I am a retired electronics engineer trying to fix my black screen Toshiba LCD/LED 40" TV. I have watched various videos showing how to test individual LEDs by scratching the track on either side. I have tried using 3 AAA batteries in series with and without a 100 ohm resistor but cannot get any to light. Where am I going wrong? Would 3 AAAs blow them either way? Or is it the opposite problem - insufficient current? (I have swapped the VESTEL 17IPS20 board for one known good. Mine had a very brown look around the Driver circuit.)

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    \$\begingroup\$ 3 AAA batteries without a resistor could definitely blow the LED. Using a DVM on the diode setting would be a safer way to test them. \$\endgroup\$ – EE_socal Mar 21 '18 at 17:17
  • \$\begingroup\$ One AAA battery would be safer for such a test and enough to lit a led of much higher typical voltage. \$\endgroup\$ – Fredled Mar 21 '18 at 22:45

Are you sure you're not trying to power a bunch of LEDs in series (i.e., a whole or partial column or row?) The LEDs' voltage drops will sum in series, so if you had a column of, say, 10 LEDs in series, each with a 2.5 V drop, you'd need at least a 10 * 2.5 V = 25 V source just to get them conducting.

Also, since LED brightness is current-controlled, you'd need to estimate the voltage left after all the LED diode drops and figure the resistor you'd need to deliver the required current.

For instance, if you have a 30 V source with the 10 LEDs above, you'd be left with

V_source - V_diode_drops_combined = V_leftover

30 V - 10 * 2.5 V = 5 V.

The LEDs might take around 20 mA apiece, and since they're all in series, they all get the same current, so:

V_leftover / I_led = R_current_limiting_res

5 V / 0.02 A = 250 Ω.

Also, these are diodes, which only conduct in one direction. Are you applying the source in the correct polarity?

  • \$\begingroup\$ I suppose putting a relatively unlimited current source through a diode is asking for trouble. It seems that these LEDs are like 2.5V zeners which I remember with fondness. My remaining proiblem therefore is to ascertain how these LEDs are connected on the invisible circuit board strip on which they are mounted. If I have blown any I would at least have expected a flash of short duration but I have so far seen no sign of life from any of them. My next step is to remove one strip and using a car battery with a 400 ohm in series, test four at a time. But must find the polarity first. Any clues? \$\endgroup\$ – Frank Mar 22 '18 at 19:55
  • \$\begingroup\$ Are there any notes from the manufacturer about how the LEDs are addressed or laid out? If you're trying to test a single LED, you probably damaged it when you applied the batteries without a current limiting resistor. But if you had more than one LED in series, the voltage supplied by the batteries would not have been sufficient to cause more than 1 LED to conduct (assuming a 2.5 V diode drop). \$\endgroup\$ – schadjo Mar 22 '18 at 20:47
  • \$\begingroup\$ Since the other series LEDs in the chain would NOT have conducted because the required diode drop overhead was not available, no current would be passed and no LEDs would have blown, and you wouldn't see any light or current flow. The key here is probably estimating the number of diode drops and providing a sufficiently large voltage to power them. \$\endgroup\$ – schadjo Mar 22 '18 at 20:51
  • \$\begingroup\$ The tv is a Toshiba 40L3453DB and the strips are five times 9 marked A (x2), B (x2) and C in the middle. They are Vestel 400 UNDS-02. Can you tell me what reverse votage they can take before breakdown? \$\endgroup\$ – Frank Mar 24 '18 at 8:07
  • \$\begingroup\$ I should have added that using my multimeter on diode test they all show as short circuit, both polarities. But could it really have blown all 54 LEDs together when the set failed? \$\endgroup\$ – Frank Mar 24 '18 at 8:11

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