# Alternator Inductor Coil Output Equation (?)

by Elektrostatik
Tags: alternator, coil, equation, inductor, output
 P: 20 Hello! I'm working on an experimental alternating current generator project and need specific equations to calculate voltage and current output from a permanent magnet driven inductor, so I will have an idea what to expect before I commit real money to expensive materials. the only variables in the equations should be those which I can directly control and measure : using the symbols below : S - wire cross section area ( mm² ) t - total number of turns L - number of winding layers W - turns per winding layer R - coil DC resistance (Ω) C - coil diametre ( mm ) H - coil height ( mm ) c - core diametre ( mm ) h - core height ( mm ) T - magnetic flux density ( T ) M - magnet diametre ( mm ) m - magnet height ( mm ) a - polar reversals per second ( 1600 / s ) g - gap between rotor magnet and stator coil inductor ( mm ) A - output current ( A ) V - output voltage ( V ) K - ambient temperature ( 293.15 K ) P - core permeability ( mH ) design component values : alternating polarity permanent magnets mounted parallel to the rotation axis near the rim of the rotor. rotor magnets : 30 mm Ø x 30 mm sintered N52 grade Neodymium magnet flux density : 1.5 T inductor core : 12 mm Ø x 12 mm soft iron core permeability : 80? mH inductor coil : 12100 turns of 0.106 mm Ø polysol insulated copper magnet wire 110 windings x 110 layers coil resistance : 2 Ω / m x 912 m = 1824 Ω DCR thanks in advance!
 HW Helper Thanks P: 5,357 Welcome to the physics forums, Elektrostatik! I hope you find the experience both illuminating and electrifying. As for your specific questions, it's a pity that such design equations weren't available from the same source as you found all those electromagnetic terms and units. Your request does seem a tall order, but I shouldn't underestimate the capabilities of some of the people here. Good luck with your project!
P: 20
 Welcome to the physics forums, Elektrostatik! I hope you find the experience both illuminating and electrifying.
Thanks!

 As for your specific questions, it's a pity that such design equations weren't available from the same source as you found all those electromagnetic terms and units
the terms and units are all from my head, from years of hands-on experience making guitar pickus, so I don't need the internet for terms and units, I need equations to put the terms and units I already know to practical use, since I've never built a generator before,
and even though I have formulas for calculating inductance.. knowing the inductance of an air core coil is useless
without including the permeance of the soft iron core in the formula to calculate the total inductance of the electromagnet,
and then all the other variables to calculate the final output voltage and current.

the coil induction formula I use is :

Ø - coil diametre
H - coil height
T - coil turns
µH - induction (microHenrys)

µH = Ø² x T² / (Ø x 18) + (H x 40)

which tells me absolutely nothing about generator output

PF Gold
P: 3,682
Alternator Inductor Coil Output Equation (?)

From Magnetics inc Bulletin FC-S8:

 Background Faraday’s Law continues to have the final word in magnetic design. Faraday’s Law: V = 4.44 N Ae f B 10 -8 Where: V is the applied potential (volts rms) 4.44 is the form factor for a sine wave N is the number of turns A is the core cross-sectional area (cm²) f is the frequency (hertz) B is the peak flux density (gauss) Since no practical materials have been discovered which allow extremely high current density without loss, or extremely high flux density without saturation, the only variable in Faraday’s equation that can be adjusted to significantly shrink the size of a magnetic is frequency.
for SI drop the 10^8, use meters and Teslas

4.44 is 2pi/√2
 P: 20 Perfect! Thank you Jim for the link and for the equation I appreciate that very much. This is just about exactly what I needed, although the induction gap is not included in the formula, which is an extremely important variable, and also current is not included in the output - but this is a very good start. So.. 2∏ / √2 = 4.44288 for 3 extra digits of precision in cgs units : V = 4.44288 x 12100 t x 1.13097 cm² x 1600 Hz x 15000 G x 10-8 = 14591 V or strictly in mks SI units V = 4.44288 x 12100 t x 0.000113097 m² x 1600 Hz x 1.5 T = 14591 V Since Current is a product of Voltage and Resistance (Ohm's law) I = V/R is it possible to extract the output current from the voltage potential and coil DC resistance? if so.. then given the DC resistance of the coil (1824 Ω) 14591 V / 1824 Ω = 8 A ...Surely that can't be right (?) .
 P: 20 I found this inductor core loss calculator on TDK's site http://www.tdk.com/core-loss.php unfortunately I won't have all the values to use it until I have equations for calculating inductance with a soft iron core, as well as impedance and current in the coil. from what I've found so far, soft iron is quite lossy at AC frequencies due to eddy currents, so maybe I would need to use a laminated core instead.
 Sci Advisor PF Gold P: 3,682 Well you DO have a lot of turns there.... 0.1mm wire is mighty small stuff, suitable for signal but not power. Isn't it around #30? Take apart a discarded electric motor and measure the armature wire's diameter. probably you need fewer turns of larger wire. And yes, at 1600 hz you probably want a laminated core here's a starter: http://www.micrometals.com/appnotes/...lossupdate.pdf
P: 20
 0.1mm wire is mighty small stuff, suitable for signal but not power. Isn't it around #30? Take apart a discarded electric motor and measure the armature wire's diameter. probably you need fewer turns of larger wire.
I normally work only with signal wire in voice coils, so armature wire is quite massive by comparison.

