Author Topic: Tommey Reed's Pulse Generator output test update. (Read 17697 times)

Tommey Reed · « **on:** August 15, 2009, 11:32:00 PM »

http://www.youtube.com/user/OverUnityNow1#play/all/uploads-all/0/nveNtdrxPbM

Greater energy coming out then going in.

Tom

powercat · « **Reply #1 on:** August 15, 2009, 11:49:42 PM »

Can you loop ?

cat

markdansie · « **Reply #2 on:** August 15, 2009, 11:56:23 PM »

sorry no cigar.
To be honest i cannot be bothered to even begin to tell you where you are fooling yourself. I am sure others will oblige.
However put it into a closed loop system with no batteries involved, then i might be impressed.
here are a few hints on what you did wrong.
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

Tommey Reed · « **Reply #3 on:** August 16, 2009, 12:37:08 AM »

Are you an idiot, or what, Volts and amps = power= watts
Anything else?
Don't be an ass!

markdansie · « **Reply #4 on:** August 16, 2009, 01:08:48 AM »

well I encourage you to contiue, it people like you that help make discoveries. All I ask is you seek some advice on testing procedures before jumping to any conclusions. Like I said close loop it and then you might have me convinced.
Mark

Tommey Reed · « **Reply #5 on:** August 16, 2009, 02:13:12 AM »

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

Ifï»¿ 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 · « **Reply #6 on:** August 16, 2009, 02:18:53 AM »

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 · « **Reply #7 on:** August 16, 2009, 02:30:01 AM »

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 · « **Reply #8 on:** August 16, 2009, 02:43:18 AM »

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

Ifï»¿ 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 · « **Reply #9 on:** August 16, 2009, 11:56:54 AM »

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

Tommey Reed · « **Reply #10 on:** August 16, 2009, 12:25:25 PM »

Thanks for the info, love information.

So when I discharge from the pulse generator, where did the extra energy come from the second cap?

As my test shows using a charge cap to discharge into the pulse generator, at the same time the pulser allow more energy into the second capacitor.

This is the test:
http://www.youtube.com/user/OverUnityNow1#play/all/uploads-all/1/nveNtdrxPbM

The pwm was run on a battery to control the mosfet.
Tom.

poynt99 · « **Reply #11 on:** August 16, 2009, 06:04:18 PM »

Tom,

In your video you started with 12.92V in one cap and ended up with about 7.7V in each after transfer through your PWM circuit. This is an improvement over direct transfer through a piece of wire where you would get about 6.46V on each.

The 7.7V on each does not show "extra' energy of any kind, but what it does show is less loss. With a piece of wire you lose 50% of the energy, no argument. With an inductor (your PWM circuit including the coil) this gives you less loss, but you are far far from perfect.

7.7V is an improvement, but if you try harder, you will be able to obtain about 8.9V on each cap. If you achieve this, your circuit will be running at about 95% efficiency. Sorry, still no OU.

.99

Tommey Reed · « **Reply #12 on:** August 16, 2009, 06:22:45 PM »

Ok, I'll take that. I was showing in my videos that energy from pre charge cap going into another cap was a loss. That's why I said 50% loss of transfer energy.

Tom

Tito L. Oracion · « **Reply #13 on:** August 17, 2009, 09:32:01 AM »

Quote from: Tommey Reed on August 16, 2009, 06:22:45 PM

Ok, I'll take that. I was showing in my videos that energy from pre charge cap going into another cap was a loss. That's why I said 50% loss of transfer energy.

Tom

Hi tom good day

how did you made your PWM?
i mean how could it last the switching on and off?

God bless
otits

captainpecan · « **Reply #14 on:** August 17, 2009, 10:17:59 AM »

Tom, Keep experimenting. Trust me, I've been down the road your on before. Try putting a small DC motor in series between the caps, and see how much energy it actually uses to do work. http://www.youtube.com/watch?v=vwp7podu06s fun little mind experiment that really get's you thinking!

Long story short, an inductor, and pulses do decrease losses. 99 is right in this respect, like I said, been down that road haven't we 99

. But it does not mean extra energy is entering the circuit yet unfortunately. Capacitors are lossy as hell. But it's when you start realizing you can be get work done, while you retrieve half the energy that you start seeing efficiency climb. Good luck.

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Author Topic: Tommey Reed's Pulse Generator output test update. (Read 17697 times)

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Tommey Reed's Pulse Generator output test update.

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