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  #211  
Old 02-11-2016, 06:00 PM
gotoluc gotoluc is offline
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No need, I'm out.
Yes... and erasing all your posts with it

I wonder why

Some things are mysteries though I have seen this pattern before

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  #212  
Old 02-13-2016, 09:55 PM
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wantomake wantomake is offline
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testing trifilar coil

Ive been learning while testing with trifilar coil setup as LC tank with cap bank, led light with resister as load, and two 9 volt batteries. All in series.

I started by charging the cap bank up as much as it will. About 2 volts below battery bank amount. The led lights brightly then slowly diminish to off.

The interesting part is when a load is placed across the cap bank to discharge it the LED lights very bright. Then remove the load from cap bank, the battery voltage drops then starts to build up to original starting voltage. At same time the cap bank (5F 2.7 v x 7) starts charging up also.
Over an a period of 2-3 hours the LED deminish in brightness then turns off when both battery bank and cap bank returns to starting voltage. Over period of days both charge by tenths of volts to higher amounts.

I do need to post a schematic and video, which I will if this is important.

So is this normal for a trifilar coil???

It seems this circuit in series would discharge a power source with any type load connected to it. As a LC tank would.

Please be nice I'm just a student here.
Thanks,
wantomake
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  #213  
Old 02-14-2016, 10:16 AM
shylo shylo is offline
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Hi WTM,
I'm just a tinkerer so not any help, are you charging the cap bank ,then disconnecting it from batteries, and then attaching the led's to discharge the bank , and then repeating?
A circuit diagram and video would help to clear the air.
But the idea of a few days of work being done ,and voltages rising is very interesting to me.
Hope you post.
Thanks artv
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  #214  
Old 02-14-2016, 03:48 PM
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wantomake wantomake is offline
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Originally Posted by shylo View Post
Hi WTM,
I'm just a tinkerer so not any help, are you charging the cap bank ,then disconnecting it from batteries, and then attaching the led's to discharge the bank , and then repeating?
A circuit diagram and video would help to clear the air.
But the idea of a few days of work being done ,and voltages rising is very interesting to me.
Hope you post.
Thanks artv
Artv,
No, I leave the circuit connected in series and do what I posted.

Later today will try to post schematic and video.

Yea me too just a hobbyist slash lover of FE and really enjoy it.

wantomake
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  #215  
Old 02-14-2016, 11:14 PM
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wantomake wantomake is offline
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Ok here is a short video with the schematic.

https://www.youtube.com/watch?v=9KUlqJp1dGU

wantomake
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  #216  
Old 02-15-2016, 10:00 AM
shylo shylo is offline
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Hi WTM, A couple questions, what was the starting voltage of the batteries, and is the led connected like your schematic shows or did you just draw it wrong? If it is opposite to the drawing then the led should be lit as long as there is power supply.(caps and or batteries)
The way it is drawn you are blocking flow no? So I can't see how the caps would charge in the first place.
If it is hooked like you drew it , then I think the led is running off the back spike of the motor which is being stored in the tri-fillar. But still doesn't explain how the caps get filled up?
That's just what I think good chance I'm wrong though.
Thanks for sharing.
artv
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  #217  
Old 02-15-2016, 12:49 PM
gyula gyula is offline
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Hi Wantomake,

I agree with shylo, the LED should normally be flipped in the schematic drawing so that the 17.2 V battery could charge up the capacitor bank at all, as per your schematic, that is.

If you started your tests with a pre-charged cap bank, then depending on how high voltage you charged up the series capacitors with respect to the two series batteries, the LED should be able to light whenever the voltage difference between the cap bank and the batteries is equal or higher than the forward voltage drop of the LED.

If you started your tests with a more or less empty cap bank, then the batteries started to charge the caps up via the LED, the 237 Ohm and the trifilar coils provided the the LED was flipped with respect to what shown in the schematic. This involved the LED had to light brightly during this first charge up process (no motor load yet).

Now, as you started the video, the cap bank had 16.32 V and the battery probably had the 17.13 V rest voltage, the difference is 0.81 V, this latter voltage is well below the 2.8-3V threshold forward voltage of the LED to emit a faint visible light so it remained dark.

