winding coils

[wind4man]

Well being new to the posting part so i thought i would ask an interesting question that i have not really seen covered here.

I have been looking at home brew wind generators all over the internet. They all give you plenty of detail as how you make the wind collectors and how to put the coil assy. togather. But i either missed or have not found where it determines what gage wire you should use for your alternator-stator ect.,ect.

I have seen 2 that mention which gage to use one was for 18 gage and the other was for 22 gage.

MY actual question would be how do i determine what gage of wire to use and how many turns of wire do i use.

As a make believe example 60 turns of 14 gage wire is the same 120 turns of 22 gage wire.

I have noticed alot of these coil plans fail to mention what gage wire to use so i know what to collect or can i use what i have on hand.

Thanks,

wind4man

[finnsawyer]

"As a make believe example 60 turns of 14 gage wire is the same 120 turns of 22 gage wire."

I presume you mean they would have the same volume.  If so, we can carry the analysis a little further.  The 120 turn winding would produce twice the voltage.  This would mean cut-in at about half the RPMs when charging a battery.  On the other hand, you would have twice the length of the 22 gage wire and half the cross section.  With half the cross section you will have twice the resistance per unit length.  But you have twice the length, so the 22 gage coil would have four times the resistance.  This means that when you reach cut-in at half the RPM you also get 1/4 the initial current, as it is usually the alternator or coil resistance that determines the current, not the battery resistance.  If they are not telling you the wire gage they are not giving you all the information you need.  But you can set a test coil with the rotor and from that determine whether you need heavier or lighter wire to give the proper cut-in RPM.

[Flux]

You are obviously in the early stages of gathering knowledge.

Any decent plan worth following will give you all the essential details. If you want to go it your own way from first principles then you are going to have to do much more learning.

Either follow something that works from  our hosts at Otherpower or Hugh Piggott or Windstuffnow or somewhere sensible or resign yourself to having to learn and understand how to design things for yourself.

This is not rocket science, if you are prepared to read all there is here and sort out the good stuff from the not so good then there is no reason why you can't master it, but it is far more inspiring to start with something that will work and gain constructional experience on the way.

Sorting the wire gauge is not something that can be done without working from the basics of prop diameter wind speed and a lot of other factors. If you find a design that gives you incomplete information they it is likely to lead you problems.

Flux

[kenl]

Math hurts my head, I just take a guess or follow one of the proven designs here or on Ed's site. If you end up making test coils you don't use save them for the future and tag them with the # of turns and rpm to voltage. Maybe you can use the info in the future and put them to good use.

kenny

semmed like a good idea at the time

[kenl]

 The info you seek is posted on this site though. Try searching and look for posts from 1 1/2 - 2 years ago. As I recall it does involve a lot of math. When you figure it all out maybe you could explain it to me :-)

kenny

it seemed like a good idea at the time

[TheCasualTraveler]

Wind4man, far from not being covered here it is probably one of THE most asked question by newcomers.

     First, determine the number of turns in your coils.

     You can best do this by following someones documented plan. If your doing your own design, try to find a plan that is close, re # of magnets, size of magnets, size of magnet disc(s). It is best if you then wind a "test coil" and test it with your magnet rotor(s) at the rpm's you want it to start charging.  A typical single coil will produce about 2.5 volts at the speed where it will start to charge a 12 volt battery in a typical 9 coil, star wired stator . ( 2.5 volts from 1 coil  x 3 (coils per phase) x 1.7 (star)) = 12.75 volts).  If needed, then adjust your coil with more turns (more voltage) or less turns (less voltage).

Second, to answer your question,

     " i either missed or have not found where it determines what gage wire"

     Once you know the number of turns, simply use as large a gauge wire as you can that will still keep the coil small enough to physically fit the area alotted for 1 coil. Larger wire will allow for higher current.

     There are a myriad of factors such as matching blade set to alternator that all affect your results but this is a very basic starting point.

[oztules]

 Andy wrote:

"A typical single coil will produce about 2.5 volts at the speed where it will start to charge a 12 volt battery in a typical 9 coil, star wired stator . ( 2.5 volts from 1 coil  x 3 (coils per phase) x 1.7 (star)) = 12.75 volts).  If needed, then adjust your coil with more turns (more voltage) or less turns (less voltage)."

