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Can VLDOIN be higher than VIN on the TPS51200

Jamaal_Charles
Genius 3245 points
Other Parts Discussed in Thread: TPS51200

Hello,

I can't find any relationship between VIN and VLDOIN in the TPS51200 datasheet.  For our application, we were hoping to use VLDOIN and VREF at 3.3V, and VIN at 3.0V.  Technically, this all falls within the acceptable ranges in the datasheet.  However, it seems strange to have VLDOIN higher than VIN, and I don't see this scenario in any of the examples I can find.  In our case, we need VLDOIN to be able to range from 1.2V to 3.3V.

Thanks,

Jamaal

  • 0
    Prodigy 5 points

    Hi JC.

    VIN is just the supply for the IC brains, 2.5-3.3V, very low current, very little heat.
    VLDOin is the input to the VTT linear regulator, high current possible, high heat created.

    The intent is to operate the LDO input at the lowest voltage possible and stay above the dropout voltage.
    The LDO is designed to provide a VTT DDR termination voltage that is one half of VDDQ (VDDQ/2)

    Page 22 Figure 5 of the datasheet is a good typical example.
    VDDQ is 1.8V, so is too low for VIN, so VIN is supplied with 3.3V.
    VTT is 0.9V, VLDOIN is taken from VDDQ 1.8V so you have 0.9V overhead on the LDO.

    Since REFIN is 0.5-1.8V, it sets the VO LDO output to 0.5-1.8V.
    You can have VLDOIN at anything HIGHER than the VO + dropout voltage (Figure 22 page 21).
    The higher VLDOIN is above the dropout voltage, the more heat it will dissipate.

    VLDOIN can be higher than VIN, but VO will always be lower than VIN and the IC will be able to drive the internal regulating FETs.

  • 0
    TI__Guru**** 191445 points

    There is no direct relationship between VIN and VLDO in.  VIN is menat to be either a 2.5 or 3.3V input (max range is 2.375 to 3.5 V).  It is a supply used internal to the IC.

    VLDOIN is the input to the source/sink LDO. Maximum voltage is also 3.5 V.  So VLDOIN can be in your range of 1.2 to 3.3 V.  But REFIN would would have to be less than 1.2 V.

    The TPS51200 IS generally intended for DDR termination applications.  In that case, REFIN is intended to be one half VLDOIN and VO = REFIN.  You do nat actually have to implement the tracking function by tying REFIN to VLDOIN with a resistor divider, but you do need to maintain the voltage limits.

    The input range for REFIN is 0.5 to 1.8 V, so VO is limited to that range also.  by setting REFIN to 3.3V you imply you want VO to be 3.3 as well.

    Can you tell us more about what you are actually trying to accomplish?

     

  • 0 in reply to JohnTucker
    Genius 3245 points

    Thanks John, Ed,

    Sorry, I did make a mistake, we do have a voltage divider to halve the VREF input.  The application is DDR terminations.

    Thanks again!

  • 0 in reply to Ed Walker
    Prodigy 20 points

    Hallo,

    I have a question to check if the following is possible.

    So if i use the TPS51200 in the following way :

    VIN = 3.3V

    VLDOIN = 2.5V

    REFIN = 1.35V

    Then the output will be (correct me if i'm wrong ):

    VO = 1.35V

    REFOUT = REFIN/2 (or is it juist REFIN ?)

    The only downsite to this is that i have a higher power dissipation, is this correct ?.

    2.5 - 1.35 = 1.15 V  

    1.15V x 1.5A = 1.725W dissipation

  • 0 in reply to Hidde Goemans
    TI__Guru**** 191445 points

    In a typical DDR application VLDOIN = VDDQ and REFIN = VDDQ/2 and VO = VTT = VDDQ/2.  You would then use an external divider to generate REFIN from VDDQ.

    For your application, you can supply VLDOIN with 2.5 V to generate VO = 1.35 V if REFIN = 1.35 V also.  REFOUT = REFIN.  Your power dissipation is actually lower. as VLDOIN would usually be VO x 2 = 2.7 V.

  • 0 in reply to JohnTucker
    Prodigy 20 points

    Hallo John,

    Thank you for the fast response.

    In our application we use one (1) DDR3L chip for Micron. This one got on-die termination on the DQ lines.

    So we are using the TPS51200 as a LDO to supply the VDD and VDDQ with 1.35V . 

    and use the VREFOUT to supply the VREF of the DDR through a resistor division.

    I quess this is not the way this chip is intended to use, but is it possible ? because we use only one DDR chip, so the current needed is limited.