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LM25088: 25088

Hale Tan
Intellectual 2480 points
Part Number: LM25088
Other Parts Discussed in Thread: LM5088

Hi, team!

I use LM25088(Buck) to produce -15V with TINA as shown in figure. But GND in chip(Pin 6) provides 8A current which formed a close loop with inductor, diode when switch is open.

So I want to ask that is pin 6 of the chip grounded by default?

  • 0
    TI__Mastermind 46061 points

    Hi Hale 

    Yes, the GND pin should be connected to the ground to simulate it properly.

    - Eric Lee

  • 0 in reply to EricLee
    TI__Intellectual 2480 points

    Thanks for your reply. But I want to get -15V from the chip, so it's not a typical application.

    I tried to use this chip which is connected like buckboost, as shown in figure.

    (This design is from webench)

    So ,my question is that is GND in chip(Pin 6)  already grounded by default, although it is disconnected with ground by wire?

  • 0 in reply to Hale Tan
    TI__Mastermind 46061 points

    Hello Hale

    Many components used inside the LM5088 model are  referenced to the ground symbol whose node name is '0' , and a few components inside the LM5088 model are connected to the GND pin (Pin 6). 

    For example,  

    IO         0 RAMP 25U ==> 25uA is sourced from the ground symbol whose node name is '0' 

    RD        61 GND 1MEG  ==> RD is connected to the GND pin (Pin6) 

    - Eric Lee