How to Resolve The Problem ????

I am making a half bridge Inverter Using IRFP450 , the operating freq is 3KHZ
here is the diagram
https://i1026.photobucket.com/albums/y325/cranker117/untitled-1.gif
the problem is that , i am not getting the 220V at Inverter output , i think the problem is that Q2 is Off i.e
If the gate is made to be 2 volts greater than the Source, the MOSFET will turn on. When that happens however, the voltage at the Source will increase until the gate is no longer 2 volts greater than the Source and the MOSFET will turn off.
whats its Solution ???
Plz Help me out

You need a high side FET driver that produces a voltage 10 volts above the source. They are common.

The fabric pattern and dots all over the background of your schematic make it very difficult to see.

I think the gate on the lower Mosfet is blown up because it has nearly 110V DC to 120VDC gate to source. Its max allowed voltage is only 20V.

The datasheet shows that its gate to source threshold voltage is 2V to 4V when its current is only 0.25mA so it is almost turned off. It is guaranteed to turn on very well when its gate to source voltage is 10V as shown in its on-resistance specification.

You are trying to drive the lower MOSFET when its' source terminal is connected to a negative supply rail, and the 555 timer is connected GND to ground. That will never work. As soon as the lower rail starts dropping more than ~3v below ground, the lower MOSFET will turn on, and you won't be able to turn it off.

Ask yourself, how will your circuit get the Vgs to 0v when S is tied to a negative voltage?

Also, you must stop thinking of gate voltage in any other terms than Vgs; that's voltage on the gate referenced to the source terminal. The MOSFET really "doesn't care" about the outside world too much, but if you exceed ±20v Vgs, your MOSFET will die.

thnkz for ur reply !!!
Here is the new one plz let me know if its gonna work or not ?**broken link removed**
And **broken link removed**