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# How do i determine the number of coil winding/size in a axial flux generator ?

i am building a axial flux wind turbine, i have built a small one, now want to move onto a bigger one. but i couldn't find any information on how to calculate the size of coils for the required power output. example i want to make a 48v 5kw generator, how would i calculate how many turns is required, the wire thickness, how many coils per phase. can someone please explaining the process in determining all this?

all the resorces online just give me numbers, example 72 turns of 1mm wire to get 48v, they dont tell me how to calculate it myself for different voltages or different power power output.

any help would be appreciated

thank you

## Comments

1 year ago

Predicting voltage and power from an arrangement of coils and magnets, is mostly based on Faraday's law, for predicting the induced voltage in each coil, and also Ohm's law, which tells us the electrical resistance in each winding, which limits how much current, and thus power, the coil can supply to a load.

The open circuit voltage of any particular coil is,

Vcoil = N*(dPhi/dt)

where N is the number of turns on the coil, and (dPhi/dt) is the time rate of change of the magnetic flux enclosed by those turns.

That equation by itself suggests a large number of turns N is good for making a coil deliver more voltage. While that is true, making N larger, for a given volume of space in which N turns must fit, will require wire that is thin and long, which will increase the coil's total resistance. This conundrum is explained in words and equations, in the paragraphs that follow.

Regarding coil resistance, that is what determines the maximum current a coil can supply, which is the short circuit current, just

Isc = Vcoil/Rcoil

Although in the shorted condition, zero useful work is done, and all the work goes into heating the coil itself.

A more useful equation might be something like:

Vcoil = I*(Rcoil + Rload); I = Vcoil/(Rcoil + Rload)

with this equation suggesting you want to make Rcoil smaller than Rload, to the extent you want to put more power into the load than into the resistance in the coil.

Regarding ways to predict, or design, coil resistance, if you know in advance the shape of your coil and how big it is, then you know its volume (e.g. in cubic centimeters), and you know that volume is going to be filled with copper magnet wire.

Volume = (1/f)*A*L, L = (Volume*f)/A, A = (Volume*f)/L

where A is the wire's cross-section area, and L is its total length.

This variable f, is a "packing fraction", a number slightly less than 1, which accounts for how well the wire fills the space; e.g. circles with their centers on a square grid (f=pi/4~=0.7854), e.g. circles on a hexagonal grid (f=3^0.5*pi/6~=0.9069).

Some expressions for the ratio (L/A), and (A/L), in terms of just L, or just A:

(A/L) = (Volume*f)/(L^2) = (A^2)/(Volume*f)

(L/A) = (L^2)/(Volume*f) = (Volume*f)/(A^2)

Coincidentally, those variables {A, L}, the wire's cross-section and its length, are what determine the coil's resistance. The variable rho is bulk resistivity of the wire material. In practice, this wire material is almost always electrical grade copper.

Rcoil = rho*L/A = (rho/Volume*f)*(L^2) = (rho*Volume*f)*(1/A^2)

And this not unexpected. If you use thicker wire (larger A), you will get a coil with less resistance. Also thicker wire causes the total length of wire L, in each coil, to be smaller, since Volume*f=constant=L*A.

Also thicker wire means you will get fewer turns, since the total number of turns N depends on the total length of wire in the coil, since

L = N*(average turn length)

I am trying to think of a good variable name for, "average turn length". Maybe use capital "C" like for the word, "circumfrence", plus a subscript "avg" signifying an average value over all N turns.

L = N*Cavg; N = L/Cavg

and I am expecting "Cavg" is constant for a particular coil shape and size, much the same as the total "Volume" remains constant for a particular coil shape and size.

Anyway, the point of all this math, is to show you there is kind of a trade-off between induced voltage Vcoil, which requires large N, number of turns, and making a coil with low resistance Rcoil, since more turns, for a given coil volume, make the wire longer and thinner, and that tends to increase the resistance.

If you need both low resistance Rcoil and big voltage Vcoil, probably the only way to do that is by scaling up the volume of the coil, and the magnets too. You know, make everything bigger, and heavier, and more expensive too!

What is the cliche about bigness? Go big or go home? Ha!

;-)

Or buy an alternator from someone. Although that is a SEDIFY (somebody else does it for you) approach, when you maybe wanted to do it yourself, DIY.

Anyway, I am hoping this explanation can give you kind of an overview, since you said "all the resources online..." fail to explain to you how to go about designing a permanent magnet alternator.

I am guessing that my explanation will fail you too, because it has too many abstract equations, and too much hand waving, but not enough real examples. Ha!

;-)

Reply 10 months ago

hi Jack how about if i want to know if my magnets are good enough for my coils , like for example i have 1" x 1" x 0.5 " N52 magnets , and the coil will have a core which have the same size 1" x 1" U shape one facing the north and one facing the south , how to calculate if my magnets are good or i need to get bigger magnets for my coils.

am not sure if my question is valid , am trying to design a generator and i have these magnets 1" x 1" x 0.5 " and i want to know if the core of the coil which will be facing the magnets are the same size what would be the maximum size of the coil ?

thanks a lot for your help

Reply 10 months ago

I think I can give you some hints. I mean, I can give you a kind of simple model, for the voltage I would expect to be induced in a square coil, with a square magnet moving past it.

