0
\$\begingroup\$

I homebrewed a flashing LED landing light for my high performance experimental aircraft. The details are here

Because I was afraid of electrical noise (which could impact communications or navigation equipment), I went with a linear regulator. That's been a mistake. Either the linear regulator or the MOSFET I used keep blowing (even with very large heat sinks attached). I've got to do better.

What I'm after is a system that can handle between 7 and 10 amps, either flashing (normal) or constant (for night landing operations).

I've concluded that a buck regulator is probably the only way to handle the demands.

I'd sure appreciate any input! Thanks!


enter image description here

\$\endgroup\$
  • 2
    \$\begingroup\$ Add your actual schematic into the question. The Instructables articles are usually quite long so not many of us will be inclined to follow the link. \$\endgroup\$ – Transistor Feb 9 '19 at 23:44
  • \$\begingroup\$ so, hi Dave, this is an exciting project. Problem is that you're not actually asking a definite question. If that question is "how do I build a Buck converter for the following in- and output voltage at up to 10 A", many semiconductor manufacturers do have design wizards. \$\endgroup\$ – Marcus Müller Feb 9 '19 at 23:46
  • \$\begingroup\$ (and by the way, your instructable is actually missing, as far as my scanning goes, key information such as LED type and typical forward voltage at 5/7 A. \$\endgroup\$ – Marcus Müller Feb 9 '19 at 23:51
  • \$\begingroup\$ Also, what is your raw supply voltage? 12 volts? 28? \$\endgroup\$ – WhatRoughBeast Feb 11 '19 at 5:17
  • \$\begingroup\$ And what, exactly, is the value of the current set resistor on your LM338? What version of the CXA1304 are you using? I ask, because the data sheet does not indicate anywhere near a 10 amp capability. So, why show this schematic as part of your question? \$\endgroup\$ – WhatRoughBeast Feb 11 '19 at 5:26
4
\$\begingroup\$

A simple fix for you: Don't put all your LED's in parallel. The reason you heat up so much is the high current and low voltage. The high current heats up your resistors and MOSFET, and both the low voltage and high current heat up your linear regulator.

You state in your instructable that your LEDs are burning 2W each. If the 5A current is evenly divided amongst the seven LEDs, that's 0.7A and 2.8V each.

Try stacking 6 LEDs (Using six because of the even number and the chance you don't have eight) in two parallel branches of three. For the same brightness and LED wattage, you now only need to push 1.4 A (0.7 x 2), and the voltage across the three LEDs in series will be approximately 8.4V (2.8 x 3). Two parallel branches of four would be fine as well, just don't let the required voltage exceed what the regulator can drive (Dropout voltage and current resistor voltage drop).

In this new configuration, the resistor heating will be only 30% of previous, MOSFET heating will be only 8% of previous, and the current regulator heating will be only 11% of previous!! Another possible benefit: You will pull far less electrical power from your aircraft (~30% of previous).

Calculations:
Old:
MOSFET heating: \$I^{2} \times R_{DSon} = P \rightarrow (5A)^{2} \times 0.5\Omega=12.5W\$
Resistor heating: \$I \times V = P \rightarrow 5A \times 1.25V = 6.3W\$
Regulator heating: \$I (V_{in} -V_{out})=P\rightarrow 5A*(13V - (2.8_{V_{LED}} + 1.25_{ V_{res}} + 2.5_{ V_{MOS}}) ) = 41.3W\$

New:
MOSFET heating: \$(1.4A)^{2} \times 0.5\Omega=1.0W\$
Resistor heating: \$1.4A \times 1.25V = 1.8W\$
Regulator heating: \$1.4A*(13V - (8.4_{V_{LED}} + 1.25_{ V_{res}} + 0.7_{ V_{MOS}}) ) = 3.7W\$

Note: LED brightness balancing can be tricky. Better balance would have each parallel string of LEDs with its own current regulator, but this is mostly cosmetic. Having all the LEDs on the same heatsink as you do will help with keeping the brightness even. Here is a decent discussion of different LEDs series and parallel layouts. An extra consideration is reliability due to the aerospace application. If one LED fails to an open circuit, the whole series path will go dark. The current source will try and push twice the current through the other line, which might just brighten it, or might be enough to burn it out. Having two regulators with their own series LEDs creates a completely redundant segment, like having two separate lightbulbs.

