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ron_o
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Posted - March 02 2013 :  16:47:38  Show Profile Send ron_o a Private Message  Reply with Quote
This will be a useful ( hopefully ) guide to inductors. It will follow the same format as my capacitor guide where practical applications are given and theory ( unless requested ) will be kept to a minimum.

Any questions / suggestions welcome

ron


Edited by - ron_o on March 04 2013 16:02:01

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ron_o
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Posted - March 04 2013 :  15:27:00  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 1 AC Generator Coil Design

A simple equation approximates the maximum output that can be expected from a generator coil

Vmax = ( number of coil turns x coil area x number of magnet sets x magnet strength x revolution per second ) / 2

where :-

coil area is measured in square metres

magnet strength ( per single magnet ) measured in Tesla's

revolutions per second = rpm / 60

Vmax = Vrms x square root of 2 .... reason is Digital meters display rms value not peak ( max ) value of AC voltage

Ok , now for a practical example

required Vrms = 10 volts so Vmax = 10 x 1.414 = 14.14 volts peak AC

number of magnet sets = 8 ( arranged N S N S etc i.e 8 North and 8 South )

magnet strength = 1.5 Tesla

rotor travels at 500 rpm so it complete 500 / 60 rotations per second = 8.33'

the diameter of the coil is 5 cm ( 0.05 metres ) so the coil area = 0.025 x 0.025 x 3.142 = 1.96 x 10 exp -3 approx

to determine the number of coil turns we re-arrange the equation , thus

number of turns = Vmax x 2 / ( coil area x number of magnet sets x revolutions per second x magnet strength )

number of turns = 14.14 x 2 / ( 1.96 exp-3 x 8 x 8.33 x 1.5 )

number of turns = 28.28 / 0.1959 = 144 turns approx

ron


Edited by - ron_o on September 13 2013 18:16:21
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ron_o
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United Kingdom
1052 Posts

Posted - March 04 2013 :  16:00:59  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 2 Maximum Inductance for a given length of wire

This is achieved by following the "Brooks Coil Dimensions" , which are :-

00000----------00000 -
00000----------00000 A
00000----------00000 -

|-A-|--- 2xA---|-A-|

A circular coil that has windings with a square cross section and a centre "hole" that is twice the thickness = height of the winding

Carrying on from Topic 1 which had a coil of 144 turns with a outer diameter of 5 cm

Dimension "A" will be equal to 5 cm / ( A + A + A + A ) units

A = 5 / 4A = 1.25cm

so the centre = 2 X 1.25 cm = 2.5 cm and each square section is 1.25 by 1.25 cm

we now need to find the "average" diameter of the winding

Average winding diameter = centre diameter + width of one winding

average diameter = 2.5 + 1.25 = 3.75 cm

Average turn length = average diameter X 3.142

Average turn length = 3.75cm x 3.142 = 11.78cm

Approx wire length = average turn length x number of turns

Approx wire length = 11.78 x 144 = 1696 cm

to be continued ....

ron
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ron_o
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1052 Posts

Posted - March 04 2013 :  18:26:06  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 2 Maximum Inductance for a given length of wire.. Continued

The approx wire gauge can now be calculated from :-

Wire diameter ( including insulation ) in cm = length "A" / square root of number of turns

so from the above values

wire diameter ( cm ) = 1.25cm / square root of 144

wire diameter (cm ) = 1.25cm / 12 = 1 mm approx diameter

from wire gauge tables 19 AWG would be suitable , now just to double check

19 AWG wind 26.8 turns per inch or 26.8 / 2.54 turns per cm = 10.55

since dimension "A" is 1.25 cm we can wind 10.55 x 1.25 = 13.2 turns per linear cm

The approx DC resistance will be ( tables say 19 AWG = 8.04 ohms per 1000 ft )

Ohms per cm = 8.04 / ( 1000 x 12 x 2.54 )

Ohms per cm = 2.63 exp -4

So our coil had an approx wire length of 1696 cm , therefore 1696 x 2.63 exp -4 = 0.446 ohms approx.

PLEASE NOTE although the Brooks Coil dimentions could be used as a generator coil design the thickness of the overall winding reduces the efficiency of the coil due to the distance of the outermost winding and the magnet.


ron
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TinMan
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4082 Posts

Posted - March 04 2013 :  22:39:18  Show Profile Send TinMan a Private Message  Reply with Quote
Hi Ron
Thanks for taking the time to present this to us.
I have to ask though-is this with or without a core?,or dose it not matter.

swim at 90 degrees to the current and gain speed in two directions

skype-thetinman.69
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The_Architect
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Posted - March 04 2013 :  23:22:41  Show Profile  Visit The_Architect's Homepage Send The_Architect a Private Message  Reply with Quote
I would guess the core helps to drag more of the magnetic field down for full penetration of the coil. I am kinda wondering if an occasional wrap of sheet steel like used in the core would help with this in outer layers of the coil, to get better saturation of all levels of the wire from the ones closest to the magnet to the coils furthest from the magnet.

