problem in build the inverter

build an inverter

first, i want to ask how to make the inverter from 12Vdc(mobile battery) to single phase 240Vac.Second,i want to ask why my inverter only out about 110Vac from 12Vdc,order i have used transformer 12V for primary and 240V for secondary....

how to build inverter

you may used half bridge configuration.
if you use full bridge you can get 220V output voltage.
with half bridge, if you use [6V:220V] transformer you can get 220V.

Regards,
Davood.

how to build an inverter

There may be a lot of reasons why you don't get full 240V at the output ..
For example, you may not drive the power transistors from full saturation to full cut-off ..

Can you post the circuit diagram ..
Is it similar to what you can see at the attached picture?

Regards,
IanP

thanks IanP and Davood Amerion for answer my question. For your information my circiut is similar with the circiut that have you sent to me. but now i want to know how to build full bridge.can you sent the circiut to me. The another question is how to drive the power transistors from full saturation to full cut-off. I really refer to your circiut but can't get the expected output 220Vac.The circuit have additional circiut or not to get the full output.

potetojb said:

first, i want to ask how to make the inverter from 12Vdc(mobile battery) to single phase 240Vac.Second,i want to ask why my inverter only out about 110Vac from 12Vdc,order i have used transformer 12V for primary and 240V for secondary....

Click to expand...

you have a transformer 12V->230V, the voltage applied to the primary is 12V-Vcesat.
Vcesat is the voltage drop of your transistor when it is closed (in conduction). Consider for example vcesat=0.5, the voltage applied to the primary is 12-0.5=11.5, at the output the voltage will be (11.5/12)*230=220.4
The output , according to your last post, is a square wave, so the 220.4V is the value of peak of square wave.
If you consider the mains at 230V you have a sinusoidal voltage that have a rms value of 230V and a peak value of 230*1.41=324.4V.
Many digital voltmeters don't have the true rms reading, when used in ac they read the peak value and divide by 1.41, so 220.4/1.41=156.31.
In any case if you want to have correct value at the output of your inverter, the square wave must have the same peak and rms value of the sinusoidal waveform.
If the frequency of your inverter is 50Hz (20ms) the square wave must be positive for 5ms, zero for 5ms, negative for 5ms and so on. In the attachment you can see this type of waveform