Then you connected the DC motor to the cap bank, it started to discharge the bank, hence the voltage difference increased to 3.96V between the batteries and the bank (17.12V-13.16V). This is already a decent forward voltage for the LED to bright strongly, the current is limited only by the 237 Ohm and the coils DC resistance of 14.2 Ohm, both are in series (neglecting batteries input resistances).

I think if you used a load resistor of a few Ohms to start discharging the cap bank (instead of using the DC motor), the same process would take place, the LED would light up brightly as the voltage difference would dictate.


The highest current in your closed circuit (when no motor load is connected and the cap bank is fully discharged) is determined by the 237+14.2=251.2 Ohm and the 17.13 V battery voltage, this current is roughly 17.13/251.2=68.2 mA. This current decreases as the cap bank is charging up from the empty or near empty condition.

Connecting the motor, the brushes surely produce voltage spikes across the cap bank but these spikes are mostly dumped in the bank and the trifilar coils and the batteries may not benefit much from them. A scope across the cap bank should show this of course. But normally a very high capacitor value like this cap bank represents strongly prevents any high spike amplitude across it, just like a big inductance works against any current change by default.

So I mean the voltage level of the batteries is a normal chemical effect when it increases back to its rest voltage after you switched the motor off. And the cap bank is being charged up back to its 16.3V level via the LED, the resistor and the coils. the 0.81 V difference (you say as near to 1V in the video) is set, this is the minimal leaking current in the forward direction via the LED, causing that voltage difference.

Gyula
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  #218  
Old 02-15-2016, 12:52 PM
gyula gyula is offline
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Quote:
Originally Posted by shylo View Post

....
If it is hooked like you drew it , then I think the led is running off the back spike of the motor which is being stored in the tri-fillar. But still doesn't explain how the caps get filled up?
....

artv
Hi Artv,

Notice that the LED remains lit when he removes the motor, so the LED cannot operate from the motor back spikes. See my other thoughts above.

Gyula
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  #219  
Old 02-15-2016, 01:16 PM
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wantomake wantomake is offline
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schematic

Quote:
Originally Posted by shylo View Post
Hi WTM, A couple questions, what was the starting voltage of the batteries, and is the led connected like your schematic shows or did you just draw it wrong? If it is opposite to the drawing then the led should be lit as long as there is power supply.(caps and or batteries)
The way it is drawn you are blocking flow no? So I can't see how the caps would charge in the first place.
If it is hooked like you drew it , then I think the led is running off the back spike of the motor which is being stored in the tri-fillar. But still doesn't explain how the caps get filled up?
That's just what I think good chance I'm wrong though.
Thanks for sharing.
artv
Artv,
Starting voltage:
17.36 batteries
16.56 for capacitors
No the LED type I'm using can connect either direction . I only used the electronic symbol to draw the circuit. If backwards then my bad. The motor was only used to draw down the cap bank. Other type loads have been used to temporarily draw down the voltage with same resuls. I even left the shop 4 ft lighs off because the coil can pick up voltage from them. Been fooled there before.

Just thought this circuit was interesting using a trifilar coil. I did try single and bifilar coil connections, but that wouldn't recharge the battery bank.

Thanks for your interest and thoughts.
wantomake
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Last edited by wantomake; 02-15-2016 at 01:18 PM.
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  #220  
Old 02-15-2016, 01:49 PM
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wantomake wantomake is offline
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not sure

Gyula,
Sorry didn't see your post until after my post.

The resting voltage is not what the video shows. But the LED comes with the tiny resister attached and I believe I've connected it both ways. Will recheck that.

But leaking voltage is the culprit here? I did measure across the led and coil as the batts and caps charged and it was 3 - 5 or so volts and decreasing a slow discharge until the banks recharged.

So where is the extra voltage coming from?? I've unplugged the caps and the batteries with no return to resting voltage. The caps discharge as they sit there. So that is why I thought to post this so someone with more knowledge than I could explain this.