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ron_o
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United Kingdom
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Posted - March 05 2013 :  14:57:54  Show Profile Send ron_o a Private Message  Reply with Quote
Hi TinMan & The Architect

The equations / calculations above are normally applied to "air cored" coils , however if you introduce a suitable core the efficiency will increase ( higher output from coil ). The "core" material must be able to magnetise and demagnetise very quickly because of the alternating magnetic poles ( ferrite is a good example ).

Obviously magnetic drag increases with the introduction of a core.

Introduction of laminates between the winding :- never tried it but "classical" thought would suggest ( to me anyway ) that it would not be benificial.

To the best of my knowledge the core needs to act as "one" so that the change in magnetism is uniform.

ron
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49er
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USA
4442 Posts

Posted - March 05 2013 :  20:52:27  Show Profile Send 49er a Private Message  Reply with Quote
Hi Ron

Very good job. I did almost understand it all and I will study more.

Doug
The sky is not the limit...There are footprints on the MOON.
Your only as DUMB as where your standing.
No matter where you go there you are.
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The_Architect
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USA
327 Posts

Posted - March 05 2013 :  23:11:16  Show Profile  Visit The_Architect's Homepage Send The_Architect a Private Message  Reply with Quote
I posted in the challenge thread that I have some metGlas 2714M and 2705M which I am wondering if it is to conductive, and might concentrate the magnetic field right through the coils leaving little flash out on the sides to penetrate the coils too. any thoughts on that? zero magnetic restriction,

any info on what those numbers mean and how they would effect a coil's effectiveness?

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ron_o
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Posted - March 06 2013 :  18:57:29  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Doug

Well hopefully i can continue to keep your intrest in this topic / thread especially when we look at optimum windings for SSG type coils.

ron
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ron_o
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United Kingdom
1052 Posts

Posted - March 06 2013 :  19:12:36  Show Profile Send ron_o a Private Message  Reply with Quote
Hi The Architect

I'm not familiar with the grades of metglas that you have quoted , did try a google search , but no luck. have youany additional info ( manufacturer etc )

ron

Located it ! , looks very promising

Edited by - ron_o on March 06 2013 19:29:51
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The_Architect
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Posted - March 07 2013 :  10:49:47  Show Profile  Visit The_Architect's Homepage Send The_Architect a Private Message  Reply with Quote
sure their site has all the info on it You must be logged in to see this link. has all of their materials for building inductor and em shielding material. *the stuff I have they told me could be laminated to make a core, but it is mainly made for shielding cables from higher power em fields*

I have though a question that is not pertaining to inductors or capacitors (other wise I would have asked over on the capacitor thread in this section :) ) instead if has to do with diodes. If you have any info on them, we can go to PM or start another thread, my question is pretty simple, and does pertain to a device that I have questions regarding the air capacitors though so I can always bundle it too if you want over on the capacitor thread with it.

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ron_o
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Posted - March 07 2013 :  14:50:50  Show Profile Send ron_o a Private Message  Reply with Quote
Hi The Architect

Since diodes have capacitance that can be varied by application of a DC voltage i would suggest the "Learning about Capacitors" topic , please.

ron
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ron_o
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United Kingdom
1052 Posts

Posted - March 07 2013 :  16:28:18  Show Profile Send ron_o a Private Message  Reply with Quote

Back to Vespers "coil(s)"

I tend to write my "notes about "whatever" directly / on the fly , so to speak. After re-reading my own postings i noticed this comment from Vesper :-


"Hum Coil testing alone......
Now on to the testing the RPM is 500 rpm and the coils are a max of 100 turns but as you will see and hear is the testing, what we are testing is 18 feet of wire to produce the voltage and amps"

So lets recalculate using this additional information:-

Firstly a circle offers the maximum enclosed area for a given length of wire

18ft x 12 x 2.54 = 548cm approx

Average turn length = 548 / 100 = 5.48 cm

therefore average diameter = 5.48 / 3.142 = 1.74 cm and since some turns will have a smaller dia and some will have a larger diameter ( to keep the average value ) lets say that the maximum diameter is 2 cm

coil area for a 2 cm dia coil is 3.142 x radius squared = 3.142 x 1 x 1 = 3.142 cm2 ( 3.142 exp-4 square metres )

re-arrange the initial equation to yield magnetic strength in Tesla

magnet strength = 2 x Vmax / # coil turns x coil area x # magnets x rpm/60

using the initial Vmax of 14.14 volts

magnet strength = 2 x 14.14 / 100 x 3.142exp-4 x 8 x (500 / 60 )

magnet strength = 28.28 / 3.142exp-2 x 8 x 8.33 = 15.5 Tesla ( thats one very powerful magnet ! )


Ok , now lets say the coil has 50 turns

average turn length = 548 / 50 = 10.96 cm

average turn diameter = 10.96 / 3.142 = 3.488 cm

as before lets round this up to 3.75 cm as our maximum diameter

coil area = 3.142 x ( 3.75 / 2 )^2 = 1.1 exp-3 square metres

to save my typing :-

28.28 / 1.1 exp-3 x 8 x 8.33 x 100 = 3.858 Tesla ( a more "reasonable" strength magnet )

you should by now see where this is heading........

