This is another 100 watt inverter circuit diagram. Built based on IC CD4047 and Mosfet IRF540, this inverter have ability to supply electronic device -which require 220VAC- up to 100w from 2-3A transformer from 12V lead acid battery.
In the circuit diagram we can observe that 12V battery is connecter to the diode LED and also connected to the pin8 of the IC 4047 which is VCC or power supply pin and also to pin 4 and 5 which are astable and complement astable of the IC. Diode in the circuit will help not give any reverse current, LED will work as a indicator to the battery is working or not.
IC CD4047 will work in the astable multivibrator mode. To work it in astable multivibrator mode we need an external capacitor which should be connected between the pin1 and pin3. Pin2 is connected by the resistor and a variable resistor to change the change the output frequency of the IC. Remaining pins are grounded .The pins 10 and 11 are connected to the gate of the mosfets IRF540. The pin 10 and 11 are Q and ~Q from these pins the output frequencies is generated with 50% duty cycle.
The output frequency is connected to the mosfets through resistor which will help to prevent to the loading of the mosfets. The main AC current is generated by the two mosfets which will act as a two electronic switches. The battery current is made to flow upper half or positive half of the primary coil of transformer through Q1 this is done when the pin 10 becomes high and lower half or negative half is done by opposite current flow through the primary coil of transformer, this is done when pin 11 is high. By switching the two mosfets current is generated.
This AC is given to the step up transformer of the secondary coil from this coil only we will get the increased AC voltage , this AC voltage is so high; from step up transformer we will get the max voltage. Zenor diode will help avoid the reverse current.
Warning!: Lethal potential at the output of transformer, please be careful, enclose this power inverter circuit into the plastic box.