Tommey Reed's Pulse Generator output test update.

[Tommey Reed]

Not so fast, c1xc2/c1+c2 is two caps in series

If you have charged a cap at 12v and 32000uf

You have (.5 x .032 x (12^2) = 2.304 joules

If you have to cap at 6v with 32000uf then in Series you will have 12v but less charge:
32000uf x 32000uf
------------------------- = 16000uf
32000uf + 32000uf
(.5 x (c1c2/c1+c2) x ((6v+6v)^2)=1.152 joules

Parallel is 1/c1+c2 or:
.5 x (.032000uf+.032000uf) x (6v^2)= 1.152
But the total was starting cap was:
.5 x .032 x (12^2) =2.304 joules from the start.

50% lost of energy going to second cap.

Tom

[Tommey Reed]

1. volts do not equal power
2. you did not take into account the power being fed to the transistor
You really need to lift your game.
Kind Regards
Mark

Mark,

You're right about the pwm not being powered by the cap.
The voltage drops from the main cap, this cause the pwm not to work after 9v.
That is why I used a battery to power the pwm only.

Tom

[markdansie]

thanks Tom,
by the way I do like your videos and your experiments. Thats why I encourage you. You solar inverter is a great solution and would save a lot of people money who cannot afford expensive solid state devices.
Like I said I encourage you very much...however just think through your testing procedures sometimes.
I am sure many others enjoy your posts as well and your video's
The reason I am a hard arse skeptic sometimes I have seen mistakes made by people all over the world when it came to testing their own devices. I have seen electrical engineers make mistakes as well. Please do not take offence.
Mark

[Tommey Reed]

Now if my calculation are right, then using a pre-charge cap to charge the second one has a lost of 50%.
 
c1xc2/c1+c2 is two caps in series

If you have charged a cap at 12v and 32000uf

You have (.5 x .032 x (12^2) = 2.304 joules

If you have to cap at 6v with 32000uf then in Series you will have 12v but less charge:
32000uf x 32000uf
------------------------- = 16000uf
32000uf + 32000uf
(.5 x (c1c2/c1+c2) x ((6v+6v)^2)=1.152 joules

Parallel is 1/c1+c2 or:
.5 x (.032000uf+.032000uf) x (6v^2)= 1.152
But the total was starting cap was:
.5 x .032 x (12^2) =2.304 joules from the start.

50% lost of energy going to second cap.

Did I make any mistakes?

Tom...

[gyulasun]

Hi Tom,

Your calculations seems correct, they reflect the 50% loss you also show in practice.

Maybe you have seen this but if not, here is a file uploaded by member poynt99 here: http://www.overunity.com/index.php?action=downloads;sa=view;down=209

He figured out how to reduce this 50% loss by using an appropiate coil for the charge transfer between the two capacitors.

Thanks,  Gyula