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ISL6844IUZ Datasheet(PDF) 2 Page - Intersil Corporation |
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ISL6844IUZ Datasheet(HTML) 2 Page - Intersil Corporation |
2 / 10 page Application Note 1612 2 AN1612.1 November 28, 2011 Transformer Core Selection From Figure 2, the RMS current in the transformer primary side can be calculated from: The RMS current in each transformer secondary side can also be computed from: The transformer used in this design is Pulse’s PA3374Nl. It is a gapped ferrite toroid core, which has the following parameters: •Ae = 4.3mm2 •AL = 35nH/n2 •le = 13.1mm •Ve = 56.5mm3 This section provides general guideline to calculate the number of turn and wire size. For more details on designing transformer parameters, please contact a Pulse representative. The number of turns on the primary side, Np, can be determined from: Therefore, the primary side has 26 turns. With the turn ratio of 1, the secondary side and the auxiliary primary side also have 26 turns. Next the calculate the maximum flux density to make sure that it is below the saturation limit.Where: For the operating power level, the wire sizes of the primary, secondary, and auxiliary windings are selected such that the current density in each winding is about 0.25335 cm2/A (50 circular mil/A). To simplify transformer winding, AWG#37 is used for all primary, secondary and auxiliary windings. Primary MOSFET Selection The primary MOSFET needs to be able to handle the voltage stress, given by: As a good design practice, some margin is provided to this peak stress voltage to accommodate transient spikes and for a good reliable performance over time. Providing a 30% design margin as a rule of thumb, the minimum rating on the primary MOSFET needs to be 54.6V. FIGURE 2. TYPICAL OPERATIONAL CURRENT WAVEFORMS TSW dTSW d2TSW d3TSW ILM Ipri Isec,1 Isec,2 <Iout,1> <Iout,2> Ipk Ipk MOSFET ON DIODES (D1 AND D2) OFF MOSFET OFF DIODES (D1 AND D2) ON MOSFET OFF DIODES (D1 AND D2) OFF I rms pri , I pk d 3 --- ⋅ = (EQ. 3) 1.06 0.35 3 ----------- 0.362A = ⋅ = I rms sec , I pk 2 ------- d 2 3 ------ ⋅ = (EQ. 4) 1.06 2 ----------- 0.5 3 -------- 0.216A = ⋅⋅ = N p LuH [] 1000 × A L ------------------------------------ = (EQ. 5) 23.8 1000 × 35 ------------------------------- 26.07 = = B max L M IMmax , ⋅ N p Ae ⋅ -------------------------------- = (EQ. 6) 23.8 6 – ×10 1.06 ⋅ 26 4.3 2 – ×10 ⋅ ------------------------------------------- 4 ×10 0.226T = = A wpri , 0.25335 0.362 0.0917cm 2 = ⋅ ≥ (EQ. 7) V DSFET V = IN MAX , nV out V f + () × [] + (EQ. 8) 26.4 1 15 0.6 + () × [] 42V = + = |
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