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 Extra simple - Extra fast Fluorescent to LED Lamp Mod AuthorGiorgos LazaridisDecember 5, 2012

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Boom! And there goes my lamp

Let there be light!

Long time since I've posted anything to my site, and that is because i work on something big this time which absorbs all my free time.

So, as i was working on this big thing, my spot lamp exploded! That is really a very bad situation because i need LIGHT to see those tiny SMDs and the four 1.20 meters CFL lamps above my workbench are just not enough. It is Saturday night, and finding a replacement CFL U-type lamp for my spot lamp is just impossible. Add to this that i was not really sure if the new lamp would work or not. And add to this that i had ten 1-Watt LEDs somewhere in my lab waiting to be used... The decision to hack and mod this lamp was easy to be taken...

Starting form the supply
First, let's start with the supply. The adapter is just a 1:1 transformer, 220V comes in, 220V comes out. It wont do any good for my project so it goes to the adapter bin... Instead, i will be using a 2x12V transformer. The LEDs are all rated at 350mA @ 3.6V and i plan to connect them in series. I plan to connect 8 of these LEDs in series, that will be 8x3.6 = 28.8 VDC. The 2x12VAC transformer can provide some 34 VDC, which is just enough to light all the LEDs and don't waste too much power as heat on the current driver that i will make.

 The adapter Only one 1:1 transformer inside No extra circuitry in the lamp

Where do those 34 volts come form?
If you ask "where do those 34 volts come from", here is the quick answer. When we say that this transformer is rated at 2x12VAC, we mean 2 x 12 Vrms. When this voltage is rectified to a DC voltage and smoothed with a capacitor, it will be higher because we then talk about peak voltage. The formula to convert RMS to Peak voltage is simple:

Vpeak = Vrms/0.707

So, in our case:

Vpeak = Vrms/0.707 = 2x12/0.707 = 24/0.707 => Vrms=33.9 Volts

You may wanna go to an online rms to peak converter to help you with these calculations.

The LEDs
As i said, i plan to use 8 LEDs rated 350mA at 3.6 Volts. Actually, i will drive them at around 320mA, no need to push them to their limits, right? So, i connected them in series and used an aluminum profile to secure them, providing this way extra surface for cooling. This aluminum profile is used to mount electrical equipment into electrical switching boards, such like this star-delta automation. It is perfect for this job!

 8 LEDs in series The aluminum profile for electrical equipment is perfect! I cut it to size And then i drilled 2 holes for each LED

 This is how i plan to secure the LEDs I need 16 holes for the 8 LEDs Heat-transfer paste of course Ready!

And now it is time to secure it into the lamp...

 Hot-melt glue, what else? The wire on the right-hand side goes underneath the aluminum base... I will still use the On-Off switch of the lamp. Close the cover and we're ready for test#1

The LED Driver
I mentioned before that the 34 volts of the transformer are perfect to power all LEDs without wasting too much power in the form of heat, since i will be using a simple linear constant current driver. Some basic math revel the power that has to be dissipated on the driver. The 8 LEDs require 8x3.6=28.8 (say29) volts at 350mA. I have 34 volts. The difference is 34-29=5 volts. Since i will operate the LEDs at 320mA, i will need to dissipate some 5x320=1.6 Watts of power, which is normal.

Although i've written a long theory on transistor constant current drivers, i will have to cheat for this one, simply because it is Saturday and i do not have a proper power transistor for the job. but i do have a batch of those good-old LM317 regulators! In the LM317 datasheet one can find many typical application circuit, among which is one rather simple constant current driver. Here is the circuit as found in the datasheet:

Can it be any simpler? Frankly speaking, No, this is the simplest constant current driver i've seen so far. All you need is the ye old mighty LM317 and a resistor... That's it.
Now, to calculate this resistor, that is even simpler than the circuit itself! In the datasheet the formula is this:

Iout = 1.25V / R1

So, solving for R1:

R1 = 1.25/Iout => R1 - 1.25/0.32 => R1 = 3.906 Ohms

Simple? Simple indeed! So i need a 3.9 Ohms resistor to regulate the current at 320mA (0.32 Amperes). What a luck! I just happen to have a 3.9 Ohms 5 Watts resistor in my bin! Ok, 5 watts is super overkill, but i just need light for tonight!

 A 2x12 VAC transformer... I could use also a 1x24 VAC as well. Super simple circuit: A bridge rectifier with 4 1N4007 diodes, a big 4700uF @63V electrolytic capacitor, and the LM317 with the 3.9Ohm 5 Watts resistor.

There is a saying that goes like this: The Devil is hidden in the detail...

 The first test failed! Only one LED lit... Strange....

Check out the next page to see why this happened and how i solved it.

 At 16 October 2013, 16:34:59 user santosh wrote:   [reply @ santosh]sir,i want to design an led driver to drive 25nos. of 1w led please help me..plsAt 6 June 2013, 16:31:48 user Larry wrote:   [reply @ Larry]Nice refurb. I noticed the shorting of the screws before I went to the second page and saw you solved that problem. I'm getting into the new high power LED stuff myself and found an easy solution to the washer for mounting screw problem. Use rubber hose and cut into 1/8" lengths with a pair of wire cutters. Very fast to make lots of rubber (non-conductive) washers. Use 1/8" diameter hose (or whatever size screws you are using, although all star-based LED's I have seen you must use #4 or smaller screws) Viton (400 degrees F max) or Norprene (200 degrees F max) hose. keep up the good work.At 17 January 2013, 21:30:13 user Metalshrine wrote:   [reply @ Metalshrine]@Giorgos Lazaridis You are right. I guess I was too tired that time. However, congratulations for the lamp.At 2 January 2013, 9:20:11 user Giorgos Lazaridis wrote:   [reply @ Giorgos Lazaridis]@Metalshrine 1/sqr2 = 0.707 so it is the sameAt 31 December 2012, 8:54:34 user Metalshrine wrote:   [reply @ Metalshrine]Man, I think you have a mistake at the beginning. Vrms*sqr2=Vpp. At least this is the way I know it.At 10 December 2012, 18:00:40 user spi wrote:   [reply @ spi]I have the exact same lamp and i was thinking the exact same thing. Nice work!

 HOT in heaven!

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