Coils wound in pairs using 2-in-hand 1.6mm diameter wire. Each coil has 37 turns in the one we did for the workshop.

Coils in each pair are roughly ‘in phase’ if one is flipped over as shown. (Actually there is still 30 degrees of electrical phase difference between them, so they are each 15 degrees off the total combined phase angle, which means about 3.5% loss of voltage.)

Pairs that are opposite to each other are ‘in phase’ if connected backwards as shown above. Then the phases are connected in star by linking all of the starts to a neutral (black wires).

This alternator uses 10 poles made from ferrite magnets. When magnets are fitted tightly together like this it makes sense to me to use smaller coils with smaller holes and benefit from the shorter turns in each coil. It’s fun to try something different anyway.

The same winding could work with 14 magnets. But I see no merit in doing this. Maybe somebody can? The inner turns would get more induction from smaller poles, but there would be more leakage flux.

Good Day Hugh,

what is the calculation behind 10magnets, 12 poles? how many coils if it is 6 magnets?

Vw

The calculation is a bit long winded but there is some discussion of the numbers here

There is no simple answer to “how many coils for six magnets” as this depends on the magnet shape in relation to the alternator as a whole but you could for example have 5 coils arranged in a 5-phase stator. doing a 3-phase output is harder.

thanks hugh, the magnets are ferrite and its 6″ x 4″ x 1″ in 17.2″ OD plate, outer dia gap will be 7″

Yeah i am following this thread, after this only, again thought of making use of those magnets.

copper wire thickness, is 14awg.

vw

Hi VW here is a picture of 9 coils on your 6 magnets. Connections are simple – just use 1,4 and 7 for one phase and so forth. No need to reverse any connections.

I’d suggest you get bigger disks (19 inch diameter) for this project.

Hi,

What will happen if i wind it this way but use a 12 pole as well (12 pole, 12 coil). Would it work even phasing is the same?

If you use 12 poles and 12 coils then this will be a single phase alternator. You will need to reverse the wires on every second coil. Finish of coil 1 is connected to finish of coil 2, start of coil 2 to start of coil 3 etc. It will work well but you will get some vibration/noise in the machine and there are other advantages to 3-phase.

Actual coils and poles were designed as a single phase, 1st coil connected to the 2nd, 3rd up to the 12th coil winded alternatively. Since pole modification is of my limitation right now, would winding the coils like of a 3 phase (preferably delta connection) produce more power output compared to the output of the original winding connected in series up to the 12th coil?

please send connection diagram of 12 magnets and 9 coils connected in 3 phase connection

hi

You can find a diagram at the end of this one

Hugh

http://www.6pie.com/images/flatcoil.jpg

Hi, are you having any problems when one magnet passes over both sides of same coil?

hi Lauri

That’s a good question, and in the past to be honest I would have thought that this would be a problem. When there is one magnet on both sides of the coil, it’s induction is indeed cancelled out. However this only happens during a certain part of the waveform when the voltage is low anyway. I would not say it happens when the voltage is anywhere near its peak, which is when the coil is conducting current.

That’s my explanation but whatever the reason I seem to get good results with these smaller coils.

The main advantages of smaller holes in the coils are that I can fit more coils with wider legs into the same space so more coil turns of wire, and at the same time the turns are shorter so the resistance is less. It’s a new way of thinking for me since I would have always said you should go for the maximum voltage by using a large hole in the coil but I am finding out that this way works well for alternators with magnets that are closely packed together, and the ferrite magnets are cheap enough to do that.

Hugh

respected sir

i am making axial flux generator of 16 pole AS i am refering ieee paper they have mentioned three phase stator winding

double layered full pitched ..

no of coil 48

turns per coil 7

thicknesss of the winding is 5mm

can u please help me .. i m getting confused as 16 pole and 48 coils so one pole would cover 3 coils

could u please make understand with drawing of three phase stator with pole and winding connection

sorry i had put some data misssing tht is..

outer diameter of rotor 200mm’

innner diameter of rotor 110mm

thickness of magnet is around 6mm

please help me out …

hi Mallikarjun,

the winding has overlapping coils, I would say and they are large. Full pitch coils means they span from the centre of one magnet to the centre of the next.

Distribute them evenly, and you will find that you have 3 phase groups that match the pattern of magnets at different angles. Connect the first coil to the fourth and the seventh etc. That is one phase. 2nd to 5th and 8th is the next.

You will need to reverse every other coil, so swap the wires around on coils 2,4,6,8, etc. Then connect all the coils in each phase in series and connect the phases in star or delta as you wish. I prefer star.

