Instructables
Picture of LED bike lamp
If you wish to make by yourself a powerful lamp for your night bike rides, as some beautiful lights you can buy, the best choice in my opinion is to build it with a commercial LED lamp. Since some of these lamps work with 12V AC current, I decided to use as power source a 12V DC Li-po battery. The lamp has an included rectifier which converts current from AC to DC, so using a DC source works good here.

NOTE: I made this projects specifically to have a powerful lamp to mount for a night ride. All the connections, the meter, the lamp pins, the battery terminals, are NOT waterproof, neither water-resistant. For the same reason I don't need any power switch, and I need a fast to plug lamp support. Pay attention to use a so much powerful battery pack with rain or wet hands!

 
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Step 1: The Lamp

Picture of the Lamp
bikelight05.jpg
Since commercial 12V LED lamps are now very common, you can find any number and type of them. There are lamps with a big number of standard white leds, with 3 or 4 or 5 more recent CREE leds, also with one single huge beautiful COB led.
I bought in a shop a 4 cree leds 6W lamp, but of course I can change it whenever I wish with a more strange or powerful LED spotlight. You can determine your best choice also referring to the beam angle. You can find an angle from about 30° to 60°, remember that a narrow beam will light further, but street sides will remain in darkness, the same as road signs.
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QuanQPham9 months ago

I am operating my electric assist at 24V, can I use 2x of these lights in series and power it via my 24V pack without a DC-DC transformer?

andrea biffi (author)  QuanQPham9 months ago

Yeah of course :-)

This way I can use a 60° and a 30° for a long range narrow beam and also light up the sides of the road and signs as well.

I have a 24V-20AH LiPO4 pack that I normally only deplete to about 7-10 AH so plenty of juice for the lights.

wobbler1 year ago
I like this idea.

I'm finding the conversations about the apparent "loss" in the bulbs interesting. I think the real reason is much simpler- they are designed to take less current than the potential maximum.

Several posts mention 700mA as the driving current. This is usually a maximum current for these types of LEDs and to achieve that level of current through the LED you need a significant amount (relatively) of heatsink. This lamp is rated at 12v(AC) 6W, so it is only designed to be taking 500mA maximum anyway (Watts=Volts x Amps). To cut down the heatsink sizes to the lamp sizes means reducing the heat losses and hence driving on lower currents. If you look at some of the higher power LED bulbs, you will see various external heatsinks used to solve this problem, which will get better over time as LEDs are designed ti be more efficient.

Regarding the rectifier, it has to be a bridge rectifier as a single LED will cause way too much flicker on AC unless they added a smoothing capacitor and this wouldn't be feasible due to the size of the capacitor needed. So does this cause the apparent loss in current/power? Yes and no. It will cause a power loss as stated by several posts, but it won't affect the brightness of the bulb versus AC in a designed lamp. Why not? Again, because that voltage drop will be compensated for in the design and will therefore be present both in AC and DC.

There is however an issue with driving it with 12DC versus 12AC that needs to be considered. LEDs do not obey Ohm's Law, which is for linear resistors. As LEDs are semiconductors, they have a non linear voltage-current curve. They take very little current until they turn on, then have a very steeply rising current use as the voltage rises slightly. This non-linear voltage versus current curve is the reason you need to use either a current limiting circuit or resistor with LEDs. Therefore, as these are probably using simple (cheap?) circuits to control current, they will set the max current to be used at the maximum AC peak voltage (1.414x12v=~17v), which will probably be causing the drop from the expected 500mA to the 350mA quoted.

Also, in designs like this they will need to build in an overhead for over-voltage of the supply. Halogen bulbs are more tolerant of overvoltage than LEDs and transformers are not in themselves a controlled voltage. The voltage of a transformer drops as more current is taken. So, if you place a lower wattage bulb in a transformer driven system, the driving voltage will be higher. With a lamp which is sensitive to overvoltage, designing for this means making the current draw lower than the max possible to allow for this contingency.

Finally, there are also efficiency advantages to running LEDs at lower currents. They convert current to light more efficiently at lower currents. So therefore, using the same total current, four LEDs of the same type will give more light than three of the same type, with less heat wasted and will also be more reliable due to the lower current.

Putting a greater voltage in the bulbs will definitely cause them to be brighter but measure your current whilst doing this and keep an eye on the heat in the bulbs, especially if you intend to mount them in an enclosed case.

