If you are not already familiar with the "captret" claims just google it for the claims or go here:
http://www.overunity.com/9878/captret-capacitor-and-electret/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.
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Part 2
Did a test on another capacitor in "captret" configuration monitoring current vs time 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.
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
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Topic: "Captret" Demystified (Read 1164 times)
