Here is the circuit diagram of a simple 100watt inverter using IC CD4047 and MOSFET IRF540.CD 4047 is a low power CMOS Astable/Monostable multi-vibrator IC
Here it is wired as an Astable multi-vibrator producing two pulse trains of 0.01s which are 180 Degree, out of phase at the pins 10 and 11 of the IC. Pin 10 is connected to the gate of Q1 and pin 11 is connected to the gate of Q2. Resistors R3 and R4 prevents the loading of the IC by the respective MOSFETs. When pin 10 is high Q1 conducts and current flows through the upper half of the transformer primary which accounts for the positive half of the output AC voltage. When pin 11 is high Q2 conducts and current flows through the lower half of the transformer primary in opposite direction and it accounts for the negative half of the output AC voltage.
IC1 = CD4047
Q1, Q2 = IRF540
T1 = 9-0-9 Transformer
R1 = 330R, R2 = 1K, R3, R4 = 220R
R5 = 390K, VR6 = 1K
C1 = 0.01uf, C2 = 2200uf 25V
C3 = 0.1uf 600V
D1 = LED (Red)
D2 = Diode 1N4007
B1 = Battery 12volt 6-7Ah
B1 can be a 12V 6Ah lead acid battery.
Q1 and Q2 must be fitted to a proper heat sink.
T1 can be a 9-0-9 V primary, 230V secondary, 150VA Transformer.
Do not expect much from this circuit. The is very simple one suitable for low grade applications.