Further Note: Paralleling other components is an easy way to increase the current capacity of other sections. It does come with a few considerations, but can help you reach much higher amperages with the same components. See this page for linear regulators, and this page for MOSFETs.

\$\endgroup\$
  • \$\begingroup\$ Wow - thank you so much for a really good discussion! Many great ideas. Key consideration will be to find what the voltage drop across the linear regulator is, to determine how many LEDs can be in series. Thank you Brandon! \$\endgroup\$ – mnvelocitypilot Feb 10 '19 at 14:45
  • \$\begingroup\$ Shouldn't the Rdson be milliohms? A lot has improved with LEDs in the past 10+ years since the article you cited on parallel series. It's not as much of an issue today, although I do agree they should be in series. A linear regulator should not be used for this project. \$\endgroup\$ – Misunderstood Feb 10 '19 at 20:07
  • \$\begingroup\$ The OP was using a 12V Cree XLM. I assume Vin is a nominal 12V and this is why he chose this LED. It is a very high density part with four dies. Stacking 6 of them (72V) is not practical. \$\endgroup\$ – Misunderstood Feb 10 '19 at 20:53
  • \$\begingroup\$ @Misunderstood Where did you get 12V from? Datasheet says 2.9V @ 0.7 A, which is nearly identical to what was calculated from measurement above. The aircraft supply is approximately 12V (13.6V), but that is a separate thing from the LED forward voltage. And the MOSFET RDSon is 0.5V from the datasheet. \$\endgroup\$ – Brandon Hill Feb 11 '19 at 3:29
  • 1
    \$\begingroup\$ @Misunderstood I assume the 1510 was just a misprint. There are definitely better FETs cheaply available. This one for example will take 60A with an RDSon of about 0.001 Ω at 1.45$ US a pop. There is some tradeoff with gate capacitance and breakdown voltage, but neither are important in the OP's case. I'll bet the IRF510 was simply available and/or cheap at the time. \$\endgroup\$ – Brandon Hill Feb 11 '19 at 5:28
1
\$\begingroup\$

What I'm after is a system that can handle between 7 and 10 amps

There are not many drivers for that kind of amperage.
The only one I know of is the Texas Instruments LM3409.
The inductor and variable switching frequency will help with EMI. PCB layout must be done correct to minimize EMI. Even with an inductor there are still some paths with discontinuous currents, theses are the most likely to generate EMI and should be kept as short as possible.

enter image description here


I see on the schematic you are using a Cree CXA 1304 (or Cree XML?). There are much better choices available.

The highest efficacy LEDs today is the Cree XP-G3 which is up to 185 lm/W @ 350 mA.

To increase irradiance by an appreciable amount you should use an optical lens to collimate the beam angle. Reducing the view angle from 120° to 10° will increase the irradiance on the other pilot's eyeball by over 100X. If you look at the inverse square law you will see the irradiance decreases exponentially with distance.

UPDATE

I am currently using Cree XML leds

That changes things. The LM3409 is for high current applications.

I do not see the value of Vin. This is very important.

I would guess 12V. There are many considerations for a 12V design for a string of LEDs. This is why I though you may have switched to the Cree CXA in the schematic. UPDATE I was thinking you were using the 12V, 1200 lumen Cree XHP "extreme High Power LED" which is used in some flashlights.

The reason I suggested high efficacy LEDs is because they generate LESS heat per lux. The electrical watts are converted to either light or heat. More light means less heat. The XP-G3 are 70-80% efficient.

That $6 light bulb is looking better and better as you learn about LED design. LEDs are nothing like a light bulb. Light bulbs can withstand high heat. LEDs do not work well at case temperature above 100° C. Light bulbs only need a voltage. LEDs require current regulation. LEDs are more efficient. Efficiency is not a huge consideration in aeronautical. Heat on the other hand is a huge consideration.

