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Original question:

LM3914: LED current difference over four times

LM3914: overheat

user5925812
Prodigy 30 points
Part Number: LM3914

I understand the LM3914 is obsolete but I never used it before. I have overheat problem. I use 1k minimum load resistor in series with 5k pot to GND for R1 and 100k pot for R2. I should get 6k resistance for R1 but the highest I get is 3.2k. By the datasheet calculation 3.2k equals 4mA current. The IC overheats quickly and some LEDs go to shutdown. I don't know how to resolve it, lower the current draw.

  • 0
    TI__Expert 6330 points
    Hi,

    I have notified our expert regarding this topic. Please expect a response by 1/23/19.

    Thanks,
    Aaron
  • 0
    TI__Mastermind 18550 points
    Hello,
    How are pins RHI and RLO connected in your design? Their connection will also determine how much current flows through each LED. Can you
    please measure how much current is actually flowing through the LED? Also, can you please send schematics of your design so that I can review to try to understand the cause of device overheating?

    Kind Regards,
    Liaqat
  • 0 in reply to Liaqat Khan
    Prodigy 30 points

     

    When signal input is at GND level, no LEDs lit, current draw is under 8mA.

    With input at 13.8V, all LEDs lit, the current draw jumps up to 150mA, to much

    for the IC to dissipate.

     

  • 0 in reply to user5925812
    TI__Mastermind 18550 points
    Hello,
    From you schematics and previous description: Assuming R1 is at 3.2K (as you mentioned in your post) and R2 is at 100K, I think calculations will be as follows:
    Since 120uA (Iadj) flows through R2, a voltage of 120uA*100K = 12V will be present at pin 8 (Ref Adj).
    Since reference develops 1.25V nominal voltage between pins Ref Out and Ref Adj, voltage at pin Ref out will be 12+1.25=13.25V
    Current through the internal 10-resistor divider will be 13.25V/10K= 1.325mA
    Current through R1 will be 1.25V/3.2K = 0.39mA
    So total current out of Ref Out pin will be 1.325mA + 0.39mA = 1.715mA
    Current through each LED will be 1.715*10 = 17.15mA
    Total current through 10 LEDs will be 17.15*10 =171.5mA

    Accounting for tolerances, I think 150mA that you are seeing is reasonable.

    Kind Regards,
    Liaqat
  • 0 in reply to Liaqat Khan
    Prodigy 30 points

    Thank you for taking the time to clarify the LM3914 calculations. The datasheet doesn't make it clear. I couldn't figure out how to reduce current draw, so I squeezed in 3.3k current limit resistors for each LED of the bargraph display. It runs cool now, but it is brute force approach to a problem. Thanks again for your help.