However the simple model is only approximate. So if you want believable numbers, then likely the best way to get them is empirically; i.e. by building your generator, or a smaller representative part of it, like one magnet and one coil, that can move past each other at the speed you want... and then actually measuring the open-circuit voltage induced in that coil, using an oscilloscope or similar.

Or there might exist software, for modeling electric machines like motors and generators. Although at present, I cannot just point you to an easy, free, way to do that, because I myself do not know enough about this kind simulation.

Anyway, for the simple model, I imagine a square shaped block of magnet flux lines, moving past another square shaped coil. Then I imagine the flux through the coil, Phi(t), as a function of time. Then I imagine the time derivative of that, (dPhi/dt), also a function of time.

I drew some pictures of this, with pencil and paper, and I will attach images of those drawings to this post.

For this easy, approximate, math, the voltage induced in the coil looks like two square pulses, a positive pulse followed by a negative pulse.

I also drew some plots of what a, more realistic, plot might look like, e.g. voltage across a test coil, measured by an oscilloscope. For this plot, I expect the voltage function will likely more closely resemble a single period of a sine wave; again a positive pulse as the magnet moves closer to the coil, then a negative pulse as it moves away.

By the way, I do not know exactly how strong is the magnetic field at the surface of one of these neodymium-iron-born (NeFeB) magnets, but I am going to guess it is about 1.0 tesla. This plus some other guesses, about the size of the magnet and coil (square with L=2.5 cm), number of turns (N = 100), and relative speed (v = 1.0 m/s)between the magnet and the coil...is enough to give you an example calculation using Faraday's law for the voltage induced in a coil via changing magnetic flux.

Given:

L = 2.5 cm = 2.5e-2 m = length of one side of square magnet or coil

B = 1.0 T = 1.0 tesla = uniform magnetic field strength at magnet surface

v = (dx/dt) = 1.0 m/s

N = 100 turns

Identity:

(1 tesla*meter^2*turn*second^-1) = (1 volt)

https://en.wikipedia.org/wiki/Tesla_(unit)#Definit...

https://en.wikipedia.org/wiki/Faraday%27s_law_of_i...

Calculations:

Amax = L^2 = 2.5*2.5 = 6.25 cm^2 = 6.25e-4 m^2

Phimax = B*Amax = 6.25e-4 (T*m^2)

(dA/dt) = L*(dx/dt) = (2.5e-2 m)* (1.0 m/s) = 2.5e-2 m^2/s

(dPhi/dt) = B*(dA/dt) = (1.0 T)*(2.5e-2 m^2/s) = 2.5e-2 (T*m^2/s)

Vmax = N*(dPhi/dt) = N*B*L*v = (100 turns)*(2.5e-2 T*m^2/s) = 2.5 volt

The promise of Faraday's law, V = N*(dPhi/dt), is that we can get more voltage, by using more turns (larger N), or by faster motion, v, between the magnet and coil (since dPhi/dt =B*L*v).

So if 100 turns gave us 2.5 volts, then 1000 turns will give us 25 volts, with everything else unchanged.

Increasing B, the strength of the magnetic field, would work too, similar to increasing N or v. However, I think you are already using the strongest kind (i.e. NeFeB) of permanent magnets known to exist, at the time of this writing.

I mentioned these numbers are just guesswork, right? I mean, these are just example numbers to show you how to do this calculation, which itself is only a crude approximation for the voltage seen across one coil, with one magnet moving by it.

However, I am hopeful that this example, sort of thinking about the problem, can in part, get you closer to building a working generator.

Reply 1 year ago

Jack, you are the only person I can find, in the hour I have been looking, that actually answers the question. Give that man a Bell's

Reply 1 year ago

Now the next question, is the number of turns, the total amount of turns needed, per coil or per stator?

Very new, doing a lot of reading and relearning math. Tedious process.

Reply 1 year ago

I love it!

But if I struggle with designing a motor or generator from scratch AND to match specs....

I Guess Maatz would not mind to get a slightly simpler approach ;)

1 year ago

If it would be all that easy we would not bother to buy motors or generators anymore ;)

Without knowing all the specs for the apperature it is simply impossible to tell you what you need.

Don't know what your base is either, a DC motor? A salvage generator that needs to be rewired?

For a basic DC motor you just need to check that the thing runs close to or above the rated RPM's to get the stated voltage back out - minus some losses here and there.

In the example you gave the wire needs to be able to handle well over 100A.

I would go for 120A by checking the common wire gauge tables for copper wire.

Even at less power it is a true pain to wire your own generator or motor properly - not for the beginner IMHO.