Help me with an 500W inverter design

[tatnur]

I implemented the above circuit in multisim and found the output before the transformer between collector and the ground..I used a 12V battery.. Instead of getting 12V peak to peak, i'm getting a very less value.. Only if i increase the resistances from 0.1R to approx 6k ohm, i'm able to get atleast 9V output.. What is the problem in the above circuit..? What is the function of the transistor bank in it..And what is the use of such a low resistance value when i'm able to get a considerable output for resistance in kilo ohms..?

 

[tmd_63]

OK. The IC U1 is a simple basic oscillator. The outputs Q and /Q are complimentary (ie one is on when the other is off).
When Q is on (high), this is buffered by the follower U2 and this powers Q1 when it's base is above 2.6V. This turns on Q3 which turns on Q5, Q7, Q9 and Q11 all at the same time. Each 2N3055 passes 10A into their respective 0.1R resistors. This provides some overload feedback. This drives 41 amps of current through one side of the transformer. When Q is off, then /Q is on and the other side operates driving the current the opposite way on the transformer and giving approx 220V output at 500W.
The circuit appears good and as to why your multisim is giving unusual reading, I can not understand. But I hope my circuit description helps.

 

[goldsmith]

Hi Tatnur
As it appears , your circuit is correct ,By the way why you used used those two diodes ? for more protection ?
Did you see base voltage of your transistors ? have them square wave ?
Best Wishes
Goldsmith

 

[picgak]

Hi tatnur ,
you can try the modification given below. hope it will help if so dont forget to press the helped me button.
regards ani

 

[goldsmith]

Hi Picgak
You have added a resistor tied to -12 volts but it is a push pull amplifier doesn't need minus supply .
Regards

 

[picgak]

Hi Picgak
You have added a resistor tied to -12 volts but it is a push pull amplifier doesn't need minus supply .
Regards

Hi goldsmith,
I have added 10 ohms to +12v supply not -12 on both sides to get sufficient base drive, for 500w he may require 4.7ohms instead of 10 ohms
regards ani

 

[goldsmith]

Hi Ani
I thought it is minus . sorry for that . but i think it still is a problem , because when power transistors are going to be on , you will have 1.2A through that resistor , because it will connect positive supply to the ground ( via that resistor ) and it is additional current . for my opinion Driver transistor will create sufficient bias to drive power transistors if we select one of them with high beta , such as darlington pairs that are available with beta rating around 10000 or higher .
Respectfully
Goldsmith

 

[picgak]

Hi goldsmith,
2N3055 has very low gain I think nearly 10 to 15 so the power transistors connected to the transformer requires large base current. I think it is better to replace the entire power transistors and the darlington pair by two nos of IRFz44 in parallel it will cheap and efficient and it doesnt require LM358. just a small mofet drive circuit is required.
regards ani

 

[goldsmith]

Hi Ani
Yes you are quite right . current gain of power transistors can't exceed from numbers around 10 or 35 or a bit lower or higher .
and about replacing driver with a mosfet like IRFz44 , yes your are right again . it can handle large value of currents . but mosfets can't start to conduction from 0.6 volts . they will have different threshold . for example 3 volts . hence input should be higher to have the same result . or a bit DC biasing will be required . but in a realistic mode your suggestion is very good .
Best Regards
Goldsmith

 

[Tahmid]

For a realistic circuit for 500W, the entire circuit should be changed for multiple reasons, efficiency being a key factor. The driver stage should be replaced with MOSFETs. There should be output regulation under line/load variation. A PWM controller stage should be used. Features such as low-battery protection and overload protection should be implemented.