I am using a constant current LED driver from Infineon - TLD2311 (datasheet)

This is a 3 channel driver with a maximum of 120mA (configured to give max) per channel driven with a 12V supply. The LED I'm using has forward current (If) of 120mA and forward voltage drop (Vf) of 3V. 3 of these are connected in series per channel to have a total of 9 LEDs.

So, are these following understanding correct:

  1. Each LED gets only about 40mA as there are 3 in series and max current per channel is 120mA
  2. The voltage the output channel will be about 9V. (Vf = 3.0 * 3Nos in series = 9V)

Now, if I add one more in series (as I want to increase the total brightness) to have 4 per channel then are these correct:

  1. Each LED will now be getting only about 30mA which actually dim each LED further.
  2. Since the supply voltage to the driver is 12V, having 4 LEDs in series will make the voltage drop across the LED to be 12V which is equal to the supply voltage and hence the IC won't be able to drive the LEDs at all.

Please share your inputs.

  • \$\begingroup\$ No that's not right. Everything in series gets the same current. \$\endgroup\$ Commented May 14, 2023 at 15:05
  • \$\begingroup\$ @user1850479 But doesn't that mean that the driver is outputting 360mA and *3 channels that's about 1A in total for the whole IC? I'm not sure if that's true. \$\endgroup\$
    – Abhijith
    Commented May 14, 2023 at 16:01
  • \$\begingroup\$ You said it's outputting 120mA per channel, so that would be 360mA total. \$\endgroup\$ Commented May 14, 2023 at 17:42
  • \$\begingroup\$ @user1850479 I meant, if all resistors in series were getting 120mA like you said then per channel output would be 360mA while the max the IC can do is only 120mA per channel. \$\endgroup\$
    – Abhijith
    Commented May 14, 2023 at 18:18
  • \$\begingroup\$ Current does not add in a series circuit. \$\endgroup\$ Commented May 14, 2023 at 19:00

1 Answer 1


With 120mA from the constant current LED driver entering the node of series LEDs, each LED will have that same current flowing through them, as described by Kirchhoff's Current Law. With the forward voltage across a single LED at 120 mA as \$V_{f(120mA)} = 3V\$, then yes the output of the driver will be 9V -- \$V_{f(120mA)}\$ may not be exactly 3V, so the voltage at the output of the driver may not be exactly 9V, but the goal is to maintain that 120mA of current. Of course, all of this assumes that the supply voltage is high enough to accommodate the forward voltage drop of three series LEDs and the power dissipation of the output power stage of the LED driver.

With 4 LEDs instead of 3 LEDs, the supply voltage would need to be greater than \$4V_{f(120mA)} + V_{PS(CC)}\$, where \$V_{PS(CC)}\$ is the voltage drop over power stage during current control (see 9.2.6 of the datasheet). In this case, your supply voltage will need to be greater than 12V.

  • \$\begingroup\$ I can take care of the supply voltage problem by giving a higher voltage. Regarding the current consumption, if I understand correctly, for the LED to glow in max brightness, then at 3V I have to supply 120mA. Doesn't this mean for all 3 LEDs to glow at max brightness, per channel should be able to supply 360mA? \$\endgroup\$
    – Abhijith
    Commented May 15, 2023 at 5:21

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