Andy, I think you forgot the 1.414 to get the DC volts.

You could of course rectify the single coil and then you can ignore the 1.414 part, but the diode voltage drop would have to be considered (what sort of diode.. shotky, silicon, germanium etc.) in that case, and a capacitor put in parallel with the output..... better to measure the AC volts of the test coil.

So the 2.5vac coil x3 (for phase) x1.7 (for star) x 1.414 (for DC.Volts)= 12.75 x 1.414 = 18v DC.

I think 18vdc is a serious cut in voltage for a 12v system.

Perhaps

1.8vac for a single test coils gives 1.8 x 3 x 1.7 x 1.414 = 13 volts DC.

.........oztules

[Flux]

Yes, allowing for a bit of diode drop and a charged battery then about 2v ac per coil for cut in at 12v.

This is all fine if you know the rotational speed you are aiming for. It is also ok to say use the thickest wire that will fit. That will indeed give you the best alternator with a cut in that you chose with the magnets chosen.

If the prop can't manage the cut in speed at the wind speed you want to cut in at you have a problem. If the power that alternator can produce is badly mismatched from what the prop can produce you are still in trouble.

This was the issue that I tried to emphasise at the start. You need to know what the prop can do in terms of cut in speed and you need to know its expected power out at furling speed. You just cant arbitrarily choose some magnets and try to build an alternator.

If you happen to have chosen the right amount of magnets and a suitable cut in speed without reference to the prime mover ( prop) you will be extremely lucky.

A test coil will get your cut in speed but you need to know about lots of factors before you even get to the test coil stage if you want to make something that works reasonably well.

Flux

[finnsawyer]

"Andy, I think you forgot the 1.414 to get the DC volts."

No he didn't.  It doesn't apply.  He did ignore the forward diode voltage drops.  When the peak value of a lead to lead voltage in the star connection becomes equal to the sum of the battery voltage plus twice the forward diode voltage (about 1.2 volts) a current will begin to flow into the battery.  This will constitute a small dc current pulse, as current flow will always be in the same direction.  The action of the diodes prevent any substantial reverse current.  And each phase will provide its own small contribution.  This is cut-in and the RPM at which it occurs will depend on the state of charge of the battery, as the battery voltage can vary considerably.  As the RPM of the alternator increases the amplitude and duration of these pulses will increase.  But each pulse is determined by the part of the lead to lead voltage that exceeds the sum of the battery and diode voltages.  There will be some small change in the diode voltages, but it is the battery voltage that will show the greatest change.  The 1.414 factor as well as the RMS value of the lead to lead waveform do not apply in this case.  That is not to say that you can not calculate the power into the battery if you know the shape and duration of the current pulses.  Since the ripple voltage of the battery is usually negligible all one needs to do in practice is measure the dc current into the battery with a meter and multiply by the battery voltage.

By the way, these considerations also apply to single phase rectification into a battery.  The peak value of the single phase voltage must also exceed the sum of the battery voltage and two diode voltage drops (full wave rectification).  Positive going voltages and negative voltages have the same effect on the current into the battery, again thanks to the action of the diodes.

[TheCasualTraveler]

     Thanks for keeping me straight folks. When Ed gave me that formula for voltage he added, "a little higher when rectified to DC" but I forgot that part and didn't know the 1.4 number till now.

     As I said there are a myriad of factors affecting output, I was trying to keep it as simple and basic as possible so he could get to some hands on stuff. Too much theory without some building kills my spirit. However, not getting the numbers right can lead to a missmatched stator.

     So, I stand corrected and learned something too.

[oztules]

Finn says

"No he didn't.  It doesn't apply.  He did ignore the forward diode voltage drops.  When the peak value of a lead to lead voltage in the star connection becomes equal to the sum of the battery voltage plus twice the forward diode voltage (about 1.2 volts) a current will begin to flow into the battery."

