COB LEDs get hot. It is true that they are more efficient than CFL lamps (compact fluorescents – the ones that take forever to warm up), and way more efficient than incandescents (either the old light bulbs or the hotter halogen bulbs). Even so, some 70% of the electrical energy going in comes out as heat. Since the heat from LEDs doesn’t radiate as infra-red, like an incandescent bulb, they need to be attached to a heatsink of some description to remove the excess heat.
Cooling the LEDs used for the candle effect
It’s all about surface area – the more the better. The space available is basically a long, thin tube, since the lamps will be inside the frame at the bottom of the art work. So an individual heatsink attached to one lamp must be less than 25mm in height or depth. However, the width of a heatsink is limited only by the width of the frame – 1.2 metres.
For the previous project I used a small heatsink on each LED. However, that had the disadvantage of raising the LED closer to the edge of the frame, making the light source more visible. High power LEDs are unpleasant to look directly at, even from a distance, and looking directly at them from close up can cause eye damage (the recommended minimum safe distance for a 10 Watt LED is 2 metres). Therefore, I wanted to place the LEDs well inside the frame so that they were not directly visible. However, there’s no room to do this if they have a big heatsink underneath them.
You can see from the picture how the heatsink makes the LED stick out (the artwork is on the left and the front of the frame is on the right). It would be better if the LED could be attached directly to the inside-bottom of the frame, but it’s made of wood which doesn’t conduct much heat away.
It seemed the best answer was to have one long, thin heatsink extending the width of the frame, rather than individual heat sinks for each lamp. That way the heat will be more distributed, with the whole heat sink available to dissipate the heat from the lamps. So the solution I chose for this project was to replace the inside-bottom of the frame with a metal strip and attach the LEDs to that. In fact, what I did was to remake the frame with a gap underneath and then attach the metal strip containing the LEDs and electronics.
Finding a suitable metal strip wasn’t easy. First, I looked on eBay, but I couldn’t find anything suitable. Then I started looking around the DIY stored for something I could repurpose. In B&Q I found these. The 25x25mm right-angle piece looked promising. I could screw it inside the front of the frame so that the horizontal face was just above the bottom of the frame, then attach the LEDs to that. The LEDs would be as deep inside the frame as possible and the metal strip would act as a long communal heatsink. The only thing was that the strips were about £7 for 1/2 metre, which was expensive and a bit short. Then I checked in Homebase and found they sell identical metal strips, but theirs are 2.4 metre long for £10.
However, when I went to attach the LEDs I discovered a slight problem: the 3 Watt LEDs are too big for the heatsink (and they would stick out enough to touch the artwork). So I decided to use 10 Watt LEDs instead (the square one in the picture) and simply use a lower voltage so they only run at 3 Watts. They are the same price (about a pound each). The voltage necessary for this is slightly lower than that needed by the actual 3 Watt LEDs; this is achieved by using a series resistor (see previous post).
The first task was to cut a 1.2 metre length of the right-angled metal strip (i.e. the width of the frame – yes, it’s a big painting).
To attach the LEDs I sent off for some 2.5mm bolts, nuts, washers and shake-proof washers. I used two per LED (diagonally opposite). So next, I marked and drilled the two holes per LED. Since the strip was so big, I decided to mount all the electronics on it as well, so I drilled some holes to mount the two PCBs (see next post). I also drilled a number of holes in the other side to attach the strip to the inside-front of the frame.
Then I sprayed the outside faces black (in the garden as the fumes are surely toxic). Black surfaces radiate much more heat than silver, and the idea was to radiate the heat down, out of the frame and away from the artwork. I tried a few types of spray paint, but settled on a £1 can of black car spray paint from my local low-cost household shop.
The candle LEDs will only be running at a low power – about 300 mA maximum (i.e. around 3 Watts), and most of the time they will be flickering or even not on at all. I tested one LED at 300mA on a 20cm piece of the metal strip and it reached a maximum of 50 degrees centigrade, so I think the metal strip solves the heat dissipation problem nicely, as well as allowing the LEDs to be hidden well inside the frame and providing a chassis on which to mount the electronics. Yes, I was quite pleased with myself 🙂
Cooling the LEDs for the Flash
The 10 Watt LED is, of course, going to produce around three times the heat of those running at 3 Watts. So, how to keep it cool? This LED needs to fit in the side of the frame, so there’s less room for a metal strip or a heatsink. With the candle LEDs I simply removed the bottom of the frame to allow for the metal strip and a gap between that and the artwork that would allow air to flow for cooling, but I couldn’t remove the side of the frame, that would look weird!