Ok, thanks then.
wantomake
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  #221  
Old 02-15-2016, 02:02 PM
shylo shylo is offline
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Hi Gyula, Thanks for the explanation, that makes more sense than what I was thinking.
WTM, every led that I have ever used has always needed to use the correct polarity.
Your coil can pick-up voltage from your lights with no connection? Can you connect a cap to the coil leads and the cap will charge up?
If so that would be awesome.
artv
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  #222  
Old 02-15-2016, 03:25 PM
gyula gyula is offline
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Hi Wantomake,

Okay on the 17.36 V battery and 16.56 V cap bank voltages, the difference is 0.8 V. On leaking current via the LED I mean the following: the cap bank from its 16.56 V level slowly self discharges when left alone (disconnected) and the battery is is able to supply this current via the LED when the cap bank is in the circuit (this is just some microAmper maximum) while the voltage drop in the forward direction across the LED is just 0.8 V, and this small forward bias is normally not enough to emit visible light. This tiny microAmper is what I meant on leaking current via the LED to supply the self discharging loss of the cap bank.

When you discharge the cap bank, the 0.8 V difference invariably increases because the cap bank voltage reduces as your voltmeter showed, this must be the 3 - 5 V or so you measured in the process. (The battery voltage nearly remains the same.) The difference increases as the cap bank loses charge due to the motor or other load and the difference decreases when the load is removed from the cap bank and the battery starts recharging it.

So what do you mean by extra voltage? If you mean the recovery of the unplugged batteries (some ten to some hundred mV) I think that is normal for both the rechargable and the alkaline batteries, depending mainly on their age / usage. The capacitor bank is also able to recover some hundred mV when they were charged up earlier but got discharged from say 16.5 to 14 V and you disconnect them completely (the dielectric material in the capacitors has a 'memory' effect, they have been stressed by a higher voltage and then this voltage disappeared by the discharge). When the cap bank is included in the circuit, the batteries are able to charge them up via the LED, starting from the 3 to 5 V difference till this difference settles at 0.8V and the LED becomes dark. If you still have questions, ask.

Gyula
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Last edited by gyula; 02-15-2016 at 03:35 PM. Reason: correction
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  #223  
Old 02-15-2016, 03:27 PM
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wantomake wantomake is offline
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my bad

Artv,
What I should've said in above post is that I did turn the lights off to test whether the coil picks up any residual voltage from the lights. This one doesn't. This LED is from Christmas light cord and I drew the circuit wrong so my mistake there.

I've been learning more about coils to finish the Lockridge motor/generator that sits on my bench for last year or so. I believe a capacitor coil mechanism is the answer to the energy storage needed to complete this project. It's been very interesting but frustrating at times.

Anyway, still learning.
wantomake
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  #224  
Old 02-15-2016, 04:05 PM
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wantomake wantomake is offline
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Too many questions

Quote:
Originally Posted by gyula View Post
Hi Wantomake,

Okay on the 17.36 V battery and 16.56 V cap bank voltages, the difference is 0.8 V. On leaking current via the LED I mean the following: the cap bank from its 16.56 V level slowly self discharges when left alone (disconnected) and the battery is is able to supply this current via the LED when the cap bank is in the circuit (this is just some microAmper maximum) while the voltage drop in the forward direction across the LED is just 0.8 V, and this small forward bias is normally not enough to emit visible light. This tiny microAmper is what I meant on leaking current via the LED to supply the self discharging loss of the cap bank.

When you discharge the cap bank, the 0.8 V difference invariably increases because the cap bank voltage reduces as your voltmeter showed, this must be the 3 - 5 V or so you measured in the process. (The battery voltage nearly remains the same.) The difference increases as the cap bank loses charge due to the motor or other load and the difference decreases when the load is removed from the cap bank and the battery starts recharging it.

