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Can anyone explain the PARALLEL MOSFET for PWM dimming ? I have read it in LM3406 datasheet.

Other Parts Discussed in Thread: LM3406

  • Hi Luong Hai,

    The differences is not easy to see until you think about what is happening with the inductor current.

    In the datasheet example the LED's are temporarily bypassed such that the LM3406 can attempt to main the constant current at the CS pin.

    If we put Q1 in series with the LED's a few things happen:

    1.)  The LM3406 would see the inductor current as zero and therefore try to raise VOUT to force current into the LED. This could lead to VOUT approaching VIN.

    2.) When Q1 is engaged the next PWM cycle, there can be much more current flowing through the LED's. This could potentially damage LED's and will likely cause a small color shift in the LED.

    3.) If VIN is very high, VOUT could violate VOUT max. During normal operation The internal FET is sourcing current into the inductor and when it turns off the current in the inductor is sourced through D1 to ramp down the current. This maintains the Volt-Amp balance across the inductor.

      a.) If you suddenly disconnect the load, this current has nowhere to go, so VOUT can actually rise close to or above above VIN

    So my recommendation is to follow the example listed in the datasheet.

    I hope this helps.

  • Thank sir very much. :)
    Can i ask one more question.
    Is there a same problem or any if i do this with other constant current such as AMC7135 circuit driver?
    Schematic example: www.mars.dti.ne.jp/.../lantern.gif
    Thank again.
  • I am not familiar with the AMC7135, but this is a linear device so the power loss is ILED*VDROP. The power loss in the LED's is the same for both solutions VLED*ILED. It does not have an inductor in the circuit so you would not need to worry about issues with voltages going higher than expected.

    The LM3406 can achieve over 95% efficiency so this helps with the thermals and overall size of the solution. (yes with the AMC7135 you would not need an inductor, but you likely would need a much larger heatsink).

    I hope this helps.