I hope this helps? Hugh

sir

i taught that in double layer the winding would be laid side by side first phase R and R’ would be connected 6 and 7 of R and R’

and other phase would be connected to Y and Y’ would be connected to 8 and 9 of Y and Y’

AS it is axial ..

then

6×48/48 =6

so R – 1+6=7

R’-2+6=8

so on for other phase ..

please help me if i am wrong sir

and the above i stated will be for 8 pole and

magnet would be of NdFeB magnets in trapezoidal shapes with curvatures

magnet dimension would be of

outer diameter-60mm

inner diameter -47.5mm

height – 39.5mm

thickness -5mm

so the

rotor diameter would be 220mm

rotor inner diameter would be of 140mm

the stator coil would be connected as i stated above … does it work …sir

Sorry Mallikarjun but I can’t follow all this. Please email me with a full description document of some sort.

hi this mallikarjun i m making a generator ,”axial flux permanent magnet generator ” of 230 watts using Ne-Feb magnet of 42 grade ,

i have designed a rotor of outer diameter 236mm and inner diameter 116 mm

magnets shape is trapezoidal of 60mm in height and 47.5mm x 39.5mm top and bottom in length ..

i m using 8 pole rotor single sided not dual rotor so..

i need design the stat-or using toroidal coil winding three phase star connected single layer ..

i dont how to calculate the number of coils need and and number of turns per coil and thickness of wire to be used

for toroidal coil … pls do help me out ..

hi Maddy,

I’d suggest you use 24 coils on a toroidal core. Watch out for high thrust loads on the bearings. Before I can suggest a number of turns I’d need to know 3 things:

Type of magnet (NdFeB grade what or ferrite?)

Operating voltage

operating speed (rpm)

YOu can use trial and error instead and just wind some test coils to collect data and work from that. It’s the safest way in the end. But I can do some calcs if you give me all of the necessary information.

Hugh

thank for ur reply .. we are using neodymium magnet or rare earth magnets of grade 42 … not ferrite

magnet to withstand maximum temperature is represented as “grade” in my case 42 grade withdstand upto 180 degree Celsius..

Operating voltage.. 100v

operating speed (rpm) ..400rpm

please let me know any more information need …

There is a section at the end of my recipe book that explains how to design alternators. The toroidal one is a bit different because the coil presents only one leg to your single magnet rotor, but the equations will work in an adapted format.

total flux = magnet area A sq.m x flux density B (Tesla)

turns n = turns per coils x coils in series/phase

revolutions per second = rpm/60

Average voltage (which is 100/2.72 for star connected AC)

= total flux x number of turns x revolutions per second

= A x B x n x rpm/60

n=60 x V /(A x B x rpm) = 6000/(2.72 x 0.02 x 0.7 x 400) = 400 turns total

If this has 8 coils/phase I would say 400/8 which is 50 turns per coil.

There’s a bit of guesswork in there because I don’t really know the flux density.

sir dont we use average E=4.44* total flux x number of turns x revolutions per second

please let me know if i m wrong

sir

for star connected we use 100/1.732 but not 100/2.732..

i think we made missed typed

total flux = magnet area A sq.m x flux density B

=2160 x 10^-6 x1.29=2.8mwb..

Average voltage (which is 100/ 1.732for star connected AC)

= 4.44 x total flux x number of turns x revolutions per second

= (4.44 x 2.8 x 10^-3x number of turns x 400)/60

= 696.6090 turns total

8 coils/phase =696.6090/8 which is 87 turns per coil.

so i can take 88 turn per coil …

so can i use this one sir..

please let know sir if i have made any mistake ..

hi this Maddy,……. can u please the coil winding connection for the stator …as it 24 coil .. 3 phase 8 pole machine .. that is 8 coil per phase .. as the inner and outer diameter are 236 and 116.. please can u shown the diagram how it can be connected ….

sir let we know the above calculation i have made that are correct or not .. as i’m student there is no noe to guide me .. please assist me..sir

Maddy

hi Maddy,

I quote from the bottom of page 55 as follows:

“(Average voltage is a little lower than the RMS value that we commonly use to measure AC voltages.)

So the output voltage from the thee-phase winding will peak at 2.72 times higher than the average voltage for one phase. (2.72 = 1.73 x 1.57). ”

We are looking at the peak voltage and not the average so that is where the 2.72 comes from.

I hope this helps.

