# DIY LED Lamps

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## Introduction: 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 6: Place It and Use It Participated in the
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