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low ripple linear regulator

Mark Liang
Intellectual 740 points
Other Parts Discussed in Thread: LM1086, LM1084

Hello,

I have a +15V, 8.5A input, and I am looking for a low ripple linear regulator, which gives outputs at 12V 3.5A and 5V 1.5A.   

Does anyone have any recommendation of part?  If two outputs part is not avaiable, any recommendation for two devices, one for 12V, and one for 5V?

Thanks.

Best regards,
Mark

  • 0
    TI__Intellectual 1106 points
    Hi Mark,
    if you are ok with a two chip solution (you will need to manage thermals though), you could look at the LM1084 for 12V@3.5A and the LM1086 for 5V@1.5A respectively. Both devices are pin compatible and are available for free sampling from our website.

    regards,
    GJ
  • 0 in reply to GJ
    Intellectual 740 points

    Hi GJ,

    Thank you very much for the reply!

    I try to find out what kind of heat sink for LM1084.  The Vin = 15V, Vout =12V, IL <= 4A.  From the formulas on page 20-21 of the datasheet, I have the following:

    LM1084 Thermal considerations:

    PD=(VIN-Vout)*IL + Vin*IG = (15-12)*4 + 15* IG = 12

    ƟJA (max) = TR(max) / PD = (TJ(max) – TA(max)) /PD

     

    ƟJA(max, control section) = (125 – TA(max))/PD = (125 – 40)/13 = 6.54,    where TA(max) = 40

    ƟJA(max, output section) = (150 – TA(max))/PD = (150 -40)/13 = 8.46,     where TA(max) = 40

     

    (θHA (max)) = θJA (max, CONTROL SECTION) - (θJC (CONTROL SECTION) + θCH) = 6.54 - θJC (CONTROL SECTION) - 0.2 = 6.54 – 0.65 -0.2 =  5.69   for KTT/NDE package

     

    (θHA (max)) = θJA(max, OUTPUT SECTION) - (θJC (OUTPUT SECTION) + θCH) = 8.46 - θJC (OUTPUT SECTION) – 0.2 = 8.46 – 2.7 – 0.2 = 5.56   for KTT/NDE package

    When I use θHA (max) =5.56 with Fig 30 on page 21, it seems that it does not need any PCB Heat sink area at all.  Is it right?

    Best regards,

    Mark 

  • 0 in reply to Mark Liang
    TI__Expert 3920 points

    Hi Mark,

    You should not use θHA with Figure 30. Figure 30 is for using copper on a PCB, not for a heatsink. Notice that the line in Figure 30 never goes below your θJA max of 6.54, so PCB copper is not sufficient to dissipate the heat. You need a heatsink.

    By your calculation, you need a θJA of 6.54 or lower to meet your thermal requirements.

    In order to meet this, you need to pick a heatsink that has a θHA lower than 5.69.

    Hope this helps,

    Tim