Hugh

thank you sir.. for your kind reply

can i have some sort of pdf or copy of your book .. for reference purpose.. that would be very much helpful for us ..

thank you

maddy

My recipe book is available for only $5 on Kindle.

http://www.amazon.com/Wind-Turbine-Recipe-Book-ebook/dp/B003XVZADA/ref=sr_1_9?s=books&ie=UTF8&qid=1280727133&sr=1-9

Anybody can download it and if you don’t have a kindle you can also download a kindle reader for your computer from amazon too.

sir i do hav this edition

Hugh Piggott A Wind Turbine Recipe Book The Axial Flux Windmill Plans Jan 2009 Metric edition

i want to know more about 8 pole 6 coil diagram and

calculation part .. i want make this one

please help ..

sir

when u would be online .. that would more help full for us to chat with you and clear your doubts..

hi Maddy,

I am online most of the time I guess. I try to answer your questions. You can also email me. I don’t know what your doubts are this time. I am busy, so please ask simple, clear questions. If you tell me ‘I want to know more’ then I will not take time to answer. You have to tell me exactly what you do not understand.

Hugh

Hello !

How important is the weight of the Tail Vane and the angle of 20 deg at the hinge?

Regards,

Bipin

The weight and the angle of the tail will both be important in determining how the turbine furls. if they are too small and light then the output may be a relatively low, especially as the wind gets stronger. On the other hand if you increases them beyond a certain point you will burn the stator or cause some other problem.

Hi Hugh, great website. Is there an optimum size of coil winding compared the the dimensions of a neo magnet? For example if I have a magnet that is 30mm in diameter and 10mm thick. I understand that the coil will be wound to have an inside diameter to match the outside diameter of the magnet. So in theory the magnet could just fit inside the coil. What I dont understand is should I make coils that are tall and thin or coils that are short in height but thicker in diameter. For example, say I need 100 turns ,should I make the coil 10 turns high and 10 turns/rows wide or only 5 turns high and 20 turns/rows wide, or 20 turns high and 5 turns/rows wide? Does it make any difference?

Should the coil be the same height as the magnet?

hi Marc,

to be honest there is no simple answer and no ‘right’ way but some will work better than others. The spacing of the magnets will be just as important as their shape in determining the optimum coils shape.

Often a coil thickness similar to magnet thickness gives optimal results. Total gap between magnet faces similar to total thickness of the two magnets is a good rough idea for optimising the use of magnets.

I like to space the neos widely since they are expensive but if you want to maximise the power from a given disk you should cram them together, and this is what I do with ferrites. Then the hole in the coil can be smaller and this works well.

I hope this helps

Hugh

ok sir thank you… for your kind reply

reply for question 3

Normal we used for star connected

stator coil space = (3.14Dm/no of coils )

Dm mean =(inner diameter of rotor + outer diameter of rotor )/2

And stator coil space multipied by factor .85 we will get space for each coil ..each coil are placed at angle of 15 degree ..

and magnet were placed at angle 45 degree in the rotor

average voltage)=4.4 ( Phimax) x f xTph where f is frequency and Tph is the total number of turns per phasex.936

( 100/1.73)=4.4 ( Phimax) x f xTph where f is frequency and Tph is the total number of turns per phasex.936

(Tph)total number of turns per phase=103.908

Tph=ZxS/3

103.098×3/24=Z

Z=12 turns

instead of 12 we made 18 turns

we made this with 18 turns for each coil but we could get the voltage as the air gap between the stator and rotor was more than 5mm as stator mould was a bit more out of shape ..may be the no of turns were less

so couldnt complete the project so we trying with different stator coil design

.

Let us now consider the design of 6 coil and 8 magnet

magnet dimension =43.5X60

thickness=5mm

Average voltage= 2xtotal flux x number of turns x revolutions per second

magnet area A sq.m=.002610

no of magnets=8 so..

Total magnet area A =.00261×8=.02088

flux density=N42=(Br) rem-anent flux density =.65

flux=.02088x,065=0.01357

Average voltage= 2xtotal flux x number of turns x revolutions per second

(60/2.72)=2x.01357 x Nx 400/60

N=121.91 approximate 122

for two coil per phase N will be 61

but in your calculation sir \

Question

B you have taken as .3 as flux density=N40=(Br) rem-anent flux density =.63 do we need to take it as .3 itself any reason for that

since i m using the single sided rotor not a dual rotor ..so the magnetic flux cutting the coil would be less so do we need increases the number of turn ?? if so mean.. by much ??

what is the air gap that we have to maintain between rotor and stator would

waiting for your timely reply

Thanking you

Mallikarjun(maddy)

hi Maddy,

I do not understand much of the above since it is mostly numbers and no clear explanation of where they come from nor where you are going with them. But I can expalin why the example in my book uses B = 0.3. This is because the example is the 1200 turbine which has only one magnets rotor (single sided). In my experience this has B=0.3 T.

