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TIDM-BIDIR-400-12: Questions about the target customer and the estimated applications of TIDM-BIDIR-400-12

Takumi Kimura
Prodigy 100 points
Part Number: TIDM-BIDIR-400-12

Hello, Team,

 

First, I’ d like to show my gratitude towards the team who supported me to consider prototyping the converter kindly last year.

 

I’d like to ask about the target customer and the estimated applications of the converter.

 

As you know, nowadays, various types of bi-directional DC-DC converter are developed all over the globe. Especially, I often see some converters whose nominal low voltage outputs are 48V are released. However, I’ve hardly seen the converters whose nominal low voltage outputs are 12V, like TIDM-BIDIR-400-12.

 

I assume that this is attributed to the needs in the world or something. Also, according to the design guide, I assume that the reason why the nominal low voltage output of the converter is 12V is to charge the battery.

 

Based on the above expectations, I’d like to ask you the target customer and the estimated applications of the converter, in other words, the purpose of developing the converter in detail.

 

Also, if you know, I’d like to ask the reason why the converters whose nominal low voltage outputs are 48V are mainly released.

 

I appreciate your every support.

Best Regards,

Takumi

  • +1
    TI__Mastermind 42015 points

    Takumi-San,

    the applications that are used for TIDM-BIDIR-400-12 are listed here: https://www.ti.com/tool/TIDM-BIDIR-400-12#applications 

    Historically 12V has been common for many electronic systems. I do not have a full history as to why this is, but it is a relatively safe voltage level which can provide enough electromotive force for many applications. And of course we cant forget that 12V was easy to obtain by historic battery chemistries. 

    For smaller wattage loads 12V can work well, but when you start to look at 2 or even 3kW loads this would require 250A at 12V. This is where 48V systems begin to make more sense. When using 48V you can reduce the size of the conductor for a given load because the current flowing will be smaller.

    Regards,
    Cody 

  • 0 in reply to Cody Watkins
    Prodigy 100 points

    Dear Cody,

    Thank you for answering my question.

    Thanks to your detailed explanations, I could get convinced.

    Thank you for everything.

    Best Regards,
    Takumi