Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[neptune]

Hi Rosemary, Poynt 99 does not seem to be about at the moment , so can I offer my opinion on what he is saying ? The Fet is acting as a closed switch , and its resistance is assumed to be negligible . So we just have 2 resistors connected in series across a 60 volt battery , We have the load resistor , 11 ohms . And we have the shunt , at one quarter ohm . Imagine instead that the load resister is 10 ohms .and the shunt is one ohm . Now , if we measure the voltages across each resister in turn , we find the the voltage across the load is ten times the voltage across the shunt .That is how he arrives at his voltage figure , being the voltage across the shunt .So going back to the original values , the load will have 44 times the voltage across it than the voltage across the shunt . There will be approx 58.6 volts across the load and about 1.3 volts across the shunt . 58.6 plus 1.3 = 59.9 volts . Near enough for me , and I failed my maths exams .@Poynt99 feel free to tell me if I am wrong .

[Rosemary Ainslie]

Hi Rosemary, Poynt 99 does not seem to be about at the moment , so can I offer my opinion on what he is saying ? The Fet is acting as a closed switch , and its resistance is assumed to be negligible . So we just have 2 resistors connected in series across a 60 volt battery , We have the load resistor , 11 ohms . And we have the shunt , at one quarter ohm . Imagine instead that the load resister is 10 ohms .and the shunt is one ohm . Now , if we measure the voltages across each resister in turn , we find the the voltage across the load is ten times the voltage across the shunt .That is how he arrives at his voltage figure , being the voltage across the shunt .So going back to the original values , the load will have 44 times the voltage across it than the voltage across the shunt . There will be approx 58.6 volts across the load and about 1.3 volts across the shunt . 58.6 plus 1.3 = 59.9 volts . Near enough for me , and I failed my maths exams .@Poynt99 feel free to tell me if I am wrong .

Ok.  Thanks Neptune.  I'm still not sure why one doesn't take the actual voltage reading.  But it's close - so.  I get it. A kind of check?

Now I'd be very glad if Poynty could print those pictures.  I've finally opened them - but would be glad to get them up here - in case anyone, like me, struggles.  Can you oblige us Neptune?  Someone?

Kindest regards,
Rosie

[woopy]

Hi Mag and Rose

i made a quick test with a MO high voltage cap with 0.95 micro F, so it discharge fast enough to see what is going on here.

The battery is at about 4.6 volt and goes to a BAT 43 schotky diode (very few lost in voltage) than to the 220mH inductance(primary of MOT)  ant than to the cap and back to the negative.

I enclose a scope shot .
when i close the circuit the voltage jump to 7.6 volts than the cap descharge down to 4.6 volt (that is the battery voltage, and stay at this value until i open the circuit and the cap goes on descharging.

The best result i got today is a 1.7 time the voltage supplied. In this pix it is 7.7 volts(the freewheeled voltage ) / 4.6 volt (the supplied voltage) = 1.67

Not bad at all.

And in this config (but with a much stronger diode a BYV26D ) , by shorting the cap i can get really high voltage (MORE than 400 volts) impressive.

Now how to use this effect ?

Will test your new circuit ASAP, any idea for the resistor value?

good luck at all

Laurent

[penno64]

Last time - I promise

posted feb 19 in shorting coil thread -


********************
Hi All,

How pertinent is this guys video now ? (1 and 2)

http://www.youtube.com/user/NRGFromTheVacuum#p/u/10/2cUS03yNl40

Looks like it's been under our noses all this time and we couldn't see the forest for the trees.

Kindest Regards, Penno
________________________________________________________

Penno, (Garry)

[nul-points]

Will test your new circuit ASAP,

Laurent

hi Laurent

nice testing - but you may want to reconsider Mags idea of a diode across the battery/switch - as shown, it's forward biased and if the switch gets closed for too long, then the 'magic smoke' is likely to to get released from either the new diode or the battery! 

interesting to see that shorting the capacitor has such a strong effect - especially when the current fashion seems to be 'shorting coils'!!!

cordialement
sandy


http://docsfreelunch.blogspot.com/