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I have an old notebook which the backlight is not working. In order to know the reason why, it would be interesting to measure the output voltage 'Vout' from the inverter. Unfortunately, there are some technical difficulties to do so:

  • The inverter exact specifications is unknown, however the best guess so far is Vout ≈ 650 Vrms @ 50KHz
  • The multimeter frequency range is unknown, the manufacturer specifications only mention MAX 750V~

Questions

  1. Is it reasonable to expect Vout ≈ 650 Vrms @ 50KHz? At the lowest dim, this Vout should also be lower?
  2. Considering the high frequency output from the inverter, is it possible to measure Vout with the aforementioned multimeter?
  3. Considering that there are no spare parts available for testing, any other idea to check whether is an inverter failure or a screen failure?

Additional Notes

Notebook Specifications
Toshiba Satellite 1135-S1553

  • Intel® Celeron® 4 processor at 2.20GHz
  • Intel 852GM Chipset
  • 1Gb RAM
  • 32MB internal Integrated Intel® 852GM video memory
  • 15.0”TFT active-matrix display; Internal display supports up to 16M colors at 1024 x 768
  • 75W external AC Adapter, 100-240V / 50-60Hz frequency (Universal) input voltage, 19V x 3.95A Output
  • 8-cell Battery Pack, rechargeable, removable Li-Ion battery, 14.8V x 4300mAh

 

Inverter Images

 

Multimeter Description
Digital Multimeter DT830D

  • Low cost multimeter which can be used to measure resistant, AC, DC Voltage, DC current, transistor, diode.
  • Safety ensure: Overload protection function
  • ACV: 200-750V ±1.0%
  • From experience. It’s 45% chance the cable, 45% chance the inverter, and 10% chance anything else. You can probably get the cable and inverter for next to nothing on eBay. – Appleoddity Jun 5 '19 at 4:13
  • @Appleoddity > "You can probably get the cable and inverter for next to nothing on eBay." Here in Brazil such kind of e-commerce doesn't work that well. I need to be more sure that the failure is due the inverter + cable. – Mark Messa Jun 5 '19 at 4:26
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The high-frequency, high-voltage will not be read correctly on the multimeter and might well damage it. You could make an RF probe with voltage divider, as shown below, to use your multimeter to measure HV RF:

HV RF probe

Try ~1 megohm (MΩ) for R1, 100 kilohm (kΩ) for R2, 0.1 microfarad (μFd) for C1 and almost any diode (e.g., 1N4007) for D1. This would give ~100:1 (well, 99:1, but that's well within component tolerance) ratio, so on the 0-20 VDC (not VAC) range, 650 VAC would read about 6.5 VDC.

That said, all you really care about is go/no go for the inverter; they don't get "weaker" with age. If you place a fluorescent lamp across the terminals, it should light. Any bipin fluorescent lamp, such as a 15 cm tube tube from a camping lantern, Circline, or even a 40 Watt tube, should light. Just connect a wire to pins at each end to each side of the inverter output. If the lamp lights, the inverter is OK.

Bipin Fluorescent Lamp

Since each pair of pins connect across filament, F, connecting to either pin on each side (or both) is sufficient. On the Circline, connect to any two pins that read open-circuit on the Ohm scale of the meter. N.B.: Do not connect the two output wires of the inverter across the filament of the lamp, i.e., two pins on the same side.

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  • > "all you really care about is go/no go for the inverter" You are 100% correct. – Mark Messa Jun 4 '19 at 21:17
  • > "Any bipin fluorescent lamp, such as a 15 cm tube tube from a camping lantern" Good to know. Indeed this is a very nice alternative. Just doing the "devil's advocate" (I'm a little bit concern about the safety of the notebook): Mind to explain a little bit more about the technical details of that? Why an inverter designed for low power CCFL (<4W) could safely turn on a 40W tube? – Mark Messa Jun 4 '19 at 21:30
  • AFAIK, the output of the inverter is two wires. However, bipin lamp has 4 pins in total (two pins in each base). Therefore, how should I connect the two wires from the inverter into the 4 pins of the lamp? – Mark Messa Jun 4 '19 at 21:49
  • @MarkMessa, current is limited by the inverter (or ballast), not by the fluorescent tube. Since the inverter only supplies a few watts, a longer tube actually draws less current when starting.(Once warmed up, there is a constant voltage drop, limited by the ballast/PS.) – DrMoishe Pippik Jun 4 '19 at 22:03
  • > "The high-frequency, high-voltage will not be read correctly on the multimeter" Ok, but following your principle of *go/no go, do you think it is possible to at least check the following: 1) Is there voltage present? Y/N, 2) Is it >100V? Y/N – Mark Messa Jun 5 '19 at 1:08

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