And yes if you have taken B=0.6T then you will not have enough turns and you may have to double the number of turns to get the right speed. This will make the resistance four times higher.

Hugh

i want to know 6 pole single phase generators connection plz tell me about it with dia gram also

if you want single phase then the simplest scheme is to use 6 coils for the 6 poles. However the coils are not actually all in the same phase if you simply connect them in series. You will have to reverse every second coil. So the connections are:

Start of first to output wiring

finish of first to finish of second

start of second to start of third

finish of third to finish of fourth

…

finish of fifth to finish of sixth

start of sixth to output.

I am planning to design one alternator with 100mm X 12.5mm X 12.5mm

with 10 pole on a stator with rpm 900. What should be no. of coil, wire for coil, size of coil

for output of 50hz 230volt 4-5kva.

my magnets were N35.

Please let me know.

hi Rajeev

This type of alternator is unsuitable for producing 230V 50Hz. Also the 10 pole will produce 75 Hz at 900 rpm. I also doubt that you can get as much as 4 kW from this size alternator at this rpm. It’s much too small. sorry

Hugh

1 KW @ 220 Volts & 750 RPM with 10 N35 Magnets , Pls suggest what will be the best possible values of Number of Turns & coil dimensions.

you could try 400 turns and connect all four coils in series for each phase. It’s a rather stupid shape of magnet to start with though. YOu might get 1kW at 750 rpm 62 Hz or it might burn out. YOu’d need a large rotor to make room for thick enough wires.

I am trying with 600 rpm to get 50Hz, but I want to know the dia of winding wire for 400 turn as advised by you

Hi Rajeev,

You need to look at the rotor diameter and figure out the coil spacings and hence the cross-sectional area available for wire and hence the wire size from there. It’s all documented clearly in my Recipe book so I suggest you have a read of that and then get back to me with any questions when you have done some reading.

Any one can help me with a single phase output AC when i use a 8 poles and 8 coils?

All result i’ve got is 2.++ V(AC) and current around 0.5 to 1A(AC). How can i increase the voltage and current up to 6V and 1A by using:

•Copper wire SWG 23

•200 turns per coils

•parmenent magnet type N48 (square shape)

•size: stator 6.5″. Gap between each magnet 1.3″..

Please anyone reply my question as soon as posible.

Thank you guys!!!

hi Safwan,

YOu get more voltage and current by turning the altenator faster. Check that you have the coils correctly wired. If you are using single phase then you connect the coils in series for more voltage. But you have to reverse the connections on every second coil as it produces the opposite voltage. So connect finish of 1 to finish of 2, start of 2 to start of 3 etc…

Hugh

you mean for first coil in clockwise and second coil in anti-clockwise and the connection finish of 1 to finish of 2 and start of 2 to start of 3(clockwise)?

If you make the connections that I suggest then the effect is to reverse the direction of the turns. If you reverse the turns (turn coil over for example) then you do not need to connect finish to finish etc. Just connect finish to start of next one if you are turning every second coil over.

Hi Sir, you know what? I’m very thanks to you about this discussion. with all clockwise direction I can produce more than 6V, now my output voltage arround 10-13V.

I use connection all in clockwise, finish of coil 1 to start of coil 2. finish of coil 2 to start of coil 3 and so on until at last coil of 8. Finish of coil 8, it was my output(-).

But sir, what is the different between both connection? All my research I do, every connection must be in clockwise and anti-clockwise (forward and reverse). And Sir, I use formula, I=V/R to get the current value, it is the right method to get the current?

Can you explain to me about this situation?

I really appreciate your help. very very very appreciate.

when making a single phase stator (which I don’t really recommend) with the same number of magnets as coils, you will have half of the coils under north poles while half are under south poles. So they will produce the opposite voltage and you have to reverse the sense of the coil, or its connections.

do you mean all in same direction of coil is wrong?

If all the coils are wound in the same direction, and you have equal magnets and coils like this, then you need to reverse the connections to every second coil, or the voltages will cancel out and a well made machine will produce little or no output.

thanks you sir for your time.. may have a nice day.. hehe

Dear Sir(Admin)

I am a student and interested in Low RPM PMG. Through Google I had been searching for some designs for 5kw PMG for river current and I got a link which led me to this discussion board and now I know what to do. Thanx for all.

But what is your idea to get constant frequency suppose at 50Hz when RPM of the turbine varies in a wide range from 50 to 200 RPM??

I will be glad to being answered by you.

Thank you sir.

Use an inverter if you need constant frequency. Rectify the output to DC and inverter again. You will also need a battery or a grid connection to stabilise the situation.