A High-power LED Torch Using a Single AA-battery

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Introduction: A High-power LED Torch Using a Single AA-battery

This high-efficiency design can power LEDs at 100mA of current and is not much bigger than the battery that powers it. The unique circuit uses 2 commonly available transistors (BC327 and BC337) in an oscillator to boost the 1.2 or 1.5 volts from a rechargeable (nickel) or alkaline battery to the 3+ volts necessary to light a LED. No fancy ferrite inductor is needed as we will be using the metal jacket around the battery as part of our coil.

Step 1: The Circuit and Layout.

The circuit uses a BC327 (PNP) and a BC337 (NPN) transistor. They are connected as a blocking oscillator at about 100kHz through the 150pF capacitor. Each cycle starts with the coil being charged, then this charge, plus the voltage of the battery, about 3.5v, is applied across the LED. The 1N4148 diode takes this voltage and stores it in the .1uF cap, which is used to drive the transistors, resulting in a much more powerful and efficient circuit.

The 22uH coil consists of 35 turns of #28AWG wire over the battery, using the metal in the shell to multiply the inductance.

A simple diagram shows the placement of the parts.

Step 2: The Parts List.

The LEDs are standard 25mA 5mm 120000mcd white LEDs from eBay. The rest are available from Radio Shack or surplus at places like http://www.AllElectronics.com.

BC327 PNP transistor
BC337 NPN transistor
1N4148 Low-signal diode or similar (Virtually any silicon diode)
0.1uF ceramic capacitor (can be anything larger)
150pF ceramic capacitor
12uF 6v tantalum cap (can be anything over 5uF)
100K resistor (controls brightness: 470k for 50mA; 68K for 120mA)
4-ft of #28 or #30 AWG wire wrap
Perforated copper boards, pre-trimmed as shown

Step 3: Making the Tube 'body'

Cut a strip of paper 11" long, 1/4" wider than the length of the battery. Apply stick glue as you wrap it around the battery. This will be stiffen up the tube.

Use the widest battery you have as the template so that the tube is loose enough for the battery to be removed and replaced.

Set aside to dry while you assemble the circuit board.

Step 4: The Assembly

Start by placing the 4 LEDs. Make sure the shorter (-) leads are on the outside. You will need to make sure they are all flat against the perfboard. Solder the shorter leads (only) to keep it secure.

Then 'fold' the longer leads over to its neighbor so that all the + leads are connected. Note that only 3 sides are connected and ends in a 'hook' which extends a bit out from the board. This will be where the coil is connected.

Continue with the BC337 and the BC327 transistors and the other components. Note that one lead (the 'E') of the BC337 and all connetions to B- are also left pointing up. See the images below. The red sleeving is used to prevent shorting - from a connection error... (:P)

When all is soldered and checked, slip the smaller perf-board, copper-side up. This is the level all the B- connections are made. You should be able to wrap the leads so they can touch each other. DO NOT solder them yet!

Step 5: A Quick Test...

If you have a 22uH inductor handy, you can test your circuit by connecting one end of the coil to B+ and the other end to the 'hook' left in Step (4).

Touch B- to the wires at the center of the small perfboard and the LEDs should all light up. SUCCESS!

On to the the final assembly bits now.

Step 6: Final Assembly

Crease and push in one end of the paper tube as much as you can to add support. Then make 2 pin-holes at the base of the crease. Push the negative end of the 12uF capacitor through one of the hole.

Solder this wire to the wire cluster in the middle of the perfboard before pushing it out the other pin-hole.

Start the coil by soldering the stripped end to the 'hook' on the main perfboard and start winding around the tube over the battery and the 12uF capacitor. Solder the other end of the coil to the + side of the capacitor.

The wire of the capacitor is long enough to reach around to the positive of the battery - that is your on-switch. Push it to touch the battery and you get LIGHT!

Do NOT look into the light - it will be bright!

Step 7: What's Next?

I hope you have lots of fun making these lights.

Here are a couple of different combinations: 4-LEDs or 1 120mA LED. The single LED is simple enough that I soldered the components directly onto the leads of the LED. Sealed in 3/4" heat-shrink tubing.

You will find much more about LED circuits at my Website: http://www.quantsuff.com Give me a visit and let me know what you are building this week!

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    111 Discussions

    A addendum to the previous posting. I've tried substituting higher value resistors to the PNP base. The LED(s) brighten just a bit but nothing to write home about.

    3 replies

    NTE components tend to be much lower performers than their "equivalents". For example, if you're using the NTE-298, it only has less than half the gain of the "real" thing! To offset that you can try (1) reducing the 100k to 75k of 82k, or (2) reducing the coil to 28 or 30 turns. Change ONE element and test.

