Today I recycled parts of many computers, and I also recycled my old 2007 laptop which have a CCFL backlight.

The inverter input have one red wire, two blue wires and four black wires. Should I just apply 5v across the red and one of the black wires?



Edit: When I apply 12V it does nothing, maybe I will need a complicated PWM signal to make it work.

I would assume the red is a '+' & the blacks are '-', and the blues are '+'(sorta) for brightness control..but you'd need to have the rest of whatever logic circuit was in the monitor for it to work.
Your best bet is probably to dump the existing inverter, and get a new basic one (should be able to find basic CCFL inverters fairly cheap on eBay - just measure how long your lamp is, and get one that'll drive lamps of that length.)

I wonder if they still make the simple ones for PC case lighting? Those just take in 12 VDC and light up max brightness.

There use to be the main supply (12..20V; assume a standard 4S cell battery and 19.5V mains adaptor), ground, brightness control line, fault output line.
Some may have a 3.3V logic supply line andeventually an mbus (ie I2C) control/diagnostic interface.

I will try with the laptop 19V from the charger and connecting one of the blue wires to 3v for simulate a 100% duty cycle control signal. The rest of the control circuit is in the laptop motherboard, so there's no way to recuperate it.

The wires for those inverters are typically :

+ supply (12V or similar)

-

ON/OFF control. In some boards it is ON when left disconnected, in some it is when shorted to -, and in some when supplied with a signal like 3.3 or 5V using a voltage divider. (Not all of them can take 12V in the signal line !)

Brightness control. In some boards it could default to some fairly bright setting when left disconnected, in others you'll have to supply in a relative voltage or PWM signal (where supplying a constant voltage would set it to max brightness)



The wire colors are something chosen by the manufacturer according to their designs, they dont necessarily line up with common wire colors elsewhere. You can identify the supply wires this way :

The - normally goes to many components, including source pins of power transistors (either directly or through a quite big in size, but low resistance value, resistor)

The + often goes to a fuse (the fuse normally looks small and similar to a resistor, but of odd color, often with "X" or something similar printed on it) right next to the power input connector

There is normally a capacitor of significant size between + and -

The few I've powered worked with just 12V for power and a 5V to the brightness control. I'm sure designs vary a bit.

You can determine which pins are common with a continuity meter. Based on coloring, my guess is red is 12V and one of the blues is brightness control. The safest bet is to find where the traces lead, but I'm going to assume you're not keen on reverse-engineering given the topic.

19V is probably too high.

If you use a "diode test" and check for wirecombinations, the GND will appear as it is acommon anode for many "diodes"...

I think I'm gonna get rid of the CCFL tube and the inverter. I tried many ways to make the inverter work but nothing. Also the tube seems to be faulty, I tried to light it up with my fluorescent flashlight circuit that outputs around 500V DC, and it doesn't lights up; it lighted up a little at one end but with the dead pink color so the tube might lost all the mercury.

Those tubes get really dark "socks" in the ends and often crack when they are EOL. If yours doesnt look like an EOL normal Fluorescent tube (T5/8/12) then it is ok

The lighting up in one end is just because your flashlight inverter (meant for hot cathode tube ?) just doesnt have high enough voltage output. The CCFLs often need anywhere between 600V..1kV just as their run voltage and lot more to strike. As the inverter output is HF, the capacitance between one electrode and external capacitive electrodes (the wire going along the tube) were sufficient to light up just that part of the tube in electrodeless way at low power