These boards create around 3000Vpp to power CCFL lamps.
I have scope probe 100:1, however, on some inverter boards it does not start properly if I measure voltage on the outputs...
Also is it possible to use some kind of load instead of lamp in order to isolate faulty lamp since inverter board does not start if at least one CCFL lamp is defective.
??
These boards create around 3000Vpp to power CCFL lamps.
I have scope probe 100:1, however, on some inverter boards it does not start properly if I measure voltage on the outputs...
Also is it possible to use some kind of load instead of lamp in order to isolate faulty lamp since inverter board does not start if at least one CCFL lamp is defective.
you need proper probe, DIY cap divider for AC measurments should work
http://www.repairfaq.org/sam/hvprobe.htm
They only produce 3kv to start the lamps - once the lamp has struck, it's more like a few hundred. The initial HV is produced by series resonance with the output cap and the transformer secondary inductance. Available current at 3kv is minimal, so any capacitive loading will reduce it significantly.
Why do you want to measure it?
Yes, according to theory, it firing up the lamp first and than goes back to "normal".
However, when I measure with scope it does not change like that, all I see is the same sine signal. If probe moves a bit it will trigger protection circuit. On some TVs if I connect the probe it goes into protection mode right away..
I think I need some higher impedance probe for scope or something.
I fix TVs so by measuring each inverter board output I diagnose where the problem is, if waveform is distorted it mean that one channel is defective, if it has amplitude change it usually means that CCFL lamp is defective...
You'll need a probe with very low capacitance - I'd suggest a few 10M resistors in series, with the probe ip being the short lead of the last resistor.
Watch the voltage rating of the resistors. Standard 1/4W resistors are usually rated for 500V. Depending on the actual voltage across the resistors, their voltage coefficient may also be significant. Watch the dissipation, 1kV over 1Mohm is already 1W.
Compact fluorescent lamp ballast has high voltage and current and no isolation for the technician between the AC power outlet and fluorescent lamp ballast. Isolation transformer will help between the AC main and compact fluorescent lamp ballast. The voltages are often 10,000 V and the current is high to ionized the argon and mercury vapor gases, homogenous mixture noble gas and toxic mercury gas to an excited state and falls back to emit a photon of ultraviolet light to struck a phosphorus coating to convert the ultraviolet to visible light. Ceramic resistor as mention by others would make a good load to measure the current. Yes, a high voltage proble like those used to measure a fly back transformers in the past, when cathode ray tubes were used as television and computer display screens are a must or a clamp on amp probe Using Ohms law to figure out the voltage, by knowing the current on both sides of the high current ceramic resistor, really large size resistor. This is my thoughts, but seek someone with more experience, if you have questions. Electronic fluorescent ballast have little isolation from the AC power outlet to protect the individual doing the measurements.
Lawsen
Note that most HV probes are only rated for low-energetic (CAT I?) circuits, like the HV supply for CRTs. These can only supply a tiny amount of current due to their design. I would be hesitant to use it on circuits with little isolation from the high-energy mains circuit.
Yes, it's possible that a high voltage probe could and connect the test equipment directly to the high energy mains supply.
What about simply using a ferrite cored transformer?
It should be easy to make one with a bandwidth covering 20kHz to 200kHz. Calibration could be achieved at lower voltages, for example using a signal generator set to 20kHz 20V.