    Researching "equivalents" is an exercise in insanity!!! For the BC337/BC327 pair you constructed your circuit with cross to NTE123AP/NTE159 which is what I used. BC337/BC327 supposedly cross to BC637/BC627 but with NTE they cross to NTE297/NTE298, and don't get me started with BC337 crossing to 2N5818 which cross to a NTE128/NTE129 pair. With transistors, WHAT BRAND IS THE REAL THING and which cross reference is the correct one???!!!

    Inductors have me somewhat guessing. The transistor pair's switching, I'm led to believe, continually build and collapse the inductor's magnetic field leading to an increase in voltage. Won't decreasing the number of turns decrease the magnetic field strength and thereby decrease the voltage output?? Is decreasing the number of turns simply decreasing the inductor's resistance and the resulting decrease in inductance irrevalent?? Inquiring minds want to know! :-)

    Most general purpose transistors (BC3x7, 2N2222 etc) are available through sales channels as well as eBay. Then you would be sure of getting the correct performing components.

    Yes, larger inductors do build up a greater charge, but it also takes time to do so. Lessening the inductor may yield a lower charge, but it does it at a higher frequency, so the resulting light may be greater.

    Not sure if you're still responding to questions about this circuit. If you are, the circuit functions as designed but I can't get near the current from it as you seem to be getting. All components have identical values using NTE brand equivalent parts. The choke has 35 turns of 26 gauge (had no 28 or 30 on hand) around a label-less AA NiCd cell. I only use the NiCd case for the inductor. I've powered it with either a NiMH or alkaline cell. The LED's are 25mA max, 2.4Vf max. Each LED added to the circuit just dims them all slightly. Even one LED isn't all that bright. I can't accurately measure the amperage thru the LED(s) as the oscillating current causes the VOM to read an average. Thoughts?

    0
    user
    DuchoD

    1 year ago

    Say, If I use a different resistor for example like 1 ohm and connect a 3.5V 1watt led diode do You think it will work??

    1 reply

    No -- the resister is there to help the transistors work, and doesn't have a direct function in the LED's brightness. TO use a 1W LED, the transistors will also need to be changed to handle the current to 'charge' the coil.

    Also, most AA batteries will not support the 1000 mA draw to power the light.

    What happens to your website?
    It's not accessible anymore.

    Which component in your schematic determines the voltage across the LED. For example, instead of a white LED if I wanted to power a red LED what would I have to change. Thanks.

    1 reply

    One way is to increase the # of turns by 10 to 45, but that would make for quite a large winding!

    Alternatively, the charge on the coil is approximately proportional to (V^2 * I) so if your V is 2.5 instead of 3.3, then adding two or three more LEDs in parallel would compensate for the difference.

    I have finally made it to work, can anybody suggest me how to increase the power of led, using bc337 intensity of light not as much as 3xAA battery torch i have, will using 2sc2500 make it powerful? I have used 50k resistor instead of 150k

    hello sir i have used every thing u have mentioned but still can't make it to work pls pls pls pls help

    6 replies

    If your parts are correct, then the problem would be your assembly, so some photos and explanation of what you have done would be very helpful here!

    Also, read through the other comments below and maybe you'll see something that would help pinpoint the problem.

    Thank you for your reply, I have uploaded 3 images

    1) Schematics on breadboard layout (top)

    2) components image.

    3) layout on breadboard.

    i have tried changing the diode, led and battery polarity but nothing happening,

    but there is faint light spark if i short collector and base of BC337.

    Pls pls pls reply,

    joule_001.pngjoule_002.JPGjoule_003.JPG

    Using your layout and the specified components, I have no problem getting the circuit to work. This would suggest that some of the components have been damaged in prior tests. In particular, I would try a different 337 because if you ran the circuit without a LED attached the back-EMF from the coil would be enough to 'pop' it. Also check/change the diode for the same reason.

    140926066a.jpg

    Done everything but nothing happening, also tried your other circuit still not working, i think some of the value i am using is not correct. I don't have access to l-c meter that's why i am unable to say if my cap and inductor is of correct value.

    Boost Basic1.jpg

    All this does is duplicate the action of the 337 -- by shorting the leads, you are charging up the coil, which discharges to light the LED when you removed the short. At least you know the coil is not open.

    From your pictures, the coil, R and caps appear to be correct. Couldn't see the numbers on the transistors or diode but one or more are likely to be defective.

    hay quick question i don't care if you think its stupid but what would happen if you where to apply 3.5 to 4v to this circuit what would the out put voltage be.

    1 reply

    It will be zero.

    The additional drive at the base of the transistor will not allow it to turn off and it becomes a short circuit, quickly draining the battery until it is destroyed.