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Car power to constant current 3.6A 50v boost converter

Discussion in 'Power Electronics' started by CommanderLake, Mar 14, 2019.

  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,132
    2,663
    Jan 21, 2010
    No, it's fine. I have it memorised to 50 decimal places (the things you do at school!) and he seems to have it correct. The rounding is correct too because the next digits are (from memory) 0288419796939937511

    For my next trick I will use that to calculate the volume of the observable universe to within the volume of a helium atom.

    :)
     
  2. CommanderLake

    CommanderLake

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    Oct 2, 2012
    I use windows calculator and dont bother trimming the extra digits of precision.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I'm sorry, an oversight by me. Looks definitely funny. One could replace Gcs by an expression including Rcomp and solve for Rcomp but first of all I'm not even sure that's the right way to do it and second that would negate the meaning of design equations.
    This component is obviously outdated and replaced by more modern ones. You're probably really better off selecting another, more modern driver circuit.
     
  4. CommanderLake

    CommanderLake

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    Oct 2, 2012
    It cant be that old, the datasheet date is 2016!
    One could test the equations with the component values on the evaluation board.
     
    Last edited: Mar 17, 2019
  5. CommanderLake

    CommanderLake

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    Oct 2, 2012
    I've done the mathematicals with the values from the evaluation board where Rcomp is 2k and Gcs=0.0001 comes closest:
    (24.5 + 0.8) / 0.8 * (5000 * 0.000023 * 2 * 3.1416 / (440 * (1 - 0.714) * 0.0001)) = 1816~

    or if i use an absolute value for GEA its 100:
    (24.5 + 0.8) / 0.8 * (5000 * 0.000023 * 2 * 3.1416 / (0.00044 * (1 - 0.714) * 100)) = 1816~

    and actually there is a 100 ohm resistor on the CS pin!!!

    Ah but thats the resistance I used and not the conductance.
     
    Last edited: Mar 18, 2019
  6. Harald Kapp

    Harald Kapp Moderator Moderator

    8,931
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    Nov 17, 2011
    I'm truly sorry but at the moment I am unable to dive deeper into this matter.
    From my point of view a compoent that is not clearly understandably documented. You might try to contact the company on this page. to get more info on this issue.
     
  7. CommanderLake

    CommanderLake

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    Oct 2, 2012
    Just tested powering the led using my bench supply and a Sinergex puresine 350w inverter and it takes 340w into the inverter to power the led to 144(3.2A) of its maximum 172w(3.6A), thats a pathetic 42.3% efficiency!
     
  8. CommanderLake

    CommanderLake

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    Oct 2, 2012
    At last I got their attention, this is the reply:

    Hi David,

    Thank you for your web enquiry.
    Our FAE (Gary Church – in copy) agrees with your findings and is working with the designers to fix the problem.

    Gary will advise ASAP.

    Kind Regards
    Eric.

    Eric Spooner
    Business Development Manager

    Monolithic Power Systems
    +44 7780 xxxxxx

    [Mod edit to obscure phone number]
     
    Last edited by a moderator: Mar 28, 2019
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Hey, that's good news. Keep us up to date.
     
  10. CommanderLake

    CommanderLake

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    Oct 2, 2012
    I finally have the solution:

    Hi David

    Sorry this has taken a while but I now have an answer. The RLED_AC figure is the AC impedance of the LED which typically is 10-15% of the resistance you calculate by using VLED/ILED. I have also attached a spreadsheet which we have which calculates the values of the Rcomp etc. Please note that Gcs is the reciprocal of the sense resistor value. Please let me know if you need any more help with this.

    Kind Regards

    Gary​

    I'm trying the spreadsheet in Google docs now.
     
  11. CommanderLake

    CommanderLake

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    Oct 2, 2012
    I dont want to use PWM dimming because the circuit looks inefficient and requires many extra components so I had the idea of amplifying the feedback signal with an op-amp and using a pot to adjust it like this:
    Capture.PNG

    But there are so many op-amps to choose from, how do I choose the best one for my purpose, the feedback signal is about 280mV.
     
  12. Harald Kapp

    Harald Kapp Moderator Moderator

    8,931
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    Nov 17, 2011
    The circuit may look inefficient, but PWM is the most efficient way of dimming.

    Messing with a feedback loop is a tricky thing. You can easily mess up the stability of the control loop by changing the phase shift of gain of the feedback stage. Without good knowledge of what you are about to do I do not recommend this. Better stay with the recommended circuit. It should be proven and work well.
    But: there's no argument against fiddling with the feedback loop if you're so inclined. Just be prepared for a few setbacks.

    There is never a "best" opamp (maybe almost never). You need one that
    • is designed to operate with the supply voltages available in your circuit
    • is designed to work with frequencies exceeding your highest signal frequency (which should not be a problem here)
    • which has inputs and outputs that work at the signal voltages required. Depending on your circuit that may require so called rail-to-rail opamps which means they can accept voltages between GND and Vcc at the input and can deliver the same at the output. Note that both inputs and outputs should be rail-to-rail.
    • is available to you (in store, not excessively prized)
     
  13. CommanderLake

    CommanderLake

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    Oct 2, 2012
    Have you simulated the dimming circuit in the EV4008 datasheet? The dimming mosfet with a capacitor on the gate is allowed to drift into its linear region increasing heat dissipation.
    Garry seems to think my idea will work and I need to be able to fine tune the current limit so the LED is not driven past its limit.
    As you say I wouldn't want to mess up the stability of the feedback loop so I wouldn't want a slow op-amp and it has to be responsive in the 70-280mV input region(with 4x gain) without a negative rail.

    How about an instrumentation amplifier?
     
    Last edited: Apr 6, 2019
  14. JMW

    JMW

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    Jan 30, 2012
    I realize your desire to learn. Me, I'm a follower of KISS. I would go to Wally World, purchase 4 U1 batteries with associated cables. They will run your lamp for an hour or so
     
  15. CommanderLake

    CommanderLake

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    Oct 2, 2012
    Still trying to interpret your message but it would still need current regulation given sufficient battery voltage.
     
  16. CommanderLake

    CommanderLake

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    Oct 2, 2012
    How's this for inspiration:

    75C2D7A1-5B09-4805-B5C3-4644124D5A99.jpg

    8F878466-303D-48E6-ABD9-BFEC287D267B.jpg

    A friend visited with his Mavic pro.
     
    Last edited: Apr 7, 2019
  17. Harald Kapp

    Harald Kapp Moderator Moderator

    8,931
    1,766
    Nov 17, 2011
    Pretty bright.
     
  18. CommanderLake

    CommanderLake

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    Oct 2, 2012
    I've decided to get an OPA196 and use this circuit:
    Capture.PNG
    It will allow 10:1 current regulation and has the advantage of redundant potentiometers in case one has a problem while minimizing the load on the OPA196 output to maximize its rail to rail output capability for use with a single supply.
    One pot will be a trimmer to adjust the maximum current while the other will be a volume knob type adjustment for dimming.
     
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