I don't understand why manufacturers always exceed voltage to backlight TV (LED strips.) For example, I'm fixing a TV with some LEDs burned out.

This is the backlight:


There are two LED strips. Each strips has 9 LEDs, and each LED consumes 3 volts. All LEDs and strips are in series.

It would be: 3 volts * 9 LEDs * 2 strips = 3 volts * 18 LEDs (2 strips are in series) = 54 volts.

Just 54 volts is enough to "on" the backlights.

Why does the SMPS of the TV deliver 133 volts D when I test the voltage to the backlight?

It's 80 volts above capacity of all LEDs together. Why don't the LEDs burn out instantaneously?

  • 5
    \$\begingroup\$ Are you measuring the voltage with the LEDs on? or are you measuring the open-circuit voltage. The LED power supply quite likely is a constant current source, not a constant voltage source. \$\endgroup\$ May 26, 2020 at 22:37
  • 3
    \$\begingroup\$ Because you drive LEDs with a constant-current supply, and it will deliver the right current for the LEDs regardless of their voltage. \$\endgroup\$ May 26, 2020 at 22:44
  • \$\begingroup\$ Thanks Solomon and Brian . I'm measuring open circuit. But why the SMPS not just deliver 54 to 60 volts ? \$\endgroup\$
    – NIN
    May 26, 2020 at 23:01
  • 1
    \$\begingroup\$ @NIN Because the you can't guarantee that the diodes or the supply are exactly any voltage. if the supply doesn't add up to exactly the voltage drop of all the diodes it's going to burn out. \$\endgroup\$
    – DKNguyen
    Nov 27, 2022 at 21:44

2 Answers 2


Isn't 133V the open-circuit voltage? When LEDs are working, voltage should be lower.

LEDs are driven by constant current sources, it's far better than using voltage regulation, because LEDs have widely varying forward voltages and they tend to have negative temperature coefficient : Overheating reduces voltage which raises current with a constant voltage regulator which leads to even more overheating until one LED in the chain die.

Here is for example the datasheet of a LED driver used in a LG TV I've repaired :


"Current mode DCDC converter"

  • \$\begingroup\$ Wao. I'am experimenting and all you were very helpfull. In fact now I see this source have 450mA. . And backlight have 18 LED's. . . . So (450mA / 18 LEDs) = 25mA per LED. . . Very helpfull \$\endgroup\$
    – NIN
    Jul 20, 2020 at 5:42

Whoa! you're totally misunderstanding how this works.

Each LED will drop a slightly different voltage, so they are wired in series and driven at a contant current which flows through ALL of them. The open-circuit voltage of the supply will always be higher than the on-load figure, because that is the way the current source needs to work. I haven't looked at the data sheet which is mentioned in the original post, but 450mA as the drive current sounds like a lot to me, but the LEDs the tv manufacturers use are heavy duty units. If it is correct, taking 3V as the drop in a diode, 3 V x 0.45 A = 1.35 Watts of dissipation per diode. You can compare 18 x 1.35 to the sort of rating of a filament bulb replacement, so you would be looking at the light output from a 24 Watt LED light bulb - around the same as the amount you would get from a 150 Watt Halogen bulb, so quite bright. What tends to happen in these TVs is that if some of the diodes fail and go short-circuit, the voltage across the chain goes down, raising the voltage drop in the current source. That overheats and shuts down, protecting the rest of the set, at the expense of you losing the picture. If you are really lucky, you could replace the individual LEDs and make things work again, but getting hold of the correct LEDS is not easy, so the best option is to replace the damaged strips.


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