the 1212’s due to the low price and drive a bunch softly, not sure how much efficency I will gain in the 1212’s going 1000ma or under as I’m too lazy to do the math and figure it all out. but might jump up to the 50 volt chips, I would like to go more efficent but those bigger chips get pricey. Those luminous cobs look interesting as well, and the price is right. Might just do a dual cob with the biggest citizens I can get my hands on, got my box of heat sinks that I ordered yesterday, showed up the next day, they are very nice, over 1000 square inches a piece, because there are actaully 17 fins. I might do another order of them before they sell out and have heat sinks for every future project I will have. probably wait till next fall to get all the parts around for a second light, got to figure out how much space in the spare bedroom that I’ll be allowed to occupy next winter.