Finally, what would you actually recommend as a good angle to actually start with after using these bulbs?
andrea biffi (author)  wobbler1 year ago
Thanks! This is really interesting, I hope your explanation will be right, as you said that means a longer duration of the battery and a better efficiency of the lamp! I will make a photo-comparison between this lamp and a 220V 9W one.
As beam angle I have a good experience in using 36°, 60° and 10° halogen 12V bulbs in bike night rides. 60° are very useful in woods, as you can see high branches and side obstacles. I suggest 36° for traditional roads, and if you want to enjoy yourself mount a 10° lamp on the helmet, you can light far mountains too, although you have to be VERY CAREFUL to not point to mates or car, so it's better to turn it on only in isolated tracks.
gpavlovsky1 year ago
Andrea, congratulations on building this light! And thanks for the link to the battery meter, it's very nice (I love numbers and stats).
I don't want to sound critical but the safety aspect have to be mentioned.
To me, this light only seems suitable for road bikes on good roads, riding in dry weather. I wouldn't trust that bracket holding the bulb (at least not for a mountain bike and not for bumpy roads).
I also don't see any weatherproofing or a safety fuse. The battery connectors you're using might be water-resistant but probably not waterproof, that battery meter doesn't say anything about weather resistance (so it's most probably not), and your light bulb connector is completely exposed to water incursion. I hope you're aware that these kinds of batteries can supply very high currents, so in the event of a short circuit the battery or the wiring will most likely burst into flames (or even explode). Do you have an on/off switch? If not, do you turn off by unplugging? If the connector is unplugged, it's one more point water can get inside. Anyone making similar projects should keep safety in mind. Bike lights are not the same as R/C toys, which are used in dry weather, and have to keep the weight as low as possible so are willing to sacrifice safety for weight savings.
Would you care to mention the costs of building this project? Because these days you can buy really cheap purpose-built bike lights from China (e.g. on Deal eXtreme), which come as a set, with proper handlebar mount, with 2+ brightness modes, full waterproofing, the battery comes in a neat-looking pouch with mounts for the frame tube, level of discharge is usually indicated by a LED that illuminates the mode switch.
andrea biffi (author)  gpavlovsky1 year ago
Hi gpavlovsky, thanks for the compliments. You're right about the safety, this is actually a project I did specifically to have a powerful lamp to mount for a night ride. All the connections, the meter, the lamp pins, the battery terminals, are NOT waterproof, neither water-resistant. This could be a problem for long travels, but it's good enough for my city rides. For the same reason I don't need any power switch, and I need a fast to plug lamp support, I'll try it on bad roads to know much about stability. I'll add some better detail in the intro..
About the cost of the set it's not high at all (the lamp is about 15$, and the battery is an all purpose battery you probably already have if you like hobby-modeling). Of course we all know that nowadays you hardly save money making a tool or a gadget, but you could love the making more than the buying ;-)
Your reasoning about the safety issues is exactly how I thought, and all is fine as long as you understand the risks and avoid rainy days, puddles etc.,
But you might want to mention the electrical safety issues in your instructible, since others might want to follow/copy your project without realizing the dangers it might bring in the wet.
And if I were you, I would at least add a sealed ATM (mini) automotive fuseholder with a 5A fuse (or smaller if you can find one), to avoid a fire and/or melted wiring. It's cheap and compact and should be easy to find in automotive shops. Install the fuse on the + wire from the battery, as close as possible to the battery. I use these:
http://www.easternbeaver.com/Main/Elec__Products/Fuseholders/C-Bfuseholderlong_9089.jpg

Just want everyone to stay safe :)
I totally understand you, the fun in building something is quite often a good enough reason to do it yourself instead of buying. I think the battery is the most expensive part of your build, thus if you already had a spare one it makes sense to put it to good use. But if your budget isn't super tight, I'd still get some kind of waterproof automotive/motorcycle LED light designed for 12VDC operation (make sure it would work on 11V or even lower). If you want waterproof connectors, I'm a big fan of automotive AMP SuperSeal connectors (available from 1 to 6 wires) available from Mouser or NewArk. 
andrea biffi (author)  gpavlovsky1 year ago
Thanks! Those AMP automotive superseal connectors are great!
You're right about the fuse, I used to connect it every time when I was younger and I built some 12V night lamps for my mountain bike, powered with lead sealed batteries. I have to add this also to the note I inserted in the first step.
scubaderf1 year ago
Nice Instructable.
The explanation you are after could be something like this.
LED's are diodes so will only allow current to pass in one direction.
As the lamp you used is an AC one then it has a clever device in the head (rectifier) that allows it to use the supplied current (AC) in both directions.
As you are only feeding it with DC (one direction) then the rectifier could be working against you??
Not 100% sure that this is the reason as I think the rectifier should just allow the DC to constantly flow. but it could be.
andrea biffi (author)  scubaderf1 year ago
I suspect too that this is not the reason... but it could be, I frankly hoped that someone skilled in electronic would have given an instant answer... I ought to post my question in the science forum section or in the answers page. Thanks anyway!
The way I understand it, the rectifier essentially 'expects' only AC, so switches (at however-many-Hertz) to match that. Therefore, since you are only supplying DC, the rectifier has nothing to rectify half the time, so halves the lamp's current draw. Effectively, you are 'pulsing' your current to your lamp at half-speed.

