Sam

[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

[EGuevarae]

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 ...

[PlainBill]

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

[EGuevarae]

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

[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

[PlainBill]

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