Hi,

Where am i doing wrong?

Best regards

 

Where am i doing wrong?

Good question.
Why do think you are doing something wrong?

 

Good question.
Why do think you are doing something wrong?

I ran it without load q2 q3 mosfet burned out

 

Two things:
You have just about half a milliamp drive to the MOSFETs, half an amp would be better.
Q1 and Q3 need their gates to be driven at least 8V higher than the source, which means taking the gate voltage above the positive supply.
Have you though of using an IC to drive the MOSFETs - they are called (guess what?) Half Bridge Drivers and you’ll need two of them.

 

The gates of Q1 and Q3 have no zener diode protection from the high voltage. Vgs will be excessive.

 

The gates of Q1 and Q3 have no zener diode protection from the high voltage. Vgs will be excessive.

The 470k resistor and the internal cap makes for a very slow turn on. As the Gate voltage gets to the Turn On Voltage the transistor will turn on and thus limit the voltage to just a little more than the turn on threshold. So I think there will never be any current in the Zeners you want to add.

Next: Normally to turn on a MOSFET you apply a G-S voltage that is 10 to 15 volts. In this case, (for the top two transistors) as the part turns on the voltage that is supplying Gate current goes away. It is very likely that the D-S voltage will never get less than 3V and maybe never close to more than 4V. The bottom transistors close very well because there is a good voltage on the Gate. The top transistors do not close well and heat up and die.

Remember it is not the Gate voltage that turns on the top transistors but the Gate to Source voltage. You need a supply that is above 330V for the Gate voltage. Read up on Half Bridge Drivers.

In the schematic the IC has a "amplifier" from Vin to LO. When Lin is above 2 volts LO is at Vcc. When Lin is at 0V LO is at 0V. (Vcc is normally at 10 to 15V)
When the Bottom MOSFET is on current passes from VCC through the diode and charges up the Capacitor on VB. Voltage on the capacitor is one volt less than Vcc. This capacitor powers the amplifier that rides on the "Load". When HIN is high HO is pulled up to VB. VB is about 11 volts above "to load". When the top transistor is on the voltage on VB is 11 volts above "to load".
The Gate resistors are 10 ohm or something near that. (4.7, 22)

 

The 470k resistor and the internal cap makes for a very slow turn on. As the Gate voltage gets to the Turn On Voltage the transistor will turn on and thus limit the voltage to just a little more than the turn on threshold. So I think there will never be any current in the Zeners you want to add.

Next: Normally to turn on a MOSFET you apply a G-S voltage that is 10 to 15 volts. In this case, (for the top two transistors) as the part turns on the voltage that is supplying Gate current goes away. It is very likely that the D-S voltage will never get less than 3V and maybe never close to more than 4V. The bottom transistors close very well because there is a good voltage on the Gate. The top transistors do not close well and heat up and die.

Remember it is not the Gate voltage that turns on the top transistors but the Gate to Source voltage. You need a supply that is above 330V for the Gate voltage. Read up on Half Bridge Drivers.
View attachment 251603
In the schematic the IC has a "amplifier" from Vin to LO. When Lin is above 2 volts LO is at Vcc. When Lin is at 0V LO is at 0V. (Vcc is normally at 10 to 15V)
When the Bottom MOSFET is on current passes from VCC through the diode and charges up the Capacitor on VB. Voltage on the capacitor is one volt less than Vcc. This capacitor powers the amplifier that rides on the "Load". When HIN is high HO is pulled up to VB. VB is about 11 volts above "to load". When the top transistor is on the voltage on VB is 11 volts above "to load".
The Gate resistors are 10 ohm or something near that. (4.7, 22)

Will it create a problem when running without load?

 

Will it create a problem when running without load?

Yes. The optos are too slow, the drive current is too low. You will get a situation where some MOSFETs turn on faster than others turn off, which puts a dead short across the supply.

 

Will it create a problem when running without load?

The fact that the top MOSFETs are not turning on well should not effect heat at zero current. (no load)
Lan) is talking about a case where both top and bottom transistors are on. Because the optos turn the transistors off at 10mA and the resistors turn on at very low current (transistors on = slow, off= fast) there should not be a problem. Unless you are driving the parts at too high a frequency. What frequency or time?