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AN2495 Datasheet(PDF) 14 Page - STMicroelectronics |
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AN2495 Datasheet(HTML) 14 Page - STMicroelectronics |
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14 / 36 page ![]() Base driving circuit design AN2495 14/36 Doc ID 13127 Rev 5 After selecting the ferrite ring diameter, the minimum primary turns is determined to avoid core saturation from the preliminarily fixed turn ratio N with 0.2. By applying Faraday's law and imposing the maximum flux Bmax equal to Bsat/2: Equation 21 where Bsat is the saturation flux of the core which depends on the magnetic permeability. During the conduction time, the junction base-emitter of the ESBT can be seen as a forward-biased diode. To complete the secondary side load loop, the voltage drop on both diode D and resistor RB must be added in series with the base of the ESBT. The equivalent secondary side voltage source is given by: Equation 22 Since the magnetization inductance cannot be neglected, only IP, a fraction of the total collector current, is transferred to the secondary. As a result, the magnetization current has to be first as low as possible. Meanwhile, the value of the magnetization inductance must be taken into account for the proper calculation of transformer primary turns and turns ratio. The magnetization voltage drop, that is, the voltage at the primary of the current transformer, can now be easily calculated: Equation 23 The magnetization current will be: Equation 24 The number of primary turns should be increased if IMmax is relatively high. The core must have a window area large enough to hold all primary and secondary windings. Otherwise it is necessary to choose a bigger core size. Once both core material and size are fixed, the turn ratio must be adjusted to get the desiderated IC/IB ratio according to Equation 25 below: Equation 25 where IMmax is the maximum magnetization current. V 1 N TP d ϕ dt ------ • N TP A e • ΔB ΔT -------- • N TP 2 V 1 T onmax • A e B sat • -------------------------------- • = ⇒ ≅ = V S V BEo n V D – V RB 2.5V ≅ – = V 1 V S N 1T N 2T ---------- • 2.5 1 5 --- • 0.5 V () = = = I Mm ax V 1 TONm ax L TP ---------------------------- = N eff I P I B ----- I Cma x I Mm ax – I C 5 ----- ------------------------------------ = = |
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