12V Emergency Light Circuit Using 3525 IC For CFL/LED Bulb | Mini Inverter Circuit

This circuit is designed to run any regular CFL or LED bulbs up to 32 watts. You can try using a higher wattage bulb or changing the transformer. The power supply voltage (Vin) is usually 12 volts. The output of this circuit is fed to the transformer through two MOSFETs. This circuit is very simple and does not require any special components. It is easy to understand and build. 

Circuit Diagram

Parts Required

Resistor(1/4W):
R1- 4.7K
R2- 18K
R3- 4.7K
R4- 22E
R5- 22E
R6- 10K
R7- 10K

SG3525A Pinout

IRFZ44 Pinout

Circuit Working

The circuit is pretty simple and is based on an SG3525 controlling pulse width modulator IC that runs a couple of MOSFETs and a push-pull step-up transformer. The capacitor at pin 5 and the resistor at pin 6 set the on-chip frequency. The frequency of the output is cut in half by an internal flip-flop, and the flip-flop's output drives buffers that can source or sink more than 200mA. The two MOSFETs Q1 and Q2 are driven by the buffer's output at pins 11 and 14. The high output current is perfect for quickly charging and discharging the MOSFET gate capacitances, which makes for fast switching times. This means that the MOSFETs work well and lose little power. The step up push-pull transformer is powered by MOSFETs. AC energy is sent from the transformer to a bridge rectifier, which turns it into DC voltage. The transformer is connected to a bridge rectifier to convert the AC voltage to DC voltage.

Prototype

Connection Diagram

Caution: The output from this inverter is dangerous. So, handle with care. Never work on the circuit with power applied. The finished project should be installed in a plastic box.