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Calculation method of low frequency transformer
- Jul 27, 2017 -

The first method of calculation:

(1) Transformer steel section: 3.2cm*3.4cm*0.9=9.792cm^2 (2) Calculate the power of the transformer according to the steel section: p=s/k^2= (9.79/1.25) ^2=61.34 Watts (fetch 60 watts) (3) Calculate coils per volt According to section: w=4.5*10^5/bms=4 5*10^5/(10000*9.79) = 4.6 turns/volts (4) Primary turns: 220*4.6= 1012 turns (5) Primary current: 60w/220v=0.273A (6) Primary line diameter: D=0.715 square root 0.273=0. Panax Notoginseng (mm) (7) Secondary turns: $number (51*4.6*1.03) =2*242 (Turn) (1.03 is Buck, two-stage 51v=2*242) (8) Secondary current: 60w/(2*51v) $number. 59A (9) Secondary line diameter: D 0.715 square root 0.59=0 (mm)

The second method of calculation:

E-shaped iron core with the middle tongue for the calculation of tongue width. Calculation formula: Output power: P2=ui takes into account the loss of the transformer, primary power: p1=p2/η (of which η=0.7~0.9, the general power of large value for the calculation of the number of turns per Volt: N (per volt turns) $literal. 5x10 (5 times)/bxs (= Silicon steel sheet permeability, generally in 8000~12000 Gauss, good silicon steel Select large value, conversely take small value. = Iron core Tongue area, unit is square cm, such as the quality of silicon steel generally can be selected 10000 Gauss, then can be simplified to: n=45/s calculation of secondary winding number, considering the transformer leakage inductance and conductor copper loss, the need to increase 5% winding allowance.   No allowance is added at the beginning. For wire diameter by current: i=p/u (I=a, p=w, equals) is selected in the line diameter of mm≈2.5~2 6A per square meter.

The third method of calculation

The first thing to say is that the sectional area of the transformer is the cross-sectional area of the coil. If your core area (the coil is in position) is 32*34=1088mm2=10.88cm2 I don't have time to calculate for you.   Give you a reference, hope to be helpful to you: Simple calculation of small Transformers: 1, for each volt turn number =55/core section for example, your core section = 3.5 * 1.6 = 5.6 sq. cm, each =55/5.6 = 9.8 turns 2, the coil turns number primary n1=220╳9.8=2156 secondary. 05 = 82.32 Preferable to 82 turns   1.05 of the secondary turn-around calculation is to consider the pressure drop of the load 3, the diameter of the conductor you did not specify how many volts you require to output the current is how many Ann? Here I assume the 8V. Current is 2 Ann. Output capacity of Transformers =8╳2=16 Volt-ampere transformer input capacity = transformer output capacity/0.8 = 20 KVA Primary current i1=20/220=0.09 Safety conductor Diameter 8√i primary. d1=0.8√i1=0. d=0.09 = 0.24 mm 8√0 conductor diameter secondary. d2=0.13 mm 8√i2=0

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