nec lcd 2010x backlight board schematic
Marcel, from your postings, it looks like you are not having a lot of knowledge of electronics. If that is the case, it's hard for me to explain to you how the circuit exactly works, in order for you to understand it. Besides I feel you are wasting my and that of others their time. From the things you experimented with, it looks more like you did some random testing, instead of first understanding the circuit structured. Also in your case, working measuring equipment is a must if you want to 'get your hands dirty'. How can someone understand what is going on, if you can't measure anything? How should the community know then what's your problem?
Anyway, from the schematic, I see comparator U1C and U1D, which play the essential role in measuring lamp current. Initially, when Q12 is triggered via a positive puls across C26, Q3, Q4b and Q4a will be triggered as well. With this, the comparator LM339 itself will be powered and being able to compare voltages. The voltages of the 'Measure lamp current' circuit (see at the bottom of the schematic) are compared against a voltage which is offered at the negative input, pin 8 and 10 of U1C and U1D. Via R10, zener D4 regulates the voltage at 6.2 V, which is divided with R17 and R18, which gives a voltage of: 6.2*(100k/(100k+270k))=6.2*10/37=62/37=1.7V. For the comparators in order to remain powered, it makes sense, the output of U1C and U1D, pin 13 and 14 must remain 'high' (open collector output!). With that R34 will guarantee through Q13 that this condition will stay. However, as soon as the output of U1C and U1D becomes low, the comparators will be switched of, via Q13 and Q3.
The 'Measure lamp current' circuit should provide a voltage on the comparator positive input to be at least MORE than the previously calculated 1.8 volt. You can see at R30, I wrote 0.27v, which must be incorrect. It means that the voltage at the anode of D12a should be 0.27v+0.7v. Since 0.97v (+input)<1.8v (-input), comparators U1C and U1D will shut down themselves via Q13.
The 'Measure lamp current' circuit has only one importance, that is to protect the transformer of building up high voltage, when lamps are open. In that situation, the voltage across R30, R29, R.. and R.. (I did not draw that part of the schematic, as it is identical to that of CN5 and CN4) will get lower, because the lamps are contributing LESS to the voltage across R28, R25, etc.
One thing for sure is that the voltage across R28 etc. should be higher than 1.7v (Due to D12a/b and D11b,D9b). With that the voltage across R30, R29, R.. and R.. will raise as well. Basically the voltage across R28, R25 etc. can be measured as well at the positive inputs of the comparator. So if you measure in lamp 'warm condition' 1.7v or less across R28, R25, R.. and R.., or pins 9 and 11 of the comparators, you know the problem has to do with either the CCFL's or as well something with the royer-oscillator.
Unfortunately I can't supply you with accurate measurements which voltages you should expect in normal 'warm' operation. I've seen that other users here mentioned 1.55 volt at pin 8 and 10 of U1C/D, which is a bit different than the 1.7v I mentioned. I've read at a lot of sources, what lamp currents can be expected. Some sources mention 4 mA. If these CCFL lamps are also in this inverter, the voltage across R28, R25, etc. will be roughly 2.5v.
Regards, Marc