Your Basic Coil

[gyula]

In your studies with the Bifilar coils inductance did you also slightly
twist the 2 bifilar conductors like we were instructed with the SSG?
I would be interested to know if your measurements
(if you did that experiment) showed more or less inductance
for twisted bifilar vs just putting conductor side by side without
twisting them.
....

Hi BroMikey,

Well, I mostly twisted the wires for building transmission line transformers back then and never compared inductances with bifilar coils where the wires were guided side by side without any twist.

Just out of curiosity, I have built a single layer, air cored bifilar coil from enamelled copper wires putting them side by side (no twist) on a 3 cm OD paper bobbin, wire length was 2 x 210 cm from wire OD=1 mm (AWG 18), it gave 21 turns. See attached picture 1.
Measured inductance between Start 1 and End 1 is 8 uH and also 8 uH between Start 2 and End 2.
Measured inductance between Start 1 and End 2 (the two windings were connected in series aiding phase) was 30 uH.
The ratio is 30 uH/8 uH= 3.75 this is close to 4.
Measured capacitance is 270 pF between either Sart 1 and Start 2 or between End 1 and End 2, (the not being measured wires were freely floating, of course).

Then I unwound this coil and twisted the same two wires, roughly by 2 twists per inch and rewound it onto the same bobbin, to have 21 turns again. See attached picture 2.
The measured capacitance was only 230 pF between Start 1 and Start 2 or between End 1 and End 2. But this smaller capacitance versus the above case stands to reason because the angled wires could not have as many facing surfaces as they had in the side by side, very close guided to each other case.
The received inductance between Start 1 and End 1 was again 8 uH as was also between Start 2 and End 2. Likewise, the measured inductance was 30 uH between Start 1 and End 2 (winding were in series aiding phase).

IF somebody has confusion what a bifilarly wound coil connected in series aiding phase means, well it means that the end of the first wire is connected to the start of the second wire if you label the starts of the two parallel guided wires as start 1 and 2 and you label the ends of the two wires as end 1 and end 2. This means also that you have to prepare in advance two wires of appropiate length from two different rolls and always guide them in parallel while making the bifilar coil. (Of course you can cut twice the needed length of wire from one wire roll too and then guide the wires also in parallel while winding.)

The series aiding phase connection is the same you show in the above picture.

So the answer to your question is that with this single layer air core bifilar coil there is no difference in inductance between a twisted and the side by side guided (not twisted) winding styles. IT is possible that for a multilayer bifilar coil the twisting introduces some change in inductance because the capacitance between the twisted wires will be different from that of the side by side wires. The capacitance between the two wires is transformed in parallel with the bifilar coil so it has the tendency of reducing the overall inductance. Notice also that using ferromagnetic cores the capacitance also changes between the two bifilar wires (usually it increases the capacitance between the two wires).

Further measurements on my bifilar coil:

1) Start 1 and Start 2 are connected together and End 1 and End 2 are also connected together: inductance was 9 uH i.e the same as any one of the individual windings. Such connection reduces copper loss only, as if you had used thicker wire for the coil.

2) Start 1 and Start 2 are connected together, the inductance between End 1 and End 2 was less than 1 uH (resolution limit of my L meter cannot let it see more precisely). This connection represents a non inductive bifilar winding shown in the wiki link.

Regarding the counter-wound measurements you refer to in the wiki link, I believe in that case one winding is wound say in clockwise the other is in counter closckwise direction, I did not make such windings.
Here is a good explanation for two mutually coupled coils which can help estimating the answers:
Mutual Inductance of Two Adjacent Inductive Coils

Gyula

[BroMikey]

Hi BroMikey,

Just out of curiosity, I have built a single layer, air cored bifilar coil from enamelled copper wires putting them side by side (no twist) on a 3 cm OD paper bobbin, wire length was 2 x 210 cm from wire OD=1 mm (AWG 18), it gave 21 turns. See attached picture 1. .....................

Gyula

Thank you that answers my question, great teaching for simple
step by step understanding of coils and their possible variations.

[gotoluc]

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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

Luc

[wantomake]

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

[shylo]

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

[wantomake]

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

[wantomake]

Ok here is a short video with the schematic.

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

wantomake

[shylo]

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

[gyula]

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

[gyula]

....
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

[wantomake]

schematic

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

[wantomake]

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

[shylo]

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

[gyula]

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

[wantomake]

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