Effects of Recirculating BEMF to Coil

[wattsup]

@gotoluc

Again nice video.

I would be curious to see what would happen if you put the output bulb in series with a small transformer (110vac primary and 12vac secondary) secondary and then put the primary back to the battery terminals via a diode. This way you could light the bulb and try and send some juice back to the batteries to see if the voltage drop will be less or even increased voltage.

Or put a diode on the bulb positive pointing to the positive of battery and put the bulb negative to the battery negative without a transformer.

wattsup

[gotoluc]

Hi Luc.

Stellar video as usual.

One thing I recommend be differentiated is the difference between "energy" and "current". Do you remember the following relationships?:

P = V x I
E = P x t

I think what you meant to say was that the series string of 9V batteries has the same "current" capability as a single 9V battery, but the voltage is raised to a higher value.

Take a look at that first equation again. You'll notice that since your voltage has increased, you will have a corresponding increase in available power, and hence energy.

So what you have shown is similar to your other recent experiments with the same coil. In this case the load power (bulb) is evenly distributed among all the 9V batteries there, so indeed you should see a much more gradual decline of any one battery's terminal voltage.

Are you beginning to get the idea that it's pretty handy to be aware of and be able to utilize a few simple equations such as the first one there? That one equation explains what you are seeing.

.99

Thanks for the reply .99

Should the 9 batteries in series last 9 time longer then using the same load on a single battery for it to be unity?

Luc

[poynt99]

Luc,

Battery chemistry is a little odd and they don't work in perfect linearity.

However, your statement is fair enough.

Definitely, the 9 batteries will last quite a bit longer than "one" if used to drive the same power into the bulb for both cases.

.99

[Kator01]

Hi Luc,

Hi Kator01,

I make no claim as I do not feel qualified

Your conclusion seems that you are.

Luc

Now this is an interesting statement. We do not talk about each other´s qualification, right ? We do talk here about efficient and correct means to exactly measuring input-power.

I have done hundrets of experiments with a similar circuit, the only difference was

1) I did not use a magnet as the magnet changes the inductance in a periodic way because the magnet is phyically vibrating like the coil in a speaker. In this way there are no stable parameters in the coil

2) I did not use a load in the recycling-circuit driven by the diode.
    instead I loaded a second capacitor and thus could exactly compare the
    input versus the output. by comparing the energy stored in the second
    elko.

3) I was using very high quality Krummer-capacitors ( 200 V, 33 000 myko Farad) which can take a few Kiloampre current when fully shorted.
http://www.krummer-kondensatoren.de/

In order for all of us to become more qualified in this specific topic ( I include myself here ) it would be a good idea to know the exact power which enters this circuit in order to get the clue why you present this here. This can be easily done by ( a slight modification of what I have proposed in my last post ) by feeding your capacitor at the left via a 0.5 Ohm resistor and measure the voltage-drop Ud across the resistor and calculate the current according to I = Ud / R. Then multiplicate this current which enters your capacitor-set with the voltage directly at the plus-terminal of the cap feeding the main switching-circuit.

Another good idea would be to use a 10 000 myko-Farad Elko for your supply-capacitor in order to eliminate any voltage-ripples so this measurement and the calculation can be done according to the above simple formula without any major error.

It is just pretty easy to do this. This will give you the exact amount of power flowing into the circuit.

Also when I watched your video I could see that the laser-light-point of your infrared-thermometer was not on the resistor but somwhere above it. Usually you have to measure at a distance of about 20 centimeter in order to get the two laserbeams overlap each other in one point. Only then will you get the correct reading.
Now using a infrared-thermo-device ( which is digitally controlled ) in such a vinicity of a strong oscillator is another big questionmark regarding correct
readings.

I hate to say this luc, but I have gone through many measurement-nightmares myself but  I had 2 very good professional teachers in my life which gradually upgraded my knowledge which is far from beeing considered an expert-knowledge. But I try my best.

Best Regards

Kator01

[poynt99]

Kator,

I have suggested (based on a recommendation from aethertech) something similar for Luc to obtain an easy Pin measurement. My post and diagram is at EF:
http://www.energeticforum.com/61265-post140.html

Here is a recent post from Aaron again espousing his nonsense:

Luc,

I don't know the status of your testing at this very moment but heard about the recommended test with the meter. It gobbles the gain. Your circuit and demonstrations are easily OU. The battery is needed to show it.

Would you care to explain WHY "it gobbles the gain" Aaron?

Good grief!

.99