ron

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The_Architect
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USA
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Posted - March 09 2013 :  08:32:58  Show Profile  Visit The_Architect's Homepage Send The_Architect a Private Message  Reply with Quote
I have to thank you for all of this Ron_O, this came just in time for me, and I may have some inductor coil questions for you soon as well :)

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ron_o
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Posted - March 09 2013 :  14:40:32  Show Profile Send ron_o a Private Message  Reply with Quote
Just ask them , i will always try and answer them

ton
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ron_o
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United Kingdom
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Posted - March 09 2013 :  18:45:42  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 3 Electromagnet :- Optimum dimensions for Maximum Core Field Strength

if the diameter of the core is "c " units then the height(length) of the winding will be 3 x "c" units and the depth of the winding will be 2 x "c" units ( ie the overall width of the coil will be 2 x "c" + "c" + 2 x "c" ).

Windings settlement

the overall depth of a winding where layers "bed" into each other is given by :-

winding depth = ((( number of layers - 1 ) x wire diameter x 0.866 ) + wire diameter )

ron
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ron_o
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United Kingdom
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Posted - March 11 2013 :  20:10:22  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 3 Electromagnet :- Optimum dimensions for Maximum Core Field Strength PART 2

This example uses the equation in part 1 of this topic

Let dimension "c" = 3/4 inches , therefore the height of the coil will be 3 x 3/4 inches = 2.25 inches and the winding depth = 2 x 3/4 inches = 1.5 inches

using 18 AWG enamelled copper wire which winds at 23.9 turns per inch we get :-

23.9 x 2.25 = 53 turns ( call this "b" )

and

23.9 x 1.5 = 35 turns ( call this "d" )

total number of turns = 53 x 35 = 1855

average turn length = mean diameter x 3.142 ( note mean diameter is 2,25 inches in this example )

total wire used = average turn length x number of turns = 2.25 x 3.142 x 1855 = 13112 inches = 1093 feet

from wire tables 18AWG wire has a resstance of 6.386 ohms per 1000 feet ,

therefore coil dc resistance = 6.386 x 1093 / 1000 = 6.979 ohms ( approx 7 ohms )

ron

i have not taken windings "bedding" into each other , see next section

Edited by - ron_o on March 11 2013 20:12:32
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ron_o
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Posted - March 12 2013 :  17:15:50  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 3 Electromagnet :- Optimum dimensions for Maximum Core Field Strength PART 3

Sidestepping "slightly" lets now look at the inductance of this "special" coil , using Wheelers equation we get :-

Inductance ( microhenrys ) = 0.2 x 2.25^2 x 1855^2 / ( 3 x 2.5 + 9 x 2.25 + 10 x 1.5 )

Inductance ( microhenrys ) = 82953

Inductance ( millihenrys ) = 83 approx

This value of inductance is of course for an air core , introduction of a suitabe core ( ferromagnetic ) will increase the inductance

ron
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ron_o
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Posted - March 13 2013 :  16:50:06  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 3 Electromagnet :- Optimum dimensions for Maximum Core Field Strength PART 4

How strong is our "special" electromagnet ?

Current = 1.7 amps , air gap = 0.001 metres

F(Newtons) = ( # turns x Amps )^2 x 4 x Pi exp-7 x core area(m^2) / 2 x air gap^2

F(Newtons) = ( 1855 x 1.7 amps )^2 x 4 x 3.142 exp-7 x 0.000285 / 2 x .001^2

F(Newtons) = 1806.6

this correspond to a lifting capacity of 184 Kg

ron
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Luciano
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213 Posts

Posted - March 22 2013 :  01:40:58  Show Profile Send Luciano a Private Message  Reply with Quote
Great job ron. Very impressed by your calculations.
1 remark and one question.
Remark: After reading this, and one of the posts on page 1, i just tried to "wrap" a small air core coil into a sheet of steel ( i had some 0.1 mm steel foil lying around)
as my setup is not running yet, i just measured with my lcr meter. Inductance went up from 1.34 to 1.48 mH with the wrapping (1 layer) outside the coil.
I will definetly run some tests when my machine is up and running.

And here my question: Is there a way of calculating the ideal size of a coil for maximum inductance when you wind bi, tri or quadfiler??
and talking bifiler, induction wise speaking, is it better to have a bifiler or a 2 layer coil (if you calculate the width and height of the layered windings to get optimum inductance)?

Sorry, but the more i read, the more questions I have. And i posted this, while i was still reading on page 1. LOL

Luc

Edited by - Luciano on March 22 2013 01:46:54
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ron_o
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Posted - March 22 2013 :  18:54:48  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luciano

The inductance of any inductor will be increase by the presence of a ferromagnet material , conversly the presence of a brass core will lower the inductance.