Here in Europe - all wire gauges are the same as wire diametre in mm, not including insulation,
but I believe 0.106 mm gauge is somewhere around 37 AWG.

With power inductor wire, it seems that bigger is better, and with fewer turns.
Around 0.5 - 0.6 mm gauge is what I'm finding in motors, so I will design a new coil
with wire in this range.

I'm not sure if I can find a round laminated core,
and I'm not so excited about using a square core.

 P: 20 so now I've made some extreme design changes to the inductor new values : 10 mm x 10 mm soft iron laminated square core (chamfered or slightly radiused edges) 36 turns of 1.5 mm Ø polysol insulated copper magnet wire, wound directly to core resistance : 0.01 Ω / m x 2.88 m = 0.0288 Ω DCR V = 4.44288 x 36 t x 0.0001 m² x 1200 Hz x 1.5 T = 28.7 V if this voltage is correct, then after rectification, it's well within the limits of the LT1083 voltage regulator I plan to use for 12 V regulated output I would still need to calculate the current before going any further
PF Gold
P: 3,682
 I would still need to calculate the current before going any further
your V/Ω would be upper bound.

Be aware you'll get somewhat less because the current in your armature tends to oppose the flux from your magnet.

I dont know how to calculate that. But this link suggests that these rare earth magnets' equivalent amp-turns per meter is quite substantial, well in the tens of thousands... with just 36 turns and modest current my intuition says you'll be okay.. but it is a guess.

http://www.kjmagnetics.com/specs.asp

in fact it's an interesting site, nice glossary and tables.....

http://www.kjmagnetics.com/

you know to keep air gap as small as possible and have a complete magnetic circuit.
P: 20
 your V/Ω would be upper bound.
I wonder if using a bifilar coil would be a solution

it depends on whether or not the strands of a bifilar winding are counted separately per turn
 Sci Advisor PF Gold P: 3,682 i stumbled across a transformer tutorial you might enjoy. It's aimed at 60 hz but shows thought processes... http://mysite.du.edu/~jcalvert/tech/transfor.htm
P: 20
Thanks Jim

The transformer tutorial has quite a lot of useful equations and information that apply to any type of inductor.

Some of which has actually helped solve some core eddy current and loss issues

there is also a current capacity issue

the current capacity of copper wire is 3 A / mm² , so 1.5 mm guage wire can only handle 5.3 A

 Be aware you'll get somewhat less because the current in your armature tends to oppose the flux from your magnet.
There is actually no armature in the design

Sixty N55 Neodymium magnets with a mass of 159 g each will be mounted in force fit holes around the perimeter of a polycarbonate flywheel, parallel to the rotation axis, the 28 kg flywheel has 35% of it's mass at the perimeter
and stores kinetic energy with a moment of inertia of around 16 kg/m², an angular velocity of 125.6 rad/s ( 20 rev/s),
so an angular momentum of 2009 N·m·s

at this rotation speed, the 60 magnets produce an alternating frequency of 1200 Hz

120 inductors are to be fixed on 2 stationary ring mounts, 60 on each side of the flywheel
aligned with the 60 flywheel magnets, so that the magnet poles pass directly over the inductor coils coaxially (orientated face to face)
the inductors on the opposite side are mounted 90° out of phase for full DC rectification,
with each inductor having it's own bridge rectifier and each fed into it's own voltage regulator.

So this is not like any existing generator
 Sci Advisor PF Gold P: 3,682 make sure your magnetic circuit is a closed loop - only air part should be the airgap you need for mechanical clearance. Kirchoff's Law for Magnetic Flux, if you will allow me that misuse , says the flux must get back to opposite end of your magnet. You want it to get there through iron not air. I suppose that's what your ring mounts do? Sorry for the endless questions, just it's so hard to paint or imagine a mental picture from words only... "Science is language well arranged" - Lavoisier old jim
P: 20
 make sure your magnetic circuit is a closed loop - only air part should be the airgap you need for mechanical clearance. Kirchoff's Law for Magnetic Flux, if you will allow me that misuse , says the flux must get back to opposite end of your magnet. You want it to get there through iron not air.
If you mean the air gap between the poles of the flywheel driver magnets and the tops of the inductor cores,
the mechanical clearance gap would be no more than 1 mm.