You might be able to open the lamp housing and bypass the rectifier, but you'd have to do that each time you changed lamps - whether for aesthetic or practical reasons.

A DC-to-AC converter ought to allow you to power such a lamp (or 3), but you'd be converting battery power into mains, to be converted straight back into DC... - not exactly efficient!

And I'm not sure what size battery you'd need to provide enough AC for your lamp to work at its full potential for as long as you might need, but I don't think it'd be worth your while.

The only other alternative I can see is to replicate (maybe with a 3D printer?) the look of your lamp, but build it without the rectifier. If you look at http://led.linear1.org/led.wiz , you can see what resistors you need for however many LED's (whether 'ordinary', High Power or Cree) you want in an array.
Actually all bulbs will run on dc current. If you look at house light (incandescent) with a fast enough camera they blink. LED's will run on AC but they will only be on for 1/2 the time a standard bulb will be (look at a sine wave to explain)

The bulb looks like a replacement bulb for a track lighting system. In that case most track lighting systems have a small transformer pack that mounts in the ceiling. Depending on manufacture some of these lights actually run the bulbs on a DC current. In that case there will be a diode bridge in the transformer.

A diode brige simply converts the sine-wave(+ and -) of AC current into a proportionate DC pulsating-wave(+ only).


As for the reduction in amp draw. All bulbs (incandescent and LED) have a lead-in wattage(Volts*Amps=Watts)  that has to be overcome before the bulb emits light. Soon as that lead-in wattage the wattage required to continue emitting the light is reduced.

When powering the bulb on either AC or pulsed DC current every time the power supply drops bellow this "continuance wattage" the bulb must be exposed to its lead-in wattage before it can start emitting light again and so it requires more wattage over a given time to emit light.

On a pure DC power source(battery or normalized DC supply) the supply can maintain the "continuance wattage without drop-outs and so it requires less wattage to maintain the light.
andrea biffi (author)  m93654281 year ago
Hi, actually incandescent lamps don't blink, filament glow is a physical continuous effect, but NEON and CFL lamps blink.
Thanks for the explanation for the lower consumption, don't you think that the exact half value is connected with some other reason?
In truth this is correct, however, they do blink. As the voltage across the affiliate returns to zero the filament is no longer being heated and will decay in its intensity(lumen) by a nearly imperceptible value(probably in micro decimals) however it does blink. That statement was more directed at those lights that use only half the ac wave curve(aka non bridged).
This is not correct, a bridge rectifier is a "passive" component, as electron_plumber pointed out quite rightly though, most AC voltage specs are RMS specs which means that you can expect the Rectified output to be 1.44 times the ac voltage input to the rectifier. If however you had an AC supply of say 10 volts PEAK AC, you would expect the dc rectified voltage to be approx 8.6V DC because of the voltage drop across either two of the four diodes on any given half cycle.
Thanks for that. I did/do know I wasn't correct, which is why I phrased that first paragraph the way I did.
The most common way to convert AC to DC is to use a simple 4 diode circuit known as a full wave bridge rectifier:

http://en.wikipedia.org/wiki/Diode_bridge

The bridge effectively flips the negative side of the the AC wave positive, so you end up with pulsating DC. If you want it to be a nice, constant DC voltage, you need to smooth it with a cap. For an LED, the pulsing isn't a problem. Note that the pulsing will be at 120Hz rather than 60Hz because of the flipping.