A "rough" method for a trifilar coil is as follows:-

C(trifilar) = squareroot ( total number of turns ) x diameter of one piece of wire x 1.75

example 144 turns using 1 mm dia wire

C(trifilar) = squareroot 144 x 1mm x 1.75 = 21 mm


Again a "rough" calculation for quadfilar , use same equation but replace 1.75 by 2

On bifilar / 2 layer winding question .... very little difference on small coils so its which ever is the easier to wind

ron
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The_Architect
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USA
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Posted - March 22 2013 :  22:27:28  Show Profile  Visit The_Architect's Homepage Send The_Architect a Private Message  Reply with Quote
if brass lowers the inductance, is it expected then that bismuth would further lower this?

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ron_o
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Posted - March 23 2013 :  17:27:49  Show Profile Send ron_o a Private Message  Reply with Quote
Good question ( now i wonder why you should mention Bismuth , as if i cannot guess , lol ).

The general theory is that metals which are good conductors of electricity ( but are non-magnetic ) will lower the inductance of an inductor. Bismuth is a bit of an "oddball" , it is a metal but it's dimagnetic properties give it some very intresting properties ( superconductors ) . If it did not have such dimagnetic properties i would say , based on its electrical conductivity that it would not be as good as brass but in light of its other properties i would guess that it would be better than brass.

ron
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Luciano
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213 Posts

Posted - April 17 2013 :  17:31:53  Show Profile Send Luciano a Private Message  Reply with Quote
Sorry if i come up with another question, but i was wondering if i could use your calculation for the optimal electro magnet above and reverse them.
if i know the newton force of a permanent magnet, i could calculate a coil (core diameter, wire gauge, windings etc...) to be able to exactly produce a counter force so it would neutralize the permanent magnet?
the idea is:
A: permanent rotor magnet is attracted to coil core and moves the rotor in direction of the coil...
B: when it reaches the core the coil is just enough energized so it would newtralize the attraction.
C: the magnet continues its journey to the next coil, who's core starts attracting it again..
Dont know if it works, but i thought by reversing the calculations above, and making the core the same diameter as the permanent magnet, it would be possible to calculate a counter magnet coil, that would just neutralize the attraction from the core. the attraction to the core would actually make the rotor turn.
(maybe the idea is not so clever, but i will give it a try IF its possible)
Luc
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ron_o
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Posted - April 17 2013 :  19:06:10  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luciano

Yes you could use the equation "in reverse" . I would prefer the word "balance" out the permanent magnet.

I know exactly what you trying to achieve , i have done the calculations myself .

You use a permanent magnet to bias the core so that the rotor magnet does not "see" the iron core of the electromagnet.

ron
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ron_o
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Posted - April 17 2013 :  19:09:51  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luciano

Look at my topic ( in this section ) "Electromagnet Calculations" , they may be of use to you.

ron
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ron_o
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United Kingdom
1052 Posts

Posted - July 01 2013 :  15:16:33  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Single layer coil

The inductance of a single layer coil ( in microhenrys uH ) is given by the following equation:-

Inductance(L) = 0.2 x N^2 x d^2 / ( 3.5 x d + 8 x l )

where:-
N = number of coil turns
d = diameter of coil ( this should be measured to the centre of the wire in inches )
l = length of winding in inches

ron
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ron_o
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Posted - July 01 2013 :  15:41:52  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Transformers

Following relate to sine wave drive waveform

Firstly a few facts:-

1) the voltage ratio is equal to the turns ratio

2) the current ratio is inversly proportional to the turns ratio

So what do these two statements mean ?

1) voltage ratio , if the primary winding has 100 turns and the secondary winding also has 100 turns then ( ignore losses at the moment ) a 20 volt AC supply on the primary will produce a 20 volt AC supply on the secondary ( used for isolation purposes etc )
And likewise if 10 amps of current was available to "drive or power " the transformer then ( ignoring losses ) 10 amps of current could be drawn from the secondary.

Now lets alter our design this time the primary remains at 100 turns but the secondary now has only 50 turns ( once again ignore losses ). So now if we drive our transformer with 20 volts AC we now get 50 / 100 ( turns ratio 2 : 1 ) times the input voltage and thus we get an output of 10 volts BUT since the current follows the inverse rule we can now draw 100 / 50 x 10 amps = 20 amps

Of course in the real world losses due to resistance of wire and flux linkage losses between primary and secondary cause a reduction in output from the transformer

3) iron or a suitable alloy within the magnetic path of a transformer greatly increases the flux density ( this is due to an increase in relative permeability )

Primary inductance Lpri = Uo x Ur x Np^2 x A / l

to be continued

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ron_o
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Posted - July 01 2013 :  18:14:29  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors : Transformers Pt2

Lpri = primary inductance
Uo = permeability of free space ( asin an air cored coil )
Ur = permeability of core material
Np = number of turns on primary
A = cross sectional area of coil ( in square metres )
l = length of coil ( in metres )

Likewise for inductance of secondary

The overall inductance is directly proportional to the relative permeability of the core material


the Mutal inductance = K x square root (inductance primary x inductance secondary )

Flux density = Uo x Ur x Np x Ip x A / l ( answer in Webbers)

Ip = current in amps in primary winding

Np = number of turns in primary

K is the "coefficient of coupling" it as a value that is always less than one

rms value of secondary voltage (esec) = 4.44 x N x flux density x frequency in Hz volts

ron

forgot to say what "K" was

Edited by - ron_o on July 02 2013 17:49:17
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ron_o
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Posted - July 02 2013 :  15:41:06  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Solenoid Coil , inductance calculation ( air cored )

The following calculation is based on a coil consisting of 200 turns of 18AWG enamelled copper wire close wound on a 1 inch diameter former.

