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Overunity Machines Forum



Captret - Capacitor and Electret

Started by ibpointless2, October 19, 2010, 06:49:51 PM

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0 Members and 6 Guests are viewing this topic.

MileHigh

QuoteThe longer you stay in school ... the more you become a indoctrinated fool

Only on the Bizarro planet, where the objective is to become dumber the more you go to school.

I guess that you don't have direct access to the experiment and the data.  Where are you getting your information from?  I can't prejudge a specific experiment without access to the correct and complete information.

What I can say is that if a decent-sized 12-volt battery is connected to a string of LEDs in series, and the setup is just right, then the battery and the LEDs can engage in a "dance of death" and it's possible that the LEDs can remain dimly lit for a very long time, perhaps even months.  That could explain what was observed.

For the argument that the battery is being "recharged," I have seen countless experiments where the battery is clearly discharging but the experimenter notices a slight increase in the battery voltage during the test.  They then proudly proclaim that the battery is being "recharged."  It's possible that the same thing is happening in this case.

Vortex1

The operation of the so misnamed "*captret" is easily explained if folks would just use a milliamp meter to measure the current drain of the battery. The battery is not being charged, rather the load on the battery is progressively decreased so the voltage rises.

If the battery were being charged, the milliampmeter would reverse sign. This is not the case.

What happens is this:

The "captret" configuration of using the outer case and the negative electrode presents initially a very leaky capacitor to the series circuit, more like a capacitor with a variable resistor. This initially creates a lot of current flow and slightly warms the battery activating the electrochemical output.

As the oxide layer is formed between the case and the negative electrode, the leakage decreases from initially about 300mA to the range of 1 to 5 mA. Even less leakage can be had if left on for a very long time, so an LED can be lit on less than 1 mA for a very long time.

Because of the reduced load on the battery, the battery voltage is seen to rise. This is a simple load line as the battery is not a perfect voltage source, and much less so when the battery is near the end of it's life.

There is no mystery here, just poor measurement technique and mis-observations of what is occurring.

You could  get the same "effect" with a variable resistor instead of the "captret" Use two nearly dead 9 volt batteries with about 7.5 volts each in series.

Adjust the potentiometer of the simple series circuit to simulate an initial 300mA current drain then over a few minutes reduce it to 1 mA current drain. Do this while observing battery voltage.

You will also see the battery voltage appear to rise due to reduced loading. This also is not recharging as the current meter never changes sign.

* This is a misnomer as the so called "captret" is actually a very poor electret in that it has way too much leakage to be any good, orders of magnitude more leakage than a good electrolytic or film capacitor. It's recovery is limited by this.

Vortex1

Did a test on another capacitor in "captret" configuration monitoring current from a 10 volt source.

Initial current:        equivalent leakage resistance:

100mA                     100 ohms

One minute later:

10mA                       1000 ohms

Two Minutes

1 mA                        10,000 ohms

Three minutes

0.1 mA                     100,000 ohms

As the oxide layer forms the leakage resistance progressively goes up as current goes down.

At no time did the current ever reverse, even after one hour when the leakage was less than 0.1 mA. So there is no battery charging effect.

As I said in the previous post the "captret " acts like a variable resistor changing resistance upward over time hence presenting a much lower load on the battery from the initial connection.

The reduced current over time causes the battery voltage to increase giving the illusion that the battery is being charged....it is not.

A high brightness led fed from a 9 volt battery will be lit for a very long time  when operated at less than 0.1 mA.

Paul-R

Quote from: TinselKoala on June 26, 2014, 09:54:55 AM
I want to see some data that demonstrates this claim that "bedini motors run faster during the full moon."

Data that supports that claim would have to show RPM readings taken from a Bedini setup

The SG is not a motor; it is a generator of radiant energy. The RPM is not relevant. The wheel's motion is to trigger the transistor.  It is the rate of charging of the battery that counts, taking into account the rate of discharge of the drivng battery.

MileHigh

QuoteThe SG is not a motor; it is a generator of radiant energy.

It is not a generator of "radiant energy," that is complete nonsense.  The SG is a very crappy battery charger in terms of power-in to power-out efficiency.  It is typically only 30% efficient and 70% of the source battery power is lost as waste heat to make the wheel spin.

In the entire world of electronics, and electrical technology in general, the output from a discharging coil is never referred to as "radiant energy."

In your mind you probably think that there is something "amazing and different" about the "radiant energy spike."  Well the truth is that that is complete and total nonsense.

My suggestion to you is to learn about how a coil works so that you can understand why and under what conditions a coil can generate a high voltage spike.

It's time for you and many others like you to demystify this Bedini nonsense where he calls the output from a coil, "radiant energy."  That makes it sound like it comes from "somewhere else," possibly even the "vacuum."  The source of the energy for the voltage spike is the battery.

MileHigh