Combining radiant output, 0.2 + 0.2 = 1.5

[sucahyo]

2 x sting oscillator + modified joule thief, twice more efficient

The new sting oscillator is weirder than before lol, the HV secondary output must be shorted. Diode is replacing potentiometer. This circuit need to be activated with touch if not combined.

Edit: Fixed image:



YouTube - 2 x sting oscillator + 1 x MJT, twice more efficient



At 1.87A at 12V input, charge 12V battery at 380mA.

1 x sting oscillator + modified joule thief efficiency = 11%
2 x sting oscillator + modified joule thief efficiency = 20%

Adding another sting oscillator improve the efficiency twice previous value. Maybe adding another sting oscillator will double again its efficiency, who knows. But this is my limit, I run out of parts.

Besides, it is rather scary to hear ringing noise to come out from the battery....


New oscillator component:
Transformer is formerly modem 9V/240V adaptor 1A. still had its core
transistor is 2SB507 PNP and TT2188 NPN.
Diode is FR3002
remember to short the secondary 240V part.

[sucahyo]

Circuit correction regarding capacitor position

I just realize that I already alter the position of the capacitor of the FWBR. I think might as well remove it all except some.

You need to experiment your self which position give the highest output. Correct position increase the output voltage twice!
edit: fixed diode connection




Here is how to combine 4 radiant circuit. P1 to P4 is the coil collapse point, between coil and transistor.

[Sephiroth]

ok, this has been bugging me from the start... there should be no current output from this circuit after a few seconds because of the capacitor next to your transformer.... that will fill up almost instantly, and as far as I can see there is no way for it to discharge...

What is the voltage reading across that cap?

[sucahyo]

What is the voltage reading across that cap?

16V. You have to remember that coil collapse current is an alternating current. This show as ripple. You can see mosfet heating page for example of the ripple.

I also remove the capacitor at other locating except one. It seems all capacitor reduce efficiency.

Removing the MJT improve efficiency:
YouTube - Better efficiency without MJT
Conventional efficiency = 38% untuned, 71% tuned. However, the last number may not be true for such high efficiency when charging a battery. The amp meter may change the circuit behaviour.

It can only be verified by using two digital meter. So, anyone can replicate this circuit and have two digital amp meter, please replicate the experiment.

New circuit now only contain two SO:


Uploaded with ImageShack.us

I also test circuit efficiency when a fan is put in series with the charged battery
YouTube - 2 X sting oscillator output powering fan and charging

input = 1.05A at 12V, output = 120mA charging 7Ah 12V and a fan in series
charger COP = 11.4%


Whoops, I just notice a major mistakes in my drawing. I will fix any image connected to MJT one.....

[sucahyo]

Actually simple



should be around 40% in conventional COP.

video and info later...

[sucahyo]

Just found out today that removing all the capacitor do not increase power consumption. However, now the output is not greater than the sum of individual run.

Single sting oscillator with torroid coil with 500mA at 12V charging at 10.5V at 170mA.

COP = 29.75%.

BTW, I am not a newbie at radiant oscillator / bedini charger. I know enough electronic to be able to calculate power from measurement. If you think you have better efficiency please post your measured input current and output current charging a battery. Also mention the source voltage and charge at what voltage. This way we can all compare and find a way to improve our circuit.

The battery never get heated up during 5 hours of charging with 170mA.

[Sephiroth]

I'll take you up on that challenge

Is there a prize?

[juju]

=D

the prize is a big hug!

eheh

sucahyo your output 10,5V is voltage on load? how you measured it?

so you have 6 watts in, 1,8 watts out...

my circuit input is 6V - 0,070A .... 0,42 watt

the output is 3.5V (with 16 led panel) on load, and 0,040A, 0,14 watt

im getting out third part the energy im putting in. 0,14 x 3 = 0,42 watt

but i dont know how to calculate cop

[Sephiroth]

the prize is a big hug!

eheh

sucahyo your output 10,5V is voltage on load? how you measured it?

so you have 6 watts in, 1,8 watts out...

my circuit input is 6V - 0,070A .... 0,42 watt

the output is 3.5V (with 16 led panel) on load, and 0,040A, 0,14 watt

im getting out third part the energy im putting in. 0,14 x 3 = 0,42 watt

but i dont know how to calculate cop

To turn an efficiency measurement into a COP you take the percentage efficiency and knock it down two decimal places, so 33% efficient becomes 0.33 COP.

