Re: 52" Sharp LCD Transformer Backlight Bypass?
http://www.youtube.com/watch?v=oE6K4xBVweA&feature=plcp
This is how it works:
The SONY and SAMSUNG uses the current sensing transformer, the primary is in series with the lamp, the secondary feeds the rectifier diode, the more the lamp draws the current the more output feedback on the secondary side will be fed back to the inverter board IC UI101 Isense pin 14, normal output from the sensing coil is 0.2VAC, it trips at 1.4VAC when lamp draws too much current.
As you can see in the diagram, there is no current sensing circuit on the MOSFET's (QI210, 202) that drive the Inverter HV transformer at all, the only protection it has is the FUSE 3.12A fuse, if one lamp draw too much, it will not be enough to blow the fuse, it will not be a good design of the circuit cannot detect one bad lamp, that is why eash lamp is connected to the individual current sensing transformer.
The other sensing circuit for open lamp connection/ or lamp not drawing current after firing up turn on sequence, done by dual diode di101 (you can see them on teiny's inverter board).
I did not have a chance to probe and analyze the LD pin and the rest of the board yet, it is also used for OV protection.
The inverter circuit has to be able to detectdown to one lamp overcurrent draw, it cannot use just the fuse to protect the MOSFET's since one lamp draw too much current will not be enough to blow hi amp fuse, but the wiring for the lamp can badly burnt up so it must have a way to detect lamp current correctly.
So inverter board needs to have detection circuits to:
1) If the lamp connector is not in place, it will have to shutdown the circuit otherwise you will get nice big arcing at the connector.
2) Ability to detect the current draw of the bad lamp.
3) Ability to detect that after the high kick start voltage is applied to the lamp, the lamp will stay conducting after fired up.
http://www.youtube.com/watch?v=oE6K4xBVweA&feature=plcp
This is how it works:
The SONY and SAMSUNG uses the current sensing transformer, the primary is in series with the lamp, the secondary feeds the rectifier diode, the more the lamp draws the current the more output feedback on the secondary side will be fed back to the inverter board IC UI101 Isense pin 14, normal output from the sensing coil is 0.2VAC, it trips at 1.4VAC when lamp draws too much current.
As you can see in the diagram, there is no current sensing circuit on the MOSFET's (QI210, 202) that drive the Inverter HV transformer at all, the only protection it has is the FUSE 3.12A fuse, if one lamp draw too much, it will not be enough to blow the fuse, it will not be a good design of the circuit cannot detect one bad lamp, that is why eash lamp is connected to the individual current sensing transformer.
The other sensing circuit for open lamp connection/ or lamp not drawing current after firing up turn on sequence, done by dual diode di101 (you can see them on teiny's inverter board).
I did not have a chance to probe and analyze the LD pin and the rest of the board yet, it is also used for OV protection.
The inverter circuit has to be able to detectdown to one lamp overcurrent draw, it cannot use just the fuse to protect the MOSFET's since one lamp draw too much current will not be enough to blow hi amp fuse, but the wiring for the lamp can badly burnt up so it must have a way to detect lamp current correctly.
So inverter board needs to have detection circuits to:
1) If the lamp connector is not in place, it will have to shutdown the circuit otherwise you will get nice big arcing at the connector.
2) Ability to detect the current draw of the bad lamp.
3) Ability to detect that after the high kick start voltage is applied to the lamp, the lamp will stay conducting after fired up.
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