• State Resolved
  • Locked Locked
  • Replies 5 replies
  • Answers 1 answer
  • Subscribers 63 subscribers
  • Views 429 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

LM3410

LM3410XBSTOVPEV: How to estimate switching current, and why is this board limited to 3.3 V when the part is 2.7 V rated on the input?

Dan Holt62
Prodigy 30 points
Part Number: LM3410XBSTOVPEV
Other Parts Discussed in Thread: LM3410

What exactly is causing the 3.3 V limit on this board rather than 2.7 V min.?  It doesn't look to be the caps, etc.  Is it because the switching current will be too high with the input at 2.7 V and the output at 12 V given that the board is supposed to deliver 190 mA to the LED?

Also, how did Victor come up with the below 4A?

----------------------

  • 0
    TI__Mastermind 28655 points
    Hello Dan,

    There are equations for the boost converter in the datasheet but if you simply take 12V at 1A that is 12W. 2.7V input would need to be over 4A for the same power, (12W/2.7V=4.4A). This doesn't include losses or current ripple in the inductor, just a simple calculation.

    If the output is 0.19A I don't see why it won't run down to 2.7V.

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Prodigy 30 points
    1. I got it. Power in equals power out if assumed 100% efficiency / no losses. I thought there was more to it than that. THANKS!

    2. I don't either. Could someone explain? If you look at the very first page of the description of: LM3410XBSTOVPEV. it specifically says it can't go below 3.3V. It even states it again in the User's Guide / Datasheet ( look for the '{{{{ }}}}' ):

    Description
    This evaluation board showcases the LM3410X as a boost LED driver. It is designed to drive four, on-board LEDs (VOUT = 11.4V) in series at an average LED current (ILED) of 190mA. The circuit can accept an input voltage of 3.3V-5.5V. The switching frequency of the LM3410X converter is 1.6MHz allowing the use of small surface mount inductors and chip capacitors. This evaluation board also features the PWM capability of the LM3410 by enabling the user to apply a periodic pulse signal to the DIM terminal of varying duty cycle. This is a 2-layer board using the bottom layer as a ground plane. {{{{The above restrictions for the input voltage are valid only for this evaluation board.}}}}

    Features
    Operating Conditions

    VIN = 3.3V to 5.5V
    VOUT =˜ VF x 4 + VFB =˜ 2.8V x 4 + 0.190V =˜ 11.4V
    ILED =˜ 190 mA

    Thanks for the VERY FAST answer!
    Regards,
    Dan
  • 0 in reply to Dan Holt62
    TI__Mastermind 28655 points
    Hello Dan,

    It's noted in the documentation so it is the recommend operating range.

    It's not duty cycle limit. I can only think it's a board limitation or the limit was placed to make sure any voltage drops on wire connections to the input. Also have to watch the input current ripple on the input capacitor (looks like only one is installed).

    Best Regards,
  • 0 in reply to Irwin Nederbragt
    Prodigy 30 points

    My prototype is working great.  Thanks for all your help.

    I did have another question.  Attached I have waveforms from my o-scope of the input voltage and output voltage as well as a picture showing the efficiency (~78%).  Would increasing the capacitance on the input and / or output make the efficiency any better?  I'm not complaining about the 78% as that is pretty good, but I was wondering if I could make it better.  All I modified on the board was changing R1 to 6.49 ohms (30ish mA), adding a 100k pull-up to DIM, and adding a 0 ohm to be able to use the DIM if I wanted (not used currently in my design).  Input ripple is 36 mV.  Output ripple is 148 mV.

    Happy Holidays,

    Dan

  • 0 in reply to Dan Holt62
    TI__Mastermind 28655 points
    Hi Dan,

    Increasing the input and output capacitors probably won't make much difference efficiency wise. You can try doubling them as see. It just depends on how much is being lost in the resistive portions of the circuit (RMS losses). Average losses such as diodes won't really change much. To find out what would improve efficiency would require going through the entire design to see where the losses are. Most will probably be in the power components, switching loss, inductor losses, etc.

    Glad it's working for you.

    Best Regards,