Enter new values on the lines with the light brown background. Press the carraige-return key within one of these fields to calculate the parameters on the lines with the blue background.
Dialectric Constants | ||
---|---|---|
Material | Min | Max |
Air | 1 | 1 |
Styrofoam | 1·03 | 1·03 |
Dry Wood | 1·4 | 2·9 |
Dry Paper | 1·5 | 3 |
The formulae used by the calculator are:
Coil Inductance | L | = 1/(4π²F_{min}² C_{max}) |
Maximum Frequency | F_{max} | = 1/(2π√(LC_{min})) |
Coil Radius | R | = ³√(184000P²L) |
Coil Winding | N | = √(736000L/R) turns |
Coil Length=Diameter | D | = 2R [known as a "square" coil] |
Wire Required | W | = N√((πD)²+P²) |
Reference: http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html
L = 0·001 × N² × R² ÷ (228R + 254l) l = length of coil winding.
But for best results, coil length = coil diameter, so
L = 0·001 × N² × R² ÷ (228R + 254 × 2 × R)
L = 0·001 × N² × R² ÷ (736 × R)
L = N² × R ÷ 736000
The pitch, P, of a coil is the distance between successive turns.
The number of turns, N, of a coil is its length, l, divided by its pitch, P.
But for best results, we decreed that l = 2 × R (its diameter), so
N = 2 × R ÷ P N² = 4 × R² ÷ P²
Now substitute this in the equation for L above:
L = (4 × R² ÷ P²) × R ÷ 736000
L = 4 × R² × R ÷ (736000 × P²)
L = R³ ÷ (184000 × P²)
R³ = 184000P²L
R = ³√(184000P²L)
Now knowing the value of R, calculate the number of turns required:
L = N² × R ÷ 736000
N² = L × 736000 ÷ R
N = √(736000L/R)
The amount of wire, W, required to wind the coil is the amount of wire, T, needed for one turn times the number of turns, N. If you unwind one turn, while maintaining the pitch, its length is the diagonal of a rectangle whose sides are the circumference, C, of the coil and the pitch, P, of the coil.
C = π × D D = 2 & R
T² = C² + P²
T = √(C² + P²)
W = N × T
W = N × √(C² + P²)