Calculate resistor value for an inverter bypass of LG Flatron W1934s

[javalinex]

Hi , I have a LG W1934S monitor with a lamp that does not work , I want to use a resistor to bypass the system of investor protection.

I have seen another post of a guy, I quote here what he said :

Ohms law . At a rough guess at CCFL runs at 650 - 1000 Volts and draws 3.5 mA.

E / I = R or 800/.004 = 200Kohms

I E = P or X .004 X 800 = 3.2 watts

So a 200K ohm 5 watt resistor connector hooked to an Appropriate Should do it . Probably be a good idea to to use a resistor rated for high voltage .

No warranty at this price .

Now , I've been thinking it Should be possible to do something similar except use two resistors ( 180K , 20K ) in series to form a voltage divider so Could you actually measure the output voltage without frying a DMM .

PlainBill

The details of my monitor panel are in this picture:


I have doubts and do not know exactly what value should I choose to do the calculation

Sorry for my english

Greetings from Spain , thank you very much

[budm]

Re: Calculate resistor value for an inverter bypass of LG Flatron W1934s

That spec sheet is for the T-CON electronic not for the lamp section.

[javalinex]

Re: Calculate resistor value for an inverter bypass of LG Flatron W1934s

You're right, sorry. These are the correct data:

Thanks

[budm]

Re: Calculate resistor value for an inverter bypass of LG Flatron W1934s

So the typical power is 730V * .007A = 5.11W, so the equivalent resistance is about 104K Ohms and should be rated at 10W (twice as much for safe margin).
But since the breakdown voltage that typical resistor can handle is about 500V, so you should use may be 4 or more resistors in series then you can smaller resistor Wattage to add them up to 10W, IE. four 27K Ohms 3W resistor (if the spec indicate the breakdown voltage of 500V) will give you 12W of power handling and 2KV breakdown (start up voltage can be as high as 1700V per spec).

[javalinex]

Re: Calculate resistor value for an inverter bypass of LG Flatron W1934s

So the typical power is 730V * .007A = 5.11W, so the equivalent resistance is about 104K Ohms and should be rated at 10W (twice as much for safe margin).
But since the breakdown voltage that typical resistor can handle is about 500V, so you should use may be 4 or more resistors in series then you can smaller resistor Wattage to add them up to 10W, IE. four 27K Ohms 3W resistor (if the spec indicate the breakdown voltage of 500V) will give you 12W of power handling and 2KV breakdown (start up voltage can be as high as 1700V per spec).

Hello again, I bought 4 resistors on ebay of 27kΩ, but with these other data

Rated Power: 5 Watts.
Max. Working Voltage: 700V.
Dielectric Withstanding voltage: 1000V.

I think in total would be 20W and 2.8KV.

Is it right? Can I use them without problems?

They look like these





Thank you very much again and sorry for the inconvenience

[budm]

Re: Calculate resistor value for an inverter bypass of LG Flatron W1934s

Those will be fine. Higher Wattage will have higher breakdown voltage.