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ISL6744AABZ Datasheet(PDF) 9 Page - Intersil Corporation |
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ISL6744AABZ Datasheet(HTML) 9 Page - Intersil Corporation |
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9 / 18 page  9 FN6554.0 October 8, 2007 Transformer Design The design of a transformer for a half-bridge application is a straightforward affair, although iterative. It is a process of many compromises, and even experienced designers will produce different designs when presented with identical requirements. The iterative design process is not presented here for clarity. The abbreviated design process follows: • Select a core geometry suitable for the application. Constraints of height, footprint, mounting preference, and operating environment will affect the choice. • Determine the turns ratio. • Select suitable core material(s). • Select maximum flux density desired for operation. • Select core size. Core size will be dictated by the capability of the core structure to store the required energy, the number of turns that have to be wound, and the wire gauge needed. Often the window area (the space used for the windings) and power loss determine the final core size. • Determine maximum desired flux density. Depending on the frequency of operation, the core material selected, and the operating environment, the allowed flux density must be determined. The decision of what flux density to allow is often difficult to determine initially. Usually the highest flux density that produces an acceptable design is used, but often the winding geometry dictates a larger core than is indicated based on flux density alone. • Determine the number of primary turns. • Select the wire gauge for each winding. • Determine winding order and insulation requirements. • Verify the design. For this application we have selected a planar structure to achieve a low profile design. A PQ style core was selected because of its round center leg cross section, but there are many suitable core styles available. Since the converter is operating open loop at nearly 100% duty cycle, the turns ratio, N, is simply the ratio of the input voltage to the output voltage divided by 2. The factor of 2 in the denominator is due to the half-bridge topology. Only half of the input voltage is applied to the primary of the transformer. A PC44HPQ20/6 “E-Core” plus a PC44PQ20/3 “I-Core” from TDK were selected for the transformer core. The ferrite material is PC44. The core parameter of concern for flux density is the effective core cross-sectional area, Ae. For the PQ core pieces selected: Ae = 0.62cm2 or 6.2e -5m2 Using Faraday’s Law, V = N d Φ/dt, the number of primary turns can be determined once the maximum flux density is set. An acceptable Bmax is ultimately determined by the allowable power dissipation in the ferrite material and is influenced by the lossiness of the core, core geometry, operating ambient temperature, and air flow. The TDK datasheet for PC44 material indicates a core loss factor of ~400mW/cm3 with a ± 2000 gauss 100kHz sinusoidal excitation. The application uses a 235kHz square wave excitation, so no direct comparison between the application and the data can be made. Interpolation of the data is required. The core volume is approximately 1.6cm3, so the estimated core loss is 1.28W of dissipation is significant for a core of this size. Reducing the flux density to 1200 gauss will reduce the dissipation by about the same percentage, or 40%. Ultimately, evaluation of the transformer’s performance in the application will determine what is acceptable. From Faraday’s Law and using 1200 gauss peak flux density ( ΔB = 2400 gauss or 0.24 tesla) Rounding up yields 4 turns for the primary winding. The peak flux density using 4 turns is ~1100 gauss. From EQ. 7, the number of secondary turns is 2. The volts/turn for this design ranges from 5.4V at VIN = 43V to 6.6V at VIN = 53V. Therefore, the synchronous rectifier (SR) windings may be set at 1 turn each with proper FET selection. Selecting 2 turns for the synchronous rectifier FIGURE 6. TRANSFORMER SCHEMATIC nP nSR nS nS nSR N V IN V OUT 2 • ------------------------- 48 12 2 • --------------- 2 == = (EQ. 7) P loss mW cm 3 ----------- cm 3 f act f meas --------------- • • 0.4 1.6 200kHz 100kHz --------------------- • • = ≈ 1.28 = W (EQ. 8) N V IN T ON • 2A e ΔB • • ------------------------------ 53210 6 – • • 26.2 10 5 – 0.24 • • • ----------------------------------------------------- 3.56 == = turns (EQ. 9) ISL6744A |
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