Your Basic Coil

[wantomake]

Too many questions

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

[gyula]

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

[wantomake]

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

[Deuis]

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.

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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.

[asollid]

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?