Help Finalizing LED Boost converter

Discussion in 'The Projects Forum' started by mattaus, Apr 8, 2013.

  1. mattaus

    Thread Starter New Member

    Jan 18, 2013
    16
    1
    Hi all,

    I have designed (tried to design maybe?) a 500mA LED boost driver based around the LM3410X LED driver. The completed schematic for my driver is below. Apologies if my nomenclature is off...

    [​IMG]

    The control circuit half of the driver is based around an ATMEL ATTINY 13V micro-controller and incorporates some external parts for battery voltage detection, reverse polarity protection and thermal protection. The PWM input of the LM3410 chip (SOT-23) is driven by pin 7 of the ATMEL chip. I know the control half of the circuit works as it is currently humming along nicely in a simple linear regulator driver I built recently.

    Anyway, for the boost converter half I am relatively confident it's correct as it is more or less based around the circuit depicted in design example 4 of the LM3410 datasheet (page 28). The only difference is I have boosted my output current from 340mA to 500mA.

    A few questions and notes:

    1) I can't for the life of me figure out why they have R1 and R3 in parallel in design example 4. Why not a single resistor as per pretty much every other design example in the datasheet? I went with a single resistor and set it to provide me with my desired LED string current.

    2) I've listed the Schottky diode as needing to be greater than 500mA and 10V - this is because I have not started looking for a suitable component yet.

    3) I need help picking a suitable inductor. The maths isn't beyond me but my fundamental understanding of inductors is lacking, and I want to make sure I pick a suitable part for the job. Footprint size is an issue so if I can get some parameters to look for (and obviously an explanation as to how these values were arrived at) I can go hunting for the part myself.

    That's about all of got. If anyone notices anything wrong with the circuit, or even ways to improve it, I would be very appreciative.

    Thanks,

    - Matt
     
  2. takao21203

    Distinguished Member

    Apr 28, 2012
    3,577
    463
    What is the value of these LED driver inductors? 22uH?

    Typically a pot core SMD inductor, you can get them easily upto a few Amps.
    They are square, as the name says, the core with the coil is enclosed in a containing pot.
     
  3. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    Don't know about the rest but the reason for the two resistors is power dissipation.
    There is around 0.15W in the R1 R3 combo and the typical power dissipation for the 0805 package in the parts list is 0.125W. A single resistor is going to get very hot, but a pair should be OK.
    Even with two resistors I'd recommend making the pads extra large as more copper helps heatsinking.
     
  4. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    What is your powersupply? I would definitely add 100nF caps close to both ICs supply pins. Also, the 10uF bulk cap will probably not be enough, so maybe some 100uF added to that would be a good idea, but this really depends on how far away is the PSU and how well filtered it is.
     
  5. takao21203

    Distinguished Member

    Apr 28, 2012
    3,577
    463
    The 10uF will be helping, but uC + switcher always gives phenomena.

    Today I linked a new TFT LCD circuit with 3.3V Li-Ion dc/dc board.

    Of course, after some minutes, uC LED starts to blink erratically, and stops. Even the 3MHz dc/dc shuts down!

    Adding a large 2200uH coil from a CFL, both problems solved.

    It is maybe not as much critical for dc/dc input side, but to some degree, the same problem.
     
  6. mattaus

    Thread Starter New Member

    Jan 18, 2013
    16
    1
    Ah...see these are the things I'm still learning to recognise :rolleyes:. How did you come to 0.15W? With a string voltage of 11V and 340mA I get 3.71W... I'm clearly missing something basic? I am very tired lol. Regardless, my power dissipation will be higher so I may very well have to do something similar.




    My power supply is a single 18350 lithium-ion rechargeable cell (which is a half size 18650 which are what laptop battery packs are typically constructed from). In the past I ahve had no problems with this as a power source and half rarely had to use bulk filtering caps. Is there any reason, now that you know what power supply I am using, that you would recommend additional capacitors to my circuit above now?
     
  7. Markd77

    Senior Member

    Sep 7, 2009
    2,803
    594
    I made an error in my calculation earlier. The proper calculation for yours is:
    Voltage across resistor = current X resistance = 0.5 X 0.38 = 0.19
    power = voltage X current = 0.19V X 0.5A = 0.095W
    Example 4 in the datasheet is actually only 0.07W, a little over half the rating for a single resistor.
    I'd still go for a pair, I think a single one would get pretty hot.
     
  8. mattaus

    Thread Starter New Member

    Jan 18, 2013
    16
    1
    I think the real reason for the resistor pair is that they provide a combined parallel resistance of 600mOhm, which is what is required to properly set the LED string current at approximately 340mA. 0.6Ohm resistors are a little harder to come by than single 1 and 1.5 ohm resistors.

    I didn't figure this out either - someone else pointed it out :)

    Of course heat generated is still a concern as well so 2 resistors has 2 benefits in this case.
     
  9. mattaus

    Thread Starter New Member

    Jan 18, 2013
    16
    1
    Actually I have one more question - and this is more of a thought experiment than anything else....just a "what if".

    What if you tied the DMM pin to VIN (so the LM3410 IC is always enabled), and connected a switch with 2 different resistors in place of the resistor on the FB pin (380 mill-ohm)?

    Set the 2 resistors so that one provides the desired maximum current of 500mA, and the other provides a lower current, so for example 40mA.

    Press the switch to alternate between the 2, the feedback voltage on the FB pin changes, and a different current flows through the LED strong. High and low without the need for a micro-controller.

    Would this work? I'm reading through the datasheet now to try and see if it mentions anything like this but so far nothing.

    Thanks,

    - Matt
     
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