18 AWG winds 23.9 turns per inch so the winding length is 200 / 23.9 = 8.37 inches

The equation :-

Inductance(L) = 0.2 x N^2 x d^2 / ( 3.5 x d + 8 x l ) answer in microhenrys

L = 0.2 x 200 x 200 x 1 / ( 3.5 + 8 x 8.37 )

L = 8000 / ( 3.5 + 66.96 )

L = 8000 / 70.46

L = 113.54 microhenrys

Please note this answer will be within 10% of the actual measured value of such a coil
( this error is due to a combination of factors )

Repeating the calculation but using a different equation ( metric system :- measured in metres )

L = Uo x N^2 x A / l answer in Henrys

L = Pi x 4exp-7 x 200 x 200 x Pi x ( 1.25exp-2)^2 / 0.2126

L = 1.2566exp-6 x 4exp4 x 4.9exp-4 / 0.2126

L = 2.4629exp-5 / 0.2126

L = 1.158exp-4 Henrys ( or 115.8 microhenrys )



ron

reason for edit :- tidy up


Edited by - ron_o on July 02 2013 17:44:14
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olddawgsrule
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USA
1434 Posts

Posted - July 29 2013 :  18:03:56  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
I'm coming along slowly here, yet trying because I have a need to learn (know)..

I understand the ratio of turns and how it effects voltage and amperage.
Of which I thank you and others for your time with me.

The question becomes how to control a strong source to an 'un-needing'(?) appliance?

Problem:
My Joule ringer runs awesome on 9v's
Doesn't really matter what battery it's sourced from as long as it's down to 9v's or less.
Not where I want my battery bank running at.

So this gives me an area of wattage it prefers.

Not to increase past 9w and have at least 3w available.
And to run on a 12v battery.. Up to 200aH on my current bank..

I know there's not enough info to give direct direction.
Yet I'm thinking wire gauge size and the differences between the to may lead to using a stronger battery (source) and giving me what I hope for.





Old in age, not in mind, so
'Teach me something new'!

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ron_o
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1052 Posts

Posted - July 29 2013 :  18:34:17  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

So at 9 volts everythings fine but at 12 volts problems ..... ok lets just look at the obvious increased voltage equals current which in turn means greater voltage induced into the secondary coil ( base drive ) ie saturation / overload as been reached
Couple of ways around it... 1) reduce number of coil turns on base coil 2) reduce coupling between the two coils 3 ) thinner wire will increase resistance of coil thus lowering base drive current

hope that helps

ron
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olddawgsrule
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1434 Posts

Posted - July 29 2013 :  19:04:58  Show Profile  Visit olddawgsrule's Homepage Send olddawgsrule a Private Message  Reply with Quote
Reduce the secondary windings, easily done.

Reduce the coupling..
That I can't believe I thought of, and for once headed the right direction.
I sleeved an earlier attempt, found a higher (stronger) draw.
Thought was to go back to enameled wire on the primary and using just the electrical tape, which I use to hold the secondary winding in place, as the separator between coils.

I won't go into placement or what I've seen of spreading yet.. on the primary..
Just need to stick to the present issue.

Thinner wire:
On the secondary (base)?
Primary?
Or both??
Now here's where I would have gone the wrong direction again..
I was looking at increasing size on the secondary..

Oh ya.. It helps!!
Always appreciate your input!
And your patience...


Old in age, not in mind, so
'Teach me something new'!

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ron_o
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Posted - August 01 2013 :  17:19:36  Show Profile Send ron_o a Private Message  Reply with Quote
Hi ODR

Spreading the collector coil along the length of the output / secondary coil works best when the coil has a core , bunching the collector coil workks best for air coted coils.

Reason.... the core material tends to keep the flux lines together at the ends of the coil.
in the case of an air cored coil the flux lines spread out thus reducing the coupling effect between primary and secondary windings at the ends of the coil.

ron
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ron_o
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Posted - September 01 2013 :  16:08:43  Show Profile Send ron_o a Private Message  Reply with Quote
Inductors :- Topic 4 LR Series Circuit PART 1 :- DIRECT CURRENT

When power is applied to an inductor it initially "fights" the flow of current until eventually a steady state current flows ( normally reached after 5 time constants have passed )

Time Constant ( in seconds ) (t) = L / R

Transient Time ( in seconds ) = 5 x Time Constant

Steady state current = V / R

Example

Coil of inductance 110mH and resistance of 2.5 ohms are connected in series to a 12 volt supply

Steady state current = 12 / 2.5 = 4.8 amps

Time Constant = 0.110 / 2.5 = 0.044 seconds ( or 44 milliseconds )

Transient time = 5 x 0.044 seconds = 0.22 seconds ( or 220 milliseconds )

ron

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Luciano
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213 Posts

Posted - October 17 2013 :  00:45:29  Show Profile Send Luciano a Private Message  Reply with Quote
Hi ron,
Just a little question before starting winding a series of coils...
there are 2 possibilities and maybe you can answer my question, as i do not want to waste too much energy and copper wire testing both possibilities.