I considered a length of metal strip again. Running an LED at 3 Watts needed a 20cm strip to keep the temperature around 50 degrees. A 10 Watt LED would need something like 60 cm. In addition, mounting the B&Q-improvised heatsink vertically would reduce the effect of cool air flowing across its entire length. (albeit not cooled to the outside but pressed up against the side of the frame instead). In fact, because of the wooden cubes attached to the artwork, it was not even possible to run metal strip up the whole of the side; the best I could do was about 24 cm (as opposed to 120 cm along the bottom). This wouyld be nowhere near enough to dissipate the heat from a 10 Watt LED.
So I considered attaching a heatsink to the strip either side of the LED. What size heatsink will it need? Time for a little calculation. Any heatsink would have to prevent the temperature rising from a room temperature of, let’s say 30 degrees, to, say, around 85 degrees (a maximum suggested in various websites). That’s a rise of 55 degrees from 10 Watts, or 5.5 degrees per Watt. Looking in the RS online catalogue, there’s one that’s 50 x 32.3 x 28mm, so it would (just) fit into the frame. However, 32.3mm is almost the entire horizontal space between the frame and the artwork, so the heatsink wouldn’t have any space around it for the air to circulate, it would cover some of the artwork, and it would heat the artwork up. It would be possible o use these, probably attaching one above and one below the LED, but it’s a rather inelegant solution.
There are two kinds of heatsink; the difference is in how far apart the fins are. The heatsinks I bought for the first project have fins separated by about 5 mm and are painted black (it’s not actually paint, the aluminium is anodised, which is an electrically deposited dye). These are intended for ambient air, i.e. you need to mount them vertically and the air moves through the gaps by convection (the heated air is lighter than the surrounding air so it gets pushed upwards and is replaced by cooler air), and the black increases heat loss through radiation. The other sort of heatsink – the ones that I bought for this project, shown in the first photo next to a circular 3 Watt LED – have fins about 1mm apart and have no paint. These are intended to cool by having air forced through the fins, i.e. you are expected to attach a fan to them.
Time to think outside the box. Now, this LED is for the Flash – the representation of a sudden, unexpected, harsh and threatening release of energy. So it’s going to be lit for maybe 1.5 seconds, and then not lit again for a while, say, 60 seconds. Therefore, the average power we need to dissipate is 1.5/60 of 10 Watts, or 1/4 Watt. We need a heatsink of around 120 degrees Centigrade per Watt.That’s easier – 24 cm of right-angle aluminium can cope with that!
But now we’ve created a new problem: how to stop the LED from overheating if it ever gets left on somehow (e.g. due to a software bug). Whatever mechanism I use, it has to be independent of the Arduino, since one fault it will be protecting against is the possibility of the Arduino hanging, getting in a loop, etc.
Alternative method 1: Timer
The flash will be created by turning on the LEDs using a MOSFET switch (see previous project). My first thought was to place a timer circuit between the Arduino and the switches, so that instead of turning on the MOSFET directly, the Arduino triggers a two second timer that switches on the LED. This would address the potential problem of the Arduino turning on the LED and then, for some unexpected reason, not turning it off again. However, it would not prevent the software from repeatedly triggering the timer, so it’s not a 100% foolproof method of protecting the LEDs from overheating.
Alternative method 2: Temperature sensing
Monitoring the LED / heatsink temperature would seem more reliable. I contemplated two approaches. Firstly, a temperature sensor that interacts with the existing MOSFET switch for the Flash LED so as to shut off the power when, say, 70 degrees Centigrade is detected. This is the idea I had:
It uses an opamp (operational amplifier) to compare two voltages. Opamps are hugely sensitive analogue devices. In this case, I’m just comparing two voltages and switching the output off if the voltage on the + input drops below that on the – input. Provided the three resistors have the correct values, this will happen as the diode heats up and its voltage drop reduces (from a typical 0.6 Volts at room temperature) – the diode is placed physically next to the LED so they both heat up together.
