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NCP1200AD100R2 Datasheet(PDF) 11 Page - ON Semiconductor |
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NCP1200AD100R2 Datasheet(HTML) 11 Page - ON Semiconductor |
11 / 16 page NCP1200A http://onsemi.com 11 DRIVER PULSES DRIVER PULSES TIME TIME TIME Drv VCC 12 V 10 V 5.4 V REGULATION OCCURS HERE INTERNAL FAULT FLAG FAULT IS RELAXED FAULT OCCURS HERE LATCHOFF PHASE STARTUP PHASE Figure 21. If the fault is relaxed during the VCC natural fall down sequence, the IC automatically resumes. If the fault still persists when VCC reached UVLOL, then the controller cuts everything off until recovery. When this level crosses 5.4 V typical, the controller enters a new startup phase by turning the current source on: VCC rises toward 12 V and again delivers output pulses at the UVLOH crossing point. If the fault condition has been removed before UVLOL approaches, then the IC continues its normal operation. Otherwise, a new fault cycle takes place. Figure 21 shows the evolution of the signals in presence of a fault. Calculating the VCC Capacitor As the above section describes, the fall down sequence depends upon the VCC level: how long does it take for the VCC line to go from 12 V to 10 V? The required time depends on the startup sequence of your system, i.e. when you first apply the power to the IC. The corresponding transient fault duration due to the output capacitor charging must be less than the time needed to discharge from 12 V to 10 V, otherwise the supply will not properly start. The test consists in either simulating or measuring in the lab how much time the system takes to reach the regulation at full load. Let’s suppose that this time corresponds to 6 ms. Therefore a VCC fall time of 10 ms could be well appropriated in order to not trigger the overload detection circuitry. If the corresponding IC consumption, including the MOSFET drive, establishes at 1.8 mA for instance, we can calculate the required capacitor using the following formula: D t + DV @ C i , with DV = 2 V. Then for a wanted Dt of 10 ms, C equals 9 mF or 22 mF for a standard value. When an overload condition occurs, the IC blocks its internal circuitry and its consumption drops to 350 mA typical. This happens at VCC = 10 V and it remains stuck until VCC reaches 5.4 V: we are in latchoff phase. Again, using the calculated 22 mF and 350 mA current consumption, this latchoff phase lasts: 296 ms. |
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