I want a pickup coil, for several reasons i want to have 300 meters of 0.6mm.
for the moment i wound 3 x 100 meter coils (didnt count the windings but the other specs are

core diameter 32mm
width (depth) of coil 21 mm
diameter: 32mm core + 21mm + 21mm (windings) = 74 mm
magnet diameter : 30mm

resistance : 6.5 ohm
ah yes and measured inductance with lcr meter is around 17 mH

I put 3 of them in series (stuck to each other on one core) to fit my needs.

Now as i am starting a second series I was wondering:

Would i get better results if i wound one single one of 300 meters (same diameter but 3x the length) or should i stick to the ones i have done already?
In other words is it better to have 3 x 100 meter coils in series or one 300 meter coil?



Sorry i am not very good at drawing... the proportions are not quite correct.., but values in mm are.

Thanks in advance for any suggestions.
(I hope i was clear in my explanation)

For the moment i am very happy as i have much flexibility with the 21mm wide coils. I can use them one by one as motor coils, add another one "piggy-back" to collect back-emf or put 3 of them together as a big pickup coil.
if you say the single 300m coil is much better as pickup... i can still use the ones i already made as motor coils, but if i wind the single 300 meter coil and it turns out not as good as the others, i can throw the 300 meter coil away.

Luc

Edited by - Luciano on October 17 2013 01:18:58
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ron_o
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Posted - October 17 2013 :  06:32:51  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

There would be a slight difference in inductance
due to the effect of the former. Also a very small
change in winding self capacitance. The magnetic
field strength would be similar.

My only comment is if you intend to reuse the coils
in another project is that the smaller coils would
offer greater flexability.

ron
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Luciano
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213 Posts

Posted - October 17 2013 :  09:01:23  Show Profile Send Luciano a Private Message  Reply with Quote
thanks ron...
that is what i thought.. but i was uncertain
I will opt for the ones you say, the smaller ones.
thank you again

Luc
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Luciano
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213 Posts

Posted - October 27 2013 :  10:15:26  Show Profile Send Luciano a Private Message  Reply with Quote
Me again with all my questions... hope I'm not annoying you..
Just read about the adamsmotor (and it could also apply to other pulse motors (probably not bedini)
that to be able to get good performance (energywise)
the run coils should have a core half the diameter size of the magnets..
and the pickup coils' core should be twice the diameter of the magnets...
Also the magnets should be much longer than their diameter.

is this bs...?
I am building a small motor with 12mm magnets... so the run coils would only have a 6 mm core and the pickup coils a core diameter of 24mm.. (on my big maschine with 30mm magnets that would mean 60mm core diameter...- that sounds crazy)

I have always read that core of pickup coils should be about the size of the magnet diameter.. so the magnet would fit in snuggly.
Is their anything to the big pickup coil cores.. or is it just bs?

source of that was Patrick J. Kelly's book
You must be logged in to see this link.
right at the beginning, between page 1 and 2...

Luc
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ron_o
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Posted - October 27 2013 :  19:52:01  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

Your questions are always welcome.

"that to be able to get good performance (energywise)

the run coils should have a core half the diameter size of the magnets..

Lets think about the "logic" behind this statement..... the rotor magnet will be attracted to the ferromagnetic core of the coil so the greater the size of core the more lines of flux will be in "contact" with it. Remember that a coil without a core still produces a magnetic field , the core focuses / concentrates the flux field .
So the logic is ... less attraction = less drag

"and the pickup coils' core should be twice the diameter of the magnets..."

Think of the pickup coil as a generator coil , if you look at any basic physics textbook regarding Faraday's electromagnetic induction you will see a North and South pole magnet with a conductor drawn parallel to the magnets face ,this is the optimum orientation for electromagnet induction. Now think of a circle , if the circle is the same size of the magnet ,how much of the circle is parallel ? But if you double the size obviously much more of the circle is now parallel

Also the magnets should be much longer than their diameter."

Longer magnets have a greater depth of field , so the flux "reaches deeper" into the core of the electromagnet

I personally don't think either statement is "BS" but i do wonder about the effect of a large ferromagnetic core on the pickup coil and the drag it would introduce ( unless the pickup coil is air cored ? )

ron
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Luciano
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213 Posts

Posted - October 27 2013 :  23:38:15  Show Profile Send Luciano a Private Message  Reply with Quote
thank you !! that helps a lot... you are right.. its logic (but logic is not my strength - lol)

that reminds me i was going to do an experiment.. but was always too lazy to take a part one of my machines.
I buy my magnet wire in an industrial store (much cheaper), but it comes (0.6 mm) on big 22 kilo spools.
I always was wondering what the output would be if i put one of these spools in front of my machine. I think i'll do that experiment during the x-mas holidays...

