Hai,
I am trying to work out an inverter design on a modified sine wave inverter which is shown in the attachment... I have tested every part of the circuit. Since i do not have an oscilloscope display with me, i am not able to cop with the waveforms, but the voltage measurements were accurately done...
In the current design, the 555 stage at the extreme left side operates in accordance with the output clock signal from IC 4047. The Ic has been configured such that it generates stable 52 Hz flip flop frequency signals at its two output pins pin#10 and pin#11. Their tandem appearance was tested by two LEDs placed at the terminals keeping the frequency at 1 Hz. So the leftmost 555 gets 104 Hz frequency and converts it into a corresponding sawtooth waveform.. The Voltage of this waveform detected by the voltmeter is approximately 3.0 V only. Now this sawtooth waveform is being compared in IC2 with the high frequency square waves generated by IC1. The resulting signal is a PWM signal ( from pin#3 of IC2 ), which is then fed to the gate terminals of each MOSFETs using 1N4148 diodes.. SO the MOSFETs are switched in accordance with this PWM signal, resulting in a modified square wave at the output of the inverter with RMS value very much resembles to the Actual AC Voltage...
Now, My problem is ;
I am getting much power output ( Almost full, tested by a 100 W bulb as load ) when the PWM stage is kept disconnected . ie, the normal square wave inverter... ( 1N4148 diodes are not connected to the gate terminals ). The gate voltage shown by the voltmeter at this condion was 5.98 to 6.0 V approximately.

BUT When i am trying to connect the PWM stage (Connecting 1N4148 diodes to MOSFET gates ), the voltage at the gate terminal reduces to half ( 3.0 V ), so the output power is also getting reduced considerably.

SO What should i have to do to keep the voltage stable at the gate terminal at PWM and thereby maintaining the maximum power output ?

I have also tried using a voltage amplifier, buffer stage with Opamp etc. etc. prior to gate terminals.. But no change were detected..
I think, there might be someone to help me a lot..

 

What is the supply voltage to the lower part of the circuit?

Note that a voltmeter measures the average voltage, not the peak voltage of a switching signal.

And you understand that a square-wave output will have have a higher RMS voltage than a modified sine-wave for the same peak waveform voltage(?).

 

Thanks for replying.. All the Parts of the circuit has been powered with the 12 V battery voltage....
The transformer i am using is 230V/15-0-15 300W transformer..
I have used 4 no.s of 12 V 20 Ah batteries in such a manner that, the series combination of two 12 V 40 Ah battery banks ( 24 V 40 Ah ) was used to fire the center tap of the transformer in which each battery banks were formed by parallel combination of two 12V 20 Ah batteries.....
SO i am having a 24 V 40 Ah battery bank for the required inverting operation, therefore expecting a minimum power of 150W..

I have already told that, 100 W bulb was used as the load.
The voltmeter reading on the gate terminals when the PWM stage kept disconnected is 6V. I know it is the average voltage.. And i am getting the full power output by the inverter...
But on connecting the PWM stage that voltage becomes half the original average value. ie., 3 V and, So the power output is also less.
Therefore i am asking, why this voltage has been halved ?

Also i have an another problem too.
When single battery is used to fire the center tap of the transformer, the output voltage is about 200 V but the power output is less..
But when 24 V is used, it is more than 300 V and the power output is maximum... So suggest me a method by which this output voltage can be corrected to get a stable 230 V output independent of the loads and providing maximum power output

 

Hai..these are some images regarding the current inverter

 

Without knowing what the waveforms look like it's difficult to answer your questions.