With 7 LEDs in a small space you have a nearly insurmountable thermal management problem. If you could, but doubt it is allowable, to have the LEDs only turn on when the wheels are off the ground and forced convection would solve the problem. Otherwise you need natural convection. For natural convection you will need a massive heatsink which is not likely to fit into the space you have.

If you still want to try, you will likely need a metal core PCB.

Use every square mm of area available to place the LEDs. Look at how higher watt LED light bulbs are designed and and mounted in a heatsink. Use the heatsink from a light bulb. I like the heatsinks in the third and fifth light bulbs from the left. The fifth one is almost ideal for your project.

enter image description here

As for the LED parallel or series issue you should use a combination of the two. For a 12V project 2 or 3 sets of 4 series LED strings will work well. I prefer all in series but then you will need a boost regulator.

Another good thing about high efficacy LEDs is their max Vf is also lower (lower Vf = higher efficacy). The Cree XP G3 have a max Vf of slightly over 3V. At 700 mA the typical is 2.83V @ 25° C, but will rise 0.0013V per degree when the ambient temperature drops in flight.

Your Vin, if similar to automotive is likely to be more than 12V, like 13-14V. This is an important factor. If the LED Vf for a 4 series string of XP-G3s is close to your Vin then a regulator may not be necessary. This is where a current limiting resistor works very well. Efficiency is not an huge issue but it will still be about 90%.

Because the XP-G3 is newer and more efficient than the 4 die XHP the Vf will likely always be below 12V if your are using a regulated 12V Vin.
You could use current limiting resistors on each of the XHP LEDs. This would be more efficient and more reliable than a linear regulator and your EMI fears are eliminated. A 2 W and about 2 Ω resistor will work with either the XHP or strings of four XP-G3. Keep in mind the XHP will generate a little more heat and a little less light.

If the current regulation generates any significant heat it may be better to move the regulators off the LED board. One PCB for the regulators and one for the LEDs.


I am also using optical reflectors and they do have lenses

Another difference between LEDs and light bulbs is light bulbs are isotropic light sources and LEDs areanisotropic (directional). Optical reflectors for LEDs are different than those for light bulbs.

The XP-G3 have a view angle of 125°.

I wrote an app to compare the radiation pattern of two LED. The image below compares two OSRAM Olson SSL LEDs, one 80° (bottom emitting up) and a 150°. A light bulb would be emitting photons equally in every direction.

enter image description here


A optical reflector must be designed for each LED. Because photons are sub-atomic particles not all of them will reflect. If you look at the angle of the LED radiation pattern you can see why the reflector should be unique for each LED. Using a reflector for a isotropic point source the photon will not reflect reflect in a forward direction.

There are many optical lenses made specifically for the XP-G3 and XML.

When you design your PCB you implement thermal management as best you can, then adjust the intensity to the maximum target temperature.

The optical lens is your best chance of achieving the desired illuminance (lux) on the eyeball's light sensors by reducing the LED's luminous intensity (and heat) to a point where the heat can be managed.

Illuminance or lux is a measurement of how the photon irradiance as it strike a surface (e.g. photons per square meter).

Lumen is a measurement of how many photons are emitted from the light source, not how many photons reach the eyeball. To estimate the lumen to lux, you must use the angle.

The easiest way to demonstrate the effect of a lens is to use a lumen to candela calculator.

Using this caclator you will see that 2000 lumen at 125° will create a bean with about 590 candela. A beam with a 10° angle with result in 83,650 candela.

If you do not use a lens you could use an OSLON SSL 80 which would give you 1,360 candela from 2,000 lm.

the control tower at my local airport can see me from 3-5 miles away.

To calculate the best view angle would be to know the the full range it must be seen. As you get closer, the beam width gets smaller. The irradiance will be much more intense the closer you are so it may be a wash when comparing max distance to min.

Right angle trig will easily solve this angle business.

At 5 miles and 10° the 50%, full width at half maximum (FWHM) of the lumen flux will spread out 1.76 miles (twice side A).

enter image description here

I assume the width is also important to be seen by other pilots as close as 1 mile, hopefully before 1 mile, but...