Whilst this is true, finding the peak to peak voltage with a normal  digital volt meter requires that you use the AC scale, and multiply by 1.414 to find the peak (for sine waves). If we measure the test coil with a scope, use the graduations /scale correctly, we get a good idea of the peak voltage.... but very cumbersome.... so just use a normal ac meter and multiply by 1.414.

I don't see the sense in working in figures most people can't measure.

AC on a multimeter we have ready access to. Peak waveform measuring devices.... not so common.

Your peak discussion whilst being accurate in content is of no use in the real world where measurements are taken with field equiptment most of us have, which is AC rms, not AC peak.

If people experimented as per the thrust of your argument, without knowing what you were actually doing, there would be a real risk of stators being wound at far too high a cut in, as they didn't realise that you didn't mean AC that you can measure with your trusty multimeter, but AC that most cant.

.........oztules

[madkane]

so raly with cut in and the volts of gen need to look at 15 volt at a good cut in speed for gen  when testing for beating the rectifiers in any phase in ac

[finnsawyer]

I don't know what you are getting on about here.  When the alternator is connected to a battery you can't measure peak to peak voltage anyway either before or after the diodes.  Before the diodes the I-R drops of the coils reduces the measured voltage messing up that nice 1.414 multiplier, and after the diodes you would essentially measure dc (zero ac reading) as the battery ripple is usually quite small.  People need to understand the distinction between making measurements open circuited or with resistive loads and making measurements with a battery connected.  For an open circuit test coil one could use the 1.414 factor to get the peak value for a given RPM and from that some idea of the RPM for the cut-in voltage.  If that was the point then I apologize, but it seemed to me that there was some confusion.  We went around on this RMS thing before and I may just be overly sensitive,  But it does not apply when the alternator is connected to a battery.    

[finnsawyer]

That would be 15 volts peak voltage, but really it depends on the diode forward voltage drops, which can be different for different types of diodes.  With a high number like 15 you could inadvertently have the alternator cut-in at too low a wind speed causing stall, as the props can't provide the power.  And what do you want for battery voltage anyway?  13.6 -13.8 volts as you propose or maybe 10 volts for a discharged battery or the nominal battery voltage of 12.6 volts?

[Flux]

Don't see what all the confusion is about.

Cut in takes place when the peak of the dc goes above the battery volts. With single phase there is no really defined point, a tiny current starts at the peaks when the factor is 1.414 times the rms voltage and the current increases more and more as the volts rise.

The discussion was about rectified 3 phase, Again conduction starts at 1.414 rms but as the mean value of a rectified 3 phase is about 1.4 you will be into full conduction at 1.4 times rms. The difference between the peaks of the ripple and the basic dc value is so small that you can just use the 1.4 figure that everyone has been suggesting.

So phase volts x 1.732 gives line volts rms. Line volts x 1.4 gives the rectified dc mean.

Can't see why we need to Worry about what happens as you load the thing and volt drops occur, that is a different condition and needs to be taken into consideration to work out the current at higher speeds, it is not to do with cut in.

He certainly needs to include the 1.4 factor.

Flux

[finnsawyer]

"With single phase there is no really defined point,"

I don't know what you mean by this statement.  Single phase is just as deterministic as three phase.  I've done my share of analysis of single phase power supplies to know that conduction also starts in that case when the peak voltage exceeds the battery voltage plus diode forward voltage drops.  With single phase there will always be a phase angle at which conduction starts.  As the peak voltage amplitude increases this angle will approach zero degrees, but never reach it.  That is, single phase full wave rectification will always have time periods when no conduction occurs.  It happens that three phase rectification will also exhibit this behavior just after cut-in until the voltages rise enough such that current always flows into the batteries.  This comes about because the three phase rectified voltage also has substantial ripple.  The result of this ripple is that there will be parts of the cycle just after cut-in is reached when the actual alternator voltage falls below the sum of the battery voltage plus the diode voltage drops.  Most people are either too lazy or unwilling to spend the time researching what actually happens, so that they end up making statements that can lead to confusion.  

[Flux]

I am perfectly happy with your single phase bit. I never questioned that.