Bottom line is that putting DC into a full bridge rectifier gives a constant DC out. So this will work, with some small caveats. The brightness might be somewhat less than you expect. 12VAC usually means 12VAC RMS, which goes way above 12V for a good fraction of the time. The more serious issue is that you are passing all the current through only 2 of the diodes, and they might not be able to dissipate the heat. I have seen some of these LEDs rated for lower DC voltage than AC. I presume this is why.

That said, I have seen cheap AC LED devices that use a half wave rectifier. This uses only 1/2 of the AC cycle - the positive bits. If you run across one of these (easy to recognize, because they "twinkle" a bit), you will need to plug the DC in the right direction for it to work. The good news is that plugging it in the wrong way doesn't hurt anything, but it doesn't work.
andrea biffi (author)  electron_plumber1 year ago
This lamp behavior and consumption are not affected by the polarity of the DC source, I've tested that...
Maybe you're right about the diode bottleneck, actually double current flows in each diode, and they could be meant to letting flow the exact current for AC source.
crazypj1 year ago
LED is Light Emitting Diode, the 'diode' part allows current flow in one direction. When used on AC, (alternating current, alternates between + and -) half the wave is 'wasted' and converted into heat. Supplying DC 'saves' power and doesn't create heat (except the tiny amount lost through conversion to light inside diode) It may double the life of 'bulb'?
not it doesn't. You're right, a diode will only allow current to pass in ONE direction, when on an AC supply it BLOCKS the current in the other direction (unless the voltage is above the "reverse voltage" of the diode). No power is converted to heat OR wasted, it is blocked.

The ONLY reason a diode produces heat is because it has a forward voltage drop, that varies between types of diode (ordinary silicon, shotkey....) a standard silicon diode (eg, 1n4001) has a forward voltage drop of 0.7 Volts, this also means that it will not forward bias (conduct) if the potential across it is less than 0.7 volts. Just as a signal transistor will not turn on unless it see's 0.7 volts at it's Base.

The heat is generated because of this drop and in proportion to the current (0.7v * "x"Amps will give you the dissipation in watts)
I was thinking about bridge rectifiers rather than single diode and was just trying to help with the 'loss' of the rated bulb
I'm aware shotkey diodes have much lower forward voltage losses
LED's have various thresholds white around 3.3v, red, 2.2v
I'm a motorcycle mechanic not an electrical engineer so tried to 'explain' or add a comment that was helpful even if not totally correct BTW, there is usually reverse leakage through diodes, but, you knew that :)
mikerew1 year ago
If you look at the specs it says 12V AC...this is 12V RMS which is less than the actual peak voltage that the device sees...The device is actually seeing about 17VAC peak. each LED needs about 3.4 volts to use the 700 mA....about 13.6 volts....if you supplied 16 VDC you would likely see the 700 mA. As a side note...they usually use a PWM circuit and not a resistor to limit current due the the high efficiency of the PWM circuit (about 90%).
andrea biffi (author)  mikerew1 year ago
Thanks, I'll try an higher voltage to see how much it actually draws... I'll let you know.
lexip1 year ago
Well ElectroFrank beat me to it and gave a very full and definitive explanation as to what was happening here with the led, but I was going to say, in your 'NOTE:', that you in fact already had the clue right there when you said it only draws 350 ma when the manufacturer says it should draw 700 ma.
As yo can see, 350 is half of 700, so what ElectorFrank (and others) are saying about it only using one half of the AC wave, holds pretty true to your measurements (;->
Also, as nice as your simple holder arrangement is, I wonder whether it might pay revisit the metal fork connecting directly with the base of the bulb.
I bet the shock forces over a period of time will really put stress on that area. The idea is simple and neat, but I reckon it would be worth cradling the light in a better way.
And one last thing, it's great that it works so well being AC on a DC battery, but is/was there a particular reason why you didn't use a DC LED to begin with, as I know there are some pretty bright ones out there now?
andrea biffi (author)  lexip1 year ago
Hi, you're right, I was searching a 12V DC lamp, but when I found this I didn't bothered anymore, because I thought it was the same...
The 12V AC LED "bulb" will contain a 4 diode bridge, which will normally rectify the AC to provide DC for the LEDs in the bulb unit.
When the bulb unit is supplied with DC, only 2 of the diodes will conduct, but the LEDs will provide (about) the same brightness.   The same average current will flow through the LED.

Diagram attached to show the principle.   (Note: there are more complicated circuits in some LED bulb units.)

A nominally 12V AC supply may actually supply a slightly higher voltage, about 14V.

