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AN1889 Datasheet(PDF) 11 Page - STMicroelectronics |
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AN1889 Datasheet(HTML) 11 Page - STMicroelectronics |
11 / 33 page AN1889 Application description 11/33 Equation 10 2.2 Fly-back transformer design Once defined the turn ratio, the needed primary inductance and the peak current, to complete the design of the transformer we still need to determine the magnetic core material, its geometry, and the exact number of primary turns. The choice of correct size and material of transformer is often an iterative process that may require several try and error steps before finding the optimal choice. Standard soft ferrite with gaped core and E- type geometry is a common choice for fly-back operation. The calculation of the product of the areas AP (cross sectional active area of the core multiplied by window area available for winding) shown in the formula below can help find the dimension of the core: Equation 11 Where ∆T is the maximum temperature variation with respect to the ambient temperature, KU is the utilization factor of the window (say the portion of the window used for winding that generally ranges between 0.4 and 0.7), and Bmax is the maximum flux in the core. By the way all ferrite manufacturers report tables with the suggested core type and size for given output power and frequency. For our project the type ED2924-PC40 ferrite material from TDK has been chosen. Next step is to determine the air gap length (lg) of the core and the inductance of a single turn (AL) needed to calculate the exact number of primary turns. The core must not saturate even at high temperature and in overload conditions (like start- up or secondary short circuit), the level of this current in the present project can reach 1.6 A. By imposing the IDCmax =1.8 A for safety margin, ferrite's manufacturer supplies the following values for the selected ED2924-PC40 core: ● lg = 0.8 [mm] ● AL = 0.13 [mH] By knowing AL, the exact number of both primary and secondary turns can be easily calculated. In fact, being the primary inductance: Equation 12 Finally from Equation 3: NS=7.5 The closest higher integer has been chosen for the demo board: NS=8 At this point we can also calculate the auxiliary winding needed to supply the driver. In our case the driver used is the UC3842 that from its specification can be driven with 15 V. By applying again formula1 we can calculate the primary-to-auxiliary turn ratio: I ms primary () I P 3 ------- T onmax T S ------------------- 0.38A = = A P 10 3 LPIms primary () ∆T 1 2 --- K uBmax ------------------------------------- 1.316 cm4 ] [ = L P N 2 A L N P 150 = ⇒ = |
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