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LM5021: The question about the value of test board of "Load Detecting Power Supply" by using LM5021.

user4843850
Prodigy 110 points
Part Number: LM5021

Hi,

I want to ask a question about the test board of "Load Detecting Power Supply" by using LM5021.

In BOM,I can't find the transformer "T1" in any distributors of electronic components and its datasheet!

So I want to ask T1's value!

Ex:primary side inductance.turn ratio...

 The more detailed, the better.

thx a lot ~!m(_ _)m

  • 0
    TI__Intellectual 2910 points

    user4843850,

    The design  I found for a "Load Detecting Power Supply" using the LM5021 is documented here: http://www.ti.com/lit/ug/snvu106/snvu106.pdf

    Is this the same one you are inquiring about? Note that in this design, the reference designator for the transformer is T100. I alsotried searching for the part number on the BOM for this transformer, Pulse PA2685NL. It seems this must have been a special order part from Pulse Electronics. According to the user guide, the turns ratio between transformer bias winding and secondary winding is 2.5:1. I recommend contacting Pulse Electronics inquiring about this part number.

    I will also follow up internally to find out this information, but this may take a few days to resolve, as the design is from a good while ago.

    - Daniel

  • 0 in reply to Daniel Ruiz
    Prodigy 110 points
    WOW!!!That was a quick response.
    exactly, I am inquiring about it!
    I want to know the transformer primary to the auxiliary winding turns ratio.
    Thx
  • 0 in reply to user4843850
    Prodigy 110 points

    another question~

    I want to ask the value of  R109 as shown below

    To realize the LM5021, I simulated the statup circuit.

    But Vin of LM5021 is always below 20(V)

    Is 10.0Meg so big that Q101 can't work?

  • 0 in reply to user4843850
    TI__Genius 9130 points
    The maximum startup current of the LM5021 is 25 uA, this drops maximum 5 V across R103 (200 k). Q101 (FMMT458) minimum hfe is 100 at these current levels, so base current is 0.25 uA, this will drop ~2.5 V across R109 (10 M). So Vin must be at least (23 + 5 + 0.7 + 2.5) = ~31 V in order to be able to charge C103 to the 23 V max start level. Of course, the avialable charging current will be very small, so it will take a long time to charge C103 to the start level.

    In your model, check the hfe of Q101, if it's too low it will limit the charging current and prevent startup.
  • 0 in reply to Bernard Keogh
    Prodigy 110 points
    Bernard Keogh
    Thanks for your help!!

    Can I ask you about selection of tha value R102(1M) and R103(200k)??
    Thank you again~
  • 0 in reply to Bernard Keogh
    Prodigy 110 points

    Is the R102(1M)+R103(200k) limited by Ivin(operation supply current) as show in fig2??

    fig1.FMMT458'datasheet

    fig2.LM5021's datashheet

    continouos collector current of Q101(FMMT458) is 225 (mA).

    Ivin(operation supply current) of LM5021 is 2.5 (mA).

    so the limit current through R103(200k) have to less than 2.5 (mA).

    R102(1M)+R103(200k) >= 240V/2.5mA = 96k

    I think my opinion is wrong, and I don't know why the value of R102 is 1M and R103 is 200k.

  • 0 in reply to user4843850
    TI__Genius 9130 points
    R102 and R103 values are somewhat arbitrary.

    They are used to limit the power dissipation in Q101 and protect it from over-stress, esp under fault conditions. If the VIN pin is shorted to GND, R103 will limit the emitter current to low level, but higher than the LM5021 startup current, to allow it to start (since VIN cap will look like a short anyway during startup).

    R102 will drop a lot of voltage under a fault condition such as VIN short to GND, and this limits the voltage across Q101.

    Q101 circuit is only required to charge the VIN cap initially to the start level - it does not need to supply the VIN pin running current. After startup, the VIN bias current is supplied from the transformer aux winding, through D101 & R108.


    If this answers your question, please click the "verify answer" button.

    Thanks,
    Bernard
  • 0 in reply to Bernard Keogh
    Prodigy 110 points

    Bernard Keogh said:

    They are used to limit the power dissipation in Q101 and protect it from over-stress, esp under fault conditions. If the VIN pin is shorted to GND, R103 will limit the emitter current to low level, but higher than the startup current, to allow it to start (since VIN cap will look like a short anyway during startup).

     If the VIN pin is shorted to GND, I  think  the emitter/collector current can be limited by R102(1M) without R103.

    Ex:240V/1M=240uA

    Could you explain why this circuit need R103(200k) in more detail?

    Thank m(_ _)m

  • 0 in reply to user4843850
    TI__Genius 9130 points
    As I said above, the choice of values is somewhat arbitrary - the startup circuit will also work without R103.
  • 0 in reply to Bernard Keogh
    Prodigy 110 points

    Thanks for your explanation !