So what do you mean by extra voltage? If you mean the recovery of the unplugged batteries (some ten to some hundred mV) I think that is normal for both the rechargable and the alkaline batteries, depending mainly on their age / usage. The capacitor bank is also able to recover some hundred mV when they were charged up earlier but got discharged from say 16.5 to 14 V and you disconnect them completely (the dielectric material in the capacitors has a 'memory' effect, they have been stressed by a higher voltage and then this voltage disappeared by the discharge). When the cap bank is included in the circuit, the batteries are able to charge them up via the LED, starting from the 3 to 5 V difference till this difference settles at 0.8V and the LED becomes dark. If you still have questions, ask.

Gyula
What I mean by extra voltage is this.

Why aren't the batteries being slowly drained by the load of the led?

I understand the 'memory' of the batteries and the dialectic of the caps. But if connected in series a battery, led, and small resistor would it not discharge the battery? I thought it would? So the resistance of the combined coil and resistor plus the slow leakage of the led will keep the battery and caps from discharging??

So the led turns off after the caps are charged up to max memory and the batteries nor the caps will never drain or loose charge??

Thanks for the help,
wantomake
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Last edited by wantomake; 02-15-2016 at 04:21 PM.
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  #225  
Old 02-15-2016, 05:19 PM
gyula gyula is offline
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Well, if you let the circuit setup in the ON state as you showed it at the start of the video (no motor load and the voltage difference is 0.8V), the batteries would keep charging the cap bank for a very, very long time because the some nano or some microAmper load current for the batteries via the LED towards the cap bank is not a real load for the batteries. Several days or even weeks may pass when you could measure some mV voltage drop from the batteries under this continuous supply of charge current into the cap bank.

Remember: the two batteries are the main active source of current in your setup, the cap bank is also a load for the batteries. If you by-pass (i.e. short circuit) the LED, the resistor and the trifilar coils in your setup with a piece of wire, then the cap bank would have the same voltage across it as the batteries have but the some nano or microAmper charge current would still go into the cap bank from the batteries because of the self discharge nature of those caps. However, this tiny load current is quasi nothing for the two batteries for a long time.

You wrote: "So the resistance of the combined coil and resistor plus the slow leakage of the led will keep the battery and caps from discharging??"

My answer is: it is the batteries which would keep the capacitor bank from discharging, okay? The LED and the series resistors are simple current limiting components in this process.

You wrote: "So the led turns off after the caps are charged up to max memory and the batteries nor the caps will never drain or loose charge??"

My answer is: On the long run the batteries themselves would also discharge of course.

Let's put this process in another view: suppose you replace the LED with a 15 V Zener diode in your circuit. This diode would maintain a continuous 15 V drop across itself (this is its job, right?) while the charging current would still flow towards the cap bank from the batteries but this time the cap bank would be able to charge up to 17.36-15=2.36 V only, okay? And in case you would keep this circuit working for a long time (no load across the cap bank, just its self-discharging), the moment the battery voltage would go down below 15 V (after some months) the cap bank would discharge to near zero voltage, understand? Should you test this, use a current limiting series resistor for the Zener because it may get burnt from the intial high current if the cap bank had zero voltage in it at switch-on.

Gyula
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  #226  
Old 02-15-2016, 06:03 PM
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wantomake wantomake is offline
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very informative

Thanks Gyula,
Good information. I do have some zener diodes and will give it a try.

Plus will leave it in on position for number of days until battery starts to discharge.

Thanks again,
wantomake
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  #227  
Old 10-15-2018, 09:42 AM
Deuis Deuis is offline
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Coil output results.

I have been testing some 8 filar 23awg bobbins with iron cores.
Using some very basic methods and a prime mover I achieved the following results. Running 4 of these coils in parallel I can peak the BUS at 120VDC.
Ive been running 120 Led's while losing 20RPM but also using less current on the prime mover with a better secondary recharge rate.
I have been adjusting and tinkering with these coils and currently peaking at .6w per coil.

sd.JPG
9 coil.jpg


I have also tested an 8 filar 18awg with 2"x1"x4" laminated core.
I've only been able to get 5V out of this.
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  #228  
Old 10-26-2018, 09:40 PM
asollid asollid is offline
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I have watched all of Angus wangus youtubes. He go to great lengths to explain that coils are OPPOSITELY WOUND. Doesnt that negate Lenz in the core?
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