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Overunity Machines Forum



Joule Thief

Started by Pirate88179, November 20, 2008, 03:07:58 AM

Previous topic - Next topic

0 Members and 91 Guests are viewing this topic.

xee2

The previous circuit I posted for measuring the JT output power had a safety problem because if the resistor was ever not connected across the capacitor (even for just an instant) the capacitor would charge up to the full open circuit output voltage of the pickup coil which might be well over 1000 volts. Even small capacitors charged to over 1000 volts can be dangerous. As long as the resistor was never disconnected, the capacitor voltage would stay under 50 volts and was not so dangerous. To prevent the capacitors from charging to over 100 volts I have added the neon bulb. The neon bulb remain off until the voltage across the capacitor reaches about 100 volts. Thus in normal operation the neon bulb does not change any measurements. But now, in order for the capacitors to charge up to the full output voltage of the open circuit pickup coil, both the resistor and the neon would have to be simultaneously disconnected from the capacitor. Thus there is a much smaller chance that the capacitor will ever charge to dangerous levels. If the voltage across the capacitor ever trys to go over 100 volts the neon will turn on and prevent the voltage from going over 100 volts.


jeanna

xee2,
This whole thing is so confusing.
Putting the neon in the secondary might limit the v to 100 (or 170 spiky volts, more likely,) but the thing is it will limit it and it is the volts you are measuring to use in the ohms law based formula.
I have issues with using ohms law with ac or pulsed dc, but besides those objections, how on earth are you ever going to see a volts reading that is above what the neon siphons out?

I don't get it??

thank you,

jeanna

innovation_station

thank you guys..

now  any one with 2 wits about them can put it togather ...

any how be careful... 

so i will explain ... 

just as i have showen ..  in my most recient videos ... 

i have 2 outputs ...1 normal jt spikes used to charge a low voltage cap 2 many powerful spikes ... ment to charge output caps..

this can be configured many ways ...  meaning stright through ou put ...  this means insted of recharge ... BOTH ARE SENT TO FILIMENT BULB...

thank you

william

heres the scope ...  kids are over for a few days im cleaning up then we are building and makeing videos ...  this means ... all 3 of us ...

anyhow ..  i have those ramma chasers .. still i will use those to conrtol fets to flip flop between low and high  for a basic unit i will use the ramma toy to sub a relay or other switching means ...

i can charge caps and allow them to release or i can probally drive loads  ....  never publically tryed yet ... 

l8r
To understand the action of the local condenser E in fig.2 let a single discharge be first considered. the discharge has 2 paths offered~~ one to the condenser E the other through the part L of the working circuit C. The part L  however  by virtue of its self induction  offers a strong opposition to such a sudden discharge  wile the condenser on the other hand offers no such opposition ......TESLA..

THE !STORE IS UP AND RUNNING ...  WE ARE TAKEING ORDERS ..  NOW ..   ISTEAM.CA   AND WE CAN AND WILL BUILD CUSTOM COILS ...  OF   LARGER  OUTPUT ...

CAN YOU SAY GOOD BYE TO YESTERDAY?!?!?!?!

xee2

@ jeanna

Quote from: jeanna on December 21, 2009, 02:48:39 PM
xee2,
This whole thing is so confusing.
Putting the neon in the secondary might limit the v to 100 (or 170 spiky volts, more likely,) but the thing is it will limit it and it is the volts you are measuring to use in the ohms law based formula.
I have issues with using ohms law with ac or pulsed dc, but besides those objections, how on earth are you ever going to see a volts reading that is above what the neon siphons out?

I don't get it??

thank you,

jeanna

The diode and the capacitor make a half wave rectifier, so there is DC coming out of the capacitor. The voltage across the resistor will always be less than 100 volts because it would take a lot more current than the JT can supply to get the voltage across a 10K resistor higher than that (remember, E = I x R). The voltage increases until the JT can no longer supply more current. That is the point I am trying to find. Since the voltage does not ever get to 100 volts, the neons never turns on - it is just there for safety.

You can see what is happening on your scope by just putting a resistor across the outputs of the pickup coil and then looking at the AC voltage across the resistor with the scope. Make sure that resistor stays connected or you will blow your scope. Start with a 10K resistor. Next use a 1K resistor. You will see that as the resistor value decreases so does the voltage across it. At the limit of a zero ohm resistor there will be zero volts across it.

I hope that helps




xee2


I have made a 2B-10C-250P coil on 3.38" toroid. I sprayed the core with clear plastic and used wire wrap wire. No problems with shorts (yet). I have just started doing tests. But it does not seem to put out as much power as the coil with 80 turn pickup. More later.