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Negative output Voltage

Category: Hardware
Product Number: LTC1624

Hello,

I found a component that may be suitable for my application the LTC1624 (please see the DS in this attached). I don't understand very well how it works this electronic assembly :

Is it possible to delete M2/M3 and R3 to have something simpler. I have only +12V (comes from the battery voltage) and I need -5V with 1A to the ouput. Thank you for your helps.

Regards

Yann DOPDF   

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  • FormerMember
    0 FormerMember

You can use this schematic and adjust accordingly to satisfy your specification.

  • Hi,

    After drawing the circuit I don't get -5V but -6V. 

      

     Something is wrong but I don't know where. But after the DS normally I'll have -5V with a supply +12V 

      

    I just removed the following components : M3/M2/R3 and Shutdown. Do you have a idea where is my mistake ? Please see in this attached my simulation. 

    PJ LTC1624.zip

  • Reply
    • FormerMember
      0 FormerMember
    in reply to 206_RC
    Children in reply to 206_RC

    That is a negative output.

    In the schematic,

    The referenced ground is the VNEG output.

    The circuit I sent is an alternative as to how to generate a negative output from a buck controller in the simulation.

  • The referenced ground is the VNEG output.

    Sorry but I can't see the VNEG output in your schematic. Where it's (sorry I'm a beginner under LTspice).

    Where must I take my probe to see the negative voltage ?  

  • FormerMember
    0 FormerMember in reply to 206_RC

    In this particular connection, ground acts as the VNEG terminal of the output and the OUT acts as the positive terminal of the output.

    If you want to see a negative voltage, search on the LTSpice documentation how to measure a voltage with a certain reference.

    Refer to this documentation: AN-1083 (Rev. A) (analog.com)

  • Now I can see the negative voltage

    I looked nowhere I just guessed your explanation only. But my question is Can I take my ground as the negative value ? In my application I need to supply my AOP with a voltage symmetry with a common ground. So if I connect this point to my ground 

       

    Where is my common mass to supply my AOP with +5V an -5V ?

  • FormerMember
    0 FormerMember in reply to 206_RC

    If you can see the voltage supply, the supply negative terminal of the supply is the OUT pin. This is the true ground.

    I just used the ground symbol as the VNEG terminal of the output because some of the buck ICs has internal ground, which might cause an error when connecting ground as the positive terminal and a floating ground as the NEGATIVE terminal.

  • If you can see the voltage supply, the supply negative terminal of the supply is the OUT pin. This is the true ground.

    So a little complicated for me, after your explanation the common ground is this point name OUT right ?

    And the negative voltage is this point name 'GND' right ?

     

    So I think I'll open a new ticket to make another electronic assembly that provides +5V which I put together with your assembly (schematic here) to see if I have +5V and -5V with a common ground. This point is very important for me to supply my AOP

    Thank you for your helps

    Regards

  • Hi,

    After simulating 2 electronic diagrams +5V and -5V, I assembled the 2. And I don't get the expected result. Do you know where is my mistake ? 

    In my +5V circuit it works

      

    But in my -5V something is wrong but it works independently very well

     

    I don't have -5V, I don't get the expected -5V but +12V.

    Thank you for your helps

    Regards

    PJ : The file of simulation

    Yann DOEzone-LTC1624_+5V_-5V.zip

  • FormerMember
    0 FormerMember in reply to 206_RC

    Your GND pin should still be grounded, not -5V.

    When you connect your input between VIN and VOUT of a buck regulator, it will become an inverting buck boost regulator with GND being the VNEG AND REFERENCE of the controller.

  • Hi,

    After grounding my GND (for the -5V circuit) and putting the ref on the resistor Rload1. I have +5V and -5V 

    But I think it's something wrong for my application. I need the same ground to supply my AOP, as you know the AOP need +5V and -5V with the same ground so in my simulation where is my ground ? Because the ground for the IBB is -5V ? 

    Can I connect my ground of my AOP as this point ? (Negative of my batterie ? But it's also +5V in my schematic) 

    I'm a little confused about the mass of the component and the mass of my electronic assembly. Do you have a idea to connect my AOP with the same ground ? 

    Thank you for your helps

    Regards