By the way sucahyo, what type of meter are you using to make the measurments with? I have two types of digital true rms meters and an analogue meter I can use. I'll use the one most similar to yours

ps. Sorry to split hairs though Bedini has always stated that the extra energy only shows up in the batteries, but what Suca is measuring is the direct output. This will only give you an efficiency measurement and not a COP measurement, since no external energy should have entered the circuit at that point. I don't know if this circuit is different since it isn't a standard bedini oscillator....

[juju]

hmm

i no what you mean...

the measurements should be based on the total power/time that the battery is charging, and the total power/time that the battery can deliver after charge...

since radiant energy will only show up in the battery.

but if we are putting out current, isnt that messing with the dipole, and so breaking the effect in battery charge?

i read bedini saying that you cannot put any current in the battery to see radiant output results on it!

my voltage without load, and so 'potential voltage' is 60V.

i only tested the led panel as load, and not a battery, the charging battery will never show the real voltage onload, since you will be measuring the voltage on it, will test it later.

[Sephiroth]

I think what Bedini has said is when there is current there is no radiant...

That much is true... the radiant precedes the current... once the current starts flowing, then the radiant disappears...

This is how I see Bedini's devices:

It is theoretically possible to recover 100% of the energy that we use to form a magnetic field through it's collapse... this is in the form of standard current and voltage (watts)... so if we had an ideal inductor, we could theoretically put in 1 measurable watt of power, and get 1 measurable watt out through the field collapse.

But if we use incredibly fast switching, then there is a moment when the field is collapsing at incredible speeds because there is no current to sustain the field. Voltage is generated in a coil when it is exposed to a changing magnetic field. The faster the magnetic field is changing, the greater the voltage generated. What bedini's devices do is create a magnetic field, then the current is cut off very sharply, and the magnetic field then collapses at an incredible rate. This generates a high voltage spike that last for microseconds. This is what we refer to as the radiant, or transient spike.

However. after a very brief time, current is induced in the coil by the voltage that has been generated. We know that when current is flowing through a coil it forms a magnetic field. So once the current starts flowing in the coil it actually SUSTAINS the collapsing field and slows down the rate at which it's collapsing, so the voltage generated by the coil drops to just above the charging battery's voltage.

So as soon as we cut off the current to the coil, the radiant is generated, but almost immediately after that, current starts to flow, and this kills the radiant BUT it recovers the energy that was required to create that field in the first place....

The majority of the energy that is charging our batteries is no doubt the energy that is recovered from the collapsing field, and this is what you can measure with a good quality ammeter on the back end. But the radiant adds a bit extra from the environment...

[Sephiroth]

Two of the transistors are fried on this so it can perform better than shown...

YouTube - boring video

Input Voltage = 12.48 volts
Input Current = 1.13 amps
Input Power = 14.1 watts

Output Voltage = 12.05 volts
Output Current = 0.73 amps
Output Power = 8.8 watts

Measurable Efficiency = 62.4%

Get over here Juju!!!

[sucahyo]

Thank you guys .

sucahyo your output 10,5V is voltage on load? how you measured it?

volt meter at battery in parallel.

I think what Bedini has said is when there is current there is no radiant...

That much is true... the radiant precedes the current... once the current starts flowing, then the radiant disappears...

I wonder about that. I always found that charging speed change in proportion with output current, irrespective of the heating.

A year ago, my 555 and 2N3055 version have 300mA at output charging two nicad at an hour but with considerable heat. Now I charge the same battery with 300mA at output without any noticeable heat.

My meter is a cheap chinese one.

Measurable Efficiency = 62.4%

awesome .

If my 30% efficient circuit produce more than 2000 volt at the charging part, I wonder how high yours is without load?

With such awesome efficiency, did you notice battery heating if left charged for hours? I think two or one 1.2 nicad / nimh is a good test.

[Sephiroth]

at the C20 rate the battery won't heat up until it has reached its maximum voltage (around 16.2 volts).... and even then it is only just noticeable...

My meter is a cheap chinese one.

your efficiency might be better than you think or not... have you tried load testing the batteries?

[sucahyo]

your efficiency might be better than you think or not... have you tried load testing the batteries?

Thanks. I never do that, my battery are all SLA.

I found a way to prevent radiant destroying the SLA by adding water only recently. I manage to revive 0V SLA to 9V with load in 12 hours with 1A input. I intend to try it once I am sure that my SLA is reliable.


BTW, can your meter measure more than 2000V? how high your output voltage is without load? Put the voltmeter before turning on to prevent sparking damage.