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LDO Part No. required.....

Dear Sir,


I would like to convert -15Vdc to -5Vdc(Fixed)/500mA with following attributes.
1. Highly Efficient.
2. Inbuild Input & Output protections
3. Operating Temp. -55degC to +85degC
4. Package: small as possible.
You are requested to kindly suggest appropriate regulator.
Thanks & Warm Regards
Atul Mishra
Design Engineer


  • Hello Atul,

    TI seldom sees the request for creating negative rail from a negative rail, so we do not have a specific IC to do this.

    Of course you could use a linear regulator, and TI has many negative linear regulators, however the efficiency would be 5/15 or 33%.

    Here are 2 that would work for you.

    We do have a reference design using a switcher.

    Buck Converter for Negative Input and Output Voltages Reference Design

    This uses the TPS40210 in a buck configuration.

    Since Vout -5V must be referenced to ground, and the IC is referenced to -15Vin, the output voltage must be level shifted with respect to ground. In the schematic you will see  Q2 Q3 which perform this function.

    You can modify this design to create -5V at 500mA from -15Vin.

    Best Regards,

    Ed Walker
    Texas Instruments 
    Online Support - Systems & Tools


  • In reply to EdWalker:

    Dear Mr. Ed  Walker,

    Thanks for reply,

    Actually we don't have sufficient space to incorporate buck converter at PCB board.

    Consequently,  I went through the part no. UCC284-ep (datasheet as attached) .

    By using UCC284-ep  LOW-DROPOUT 0.5-A NEGATIVE LINEAR REGULATOR,  whether i could able to achieve high efficiency at given attributes

    Input: -15Vdc

    Output: -5V/500mA

    We required efficiency minimum 80%

    Please guide me.

    Best Regards

    Atul Mishra


  • In reply to Atul Mishra:

    Hello Atul,
    A linear regulator's efficiency is simply Vout/Vin.
    So for 5Vout and 15Vin, 5/15, the efficiency is 33%, no matter how much current is passed.

    UCC284, or any linear regulator, may not work for your design, it will be very hot.
    Power loss will be (Vin-Vout) x Iout = 10V x 0.5A = 5Watts

    A switching solution is the only way to achieve 80% efficiency for your voltage conversion.

    Best Regards,

    Ed Walker
    Texas Instruments 
    Online Support - Systems & Tools