At 11.1V, the battery is already below the nominal 12V, and the 2 conducting diodes in the circuit will drop about 1.2 Volts, and this will reduce the voltage available to the LED circuit to about 9.9V.
Therefore the current drawn will be considerably less, as a small change in the voltage across a diode or LED will result in a large change in current.

This is probably the reason for the lower than expected current.   If you built a unit from basic LEDs, you would control the LED current level by means of a resistor.

(Note: mAh is milli-Ampere-hours, a measure of battery capacity.   LED current is measured in mA.  Battery life in hours = Battery capacity in mAh divided by LED current in mA.)
LED 12V AC-DC.gif
andrea biffi (author)  ElectroFrank1 year ago
Actually the 11.1 Li-po batteries as mine gives more than 12 V when charged. I agree about the diode bridge, but I'm not sure the reason is the lower voltage..
Thanks for the notes about mAh, I rectified the text.
LEDs are DC devices. To run LEDs on AC, a rectification circuit has to be added, losing power.
By then running that AC unit back on DC, you still lose that power in the rectifying diodes.
andrea biffi (author) 1 year ago
thanks guys for your discussion about current and power consumption, I'm following it with interest.
andrea biffi (author) 1 year ago
thanks guys for your discussion about current and power consumption, I'm following it with interest.
sconner11 year ago
I wonder bike light projects are using batteries at all.
I had a light kit as a kid that had a small generator that spun by making contact with the tire.
The faster you went, the brighter the flashlight bulb in the headlight got.
I would use one of those with a rectifier and capacitor(s) to keep the LED's shining for a few moments when we're stopped at a light or stop sign.
Because bikers always obey the traffic rules, right?
We should if we want motorists to respect us in return.
donleonard1 year ago
The drop in current draw is probably caused by the diode voltage drop in the bridge rectifier. A typical voltage drop of a forward biased silicon rectifier is from 0.5 to 0.7 volts depending on the load requirements. Two of the diodes in the bridge will be forward biased so the voltage drop could be anywhere between 1 and 1.4 volts and that would be subtracted from the battery terminal voltage. The LED diode array would not receive the full battery voltage if the battery voltage was 12.25 volts as shown in the picture, but could be as low as 10.85 volts (12.25 – 1.4 = 10.85).
W7LRP
Correct. And even though a diode isn't a resistor, the basis of Ohm's Law is still relevant.
Lower voltage across the diode means lower current through it.
Also higher V will mean more current and brighter light, up to the LED's maximum power dissipation rating. Too much voltage, therefore current, will force the junction to heat up faster than the package and heat-sink can conduct it away, the junction breaks down and the LED burns out. Which is why LED and Laser circuits need limiting resistors or controlled current supplies respectively.
These store bought LED lamps have a rectifier in them obviously, but is it half-wave or bridge rectification? What Happens if you reverse the polarity on the lamp? If it lights, Bridge rectification, if not, half-wave or none at all. They may be using the LED itself as the rectifier. What about current limiting? We don't know if there's a resistor or not. We know they expect 12V AC, the designers may have relied on the consistency of household power to be reliably in range with some margin to spare. I notice the Lamp the author uses has 4 LED's in it, probably in series because most blue or white LEDs have a forward bias of up to 3.2V making the lamp 12.8V capable. A 10:1 step down transformer is an optimal 10-12.6V dependent on local AC mains being anywhere from 110-126VAC.
So I think they may not have a resistor to waste voltage and create heat. Maybe a zener in parallel just to shunt momentary surges.
A 12V battery under load is always something less than that and so should be safe without a resistor too.
Ortzinator1 year ago
I've never seen a tire like that...
can you tell us what it is?
andrea biffi (author)  gzus111 year ago
Of course, I've already searched some reference but I could't find anything. I've bought these two "Made in China" tires in Milan, where they're pretty common on new trendy city retrò bikes, as this one.
pmelnikov1 year ago
It seems it is a Li-Po battery (not Li-ion). Li-Po batteries are best for high currents (you don't need). To supply your LED lamp you'd better to use a Ni-Mg / Ni-Cd battery packs. They are much safer, cheaper, smaller and will work longer. Good luck.
(Sorry for my bad English)
andrea biffi (author)  pmelnikov1 year ago
You're right, I've to correct it. Yes, I don't need so much power, but I bought some of these battery packs, maybe I'll found a better use soon. Thanks!
Thanks for the cool instructable! Can these LEDs be used outdoors in rain? I'm sure cycling could kick up a lot of water etc from the road surface.
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