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SB350 Datasheet(PDF) 5 Page - Fairchild Semiconductor |
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SB350 Datasheet(HTML) 5 Page - Fairchild Semiconductor |
5 / 10 page AN-8024 APPLICATION NOTE © 2009 Fairchild Semiconductor Corporation www.fairchildsemi.com Rev. 1.0.1 • 9/18/09 5 (Design Example) EEL-19 core is selected, whose effective cross-sectional area is 25mm 2. Choosing the saturation flux density as 0.3 T, the minimum number of turns for the primary side is obtained as: min 6 6 6 10 900 10 1.2 10 144 0.3 25 MLIM P SAT e LI N BA − ⋅ =× ×⋅ =× = ⋅ [STEP-7] Determine the Number of Turns for Each Winding Figure 6 shows the simplified diagram of the transformer. First, calculate the turn ratio (n) between the primary side and the secondary side from the reflected output voltage, determined in step 3, as: RO P SO F V N n NV V == + (13) where NP and NS are the number of turns for primary side and secondary side, respectively; VO is the output voltage; and VF is the diode (DO) forward-voltage drop. Then, determine the proper integer for NS, such that the resulting NP is larger than NP min obtained from Equation (12). The number of turns for the auxiliary winding for VDD supply is determined as: * 1 DD FA AS OF VV NN VV + =⋅ + (14 ) where VDD * is the nominal value of the supply voltage and VFA is the forward-voltage drop of DDD as defined in Figure 6. Since VDD increases as the output load increases, it is proper to set VDD at 3~5V higher than VDD UVLO level (8V) to avoid the over-voltage protection condition during the peak load operation. Figure 6. Simplified Transformer Diagram (Design Example) Assuming the diode forward- voltage drop is 0.5V, the turn ratio is obtained as: 100 18.18 50.5 RO P SO F V N n NV V == = = ++ Then, determine the proper integer for NS, such that the resulting NP is larger than NP min as: min 8, 146 SP S P NN n N N == ⋅ = > Setting VDD * as 15V, the number of turns for the auxiliary winding is obtained as: * 15 1.2 824 50.5 + + = ⋅= ⋅ = ++ DD FA AS OF VV NN VV [STEP-8] Determine the Wire Diameter for Each Winding Based on the RMS Current of Winding The maximum RMS current of the secondary winding is obtained as: 1 RMS RMS MAX SEC DS MAX D In I D − =⋅ (15) The current density is typically 3~5A/mm 2 when the wire is long (>1m). When the wire is short with a small number of turns, a current density of 5~10A/mm 2 is also acceptable. Avoid using wire with a diameter larger than 1mm to avoid severe eddy current losses as well as to make winding easier. For high-current output, it is better to use parallel windings with multiple strands of thinner wire to minimize skin effect. (Design Example) The RMS current of primary-side winding is obtained from step 4 as 0.36A. The RMS current of secondary-side winding is calculated as: 1 10.47 18.18 0.36 6.9 0.47 RMS RMS MAX SEC DS MAX D In I D A − =⋅ − =⋅ = 0.3mm (5A/mm 2) and 0.65mm×2 (10A/mm2) diameter wires are selected for primary and secondary windings, respectively. [STEP-9] Choose the Rectifier Diode in the Secondary Side Based on the Voltage and Current Ratings The maximum reverse voltage and the RMS current of the rectifier diode are obtained as: MAX IN DO O V VV n =+ (16) 1 RMS RMS MAX DO DS MAX D In I D − =⋅ (17) |
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