Luc
L
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Luciano
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213 Posts

Posted - November 24 2013 :  03:16:27  Show Profile Send Luciano a Private Message  Reply with Quote
hi ron,
In my coil testing i ran into an effect I do not really understand..
maybe you can help me:
Here is the setup of the experiment.

8 magnet 140mm diameter rotor all north.
magnet diameter: 15mm
height: 8mm (neos)

testing 2 different coils (dont know the amount of turns as I wound them to fit a specific size)

coil 1:
0.85 mm wire
width: 20mm
diameter:42mm
core diameter 16mm
resistance:1 ohm
inductance:1.7 mH


coil 2:
7 x 0.125 mm stranded wire
width: 20mm
diameter:42mm
core diameter 16mm

resistance:28.3 ohm
inductance:34.5 mH


both welding rod core with counter magnet at the opposite end
(inductance was measured without core)

at around 3000 rpm after bridge rectifier

coil 1 gives like 3to4 volts

coil 2 gives like 18 to 20 volts

--------------------

Ok now i took some AC caps... and put them in paralel to the coils on the AC side...

tested with 2 caps: 1 F 250v and 7F 400v

with the 1F coil 1 goes down to 2 - 3 volts but coil 2 jumps to like 40 volts.
also the amps (3 watt led when connected go from 20ma to 80ma and is much much brighter)

with the other 7F cap, its the opposite... coil 2 goes down but coil 1 jumps from 4 to 10 volts, when load is connected, amps go up also..

in both cases, the rotorspeed goes down, like lenz effect... from 3000 rpm to like 2.350 rpm... (even when no load is connected, just the cap)

strange... when i short the coil speed goes up again.. to around 2700 rpm. (less than the original 3000)

AND... when i connect a load, with coil 1 ( a 3W led slows down the rotor to 2100 rpm, but a 20 watt car bulb (just barely glimming) speeds up to 2500) ... very strange.. its not acceleration under load.. because the initial speed with no load and no cap and coils in place is much higher - 3000 to 3300 rpm.

note: when i slow down the rotor a little bit (by 2 or 300 rpm) output remains unchanged... then starts to drop when rpm get lower.. but is still much higher than without the ac cap.

------------------------

ah yes, with coil 2 at higher rpm, 3800, the effect disappears, the output with or without cap is unchanged (with the 1F or the 7F cap)
I tried a 0.22F cap, there was a slight change but that was in the 0.1 volts range.

--------------------------
it reminds me a little of a paralel tank circuit, with the 2.350 rpm being the frequency (2350 x 8 / 60 = 313hz) but checking with the formula:
(using tank circuit calculator on page: You must be logged in to see this link. )
i get around 140F for coil 1 (as i dont have a 140F AC cap, i tried coil 2:
313 hz, 34.5 mH ==> 7.4943 Microfarad

BUT at 7.5F with coil 2 it is very counter productive... voltage and amps go down... at 1F it goes up.

so i figure its not a tank circuit effect... but it has to do something with resonance... just the coils inductance being different when being excited by a magnet?... in that case, how can i measure that coils inductance when it is producing electricity without blowing my LC meter by connecting at the 2 outputs.(I didnt dare hook up the lc meter when the output was 40 volts)?

My questions: what is it? and is there a way of calculating te optimal AC cap for a given coil at a given rpm? or vice-versa... ideal rpm for given cap and coil...etc..

the interesting part for me, is that not only the voltage goes up, but also the amps !
Maybe that effect is very well known, except by me... but in that case, why doesn't everybody use it.. so there must be a drawback...


Sorry if my explanation is a bit complicated, If there is some mis understanding.. i would try and reformulate it...

Luc

Edited by - Luciano on November 24 2013 05:39:15
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ron_o
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Posted - November 24 2013 :  12:31:10  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

The combination of the inductor (coil ) and capacitor is indeed forming a type of resonance circuit but you have pulsed DC not AC so the tank circuit calculator is not going to give an accurate result.

A capacitor charged by pulsed DC and then effectively discharged through the same inductor will show / produce a ringing waveform but it will not be of equal and opposite polarity
Obviously the drive speed governs the pulse frequency and pulse duration , this in turn effects the effective reactance of the capacitor which provides the amount of current flowing through the coil / capacitor circuit.
This is why the speed varies depending upon capacitor size ( think of it as a form of controlled coil shorting )

Measuring the inductance of the coil whilst it is producing power is not easy but it can be roughly measured to provide what can at best be described as a "guide" value.

You need an inductor of known value , a collection of different value capacitors and ideally an oscilloscope ( or an high impedance AC voltmeter )
The inductor is placed near to the coil under test ( but not to close to act as a load ) the AC voltage is measured using different values of capacitor wired in parallel with the "probe" coil when the capacitor(s) / coils output voltage reaches a maximum , you have a "copy" of the drive coil / capacitor combination which can now be measured using your LC meter.

Because the waveform produced is not a true sine wave you would have to take actual measurements of speed , capacitance and voltage etc and produce a graph.

