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TIDA-00704: SAMPLE BOARD

Giorgio Saccavino
Prodigy 140 points

Part Number: TIDA-00704

Dear Sirs,

I'm trying to reproduce your TIDA-00704. I've all that I need except some components (tipically capacitors, but not only) which are in allocation.

I am designing my custom battery charger that is very similar to TIDA board (the difference is that the references and some controls are digitally controlled); anyway I want to check first the TIDA board because I am a beginner in resonant topologies (allocation do not involve the production cycle).

The question is: I know that the evaluation board is'n for sale pourposes, but, is really impossible to receive a prototype?

Thanks in advance

Giorgio

  • 0
    TI__Expert 5120 points
    Giorgio, hello and thank you for you interest in Texas Instruments. As a core part of our our TI Designs strategy, we don't customarily make our prototypes available for sampling. We do however make copies of the design on a limited basis such that we can test and document the solution. I will contact the local design team in India regarding any possible availability, but I want to set expectations up front that the possibility of one being available is remote. The design was originally designed in August 2016 and some minor modifications made to the schematic/BOM in March 2017.

    Best Regards,

    John Fullilove
    Texas Instruments
    TI Designs
  • 0 in reply to JohnFullilove
    TI__Expert 5120 points
    Giorgio, hello and we appreciate your patience. I have conferred with our design team and we do not currently have any sample boards available, however, we would be more than happy to consult with you on any challenges/questions you may have with board development. Please let us know where we could possibly help.

    Best Regards,,
    John Fullilove
  • 0 in reply to JohnFullilove
    Prodigy 140 points
    Hi John,
    do not worry for the board, I'll do PCB and I search sample components to do the TIDA board.
    As i told you I'm a beginner in LLC design and so your support, is really appreciated.
    I just have a technical question:
    In the document that explain the dimensioning of the board (TIDUC71A) I noticed that the plotted gain at fsw=f0 is equal to 1. I think that this is correct when the resonant inductor is not integrated in the transformer unlike in TIDA design. I suppose that integrated transformers should take in account the secondary leackage inductance. If it is true, the gain at fsw=f0 is Greater than 1 and all other peak gains are increased.
    What do you think about?

    Thanks in advance

    Best regards

    Giorgio
  • 0 in reply to Giorgio Saccavino
    TI__Expert 3860 points
    Hi Giorgio,

    Whether we use an integrated transformer or not, f0 will be dependent on the leakage inductance (combined value) and the gain of the power stage will be 1 at f0. It will be above 1 only for switching frequencies below the resonant frequency.

    Regards,

    Salil
  • 0 in reply to Salil
    Prodigy 140 points
    Hi Salil,
    I found recently an Application note in which is described the difference of gain at f0 between Transformer+inductor and integrated transformer. In practice, in case of integrated transformer, the numerator of the gain equation is multiplied for a RADQ(Lp/(Lp-Lr)) where:
    Lr=Llkp+Lm//(n^2*Llks) ; Lp=Llkp+Lm. Indeed, if it is true, at F0 the gain is greather than 1.
    What do you think about?

    Regards,

    Giorgio
  • 0 in reply to Giorgio Saccavino
    TI__Expert 3860 points
    Hi Giorgio,

    We are assuming that the effect of secondary side leakage is insignificant in our design calculations. I agree that there can be some change in the gain. However, the change being in the positive direction, it is probably not a problem. The power stage will operate at slightly higher switching frequency than the calculated value. For an initial estimate of operating point, this may be sufficient.

    Regards,

    Salil
  • 0 in reply to Salil
    Prodigy 140 points
    Hi Salil,
    I agree with you that the gain variation isn't a big problem. Anyway, I think that to obtain a gain = 1 at F0, the difference from the physical turn ratio and tha equivalent turn ratio derived from APR (all primary referred) model of the transformer must be take into account.

    Regards,

    Giorgio