I question your bit about 3 phase. Yes it does have ripple with a peak at 1.414 x rms, but the ripple is minimal, can't remember the figure but it is something like 3%. The mean is about 1.4 x rms. So the ripple must be within the band from 1.414 and something marginally less than 1.4. For sorting out 3 phase cut in it is so nearly a smooth dc that you might as well regard it as such.

If you measure the output of a single phase rectifier with a meter you can get any answer from mean at .637 rms to peak at 1.414 depending on the meter you use and whether there is any capacitor attached.

If you do the same with a 3 phase rectifier the value only changes about 1% with an added capacitor, just to bring you from the mean dc to peak.

Flux

[finnsawyer]

The ripple for three phase is about 13% of peak voltage.  That would show up on the open circuit voltage, not the battery voltage.  That means that the alternator RPM needs to increase by 15% for the current to flow all the time.  Once the cut-in or threshold voltage is reached the excess voltage essentially appears across the alternator coils and that is what determines the shape of the current pulse as well as its duration.  

[Flux]

You must be looking at a 3 pulse 3 phase rectifier to get ripple as high as that.

A 6 pules rectifier has a mean of 1.35 ( not 1.4 as I stated)

Conduction will start at 1.414 vrms and conduction will become continuous at 1.35.

I still think that using a factor of 1.4 x rms will give a cut in that is far more accurate than leaving the factor out.

This certainly will cause confusion, I hope nobody ever reads this if they are trying to determine cut in speed. They will need to choose who to believe or perhaps the wise will choose to not believe any of it.

All the advice from DanB, Hugh Piggott, Oztules, various others including myself will need to be ignored even though it has worked for us well enough in the past.

Flux

[TheCasualTraveler]

Gentlemen, gentlemen,

     If I may, I know the discussion your having is what the experienced posters on here enjoy and by all means I encourage it. However, can I ask something that would help us newcomers?

     It occured to me that in all the projects I've looked at I've never seen where they set a target voltage they wish to obtain per coil. I've seen where they measure, adjust, rewind and finnally end up with their coils set number of turns and voltage but not a voltage target per coil. It occured to me that it would be possible to set typical target voltages for coils based on the stator specs, for example,,,

     Taking into account that individual results will vary somewhat, an average coil with average losses for diode drop, line resistance etc. the following would be the results one would aim for when testing a single unrectified coil in the actual magnet rotor setup they have built and at the speed they want cut-in.

     9 coils wired star/ regular rectified - 2 volts / coil +/- 5%  for 12 volt bank

     6 coils wired star/ regular rectified - 3 volts / coil +/- 10%   for 12 volt bank

     8 coils wired star/ jerry rigged -       for 24 volt bank

     etc, etc.

     *Please note* the above are not meant to be actual numbers but examples of a simple chart covering stators showing target volts per coil in various stators. I think a chart like this with say a dozen common stator arrangments would be a good referance tool for the beginner. Since the conditions are not known you could suggest perhaps a slightly higher voltage (for the beginner) since resistance could always be added to raise cut-in.

     Again, I do not mean this as a referance to anyone knowlegeable and clear headed enough to run all the numbers but as a guide to the novice.

     Silly or do-able?

[Flux]

You are right, a simple question got out of hand and nothing but confusion has come out of it.

Regarding your point, yes it would be possible to compile tables of volts per coil for cut in ( not possible for other than cut in).

The thing may simplify things for a few but with the number of combinations of coil numbers, pole numbers and battery voltage it would be a lengthy table even if we only considered star connection.

I still think it easier to think in terms of volts per phase rather than coil. If we know the number of coils in a phase we can just divide the total turns by the number of coils.

If someone would like to make a table of all the options then fine with me. I still say that those who can't follow the very simple basics would do far better to look at something that works from our hosts or Hugh or Ed at windstuffnow. That way you will not only get the cut in right but you will get a magnet system and winding that will match your blades.

It is perfectly possible to build virtually anything that will operate effectively at the cut in point, but it may completely stall beyond that or it may provide so little load that the blades will run away and become virtually uncontrollable and risk over speed damage or burn out.

