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Ο λόγος στασίμων 1:1 ενός κεραιοσυστήματος, δεν συνεπάγεται τον 1:1 λόγο μεταφοράς ισχύος, προς την κεραία!

Saturday, 14 January 2012

29 ραδιο-κατασκευαστές!


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Use air-core-coil resistance to estimate inductance

If the coil is tightly wound you do not need the number of turns to calculate inductance.

Peter Demchenko, Vilnius, Lithuania; Edited by Paul Rako and Fran Granville -- EDN, January 5, 2012

This Design Idea shows how to calculate the inductance of a multilayer air-core coil using only its dimensions and resistance. If you know the dimensions and the number of turns on an air-core coil, you can easily calculate the inductance. With the dimensions in millimeters, the inductance, L, in microhenries is a function of the square of the turns, as the following equation shows: L=0.008×D2×N2/(3D+9h+10g), where D is the average diameter of the coil; h is the height of the coil; and g is the depth of the coil—all in millimeters (Figure 1).
If you don’t know the number of turns, you can still calculate the inductance using the dc resistance of the coil. For this technique to be accurate requires tight and regular coil winding using enameled cylindrical wire (Figure 2). You could use the wire dimensions to give an approximate expression for the total number of turns, N: N=g×h/d2, where d is the diameter of the wire, but this Design Idea assumes that the wire diameter is unknown.
Use air-core-coil resistance to estimate inductance figures 1 and 2

Because the length of an average turn is equal to π×D, the total length of the wire is (N×π×D). The square of the cross-sectional area of the wire is (π×d2)/4.
Read more design ideasYou can express the resistance, R, of the coil as R=ρ×N×π×D×4/(π×d2×1000)=ρ×N×D/(250×d2)=ρ×g×h×D/(250×d2×d2), where ρ is the wire resistivity in ohmmeters and the resistance is expressed in ohms. Thus, you can derive an expression for the wire’s diameter squared: d2=Use air-core-coil resistance to estimate inductance overbar. You would then substitute for the d2 term in the expression for turns: N=g×h/Use air-core-coil resistance to estimate inductance overbar. You can now square both sides of the equation and cancel terms: N2=250×g×h×R/(ρ×D). Substituting the value of N2 into the first equation yields L=2×D×g×h×R/(ρ×(3D+9h+10g)). Using the value of ρ for copper wire, you get an expression for L, which depends only on the resistance and the physical dimensions of the coil: L=117.7×D×g×h×R/(3D+9h+10g).

L and R are proportional to each other yields, yielding two interesting consequences. First, for a series RLC circuit, the following equation defines the damping factor, Use air-core-coil resistance to estimate inductance C/L overbar, meaning that the damping factor is proportional to the square root of R for a given C and coil dimensions D, g, and h. Second, the quality factor, Q, of a coil with given values of D, g, h, and ρ and an angular frequency of w=2πF is a constant value: Q=wL/R=2×w×D×g×h/(ρ×(3D+9h+10g)).
    I would avoid a math error and just use an inductance bridge or RLC meter. However, it would be a good exercise for a lab experiment in a beginning electronics course. A little contest can motivate.
    Mark Seibel - 2012-10-1 08:40:09 PST
    The correct form of the second expression in previous poster is: 2x(1-(sqrt(3))/(pi)) where "sqrt" = square root "pi" = 3.14...
    Marian Stofka - 2012-7-1 07:45:25 PST
      • This comment should not be taken as a criticism, but as an attempt to increase accuracy of quite cute estimation presented in the DI.
        As the wire is assumed to have circular crossection,
        the winding always contains empty (non-conductive) space;
        even if the wire were not covered by an enamel or other insulation.
        To determine the minimum empty space theoretically;
        you can imagine three parallel conductive cylinders (wires); each of which is touchimg the rest two.
        In the center of the crossection of this configuration, there is an empty area of:

        (dxd/4)x(sqrt(3)-(pi)/2)

        In the winding with many turns two such areas can be counted per each turn.
        Consequently, the filling factor of the winding drops from the value of 1, assumed in the DI, to the value of:

        2x(1-(pi)/sqrt(3)) = 0.89734



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