At 1 mile and 10° the FWHM is 0.36 miles.

enter image description here


...the LEDs mounted with thermal epoxy onto a 1/8th inch thick x 5 diameter aluminum disk, which has a 3x2.25x0.625 medium fin density heat sink (thermal) epoxied on the back. It's in an area in the nose of the aircraft that does get some circulation.

Is there circulation when the aircraft is sitting still on the ground?

The heatsink is likely insufficient. I would estimate it can dissipate about 5 watts if perfectly mounted and oriented. I base the estimate on this heatsink. Do you have a link to your heatsink datasheet?

A picture would help. A lot. Also I need to know the Vin voltage.

I do not understand how you can epoxy an XML to an aluminum disk. Are you referring to the CXA design?

Epoxy is not the best TIM. You really want to screw down two surfaces to put some pressure on the thermal interface. You only want enough TIM material to fill in the tiny air pockets in the imperfections in the metal surface. I use one 4/40 screw for each LED. I also use ½" copper pipes for heatsinks and pump chilled water for cooling.

enter image description here

You want the aluminum disk to be as thin as possible. A thick disk is for lateral heat transfer. In the project the thermal path is from the LED to the heatsink. If you used 0.062 sheet aluminum you would increase its thermal conductance 2X. The base of the heatsink will do the lateral heat transfer. Better yet would be to eliminate the disk and mount directly to a maybe a round heatsink like the one I linked to above. If we are talking about the CXA.

The advantage to using a CoB like the CXA is they have a large thermal surface and they can be mounted directly to a heatsink.

\$\endgroup\$
  • \$\begingroup\$ inverse square is not exponential. \$\endgroup\$ – Jasen Feb 10 '19 at 3:21
  • 1
    \$\begingroup\$ Thank you Misunderstood! I am currently using Cree XML leds. When I first did this design (around 3 years ago), those LEDs were the best choice. I am also using optical reflectors and they do have lenses (although I'm sure they could be better). As it stands, when it's working, the control tower at my local airport can see me from 3-5 miles away... so, while brighter is a goal, I'm not unhappy with the current output. Thanks again! \$\endgroup\$ – mnvelocitypilot Feb 10 '19 at 14:49
  • \$\begingroup\$ @Jasen Sorry, I got confused by the "square" in the name and the superscripted 2 on the distance which I had mistaken for an exponent. \$\endgroup\$ – Misunderstood Feb 10 '19 at 22:54
  • \$\begingroup\$ @Jasen so could I say it is exponential I say irradiance will increase exponentially as distance decreases? Is there another word for inverse exponential terms? \$\endgroup\$ – Misunderstood Feb 11 '19 at 0:33
  • \$\begingroup\$ @Jasen No, I do not want to use incorrect terminology. I did not know that it is incorrect to refer to a negative exponent as exponential. I was asking what term I should use to replace exponential in my answer. By definition exponential can be used for any expression that contains an exponent. You are correct if we stick to the formal definition. It seems to me using the term exponential is acceptable unless there is a formal term for a negative exponent. The exponent is still r² not K², it's just that r² is a denominator. \$\endgroup\$ – Misunderstood Feb 11 '19 at 5:42
0
\$\begingroup\$

You will never get 7 or 10A through linear regulators. 1.5A at best and with a large heatsink. You need an individual constant current regulator or power supply (can be DC to DC) for each LED. Then for each LED a large MOSFET (I suggest a 15A one or more) attached to a large heatsink. Or two MOSFETs in parallel to divide power dissipation. The gate of the MOSFET should recieve at least 5V, even if it can turn on with less, and more is better until +-16 (read datasheet before proceeding).

\$\endgroup\$
  • 1
    \$\begingroup\$ You can certainly make a high power linear regulator if you want to, but the thermal design must be done properly. But OP already said they want to do a switcher, so... \$\endgroup\$ – mkeith Feb 10 '19 at 7:29
  • \$\begingroup\$ Thanks to both Fredled and mkeith! I have done quite a bit of thermal design, but it clearly hasn't been enough. Thank you! \$\endgroup\$ – mnvelocitypilot Feb 10 '19 at 14:46
  • \$\begingroup\$ @mkeith Yes, it's always possible, but there will be too much power loss for a plane and fire hazards. \$\endgroup\$ – Fredled Feb 11 '19 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.