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UCC29002: UCC29002:During the start-up, while the current not sharing a few time

user4717993
Prodigy 30 points
Part Number: UCC29002

HI

I  parallel two isolated DC/DC module with 50 V output voltage, 12.5 A output current capability by using UCC29002. The waveform is as follows


ch2:module 1 output current
ch3:module 2 output current
ch4:output voltage


Q1:How can I sharing modules current in the start time?

Q2:How can I reduce the current error between module 1 and module 2?

  • 0
    TI__Mastermind 34465 points

    Hi,

    The UCC29002 forces the power module into its maximum load sharing capability at startup. This is described in the datasheet section 7.3.8.
    Load sharing is disabled at startup. The reason for this is to force a high current startup.

    The UCC29002-1 does not have this feature so load sharing is enabled even at startup with this device.

    You should be able to achieve better accuracy in load sharing than you are seeing by following the tips in the data sheet and design documents for the UCC29002.

    Theuvenin connections to the load are important. Use a high value of current shunt so that the sensed voltage is accurate. PCB layout and  routing of noisy traces away from the low level current signal is important. 
    If you carefully follow these rules and hints then you should expect to achieve an accuracy of 5% or so between the current levels.

    Regards

    John

  • 0 in reply to John Griffin1
    Prodigy 30 points

    Hi John

    Thanks for your reply

    I met some problem yesterday.The waveform and sehematic is as follows   ps:





    Q1:How can I improve my current sharing that be close to each other not separate.

    Q2:Could you tall me why current separate each other slowly.

    Thanks

  • 0 in reply to user4717993
    TI__Mastermind 34465 points

    Hi

    The UCC29002 has been designed for use with high tolerance current sensing shunt resistors.

    Do you know what the accuracy of your Hall effect sensors is ?

    Regards

    John

  • 0 in reply to John Griffin1
    TI__Mastermind 34465 points

    Hi 

    It has been almost 2 weeks since the last post in this E2E thread. I will close this post and recommend that you use a current shunt resistor in place of the Hall effect device.

    Please open a new thread if you need further assistance.

    Regards

    John