you can get a good mental picture of the design if you visualise an 810 mm Ø x 30 mm disk flywheel
rotating on an x axis, with sixty 30 mm holes drilled around the perimeter, parallel to the rotation axis,
10 mm from the outer edge, with a 40 mm centre-centre spacing, forming a ring of 60 magnets,

Each neodymium magnet is 30 mm Ø x 30 mm and force fit into the 60 holes around the outer perimeter of the flywheel,
with both poles sitting flush with the left and right side surfaces of the flywheel,
the polarity of each adjacent magnet is reversed : N - S - N - S..
inducing an alternating current in the inductors as the flywheel rotates at 20 revolutions per second,
with 60 alternating poles per revolution, producing a 1200 Hz alternating current.

The flywheel magnets hover past the stationary inductors in a parallel plane,
the inductors form 2 inductor arrays, one on each side of the flywheel, one array shifted 90° out of phase
relative to the other, and each mounted on two 810 mm Ø support rings on either side of the flywheel.

Both the flywheel and inductor support rings are non-conductive polycarbonate
and have no influence on the magnetic field.

I may be having second thoughts on the iron cores.. which seem to cause more problems than they solve,
not least of which is the magnetic drag effect of 120 iron cores on the angular momentum of a magnetic flywheel.
such a drag effect could effectively act as a brake (!)

Perhaps I've been underestimating the extreme flux density of the N55 Neodymiums to induce enough current in an air coil.

N55 magnets this size actually have so much magnetic force, they're considered a 'safety hazard'.

.
 P: 141 First off all i can say that (if I have understod the design correct) you wouldn't get 1.5 T through your armature coil. Look at the B(H) curve of your magnet. This because your flux path has a very high reluctance ( mostly material with permeability of 1). Thus the leakage flux in the machine would be large. As mentioned before, like a transformer core, the complete flux path should be closed ( except the air gap between rotor and stator) by a material with high permeability. Related to your design, the inductor support rings should be made of soft iron. Or have I miss understood the design? A simple drawing (from the top) would be helpful.
P: 20
 First off all i can say that (if I have understod the design correct) you wouldn't get 1.5 T through your armature coil.
Hello,

I'm afraid you have indeed misunderstood the design, as there is no armature and no stator.

A standard generator is simply a motor operating in reverse, so rather than 'drawing' current to drive an armature..
standard generators 'produce' current when their armature is externally rotated by some mechanical means.

This design cannot operate as a motor under any conditions, there is no armature and no stator.
The flywheel is brought up to operating velocity by mechanical means, and continues to spin by angular momentum.

My previous post gives a detailed description of mechanical, electrical and magnetic design and construction.

the only conductive parts of the generator assembly are the inductor coils - which are stationary,
and the magnets - which are mounted on a rotating flywheel.

The inductor coils operate in isolation, absolutely independent of one another,
and there is no electrical or magnetic communication between them.
Each coil is individually rectified by it's own bridge rectifier, and each drives it's own independent voltage regulator.

As an analogy - you can think of the support rings as two 81 cm Ø plastic hula hoops,
which only function to hold the inductors in position as the rotating flywheel spins passed them.

.
 Sci Advisor PF Gold P: 3,682 Magnetics is not well taught. There's a magnetic circuit analogous to an electric one. Magnetomotive force (MMF) is what makes magnetic flux flow(pardon the expression) around a magnetic loop, just as electromotive force (EMF) moves charge around a Kirchoff's Law closed loop. Reluctance is opposition to magnetic flux, analogous to ohms. Flux is MMF/reluctance Air is a only decent insulator for magnetic flux, but an EXCELLENT one for electric current. So it's not intuitive to us just how much magnetic flux gets reduced by an air gap. We can feel magnets acting through air, so we forget to notice how much stronger they pull when close(not much air). Put one of your magnets in a C-clamp so there's a closed iron loop and experiment with air gap . The MMF around a magnetic circuit is the sum of the rises and drops, just as in an electric circuit, and it adds to zero. Your magnet is a strong MMF source (rise). See that magnet manufacturer link, he gave MMF in oersteds and a conversion to amp-turns/meter. Your flux will be MMF / Ʃ(reluctances) The airgaps are large reluctances Iron is a smaller reluctance by its relative permeability ratio. So you want as much of the magnetic path as possible to be in high $\mu$ material. Take apart a car alternator and observe - they spin a multipole electromagnet rotor, and the magnetic path comes out rotor point, through laminated iron core, back into adjacent rotor point. A motorcycle alternator is similar but the rotor is a permanent magnet with many alternating poles like yours, just it'a all one piece not discrete magnets. Their coils are also on an iron core. So - build yours with "hula hoops" of whatever you planned, but leave yourself ability to substitute iron and see whether voltage increases. old jim

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