Standard equations will only give a rough guide

I will get back to you with more information , just need to check out an idea

ron
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Luciano
Average Member



213 Posts

Posted - November 24 2013 :  13:48:59  Show Profile Send Luciano a Private Message  Reply with Quote
Hi ron, thanks for the reply...
I do have an oscilloscpe.. usb type, 2 channel very simple --- (i only use it to measure herz and visualise waveform)

I always thought the output of my generator coils before the bridge rectifier was AC not pulsed DC..
well i will have to goback to the basics..

"controlled coil shorting" sounds a bit like it is... but when i did coil shorting with reed switch.. the voltage went up.. (very similar effect) but the amp output went down or stayed the same (at least the watts stayed roughly the same).. here both go up.. watt output increases so that you can see it (nice side effect..)

cant wait to hear about your idea..

Luc


Edited by - Luciano on November 24 2013 14:20:47
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ron_o
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Posted - November 24 2013 :  15:43:25  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

Only way to produce true sinewave AC voltage is by using N S N S N S etc arrangement.

Have not been able to remote access my computer so will have to wait till i get back home
sorry about that.

ron
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Luciano
Average Member



213 Posts

Posted - November 24 2013 :  17:05:28  Show Profile Send Luciano a Private Message  Reply with Quote
ahh ok now i understand the AC - pulsed dc
no problem.. i can wait...

question: if i change rotor config to NSNSNSNS (magnets are only pressed in, not glued, so they can be changed without complication...)
would then tank calculator work more precisely???

Luc
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ron_o
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Posted - November 24 2013 :  19:27:05  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

Yes , changing magnets to NSNSNS will produce a pure sine wave so the calculator will be much more accurate.

Still searching .....

ron
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Luciano
Average Member



213 Posts

Posted - November 25 2013 :  00:26:46  Show Profile Send Luciano a Private Message  Reply with Quote
i have time and can wait.. ;-)

i am full working so i only get to work on the project maximum 1 hour a day..

My problem is i never know where i am going..

da basic idea was to have 2 rotors going in opposite directions... one all north.. the other all south.. in between coils in series sitting on one single core..
that was before the weekend.. when i did the cap tests.. I have the feeling that i am getting much more output this way than i would in any other case...
its the first time i got my 20 watt car bulb to glim without the rotor coming to full stop because of lenz.
and i have never seen a DC cap charging so fast after the bridge rectifier.. with the litz coil.. a 220F cap charges like in 3 seconds to 80 volts... with that AC cap, its at 100 volts instantly.. i meen 100 volts is the first figure that appears on the meter.. and then goes up to 140 with usual speed like 3 seconds..
so of course i am thinking of cap dumps etc...

But i said to myself.. before i go somewhere i dont know.. i will this time take my time.. and do it right..
I meen only start winding coils etc.. when i am sure where i am going.

so i am winding only 2 coils of each wire size i have.. do the testing with only 1 pair of coils.. and then only continue the build...

So take your time, i will only be ready to continue the tests by end of the week.
Luc
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Luciano
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213 Posts

Posted - November 26 2013 :  09:16:30  Show Profile Send Luciano a Private Message  Reply with Quote
ok.. so i tested with NSNSNSNS rotor...

positive: i get a little more electricity than with the NNNNN rotor..
(one air coil... 30 volts instead of 24)

negative:
with iron cores i can hardly make it to 2000 rpm... cogging cogging and more cogging.. (no comparison with the "back magnets" on the all north rotor)

the caps hardly have any effect.. with 2 coils in series i hit the 100 volts with the caps , here 2 coils in series go up to 70 volts, but there is no difference if i put a cap or not (allthough the cap slows the system down)
i tested 0.22 F, 1, 2, 3.5, 6.8, 7, 8, 10, 13.6, and 14 F

with 0.85mm, 0.63 and my litz coils.

no effect at all or a bit counter productive. only slightly positive effect was with the 0.63mm wire coils where coils went up by like 10%

With the All north rotor they went up like 80%.

so i basically think i will go back to my initial configuration (all N)

Luc
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ron_o
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1052 Posts

Posted - November 26 2013 :  16:05:22  Show Profile Send ron_o a Private Message  Reply with Quote
Hi Luc

I will try and answer your questions in order.

1) increase in output voltage , drop in drive current
When both the capacitive and inductive reactance are equal ( tank circuit ) the total impedance of the circuit increases towards infinity.
So in effect the circuit no longer becomes a load hence voltage up current draw down

2) Why is this not used ?
If a load is applied the tank circuit becomes imbalanced so the voltage drops and the drive current increases.

3) You were initially using pulsed DC , this introduced a "ringing" effect in your coil + capacitor combination.

4) The pulsed DC version had 8 magnets , when you use NSNSNS arrangement you only count 4 magnets ( so the capacitor value changes )
If the rotor speed remained unchanged the frequency would now become approx 156.7 Hz
the 34.5mH coil would now require 29.89uF and the 1.7mH coil 606.8 uF.

5) obviously the cogging will increase in the NSNS arrangement , this being due to the force of attraction between the rotor and the "anti cogging" back magnets

hope this helps


ron
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