# How do i calculate magnetic coils output volts ?

1. Jun 23, 2009

### smart45

i am planning to make a magnetic generator & i have a question :

if i provide a disk with 12 numbers of neodymium (1x1”x1”cylinder) (n48) magnets around it ,
moving by motor clock wise at 1500 rpm ,
( magnet specification: Br: 13.80- 14.20 KG, Hcb: > = 10.5 Koe, Hcj: > = 11 KOe )

the magnets facing 12 Coils , each coil (3”x 2”) with 782 winding turns (copper awg 15),
each coil inductance = 8.685 mH & each coil resistance = 1.051 ohms .

so how many volts can be gain from the output of the coils ?

2. Jun 23, 2009

### Bob S

Is this an air-core alternator, or is there any iron in it? How are you mounting the magnets-are they facing along the z-axis in cylindrical coordinates (r,theta,z)? What is the radius of the magnets from the axis of the rotor? Do you plan to have coils on both sides of the magnets, or only one side? I think ithe radius has to be at least 2" x 12/2 pi = 4" to get the coils in. Are your coils rectangular or trapezoidal? I think you will be generating a 1500 RPM x 6 poles /60 = 150 Hz. Do you plan to rectify it?

3. Jun 23, 2009

### smart45

dear Bob,

A- 12 coils at stator (on circle shape) facing 12 magnets at one side of the round rotor.
(this is for now but later I shall make another 12 coils also facing the other side of the rotor).
B- each coil cylindrical 3”x 2”.
C- between each coil to another 1.1”.
D- Coils with 6mm hall in the middle of the 3” round side.
(I have left a hall because the N48 magnets are very strong, But if you advise to pot iron
in the middle then I shall do ).

E- 12 magnets are distributed along the circle of the rotor (round disk).
F- each magnet 1”roud face would be facing exactly the middle of the 3” round face coil .
G- radius of the magnets from the axis of the rotor = 8”
H- radius of the coils from the axis of the stator = 8”

4. Jun 23, 2009

### Bob S

Looks good. So you are starting with an iron-free alternator, and perhaps at a later time adding some iron inside or behind the coils? At 150 Hz you might get some eddy curents in iron, so ferrite would be a better choice.

5. Jun 24, 2009

### smart45

So the question now how many volts can be gain from the output of the coils ?

6. Jun 24, 2009

### Bob S

Verrry crude estimate:
Bpeak = 1 tesla
Area = 5 x 10-4 m2 (area of magnet pole)
w= 2 pi f = 2 pi 150Hz = 942 radians per sec (at 1500 RPM)
N= 1 turn
Vrms = 0.707 x Bpeak x Area x w x N = 0.33 rms volts per turn (open circuit).
Coils should be very close to pole tips.

[Added edit] I forgot to multiply by number of coils (=12) so the answer is
Vrms = 12 x 0.33 = ~ 4 volts rms per turn at 1500 RPM. So if you can get 50 turns on each coil, the output is 4 x 50 = 200 volts rms.

Last edited: Jun 24, 2009
7. Jun 26, 2009

### smart45

Dear Bob,

Every coil has = 782 turns ( number of layers = 23 X 34 turns per layer ).

Does that mean 4 volts rms X 782 turns = 3127 volts rms ?

8. Jun 26, 2009

### Bob S

You are correct. 4 volts per turn x 782 = 3127 volts, provided the coil thickness is small compared to the magnet size and magnet separation. Coil turns that are more than say one inch from the magnet pole will not couple much. My handbook shows 15 Ga. wire has a spacing of about 16 turns per inch (for enamel coated wire), so the cross section of your coil is 23/16 = 1.43" by 34/16 = 2.1". I do not believe you will couple efficiently to all of the turns unless you have iron (laminated) or ferrite in them. 15 Ga. wire has a resistance of 3.2 ohms per 1000 ft, which with your resistance measurement of 1.051 ohms/3.2 ohms per 1000' = 500 ft per coil, or 7.7" per turn. Your coils will couple inductively, so the inductance will be more than 12 x 8.7 mH.

9. Jan 31, 2011