Flux

[finnsawyer]

For your 6 pulse rectified three phase you have 6 pulses of ripple per cycle.  That's one pulse every 60 degrees of time advance.  If phase one follows the relationship V = VoxCOS(wt), it will produce two pulses per cycle, one over a range of minus 30 degrees to plus 30 degrees for wt.  The pulse will drop from a maximum value of Vo at time t = 0 to a value of 0.866xVo when wt = 30 degrees (or minus 30 degrees).  At that time a new pulse starts and advances toward one.  The maximum drop in V is 13.4%.  Anyone who bothers to put the open circuit rectified voltage on a scope can readily see this.  I won't comment on your use of the mean or average, nor do I see any confusion resulting from the idea that cut-in occurs at the RPM when the peak lead to lead voltage equals the sum of the battery voltage and the two forward diode voltage drops and that the RPM must increase by 15% for full 360 degree conduction or full time current output to occur.  A 15% increase in RPM also implies a 15% increase in the wind speed or a 52% increase in available wind power.

Shall we agree that when evaluating a test coil using an RMS meter one would multiply the open circuit voltage reading for one coil by the number of coils per phase and by both 1.414 and the 1.73 factor to get the expected peak voltage at the given RPM?  From that, assuming linearity, one can then determine the expected cut-in RPM if one knows the battery voltage and diode voltage drops.  Of course, if you use a scope you have the peak voltage for the test coil right off the bat.  It is important to know what you are actually measuring.

I have no problem with the methods of DanB or Hugh, or what you have done that give reasonable results.  But it doesn't hurt sometimes to dig deeper.  A lot of people don't know how to analyze circuits containing diodes.  A refresher doesn't hurt.  Sometimes the threads themselves engender confusion as may have happened in this case.    

[Flux]

This has got so far from the original that we may as well carry on. The others will be well advised to ignore this lot.

You are right that sometimes it does pay to have a look at things in a new light. Many issues here are not exactly straightforward and little of them can stand examination in detail without issues arising that are different from common practice.

Now I have got to grips with what you are saying I see where you are coming from. You have obviously been involved with power supplies with capacitive input filters and this is also true with a battery. I follow your ripple figures now. I spent my life on motors and rectifier loads involving inductance and resistance. A motor can behave as a capacitor in some respects in that it takes current when the supply exceeds the emf but the inductive bit is never absent.

This accounts for our differences in the value of ripple. That is probably why you don't recognise my figures for mean dc and ripple as they are based on rectifiers with included inductance.

Now we come to the issue about cut in into a battery load and how this ties in with things we normally measure and it certainly gets confusing. When we measure volts from a coil and predict the dc voltage we are predicting what we would measure with an average reading dc voltmeter ( which normal moving coil and digital multimeters are)

This may be where the confusion started as I believe you use a scope in preference to meters. You would instinctively measure peak and the mean dc would be tricky to measure.

The average is 1.35 x vrms. We start conduction at 1.414 v rms and I agree with you that full conduction does not occur into a battery until we reach the crossover point on the waveform. Throw in a few unknown diode drops, non sinusoidal waveforms and a range of battery volts from well discharged to float and we have something that takes some sorting out.

As few people have a scope and the thing can't have any degree of accuracy because of the many variables then I think we can stick to accepted method but I am inclined to revise the factor to 1.35 from 1.4.

I hope we can agree on this at least for air gap alternators with no significant inductance, I am not entirely sure that this will apply to all motor conversions, some of which exhibit a fair bit of leakage inductance.

Flux

[finnsawyer]

As you may have figured out my interest is basically the air gap alternators.  I usually steer clear of the other types.  And yes introducing inductance can certainly change matters.  As far the scopes are concerned, I would recommend anyone doing serious work in this area at least get a low cost one.  With the low frequencies found with these alternators you don't need an expensive one.  And one can make a simple circuit for calibrating one using a 9 volt battery, a resistor, and a 5.1 volt Zener diode.  Love those Zener diodes.

One last observation.  Anyone trying to test a single coil with my alternator design should use a scope to determine the peak value.  The 1.414 relationship certainly does not apply in that case.