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32V with 200 mA current LED driver IC

Other Parts Discussed in Thread: LM3429, TPS92511, TPS92411

Hello All,

I need output 32V with 200 mA current LED driver IC for my LVDS LCD. Anybody can suggest TI part number for my requirement ?

Regards,

Azlum

  • HI Azium,

    What is VIN? If VIN is higher that 32V, the TPS92511 could be a good choice. If it is lower, the LM3429 may work.

    Note: VF of LED's varies so they may not share evenly. The below technique can help to improve sharing (there still could be variation in brightness between strings) :

    Regards,

  • Hello John Cummings,

    My input (VIN) is between 5 ~ 12V. Do you have some other solution ?

    Regards,
    Azlum
  • HI Azium,

    The LM3429 should be able to work for you. Please let me know if you have any questions.

    Regards,

  • Hello John,

    Thanks for the reply. Could you advise that, how to set desired output voltage ?

    My case,

    Input:

    12V

    Output :

    32V @ 200mA

    Can you provide the reference schematic for above requirements ? Is it availble in WebBench ?

    Regards,

    Azlum

  • HI Azium,

    The great thing about an LED driver is that it will adjust VOUT to regulate the LED current. The peak current is set by:

    The LM3429 has an EVM available that you can test.

  • If you don't need accurate analog dimming or fast PWM dimming the TPS92411 might be a more cost effective solution. It has reference designs close to what you need on the web. There is a 36V output one at 700mA, you could use the same circuit but just adjust the LED current. It may need a little tweaking (like if the 700mA application ripple current is too high for a 200mA app, you would increase the inductor value) but should be close.
  • Dear John Cummings,

    Thanks for the reply.

    For LED driver IC, the output voltage will be set automatically according to the LED current. And no need of setting the desired forward voltage (Vf).

    In other words, same circuit can be used for 32V @ 200mA, 36V @ 200mA & xxV @ 200mA (different voltage at same current) without changing any discrete components.

    My understanding is correct ?

    Regards,
    Azlum
  • The concept of LED driver IC is below:-

    Same circuit can be used for 12V @ 200mA, 21V @ 200mA & xxV (maximum output voltage) @ 200mA (different voltage at same current) without changing any discrete components unless change the LED current ?

    So how is the LED forward voltage (VF) is setting automatically ?

    Regards,
    Azlum
  • Dear John Cummings,

    Could you share your feedback ?

    Regards,

    Azlum

  • Dear John Cummings,

    I have connected 6 LEDs series. One LED current is 1A. So the LED string current is 1A since 6 LEDs are connected in series. 

    When we look into the equations, there are two parameters:- 

    1) Average LED Current. Please let know how Vsns = 100 mV ?

    2) Peak LED current.

    Here why we need to consider ILIM as 6A even though we didn't connect the LEDs in parallel ?

    Also for below configuration what will be the ILIM ?

    Above ILED is 25 mA

    What will be ILIM ?

    Regards,

    Azlum

  • Hi Azlum,

    It depends on the input voltage and output voltage relationship if they are in a boost configuration.
    For instance, if your input voltage is at 6V and your output voltage is at 36V, the boost ratio is 6.
    With such boost ratio, your MOSFET needs to get current 6 x your targeted output current.
    Therefore, the ILIM should be set at 6 A.

    The above just illustrates an ideal calculations.

    In actual designs, you need to get some margin for this current limit.
    So, if your boost ratio is at 6 and targeted output current at 1 A, it is suggested to set current limit higher than 6 A (say 6.6 A).

    With your 200 mA examples, it also depends on the boost ratio.
    I would suggest a current limit of (200 mA x boost ratio + 10 ~ 15% margin).

    Best Regards,
    Issac.
  • Dear Issac,

    My input is 12V
    Output is 32V.
    Ratio is 32/12 = 2.66.

    ILIM = 200 mA * 2.66 + 15%
    = 0.6118 A

    My calculation is proper. I guess. Can you confirm ?

    Regards,
    Azlum