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Design and Example of Low Power Transformer

Design and example of low power transformerTaking the transformer used in the article of Weiner electronic power supply technology using TL431 as high-power adjustable regulated power supply (hereinafter referred to as "Li") as an example, we will lead you to complete the whole process of transformer design step by step.In this article, we do not deduce the relevant application formulas, but hope the readers can draw inferences from this design example.

Design and Example of Low Power Transformer 1

1. Calculate secondary output power (P2)The maximum output voltage in Li is 24V. Assuming that the rated output current is 1a, the voltage drop of K790 tube is 3V, and the power consumption of voltage doubling rectifier circuit is ignored, then:P2=(243)x1=27W

2. Calculate primary power (P1)Assumed transformer efficiency = 0.75, then P1 = 27W / 0.75 = 36WNote: the efficiency of transformer varies slightly according to the output power. Generally, for transformers with capacity below 100W Between 0.7-0.8, below 100W to 1000W, between 0.8-0.9. In practical application, the lower output power shall be taken as the smaller value.

3. Calculate the wire diameter of primary and secondary coils (d)Where, I - winding working current, J - current density (usually J is 3-3.5a/mm2)Primary winding current I1 = 36 / 220 = 0.164 (a)

3.1 wire diameter of primary winding3.2 wire diameter of secondary winding:Secondary winding current I2 = 1x1.17 = 1.17a

Design and Example of Low Power Transformer 2

Where 1.17 is the rectification coefficient of secondary ac current of transformer.Since there is no 0.67mm in the specification of enamelled wire, it is taken as 0.7mm.Usually, we take the current density of the secondary coil as a smaller value to obtain a small internal resistance of the power supply and reduce the temperature rise.

4. Calculate the cross-sectional area of iron coreWe use the following formula to calculate the cross-sectional area of iron core (I think it is a relatively simple empirical formula)Where s - sectional area of iron core K - coefficient P2 - secondary power

The value of K is related to the output power of the transformer. For K below 100W, take 1.25-1.1 (the lower value is taken for the higher power), and take 1 directly for 100W-1000W. In this example, take 1.15, then:Theoretically, under the condition that the cross-sectional area of the iron core remains unchanged, the tongue width and stack thickness of the transformer iron core can be in any proportion. However, in the actual design, the manufacturing process of the coil, the symmetry of the shape, leakage reactance and other factors should be considered. Generally, the tongue width is about 1.5 2 times the stack thickness. In this example, the EI sheet with a width of 66mm is selected, and the stack thickness is 2.7cm.5. Turns of primary and secondary windings

5.1 calculation of turns per volt (W0)Where, F -- alternating current frequency (Hz), B -- magnetic flux density (T), s -- sectional area of iron core (cm2)The value of B will vary depending on the iron core material. See the following table for general silicon steel sheet materials and their values:

In this example, H23 tablets are selected, and B is 1.42, then5.2 number of primary turns (W1)W1=220W0=1160(T)

5.2 secondary turns (W2)Where, 1.2 is the coefficient of adding filter capacitor after rectification, and 1.12 is to compensate the secondary voltage drop caused by winding impedance after load. Generally, the number of secondary turns shall be increased by 5% 25% (the larger value shall be taken if the output power is small).So far, the design of the transformer has been basically completed.

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