After desoldering out [QP810S:FCPH20L60 – MOSFET], and desoldering out one end of [RM801:0E22], the inserted fuse does not blow up, make sure that, all components on the section is OK. Just Check the MOSFET for its Drain to Source leak. You can check it very easily with a multimeter set to Meg Ohms range. The resistance in between Drain To Source must be very high, almost infinite. If any leak found, replace the MOSFET. Just see, the rectified main DC voltage is fed to this MOSFET by the primary winding of [LP801], and the secondary of this transformer is again passed though one winding of starter transformer, and the other end of this winding of this starter transformer is rectified by [DF803:UF4007S] and smoothed by capacitor [CS806:68uF 65VDC], and supplied to the stand-by power switching opto coupler [PS801:WK707]. The secondary AC voltage actually generates from the secondary winding of [T801:Double Transformer]. This Double transformer is switched by MOSFET[QP801S] stated above. If there must be a voltage with this secondary winding of transformer [LP801], the MOSFET and related circuit should work properly. So this circuit can be considered as the main part of this power switching system.
If fuse does not blow up, even after MOSFET[D801S] has been replaced, check the voltage of all the pins [IC801S:4863G]. This is the booster section control IC. This power supply is so designed to work with various inputs AC voltage levels, starting from 110VAC to 280VAC.
The main voltage selector function is controlled by this [IC801S]. The voltages of all the pins of this IC are
Pin –1 = 2.9V
Pin – 2 = 2.3V
Pin – 3= 2.9v
Pin – 4 = 0V
Pin – 5 = 0V
Pin – 6 = 1.9V
Pin – 7 = 0.5V
Pin – 8 = 16.4V [+ve supply]
Check this +ve 16.5V DC at pin number 8 of this IC. If this is absent, check the resistance between this pins 16 to pin 6 of this IC. It should never show very low Ohms measure. If it shows very low resistance with an Ohmmeter, replace the IC. It ahs damaged. Shorted.
Anyway, you have to first look fault at this section, with a dead set; because, this is the section that senses the input voltage, and determines to which voltage rating should be selected further to work.
Continued on Part-5
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