This is the schematic diagram of inverter circuit for 40W fluorescent light, this is great circuit for emergency lamp. This inverter allows you to power up 40W fluorescent tubes from any 12V source capable of delivering 3A. The circuit is very simple, inexpensive and easy to build. Please use safety equipment since this circuit work with high voltage electricity.
R1 = 180 Ohm 1W
R2 = 47 Ohm 1/4W
R3 = 2.2 Ohm 1W Resistor (only needed once)
C1, C2 = 100uF/16V
C3 = 100nF
Q1 = TIP 3055 or 2N3055 or equivalent
MISC = Wire, Case, Board, Heatsink For Q1, heatshrink, AM antenna rod for coil
L1/T1 Winding Procedure:
You will need an AM antenna rod that is about 60mm (2.5 inches) long to wind T1/L1 on. T1/L1 are wound on the same core. Shrink a layer of heatshrink over the core to insulate it. Leave 50mm of wire at each end of the coils.
- Primary: Wind 60 turns of 1mm diameter enamelled copper wire on the first layer and put a layer of heatshrink over it.
- Feedback: Wind 13 turns of 0.4mm enamelled copper wire on the core and then heatshrink over that.
- Secondary: This coil has 450 turns of 0.4mm enamelled copper wire in three layers. Wind one layer and then heatshrink over it. Do the same for the next two.
How to calibrate/test the circuit:
Connect the 2.2 Ohm 1W resistor (R3) in series with the positive supply from the battery. Hook up a 40W fluorescent tube to the high voltage ends of the transformer. Momentarily connect power (power up the circuit). In the case the tube doesn’t light up immediately, then reverse the connections of L1. If the tube still doesn’t light up, check all wiring connections. When you get the tube to light, remove the 2.2 ohm resistor and the circuit is ready for use. You will not need R3 again.
This circuit is should be used for 220V fluorescent lamps. It will work with 120V units just fine, but will shorten the life of the fluorescent tube.