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Low frequency transformer

Calculation of low frequency inverter

Tongue 32MM, thickness 34mm, E width 96mm, strength required, 220 key, number of turns, number of thick lines, double 51V group, maximum strength to use thicker lines, mouth refers to EI- type transformer core The cross-sectional area refers to the middle part of the E-section (inserted between the modified skeleton in the middle of the width), that is, the width of the iron core and inserted into the variable side of the total thickness of all The E film which stack product of the simplest means Transformer skeleton in the middle of the area of ​​the square, the core transformer area refers to the coils that are partially covered: blade width × degree thickness overlap = cross sectional area, unit: C

First calculation method:

(2) Calculate the power of the transformer according to the cross section of the silicon sheet: P = S / K ^ 2 = (9.79 / 1.25) ^ 2 = 61.34 (1) Transformer silicon plate: 3.2CM * 3.4CM * ( 4) (5) Primary winding current: 60W / 220V (5) = 0.273A (6) Main coil diameter: d = 0.715 Base number 0.273 = 0.37 (MM) (7) Secondary winding: 2 * ( 51 * 4.6 * 1.03) = 2 * 242 (turns) (1.03 is depressive pressure, two -stage 51V = 2 * 242 turns) (8) Secondary coil current: 60W / (2 * 51V) = 0.59A (9) Secondary wire diameter: d = 0.715 Number of roots 0.59 = 0.55 (MM)

The second calculation method:

E-shaped core with center blade to calculate blade width. Calculation of formula: Output power: P2 = UI Taking into account the loss of the transformer, the main power: P1 = P2 / η (where η = 0.7 ~ 0.9, the general power to get a large value) : (B = thickness of silicon steel plate, generally 8000 ~ 12000 Gauss, good silicon steel selection value, and vice versa to have a small value S = iron core blade In the area, the unit is square CM ) the quality of silicon steel can often be selected for 10000 Gauss, which can then be simplified as follows: N = 45 / S Calculate the number of secondary revolutions, considering the leakage inductance of the transformer and the loss of copper wire , to increase the margin 5%. No need to add main rank. By current diameter searching: I = P / U (I = A, P = W, U = V) with the diameter of each square MM ≈ 2.5 ~ 2.6A selected. The third calculation method

It is important to note that the transformer cross-sectional area is the cross-sectional area of ​​the winding position. If your core area (coil covered) is 32 * 34 = 1088mm2 = 10.88cm2 I don’t have time to calculate it. For example, if you have a core part = 3.5 ╳ 1.6 = 5.6 square centimeters / minute of turns per volts = 55 / core section For example, your core = 3.5 ╳ 1 , 6 = 5.6 cm square So the number of turns per rotation = 55 / 5.6 = 9.8 turns 2, find the primary winding coil n1 = 220 ╳ 9.8 = 2156 secondary coil n2 = 8 ╳9.8 ╳ 1.0 = 82.32 Desired for 82 secondary windings Calculating 1.05 is considering the load when the pressure drops 3, find the wire diameter that you do not specify how much power you need to produce low quantities? Here I assume 8V. Current 2 amps. Transformer output power = 8 ╳ 2 = 16 V input power of the transformer = transformer output power / 0.8 = 20 VA Main coil current I1 = 20/220 = 0.09 Ann wire diameter = 0.8 √I Coil diameter d1 = 0.8√ I1 = 0.8√0.09 = 0.24 mm Secondary coil diameter d2 = 0.8√I2 = 0.8√2 = 1.13 mm.