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LM3414: Webbench Simulation

Amy Domino
Prodigy 190 points
Part Number: LM3414


Hello,

I am trying to run a webbench thermal simulation of the LM3414 (Vin min is 4.91 V and max is 5.09 V) . My LED (Osram LUW H9GP0) has a forward voltage of 3.45V and a current of 700mA. When I try to make a custom LED in webbench, I get an error that the design cannot be created with the LM3414 since the duty cycle is above the limit. 

Webbench calculates the duty cycle to be 102%. Isnt the DC Vled/Vin? So 3.45V/4.91V which is about 70% DC. How is webbench calculating the duty cycle?

Amy

  • 0
    TI__Mastermind 25500 points

    Hello Amy,

    Webench likely takes the worst case loss in the diode and the switch when determining duty cycle. You also need to add in the 200mV reference to the output voltage. So it is possible in the worst scenario that the duty cycle is much higher. But whether that is a problem depends on the application. This device is capable of 100% duty cycle operation. So it will still work, but if it reaches 100% duty cycle the LED current could reduce some depending on the switch Rds(on) and the inductor ESR. It sounds like in this case the current drop would be minimal. It would likely work fine but let me know if you need something else for your application and I may be able to make a recommendation.

    Regards,

    Clint

  • 0 in reply to Clinton Jensen
    Prodigy 190 points
    Thanks Clint for the quick reply.

    I have a few more questions regarding this part. My main concern here is power dissipation and heat.

    1. Can you explain why a 200mV reference needs to be added to the output voltage? So the output voltage would be 3.65V instead of 3.45V?
    2. How can I calculate the DC using worst case loss in the diode and the switch?
    3. Do the webench simulations take into account worst case RDSON of switch? What is the worst case RDSON of the switch?
    4. One graph I find confusing in the data sheet is VIN versus Efficiency. As VIN increases, efficiency goes down. I would think it would be the opposite because the higher the Vin, the lower the duty cycle and the less time there is current going through that resistor.
    5. I calculated the efficiency below using equations from a different e2e post. Does 76% efficiency look right?

    From <e2e.ti.com/.../2382885>

    1 - Pswf = 0.5 * 5 * 0.7 * 500e3 * 10e-9 = 0.0088

    2 - Pswr = 0.5 * 5 * 0.7 * 500e3 * 10e-9 = 0.0088

    3 - Pcond = (0.7)^2 * 1.8 * 0.75 = 0.6615

    4 - Pdiode = 0.42 * 0.7 * .25 = 0.0735

    5 - Pdrive = 2.05e-9 * 5 * 500e3 = 0.0051



    Ploss = .0088*2 + 0.6615 + 0.0735 + 0.0051 = 0.7577

    Pled = 3.45 * 0.7 = 2.415

    Ptotal = 2.415 + 0.7577 = 3.1727

    Efficiency = 2.415 / 3.1727 = 0.7612
  • 0 in reply to Amy Domino
    Prodigy 190 points
    Here is the link to that other e2e post -->e2e.ti.com/.../2382885
  • 0 in reply to Amy Domino
    TI__Mastermind 25500 points

    Hello Amy,

    To answer your questions:

    1. I'm sorry, that was my error. This uses a low side switch with switch current sense, so you do not add the 200mV.

    2. DC = Vled/(Vin*efficiency) is the easiest and just assume 0.85 (it should be between 85% to 90%). It gets you close enough for all practical purposes.

    3. I believe they use what is specified in the datasheet which is a maximum value only in this datasheet. So yes, they use the maximum value which is 1.8ohm.

    4. This is common for a buck regulator. Usually the diode is a little more lossy than the switch, maybe they are close in your case. But the AC losses are proportional to Vin^2 so they become more of a factor as the input is increased.

    5. Your calculations look ok. But the reality of it is that the 1.8ohm spec is a maximum. Under most operating conditions it is likely to be less than about half that. I expect your actual efficiency will be about 10% higher than calculated.

    Regards,

    Clint

  • 0 in reply to Clinton Jensen
    Prodigy 190 points
    Great, thanks Clinton!