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    Sam

    I have a 26" benq monitor/tv Model: dv2680. It has 8 ccfl's backlights and one is bad. In lieu of replacing it, can someone tell me how to fake it out or bypass it. Its viewable with 7 of them working...Thanks

    #2
    Re: Sam

    Originally posted by squiddiddler
    I have a 26" benq monitor/tv Model: dv2680. It has 8 ccfl's backlights and one is bad. In lieu of replacing it, can someone tell me how to fake it out or bypass it. Its viewable with 7 of them working...Thanks
    First of all, Welcome.
    Second, use a more descriptive thread/topic name.
    Third, look around for Dummy Load. I think PlainBill would be able to help you in that topic, as I do not have enough knowledge to serve you ...
    There are 10 kind of people in this world: those that understand binary, and those who don't.
    • ASUS ROG Maximus IX Code
    • Intel Core i5-7600K 3.8GHz
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      #3
      Re: Sam

      Originally posted by squiddiddler
      I have a 26" benq monitor/tv Model: dv2680. It has 8 ccfl's backlights and one is bad. In lieu of replacing it, can someone tell me how to fake it out or bypass it. Its viewable with 7 of them working...Thanks
      Greetings. I just returned from a day escorting my Grandson around Disneyland. He got to see Buzz Lightyear, have his picture taken with Goofy, ride on Dumbo, and play in Mickey Mouse's Toonland.

      I mentioned those because I'm not going to infer a relationship between those names and what you are trying to do.

      In general, here is the theory. Measure the voltage across and the current through a working CCFL. Calculate the equivalent resistance and power dissipation of the CCFL. I.E. If there is 780 V at 8.3 ma through a good CCFL the equivalent resistance is 94K ohm. 100K ohm is close enough. It was dissipating 6.5 watts, better use a 10 Watt resistor.

      Now the problem is your typical DMM won't accurately measure the voltage or the current. The best solution is to pick up a range of 10 Watt resistors and try them until you find the highers value that works, then go down 10%.

      Of course, by the time you have done that you will have spent more than you would pay for a new CCFL, and you won't have to be annoyed by people asking why there is a dark band across the screen.

      PlainBill
      For a number of reasons, both health and personal, I will no longer be active on this board. Any PMs asking for assistance will be ignored.

      Never be afraid to try something new. Remember, amateurs built the ark. Professionals built the Titanic.

      Comment


        #4
        Re: Sam

        Originally posted by PlainBill
        Greetings. I just returned from a day escorting my Grandson around Disneyland. He got to see Buzz Lightyear, have his picture taken with Goofy, ride on Dumbo, and play in Mickey Mouse's Toonland.

        I mentioned those because I'm not going to infer a relationship between those names and what you are trying to do.

        In general, here is the theory. Measure the voltage across and the current through a working CCFL. Calculate the equivalent resistance and power dissipation of the CCFL. I.E. If there is 780 V at 8.3 ma through a good CCFL the equivalent resistance is 94K ohm. 100K ohm is close enough. It was dissipating 6.5 watts, better use a 10 Watt resistor.

        Now the problem is your typical DMM won't accurately measure the voltage or the current. The best solution is to pick up a range of 10 Watt resistors and try them until you find the highers value that works, then go down 10%.

        Of course, by the time you have done that you will have spent more than you would pay for a new CCFL, and you won't have to be annoyed by people asking why there is a dark band across the screen.

        PlainBill
        There are 10 kind of people in this world: those that understand binary, and those who don't.
        • ASUS ROG Maximus IX Code
        • Intel Core i5-7600K 3.8GHz
        • 16gb GSKILL TridentZ RGB DDR4-3200
        • 1 M2 SSD + 2 WD Blue 1TB (Mirrored)
        • Windows 10 Pro x64
        • GeForce GT1050
          2 x Acer KA240H + 1 Vewsonic VP2130 21 (a cap replacement job )

        Comment


          #5
          Re: Sam

          Yes this is what is done to load test the lamp inverters. Even in few service manuals spells this out.

          and 96K is too low, I recall around 120K and yes use 10W sand bar because of high KV.

          Cheers, Wizard

          Comment


            #6
            Re: Sam

            Originally posted by Wizard
            Yes this is what is done to load test the lamp inverters. Even in few service manuals spells this out.

            and 96K is too low, I recall around 120K and yes use 10W sand bar because of high KV.

            Cheers, Wizard
            What??? I did a SWAG and came within 20%???? Sheesh, I'm luckier than I thought.

            PlainBill
            For a number of reasons, both health and personal, I will no longer be active on this board. Any PMs asking for assistance will be ignored.

            Never be afraid to try something new. Remember, amateurs built the ark. Professionals built the Titanic.

            Comment

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