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LM3423-Q1: LM3423-Q1

Lee Leong
Intellectual 330 points
Part Number: LM3423-Q1
Other Parts Discussed in Thread: LM3423, LM3421

Hi

Question on  Buck-Boost Application ( Refer to datasheet  "LM342x/-Q1 N-Channel Controllers for Constant-Current LED Drivers",  SNVS574F –JULY 2008–REVISED JULY 2015)

1.    Can  I use  LM3423 to replace LM3421 in the  application circuits on Pg 55 , Figure 41.

2.    I intend to use LM3423 to control  4 Lighting LEDs place  in series on  the application circuits on Page 55, Figure 41. 

        Lighting LED specs : Absolute Max V reversed = 1V, Absolute Forward current max = 1A, Vforward = 3.5V

         I intend to set the LED Average  current to 0.7A by setting up the Rsns , Rhsp and Rcsh

   i. May I know what the voltage at D1's cathode. with respect to  Q2's Drain  if my input voltage is 24V

   ii. Also since  Vin take in Dynamic range 10V to 70V.. When Vin drop  from 70V to 10V during the time when Q1 is turn ON.. Will there will be a reverse Voltage across my 4 LEDs ( Since my lighting LED        absolute reverse max volt is 1V, so , total withstanding voltage is 4V only  )Will this  damage my 4 lighting LEDs in Series?

  • 0
    TI__Mastermind 28655 points
    Hello Lee,

    1) Yes.
    2) It's a buck boost so the voltage at D1 cathode would be 24V = 4X 3.5V = 38V. The Drain of the MOSFET will be a diode drop higher (D1) when off or 38.6V - 39V depending on the diode. If the schematic is per Page 55, Figure 41 no, there will be no reverse voltage. The LED voltage sits on top of Vin so it just rides up and down with the input voltage change.

    Regards,
  • 0 in reply to Irwin Nederbragt
    Intellectual 330 points

    Hi . Thanks for the reply. I have other questions.

    For my design I intend to use 0.7A to drive my External LED. So my I average set at 0.7A

    1. How Should I choose  I limit. (I,e, Current following to Q1). I see yr application note was selected to be 4A.(AN-2010 LM3423 Buck-Boost 2 Layer Evaluation Board (SNVA415C–June 2010–Revised May 2013))

       My understanding is that  I limit should be much more than  Iave of LED. Am I right?

    2. Also refer to yr  AN-2010 LM3423 Buck-Boost 2 Layer Evaluation Board (SNVA415C–June 2010–Revised May 2013)

        Stated on yr BOM  is Part no. MSS1260-473ML ( I found from internet that this part 'specs : I sat is 3.04A , Irms is 2.5A )

         But then I limit  set at 4A . is already  exceed 3.04A. of the Inductor satuation... Is there an error on the part No. of Inductor.. I think it should be higher like 6A. etc.

  • 0 in reply to Lee Leong
    TI__Mastermind 28655 points
    Hello Lee,

    1) This is part of the design process. You have to use Vout maximum, Vin minimum, know your peak to peak current ripple and maximum output current. With four 3.5V leds your output voltage is 14V plus the output rectifier (0.7V). Your minimum input voltage is 10V. Duty cycle will be close to 60%. Without current ripple the average output current during diode conduction will be 0.7A/(1.0-0.6) 1.75A. This is also the average input current when the MOSFET is on.

    As an example: If you are switching at 500 KHz the on time will be (1/500KHz)*0.6 = 1.2 uS. if the buck-boost inductor is 22 uH the peak to peak ripple will be V = L*(di/dt). 10V = 22 uH*(di/1.2 uS), di = 0.55A. The peak current for the Ilimit resistor is 1.75A + (0.55A/2) = 2.025A. I would add margin for efficiency, voltage drops, tolerance on the components, etc and probably set it around 3A - 4A. The actual peak current will be higher due to voltage drops on the board, efficiency, etc.

    2) The I limit is set high enough not to trip during normal operation and will start to run into saturation. The I limit is protection from circuit failure, connecting the wrong load or trying to run the buck-boost with very low input voltage.

    Regards,