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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