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AN708 датащи(PDF) 3 Page - Vishay Siliconix |
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AN708 датащи(HTML) 3 Page - Vishay Siliconix |
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3 / 7 page ![]() AN708 Vishay Siliconix Document Number: 70581 www.vishay.com S FaxBack 408-970-5600 3 For the present example: Output power = 5.0 V x 1.5 A + 12 V x 0.15 A x 2 = 11.1 W If efficiency is assumed to be 70%, Input power = 11.1 W/0.7 = 15.86 W For a little cushion, assume a low-input line voltage of 85 V ac. Thus, V dc + 85 2 + 120V dc. Assuming a 20-V pk-pk input-capacitor ripple voltage the minimum voltage is = 15.86 W/ 100 V = 0.1586 A Vmin = 120 V – 20 V = 100 V. Iin = Pin/VDC = 79 mF C + I dt dv + 0.1586 A x 0.01 s 20 V 68 mF is a standard value. With 68 mF, the ripple voltage is = 23.5 V, an acceptable value. Vpp + I dt C + 0.16 A 0.01 s 68E–6F The capacitor voltage rating is calculated: Vmax + 260 V ac x 2 + 368 V dc. A 400-V capacitor is acceptable. A rating of 450-V dc is preferable if high reliability is required or significant line transients are expected. Assuming a power factor of 0.65, the RMS input current is = 0.287 A Iac + Po hac (PF) + 11.1 W (0.7)(85 V)(0.65) A 1-A bridge rectifier is more than adequate. The primary inductance value is chosen by analyzing the lowest input voltage case. For a given load, the value of the peak transformer primary current will remain constant regardless of the input voltage. Since the primary inductance is fixed, the time to ramp to a given value of current is inversely proportional to input voltage (V = Ldi/dt). Therefore, low line is where the most time is needed to ramp to the desired primary current. The duty factor limit dictates an on-time limit. After choosing an operating frequency and calculating the peak primary current, a value for primary inductance, LP, can be determined as follows: For 100 kHz, period = 10 ms. At 50% duty factor, ton(max) = 5 ms. = (0.1586 A) (4) Ipk = Iin x 4 = 0.634 A pk. For VIN (dc) = 100 V = 788 mH. L + V dt I pk + (100 V)(5E–6 s) 0.634 A The actual inductance used was 735 mH. [For high-volume production applications, the design engineer should consider the worst case tolerances for clock frequency and inductor value.] See AN707 for transformer design equations and a fully worked example. The biggest considerations for universal input are related to the additional insulation required to comply with VDE isolation specifications. The physical space occupied by the insulation typically reduces the useable fill factor to 25%. Furthermore, the increase in leakage inductance caused by large physical separation of the windings has the undesirable effects of creating large voltage spikes on the power MOSFET drain, contributing to power losses, and degrading load regulation. Barrier tape at window ends will take up a lot of useable space, so a core geometry with a long, low window should be selected to minimize wasted area. This has the added benefit of reducing leakage inductance. (See equation 6.4 of reference 4.) Wind the primary first. Apply the required insulation, and then wind the secondaries. All secondaries should be wound together with no intervening insulation, if voltage levels allow. Optimal cross regulation is achieved in this way. Further reductions in leakage inductance can be realized by using interleaved windings. First wind one half of the primary, followed by the secondaries and remaining primary turns. The multiple primaries are usually connected in parallel. The spike blanking circuit described in AN707 virtually eliminates the primary-to-secondary leakage inductance problems, at least from the standpoint of the regulation effects. In selecting a power MOSFET, the main concerns will be the rDS(on) and the drain voltage ratings. The transformer primary voltage during the off time is VP = (Vo + VD) NP/NS. Using the 5-V winding, |
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