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Inductance is calculated by the following formula: Description of the coil formula

Inductance is calculated as follows: coil formula
    Impedance (ohm) = 2 * 3.14159 * F (operating frequency) * inductance (mH), set to use 360 ohm impedance, therefore:
    Inductance (mH)=impedance (ohm)÷(2*3.14159)÷F(operating frequency)=360÷(2*3.14159)÷7.06=8.116mH
    According to this, the number of windings can be calculated:
    Number of turns = [inductance * {(18 * circle diameter (吋)) + (40 * circle length (吋))}] circle diameter (吋)
    Number of turns = [8.116 * {(18 * 2.047) + (40 * 3.74)}] ÷ 2.047 = 19 ring hollow inductance calculation formula
    Author: Anonymous Copy from: Site author Hits: 6684 article entry: zhaizl
    Hollow inductance calculation formula: L(mH)=(0.08D.D.N.N)/(3D+9W+10H)
    D------coil diameter
    N------ coil turns
    D-----wire diameter
    H----coil height
    W----coil width
    The units are mm and mH respectively. .
    Air core inductance calculation formula:
    l=(0.01*D*N*N)/(L/D+0.44)
    Coil inductance l unit: Wei Heng
    Coil diameter D unit: cm
    Coil number N unit: 匝
    Coil length L unit: cm
    Frequency inductance capacitance calculation formula:
    l=25330.3/[(f0*f0)*c]
    Working frequency: f0 unit: MHZ this question f0=125KHZ=0.125
    Resonant capacitor: c unit: PF title definition c=500...1000pf can be decided by itself, or by Q
    Value decision
    Resonant inductance: l unit: Wei Heng
    Calculation formula of coil inductance
    Author: coil inductance calculation formula posted by: Reproduced Hits: 299
    1. For the circular CORE, the following formula is available: (IRON)
    L=N2. ALL = inductance value (H)
    H-DC=0.4πNI/lN=number of turns of the coil (circle)
    AL = inductance
    H-DC=DC magnetizing force I=through current (A)
    l = magnetic path length (cm)
    l and AL value size, refer to the Microl comparison table.
The transformers shall meet the requirements specified in addition to the routine tests.
For example: with T50-52 material, the coil is 5 turns and half, and its L value is T50-52 (indicating OD is 0.5 inch). After checking the table, its AL value is about 33nH.
    L=33. (5.5) 2 = 998.25nH ≒ 1μH
    When 10A current flows, its L value can be changed by l=3.74 (check the table)
    H-DC=0.4πNI/l=0.4×3.14×5.5×10/3.74=18.47 (after checking the table)
    You can see the degree of decrease in L value (μi%)
    2. Introduce an empirical formula
    L=(k*μ0*μs*N2*S)/l
    among them 
    00 is vacuum permeability = 4π * 10 (-7). (10 minus seven powers)
    Ss is the relative magnetic permeability of the inner core of the coil, and μs=1 for the air-core coil
    N2 is the square of the number of turns of the coil
    S-coil cross-sectional area in square meters
    l The length of the coil, in meters
    The k factor depends on the ratio of the radius (R) of the coil to the length (l).
    The unit of calculated inductance is Henry. k value table
    2R/lk
    0.10.96
    0.20.92
    0.30.88
    0.40.85
    0.60.79
    0.80.74
    1.00.69
    1.50.6
    2.00.52
    3.00.43
    4.00.37
    5.00.32
    100.2

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