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AN2495 Datasheet(PDF) 17 Page - STMicroelectronics |
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AN2495 Datasheet(HTML) 17 Page - STMicroelectronics |
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17 / 36 page ![]() AN2495 Startup network design Doc ID 13127 Rev 5 17/36 6 Startup network design To allow the circuit to start as soon as the line voltage is applied, it is necessary to pre- charge both C5 and C8 capacitances. A resistor connected to the DC bus directly makes the pre-charge of C5 electrolytic capacitance. The circuit must start at a minimum DC input voltage of 250 V. The current required by the L6565 driver during the startup time determines the (R1+ R2+ R3+ R4) value. Considering that the L6565 driver needs a 0.07 mA maximum startup current, we obtain: Equation 33 Startup resistance must be lower than 3.6 M Ω to reach the best trade-off between power dissipation and time-to-start. As a consequence, before choosing the startup resistor, we must determine the C5 capacitance value, which is set according to another requirement. C5 must be able to supply the L6565 driver until the steady state behavior of the converter is established. The time from bench verification is 20 ms maximum. Given a L6565 minimum hysteresis-voltage (difference between startup threshold and undervoltage threshold) of 3.7 V, the voltage across C5 must decrease less than 3.7 V during the startup period. From the L6565 datasheet we know that maximum quiescent current after turn-on is 3.5 mA, so C5 must be chosen such that: Equation 34 C5 = 33 µF is a good choice to guarantee a good margin. Finally, we can set the startup resistance value to reduce time-to-start and simultaneously optimize standby power dissipation. The L6565 has a maximum startup threshold of 14.5 V, therefore the maximum time-to-start is approximately: Equation 35 A good choice is to put in series four 200 k Ω resistors (R1 = R2 = R3 = R4 = 200 kΩ), which dissipate less than 1 W of standby power. The pre-charge of C8 base capacitance is carries out by connecting it to the OUT pin of the L6565 through a diode in series with a 2.2 k Ω resistor. R 1 R 2 R 3 R 4 ++ + () V inm in I SU ---------------- 250V 0.07mA --------------------- 3.6 M Ω ≅ = < ΔV I Q Δt • C 5 ----------------- 3.7V C 5 3.5mA 20ms • 3.7V --------------------------------------- 19 μF ≅ > ⇒ < = C 5 14.5V • V inm in R 1 R 2 R 3 R 4 ++ + () ---------------------------------------------------- ⎝ ⎛ ⎠ ⎞ I SU – ------------------------------------------------------------------------- 2R 1 R 2 R 3 R 4 ++ + () ⇒ 808k Ω ≤ sec ≤ = Time - to - start |
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