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NCP1253 Datasheet(PDF) 11 Page  ON Semiconductor 

NCP1253 Datasheet(HTML) 11 Page  ON Semiconductor 
11 / 15 page NCP1253 http://onsemi.com 11 R1 200k R2 100k R3 100k C1 4.7uF D1 1N4007 D2 1N4007 D3 1N4007 D4 1N4007 Cbulk 22uF input mains D5 1N4935 C3 47uF D6 1N4148 Vcc aux. Figure 25. The Startup Resistor Can Be Connected to the Input Mains for Further Power Dissipation Reduction The first step starts with the calculation of the needed VCC capacitor which will supply the controller until the auxiliary winding takes over. Experience shows that this time t1 can be between 5 and 20 ms. Considering that we need at least an energy reservoir for a t1 time of 10 ms, the Vcc capacitor must be larger than: CVCC w ICCt1 VCCon * VCCmin w 3m 10m 9 w 3.3 mF (eq. 1) Let us select a 4.7 mF capacitor at first and experiments in the laboratory will let us know if we were too optimistic for t1. The VCC capacitor being known, we can now evaluate the charging current we need to bring the Vcc voltage from 0 to the VCCon of the IC, 18 V typical. This current has to be selected to ensure a start−up at the lowest mains (85 V rms) to be less than 3 s (2.5 s for design margin): Icharge w VCCOnCVCC 2.5 w 18 4.7m 2.5 w 34 mA (eq. 2) If we account for the 15 mA that will flow inside the controller, then the total charging current delivered by the start−up resistor must be 49 mA. If we connect the start−up network to the mains (half−wave connection then), we know that the average current flowing into this start−up resistor will be the smallest when VCC reaches the VCCon of the controller: If we account for the 15 mA that will flow inside the controller, then the total charging current delivered by the start−up resistor must be 49 mA. If we connect the start−up network to the mains (half−wave connection then), we know that the average current flowing into this start−up resistor will be the smallest when VCC reaches the VCCon of the controller: ICVCC,min + Vac,rms 2 p * VCCon Rstart−up (eq. 3) To make sure this current is always greater than 49 mA, the maximum value for Rstart−up can be extracted: Rstart−up v Vac,rms 2 p * VCCon ICVCC,min v 85 1.414 p * 18 49m v 413 kW (eq. 4) This calculation is purely theoretical, considering a constant charging current. In reality, the take over time can be shorter (or longer!) and it can lead to a reduction of the Vcc capacitor. This brings a decrease in the charging current and an increase of the start−up resistor, for the benefit of standby power. Laboratory experiments on the prototype are thus mandatory to fine tune the converter. If we chose the 400k resistor as suggested by Equation 4, the dissipated power at high line amounts to: PRstart,max + Vac,peak 2 4Rstart−up + 320 2 2 4 400k + 105k 1.6Meg + 66 mW (eq. 5) Now that the first VCC capacitor has been selected, we must ensure that the self−supply does not disappear when in no−load conditions. In this mode, the skip−cycle can be so deep that refreshing pulses are likely to be widely spaced, inducing a large ripple on the VCC capacitor. If this ripple is too large, chances exist to touch the VCCmin and reset the controller into a new start−up sequence. A solution is to grow this capacitor but it will obviously be detrimental to the start−up time. The option offered in Figure 25 elegantly 