The output from the arduino and the output from the opamp both feed into an AND gate that only turns on when both inputs are on (i.e. at 5 Volts). So, the Arduino’s digital output must be “on”, and the diode must be sufficiently cool, for the LED to be turned on.
I created this diagram using the free, open source software “KiCad”. It seemed like a good idea, in that you can create highly complex circuits and then transfer them to a PCB design tool to create the PCB layout, and lots more. However, I started it three hours ago, and all I can say is that the software makes Windows Vista look as intuitive as an Apple masterpiece; I think it would have been easier to persuade a call centre to give me a £500 rebate on my gas bill.
A second approach to monitoring the temperature is not as flexible – it involves a thermal fuse. These little devices melt internally at a certain temperature (you buy the one for the temperature you want), disconnecting the power. However, it’s permanent. Like an ordinary fuse, the only way to get things working again is to replace it, which could be a problem is a software fault causes the thermal fuse to activate after installation (since it would require physically accessing the hardware, rather than just distributing a new version of the software. Incorporating both over-temperature mechanisms would be the ideal; the opamp would prevent the thermal fuse from blowing and the thermal fuse would protect against a failure of the opamp circuit, including the MOSFET.
Alternative method 3: Stored energy
So far, the plan has been essentially to connect the LED directly to the 12 Volt power supply for a short enough time to produce a “flash” without anything getting too hot, and incorporating a variety of mechanisms to make sure that it can’t overheat, even if things go wrong. But what if, rather than connect the LED directly to the power – via a switch – we used some type of energy storage, so that there was only enough power to light the LED for a couple of seconds, even if the switch stayed on? That would solve the problem of overheating completely.
Let’s try some calculations. We need one amp for about 1.5 seconds. Batteries generally don’t like being used up completely in that short a time, but a capacitor could do it. How big a capacitor would we need? I consulted an online calculator (set Supply Voltage to zero for discharge calculation):
Start voltage: 12 Volts Run time: 1.5 seconds Final voltage: 10.5 Volts Resistance: 12 Ohms (The LED takes about 1A at 12V when cold)
The result? About 1 Farad. That’s 1,000,000 microFarads (uF). A regular 12 Volt 1000uF capacitor is quite big, maybe the size of a thimble, and we need 1000 of them.
Now, there are things called “supercapacitors”, which, as the name suggests, are, well, super! You can get a one Farad supercapacitor for a pound or two. But not so fast… These devices have a maximum rating of around 2.5 Volts. They differ a bit according to manufacturer but if it says 2.6 Volts on the can, 2.65 Volts is really a bad idea.
The other thing to watch with supercapacitors is their lifespan. Typically, they are rated at 1000 hours (yes, that’s 41 days). They don’t fail at that point, but their capacity is reduced to 70% and their internal resistance rises. However, that’s the worst case scenario – maximum voltage at maximum temperature (typically 65 degrees Centigrade). Reducing the temperature by 10 degrees doubles the life, and reducing the voltage applied also increases the life; it seems that if one is gentle enough with them then 10 years is quite doable.
Another thing to check is the internal resistance (or “Equivalent Series Resistance”). Some supercapacitors are intended for low current applications, such as memory backup, and have several tens of Ohms resistance. We need one with a few milli-Ohms.
We have a 12 Volt power supply (since we need 11 Volts to power the LEDs to full), but we can only put, say, 2.4 Volts across our 1F capacitor. So we would need five in series. Now, that reduces the effective capacitance to 0.2F. In other words, we need five 5 Farad capacitors to get the discharge time we calculated.
There’s something else to consider. These capacitors typically have a tolerance of +-20%, which means that I might actually get four 1.2F capacitors and one 0.8F capacitor. If I charge these in series then there will be a voltage imbalance. The 0.8F capacitor will reach 2.4 Volts before the others (since it has less capacity to fill). However, the 1.2F capacitors will continue to charge, increasing the voltage on the smaller capacitor, meaning the smaller capacitor will be charged to a higher voltage. This will at best shorten its life, and at worse cause it to fail.
