What's the working principle of this inverter circuit ?

[said az]

https://imgur.com/a/yCI0j
I found this inverter circuit online.
the timer output is at 50hz with square wave signal
I can't read it , can you explain it to me .
thank

 

[(*steve*)]

Yes. The red X marks the point at which the missing image is located in your post.

Either post the actual image here or a link to the website page. Even better, do both!

Edit: was there an old version without the link?

 

[BobK]

The 555 produces a square wave.

Q7 is an inverter, producing an inverted square wave.

Q1 and Q2 are a gate driver for Q3 and Q10 which are in parallel, and ground the upper side of the transformer primary when on.

Q4 and Q5 are a gate driver for Q6 and Q9, which ground the lower side of the transformer primary.

Thus, for one half the cycle the upper side of the transformer primary is on, and the for the second half of the cycle, the lower half is on. Which produces bipolar -12 to 12V signal on the transformer. The output of the transformer is 230V, but again a square wave.

That output is good for lighting incandescent lights, or a small heater, in other words resistive loads, and not much else.

Bob

 

[WHONOES]

Apart from what BobK says, it is not the best of circuits as there will most likely be some conflict between the two halves of the circuit where they will both be on at the same time for a short period.
A single ended circuit would be a better bet. I would also run it at a much higher frequency which would enable a smaller transformer to be used.

 

[said az]

Yes. The red X marks the point at which the missing image is located in your post.

Either post the actual image here or a link to the website page. Even better, do both!

Edit: was there an old version without the link?

Yes I edited it , the image is shown now

 

[said az]

Thank you BobK and WHONOES ! one question , why is the signal need to be inverter by the transistor Q7 Isnt it already alternative ?
when I put another transistor in the lower half at same location , it gives the same signal forms at the output :


https://imgur.com/a/DVvGW

 

[duke37]

One pair of the output transistors need to be turned on when the other pair is turned off so generating an AC drive to the output transformer.

In capture1.JPG all output transistors are turned on and off together. There will be no AC on the transformer and the invertor will appear as a short circuit. Do you have a fuse?

 

[WHONOES]

Duke37 is correct in his analysis.

 

[kellys_eye]

I would also run it at a much higher frequency which would enable a smaller transformer to be used.

Not much good if the design is specifically for powering 50/60Hz equipment....

 

[WHONOES]

Not much good if the design is specifically for powering 50/60Hz equipment....

Not so. The switching frequency is determined by U1.

 

[BobK]

@WHONOES

Read @kellys_eye post again.

The circuit is an inverter to produce 230V at 50Hz. Running it at a higher frequency would not do that.

Bob

 

[WHONOES]

Yes but, I think the circuit is an example that said az has found and is not for a specific application. If it was just to power a lamp as seems to be indicated then a higher frequency is no problem.