• State Resolved
  • Locked Locked
  • Replies 1 reply
  • Subscribers 63 subscribers
  • Views 3759 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.

LM3410: Calculating delta-IL and delta-Vout

solosailor
Prodigy 10 points
Part Number: LM3410

Hello,

I'm looking to use the LM3410 to drive a string of LEDs (the normal ones, 20ma, 3.2V, about 5 to 6 of them) from a 3.3V power source (it'll be eventually a LiPo battery). I've kind of gotten it working but struggle with some of the math in the LM3410's spec.

I know about the error in the spec in regards to calculating the output capacitance (the Vout denominator needs to be delta-Vout) but this brings me to my problem. How do I determine delta-IL and delta-Vout? There is a formula given on page 14 for delta-Vout but it requires the capacitance value (chicken-egg). So far I have simply used the two example values from Table 2 (8.2.1.1).

I have stumbled through a different forum post onto a document called "Basic Calculation of a Boost Converter's Power Stage" that says delta-IL cannot be calculated because the inductor is unknown but it can be estimated: delta-IL = (0.2 to 0.4) * Iout * Vout/Vin and things seem to match up when I use the values from Table 2 in the LM3410 spec. For delta-Vout I'm totally stuck though. The same document does mention delta-Vout = ESR * (Iout/(1-D) + delta-IL/2) but when again using the values from Table 2 I'm nowhere near the stated 250mV.

  • 0
    TI__Mastermind 28655 points
    Hello,

    A simple method to calculate delta V without the LED connected:

    Vin = 3.3V, Vout = 3.2V*6 = 19.2V
    Maximum duty cycle = 1-(3.3V/19.2V)=0.828
    Assuming not ripple current (unless it is high it will be close),
    The output capacitor has to support the load for: 0.828 * 1/Fsw, if Fsw is 1.6 MHz, ton MOSFET = 0.828*(1/1.6 MHz) = 518 ns
    The output current is 20 mA, I = C*(dv/dt) where I = 20 mA, dv = what you choose, say 0.25V, dt is 518 ns, then C = 0.041 uF. If you chose 525 KHz this becomes 0.13 uF. This assumes no ESR in the capacitor. Choosing 0.25V is choosing the peak to peak voltage ripple.

    When the LED sting is added it gets more complicated. Regardless the ripple voltage will be less because as the voltage drops the current in the LEDs drops so it won't discharge as far, as the voltage goes up the LEDs will draw more current so it won't charge as high as a fixed 20 mA load.

    When adding the LED string in you need to find the equivalent series resistance at your operating current, 20 mA. On the voltage versus LED current curve you can pick two points on each side of you operating point and calculate (V1 - V2)/(I1 - I2) = R@20 mA, you also need to multiple the (V1-V2) by the number of LEDs in series.

    6*(3.3V-3.1V)/(25 mA- 15mA)=120 ohms (these numbers may be way off what you are looking at, just to show an example).

    So if the ripple was 0.25V then the current would vary 2.08 mA in the LEDs (0.25V/120 ohms)=0.00208 mA. Since this is small (approximately 19 mA to 21 mA swing the actual voltage ripple will be close to what was calculated above since it is almost fixed at 20 mA load current. If the current swing becomes very large then the non-linearity of the LED resistance will confuse things more. Sometimes it's easier to measure at that point.

    Regards,