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LM1084: 5-Volt 5-Amp Voltage Regulator

Leandro Hernandez
Prodigy 20 points
Part Number: LM1084

Hello, I am an Electrical Engineering student and would like some help with the following problem. I am using the LM1084 as an adjustable voltage regulator to output 5V at 5A as described in the datasheet. I printed a PCB to implement this design but it seems like the IC shuts off. I am using a 10uF Tantalum capacitor at the output as suggested and a 10 uF electrolytic capacitor at the input. I was also attempting to use a power transistor to output a higher current on a second design but since I cannot get the base design to function that design did not work as well and encountered the same problem. Is there something I am missing? I am using a 24V LiPo battery at the input and using a dummy load and cannot get more than 0.5 amps. I have done as much research into it as my knowledge allows me to so I was hoping someone can provide an idea as to why this is happening. Thank you very much.

  • 0
    TI__Mastermind 37925 points

    Hi Leandro!

    A likely cause is that the LDO is dissipating too much power and is going into thermal shutdown mode. With your setup, the power dissipated at full load woud be approximately

    P = (Vin - Vout) * Iout = (24 - 5) * 5 = 95 W,

    and the junction temperature would be about

    R_theta_JA * P = 22.7 * 95 = 2156 C.

    As I do not know which package you are using, I used the more conservative figure for R_theta_JA in the datasheet to illustrate that this will cause the LDO to go into shutdown mode since the shutdown curcuitry kicks in at 150 C. Your ammeter likely reads 0.5 A because the LDO will go in and out of shutdown mode as it heats and cools, and the ammeter is displaying an average of the measurements. If you can provide some scope shots of the output, that will help to further diagnose the issue.

    If the culprit is in fact the shutdown circuitry, you might consider stepping down the input voltage so that the LDO dissipates less power while driving 5 A. You can do this efficiently with a step-down switching regulator; here is a link to help you get started in your search for one that works for your application:

    http://www.ti.com/power-management/non-isolated-dc-dc-switching-regulators/step-down-buck/overview.html

    Another consideration is the thermal dissipation properties of the PCB. The thermal properties of the device listed in the datasheet are measured using an industry standard JEDEC board which is not thermally optimized, so you can improve the thermal dissipation of your board relative to the datasheet specs by carefully designing it so that heat can effectively be pulled from the device. Can you send some screenshots of your board layout so I can give you some feedback in regards to the thermal dissipation? Here is a link to an app note that could also help you understand how to design a board to be thermally efficient:

    http://www.ti.com/lit/an/slvae85/slvae85.pdf

    If the thermal dissipation is not the problem, it could be the battery that is part of the problem. Can you provide more details about the battery you are using, specifically the 'C' rating for discharge and Ah capacity? I'm wondering if the battery is capable of comfortably driving 5 A.

    Regards,

    Nick