| To give credit where it's due, I borrowed the circuit from here:
http://www.aaroncake.net/circuits/40wflamp.htm
To build the circuit as I've constructed it, you'll need:
- 50mm piece of 10mm ferrite rod;
- 0.6mm enamelled wire;
- 0.4mm enamelled wire;
- TIP31C transistor (or any other type that will handle 3A);
- 100uF 16V electrolytic capacitor;
- 0.1uF (100nF) polyester or ceramic capacitor;
- 180 ohm 1W resistor;
- 47 ohm 1/4W resistor;
- 1N5408 diode (or equivalent 3A type);
- 30x70mm blank PCB.
Here's the artwork and circuit overlay for the PCB:
To construct the coil, first insulate the ferrite rod with a couple of strips of masking tape. Next, wind 90 turns of 0.4mm enamelled wire onto it. This should be a neat fit if you start 5mm from one end. Secure and insulate this layer with strips of masking tape. Continue winding in the same direction (ie. don't reverse the direction of your turns) to form a second layer over the top of the first, then secure and insulate it with masking tape as well. Continue until you have 5 layers. This coil will be the secondary.
Next, wind 52 turns of 0.6mm wire onto the ferrite, securing the coil as before. This coil will be the Primary.
Finally, wind 12 turns of 0.6mm wire onto the ferrite in the opposite direction to how you wound the primary. Secure with masking tape as before. This coil will be the feedback. You should now have a completed transformer would on the ferrite that looks like this:
This will neatly slot into the row of 6 holes on the PCB. Connect to a 12V supply and a fluorescent tube, and it should work. For a 20W tube increase the value of the 0.1uF capacitor to 0.47uF or so. This technique can also be used to decrease the current consumption and brightness to whatever level you find acceptable.
Enjoy!
BTH |
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