# How to calculate Junction to Ambient temperature of LED?

I'm trying to calculate the Junction to Ambient temperature for this Cree LED:

https://www.mouser.ca/datasheet/2/90/ds--XHP502-1093532.pdf

I know how to calculate this:

Junction Temp = Junction to case + Case to Heat Sink + Heat sink to Ambient + Ambient

However, If I want to know what is junction to ambient temperature (aka no heat sink), how do I calculate this from the data sheet?

• It seems you have it backwards. The junction temperature is calculated from the PCB temperature. A clearer question would be: How do I calculate junction temperature from ambient? – Misunderstood Sep 8 '18 at 12:38

This is a high power LED. It is designed to dissipate the vast majority of its heat through the solder joints. So you need to figure out the solderpad-to-ambient thermal resistance of the circuitboard onto which you intend to solder it. This can of course be calculated based on the dimensions of the copper features on your board, but the easiest method may actually be to just solder the LED to a board and measure the temperature at the solderjoint for a few different power levels.

• This is actually the best method because most hobbyists may be able to bake a board, but results may vary. One had to test any single piece. – Janka Sep 7 '18 at 20:47
• What your answer is missing is the calculation of the junction temperature from the PCB temperature. The junction temperature is what determines the actual operating characteristics. PCB temperature is not of much use without the junction temperature calculations. – Misunderstood Sep 8 '18 at 12:48
• @Janka the best method for what? The measured PCB temperature is the starting point to calculate junction temperature. – Misunderstood Sep 8 '18 at 14:41
• With a hand-baked board, you never know how good your soldering is. You always had to check the calculated temperature on the board against reality. Measuring the tab temperature of the LED is key, because you can rely on the datasheet. – Janka Sep 8 '18 at 16:05
• @Misunderstood: The author of the question seems to already understand how to calculate the solderjoint to junction temperature differential. – Timmy Brolin Sep 9 '18 at 1:43

Unlike other semiconductor devices such as transistors, attaching a heatsink to the device is not much use as it will block the light.

Page 12 of the datasheet states:

The maximum forward current is determined by the thermal resistance between the LED junction and ambient. It is crucial for the end product to be designed in a manner that minimizes the thermal resistance from the solder point to ambient in order to optimize lamp life and optical characteristics. [Emphasis mine.] Figure 1. The thermal resistance figures are only quoted for junction to solder point. This implies that junction to air (through the lens) is not significant.

You should be able to design to meet the solder pad specifications at the end of the document and follow the rating curves give. Figure 2. Current safe operating area vs ambient temperature.

• +1 As to your first statement, blocking the light would be much worse than useless since as much as half the power input is radiated out as light. – Spehro Pefhany Sep 7 '18 at 20:34
• Your answer is missing the junction temperature calculation. The datasheet characteristics are related to junction temperature. Safe operating range is nice thing to know but it does not help with estimating the effects thermal management such as radiant flux and Vf. – Misunderstood Sep 8 '18 at 12:53

calculate the Junction to Ambient temperature

I assume this means calculating the junction temperature from ambient air.

I know how to calculate this:
Junction Temp = Junction to case + Case to Heat Sink + Heat sink to Ambient + Ambient

I don't think you do. That is a nonsensical formula. And it is unclear if you are referring to temperature or thermal resistance because your terms do not have proper labels. Case to junction what? Temperature? Thermal resistance?

To calculate Junction Temperature (Tj), you need to know how much thermal power (heat) is being generated and how much of this heat is being dissipated.

If I want to know what is junction to ambient temperature how do I calculate this from the data sheet?

There is no such thing as "junction to ambient temperature"

There is junction temperature, ambient temperature and junction to ambient thermal resistance.

Very little of this information comes from the datasheet.

To calculate the number of watts to be dissipated you need to know the electrical watts being consumed and the radiant watts emitted from the LED.

The electrical watts comes from the IV curve. To calculate the radiant watts you need to convert the photometric (luminous) flux (lumens) to radiometric flux (radiant watts). To do that you need the spectral distribution. If I were to digitized the blue curve to get the relative height of the curve at each wavelength then plot the wavelengths I would get something like this: So now I have the radiant watts for each wavelength. I have to convert the watts to lumens. This is a two step process. Photons at each wavelength carry different amounts of radiant energy.

