Active power from parallel resonance. Test circuit and research.

[Jim36]

Hi all,

I have a new idea and would like people to have a look at as it is quite simple. I have seen some experienced electronic guys on here so hope to get some feed back before building.

See attached circuit diagram. From my calcs the capacitor & inductor can store just over 1 joules per cycle. The current circuit would only use 50 % of the cycles but another circuit could be added to use the full wave cycle.

The circuit should work by triac T1 firing on the first half cycle which should start the resonance circuit. In theory the parallel LC circuit shouldn't draw any current from outside its own circuit. During the second cycle T1 stops conducting decoupling from the mains and T2 should fire on the negative cycle produced by the LC tank and power dissipated through the dampening load. The cycle then starts again.

Thanks for looking, and I welcome any comments.

Cheers

Jim

[pomodoro]

Is this for OU or for some other purpose?


Cheers, Pomo 

[citfta]

If you want some help you need to reduce the size of your drawing to the size recommended for this forum.  Otherwise no one can see all of it without scrolling back and forth.  You can edit your post and remove the attachment and then insert a smaller drawing the does not go way beyond the normal width of the forum.  There are several free picture editing programs on the net you can get if you need one to help you reduce the size.  I use one called "Free Photo Converter".  Very easy to use and works great.

[verpies]

See attached circuit diagram. From my calcs the capacitor & inductor can store just over 1 joules per cycle.

1 Joule is a lot in electronics.
But why do you think that the parallel LC circuit will give up more energy then it took to energize it?

During the second cycle T1 stops conducting decoupling from the mains and T2 should fire on the negative cycle produced by the LC tank and power dissipated through the dampening load. The cycle then starts again.

Please compare the forward voltage drop of a Triac vs. a MOSFET, while keeping in mind that the power lost inside the Triac is equal to the product of the current flowing through it and its voltage drop.

[sm0ky2]

here