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LM340: PCB Heat sinking Help

John Currie
Prodigy 185 points
Part Number: LM340


I need a 5V, 300 ma from a 24 VDC supply on a PCB.  I am confused about PD calculations and theta jA.  I am calculating PD as (24-5) x .3A = 5.7 W.  Fig 32 0n LM340 data sheet shows max PD vs ambient when mounted on 1 sq in of 2 oz copper and the max wattage is 3.2 W and that is at negative ambient temps..  Less than 2W if the ambient is 25C.  How can I get the current I need when the max power is so low.  I know I must be looking at this wrong.  Can someone help me understand please?

Thanks - John

  • 0
    TI__Mastermind 23361 points

    Hi John, 

    When using this device, you need to estimate the junction temperature of the device using ThetaJa and power dissipation. Different packages have different ThetaJA numbers, and you will need to make sure that the ambient temperature of your system plus the temperature rise using PDxTheja is less than 125C. 

    Does it make sense? 

    Regards, 
    Jason Song

  • 0 in reply to Jason Song
    Prodigy 185 points

    Hi Jason,

    Perhaps a real world example will help.  I have 24 V available on my circuit board. I need 5V and 300 ma.  According to SLVA118A (eq 11),  PDmax=(Vin-Vout) x Iout =  (24-5) x .3 = 5.7 W.  Next, theta JAmax = (TJ-TA)/PDmax (eq. 7).  If I assume ambient is 35C then, (125C-35C)/5.7W = 15.78 C/W.  Is this right so far?  The LM340 data sheet says that the LM340 SOT-223 package on on 1 inch sq. of 2 oz copper PCB has theta JA of 51 C/W Fig 23).

    SLVA118A chapter 6 says that if the theta JA in the data sheet is less than the theta JA max I calulated with equation 7 then the pakage is thermally acceptable.  So, does that mean that my theta JA max has to be over 51 C/W?  My JA is 15.78 C/W.  I don't see how I can get 300 ma out of this regulator based on the above example.  It is a 1.5 A regulator.  Even if my ambient was 0 C the the theta max would be (125-0)/5.7W = 21.9 C/W.  That is even worse, so I don't think I understand what to do.  Please show me a working example of how to do this.

    Thank you - John

  • 0 in reply to John Currie
    TI__Mastermind 23361 points

    Hi John, 

    Your calculation is correct, the SOT-223 package cannot support the 5.7W that your application needs. If your application allows, you may need to create multiple stages when converting the voltage from 24V to 5V to allow heat being distributed across the stages. 

    Even the LDO is rated to 1.5A when using it, it still needs to meet the power dissipation requirement by not exceeding the recommended junction temperature. Normally the maximum current is achieved with tighter Vin and Vout conversion. 

    Regards, 

    Jason Song