For example if we compare the 450 nm (blue) radiant flux and 550 nm (green) radiant flux
1 µmol of blue photons would be carrying 0.27 watts of energy. Whereas
1 µmol of green photons would be carrying 0.22 watts of energy.

To "measure" lumens actually the radiant flux is measured then the radiant watts is converted to lumens according to the Photopic Luminous Efficacy from the Relative Sensitivity Curve for the C.I.E. Standard Observer table.

1 Watt of 450 nm blue photons converts to 27.46 lumens.
1 Watt of 550 nm green photons converts to 679.55 lumens.

So you get the number of lumens from the datasheet Then adjust the lumens to the amount of current you will be using to drive the LED Then use the above conversion from lumens to watts

The temperature is a function of the thermal resistance to ambient and how much heat is being generated at the junction. In the thermal path from the junction to ambient there are a few thermal resistances. The junction temperature decreases as the thermal flux moves from the junction to ambient. The amount of thermal resistance in the path of the thermal flux determines how quickly the flux flows measured by how much the temperature is reduced for the amount of heat generated (°C/W).

Heat flows from hot to cold. So we need to provide a thermal path from the junction to ambient with as little thermal resistance as possible. The datasheet only has one thermal resistance in this path. That is from the junction to the thermal solder pad on the case of the LED.

The next thermal resistance is the solder between the LED and PCB.

The image on the left is the copper in the PCB and the solder deposition pattern on the right.  Source: Cree XLamp XHP50.2 LED datasheet

The thickness of the solder and the solder's thermal conductivity determines the thermal resistance. You can see by the stencil patter they go to great lengths to insure the solder is as thin as possible.

The rest of the thermal resistance come from the design of the PCB and heatsink.

Thermal Resistance Model For A Typical PCB For a semiconductor the formula for ambient to junction temperature is:

Power dissipation for a classical semiconductor is the same as the electrical power used:
voltage x current = watts (both thermal and electrical).

An LED dissipates the electrical power as both heat and radiant energy (light). To calculate the thermal watts you subtract the radiant watts from the electrical watts.

This Cree XHP50.2 LED has fairly poor efficacy at 129 lm/W for the 5700K LED in this line. It will dissipate more heat than a lower density LED. This LED is designed to emit as many lumens as possible in a small area as it has 4 dice creating a Vf of 11.2V or 2 dice at 5.6V. I would want to see an efficacy closer to 180-200 lm/W in a high power lower density single die 5700K LED such as the XP-3G.

The electrical power at test current is 11.2 x 700 mA or 5.6 x 1400 mA = 7.84 watts. This LED will radiate about 47% of the electrical power as light and 53% as heat. This leaves 4.15 watts of heat to be dissipated. NOTE: the 47% is an educated guesstimate based the efficacy of other LEDs for which I know the radiant flux.

If Tambient (PCB temperature) is 60°C the junction temperature is.

Tj = (Øja x 4.1 watts) + 60°

Øja is highly dependent on the PCB layout and thermal management (e.g. heatsink, thermal vias, copper thickness and etc).

In this case the required Øja would be calculated from the 4 watts that needs to be dissipated. Then the PCB would be designed to achieve the required Øja.

So you would need to estimate the Øca (case thermal pad to ambient) for your PCB to get the junction temperature from the PCB temperature. The 1.2 °C/W in the datasheet (labeled as exposed pad in above image) is added to the thermal resistance from the LED's thermal pad to ambient.

The best and easiest, IMHO, to understand source to make these calculations is:
Thermal Design By Insight, Not Hindsight
And "by easiest" I do not mean simple.

Thermal management of light sources based on SMT LEDs

PowerPAD™ Thermally Enhanced Package, SLMA002H November 1997 Revised July 2018

I found this document calculating an LED’s Junction Temperature to be of little use but it should backup some of the calculations used in my answer.

## Bottom Line

You can get yourself a masters degree in thermodynamics and do the calculations outlined above. If you are very bright you could take a short cut and take the MIT Open Course Ware class: Lecture 1: Thermodynamics Part 1

Or you can do it the practical way.

Design your PCB using the best practice thermal management methods, build your PCB turn it on and adjust the current based on the temperature of the PCB near the LED and desired brightness. I generally do not allow the PCB temperate to exceed 55°C. This is given that I only use LED with a max junction temperature of 150°C.