DIY LED Lamps

Build low power (10W) LED lamps for general illumination.

Step 1: Parts and Tools

Components

2200uF >200v electrolytic capacitor

108 BT30WCS LEDs

Switch

Heat sinks

Plastic box

Insulated copper wire

2x 5VAC transformer or similar to achieve 1:1 ratio

2x 10 ohm resistor

1x 2.2 kilo ohm resistor

1x 100 kilo ohm resistor

1x 2n2222 NPN BJT

1x TIP48 NPN BJT

1x DF04 bridge rectifier

5x perfboards 3 x 10cm (4 x 1.5 inch)

Tools

Soldering iron

Solder wire

Glue gun

Tape

Cutters

Step 2: LED Arrays

There are four arrays connected in parallel, each with 27 BT30WCS LEDs connected in series. Each LED has a voltage drop of about 3.2v therefore the whole LED array uses 27 x 3.2v = 87v with a forward current of 30mA.

The four LED arrays in parallel draw 30mA x 4 = 120mA.

Step 3: Circuit. HIGH VOLTAGE!!

CAUTION! HIGH VOLTAGE!

We need about 60VAC to make this circuit work. A custom transformer can be made or, if you are lucky, maybe you find one with this rating. This circuit uses two transformers, first a 120 to 5v (24:1 ratio) transformer followed by a 120 to 9v (13.3:1 ratio) inverted to get a total 120 to 66v (1.8:1 ratio).

So with a 120VAC input we get 5VAC aprox, after the second transformer the voltage is 66VAC. After the rectifier and filter stage the DC voltage is in theory 66 * 1.41 = 93VCD.

The constant current thru transistor T1 will be Ic = 0.7v/R where R = R1||R2. The 0.7v are the theoretical Base-Emitter PN silicon junction voltage drop of the BJT transistor T2. For a current of Ic = 120mA the resistor value is R = 0.7v/120mA = 5.8ohms.

In this case, the transistor B-E junction has a voltage drop of 0.62v so R = 0.62/120mA = 5.16homs, this resistor value is approximated with two 10ohm resistors connected in parallel.

In the schematic, each LED (D1-D4) represents one LED array of previous step.

Step 4: Build It

Step 5: Try It

Step 6: Place It and Use It

Comments

[kcraske]

I think you should look at your schematic as the second transformer where you have the diode bridge is connected incorrectly. the secondary of the transformed should be connected a the input to the bridge with the electrolytic capacitor connected at the bridge output. as it is connected the capacitor is directly across the transformer which would result in immediate failure. could you also explain why you first reduce the voltage then boost it back up? Why not just reduce it to the required level using one transformer?

[dan95]

You are right!. I just noticed it, Im fixing it right now. About the transformers, yes, I could have built or tried to find a custom transformer but I already had those and that configuration was good enough for this design.

[JoseG11]

Yees, HEY SHOW YOUR FACE MAN¡¡¡

[tomatoskins]

I agree! This circuit would have some issues and possibly not work.

[rafununu]

This project is very dangerous, it works with a voltage which can be lethal. moreover, the current driver schematic doesn't work.