Alternatively. Remember we have a “boost” regulator? That works down to just over 3 Volts. We could put two supercapacitors in series. Let’s try calculating that:
Initial voltage: 4.8 Volts Run time: 1.5 seconds Final voltage: 3 Volts Resistance: 3 Ohms (4 times the current to compensate for 1/4 voltage)
Still one Farad! (Or rather, 2 x 2F in series – actually, it shouldn’t come as such a surprise, since the total amount of energy will be the same – minus a small amount lost in the regulator). Now, the interesting thing is, in the world of supercapacitors, 1F is quite low. One can easily buy a 10F capacitor, and 100F is not out of the question at all. So, how about if we have larger capacitors and a slightly lower charge voltage – say, 4.4 Volts, plus some more leeway in case the boost won’t work right down to 3 Volts, plus the possibility of a longer flash. How about this, a 25 Farad, 2.7 Volt capacitor for £3.20 (remember, we need two and the result will be rated 12.5F, 5.4V. That should do the trick.
Now, how to charge it? The recycle time (minimum time between flashes) is an artistic decision, so let’s put the coloured artist hat on a moment… It needs to be long enough for the flash to be a memory and the candles to become the norm. Ok, I just sat staring at a clock and imagining….. I think something like two minutes. Which gives us 120 seconds to recharge. Back to the calculator (and the white hat):
Initial voltage: 3 Volts Supply voltage: 4.8 Volts Final voltage: 4.7 Volts Resistance: 3.3 Ohms Capacitance: 12.5 Farads
This will recharge the capacitors in 120 seconds at a maximum current of 0.5 Amps. Still looking good. Now, to provide 4.8 Volts from a 12 Volt supply we need a series resistance, and since it has to be *exactly* 4.8 Volts, we need a voltage regulator. Well, the world is full of 5 Volt regulators! That would put (about) 2.5 V across each 2.7 V capacitor. However, a diode between the capacitor and the voltage regulator would drop it to 4.4 Volts (with a high value resistor across the capacitor to make sure some current flows). The calculator estimates the maximum charge current would be 0.55 Amp, which is quite reasonable.
In fact, 12.5F will give us plenty of leeway. Here’s the calculation (using 3.6 Ohms, which comes from R=V/I = 11 Volts at 1A for the LED = 11 Ohms, but the regulator will boost the voltage from a median of 4 Volts (as the capacitor discharges from 4.4 to 3.5) to 11 Volts, i.e. 11/4, which is equivalent to decreasing the resistance by 4/11, so 11/11*4 = 4 Ohms, minus 10% for losses in the regulator comes to 3.6 Ohms.):
Initial voltage: 4.4 Volts Final voltage: 3.5 Volts (in case the boost doesn't work at 3) Resistance: 3.6 Ohms
Still a good 10 seconds of flash 🙂
So, two 25 Farad supercapacitors in series with a 5V regulator and diode and resistor to charge it and a boost regulator to run the LED should work. The advantage would be no need for thermal protection of the LED; there’s just not enough power to heat it up.
There are a few downsides with using these supercapacitors though:
- With two supercapacitors in series there’s the possibility of differential charging – probably not at first, but as the capacitors age it may become a problem. We could check for this by measuring the voltage across them using the arduino’s analogue in pins, and shut the system down if anything seriously wrong is detected.
- If a serious fault develops elsewhere in the charging system (such as failure of the 5 Volt regulator), there is the possibility that the supercapacitors could go over-voltage, which is likely to cause them to fail (by which I mean explode – which would be ironic, given the nature of the artwork). This could probably be adequately prevented, as above, by regularly checking the capacitors’ voltages.
- The supercapacitors are rather large (16.5 diameter x 28.4mm length), so they won’t fit between the heatsink and the frame like the rest of the electronics.
- If we want the system to last well over ten of years then the supercapacitors need to be kept cool, so they can’t go near any of the lamps or the charging resistor.
- There would be a time delay of up to five minutes at power-on and one minute after a Flash before another Flash could be triggered. That is mostly within the range required, but it could limit the artistic effect if, for some reason, the specification were to change after the system is built.
I decided against the supercapacitors, mainly because they were too big to fit inside the frame, but also because I don’t like the idea of built-in obsolescence and I’m not sure that they would last long. So I went with a 24 cm length of metal strip as the heat sink, separate hardware to monitor the LED temperature and shut it down if necessary, plus a thermal fuse, just in case.