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LM317: Regulator output current stability and heating issue.

sachin shet
Prodigy 10 points
Part Number: LM317

Hi,

I would like to use LM317T/LF01 regulator for our application.

Our Requirement is as below:
Input voltage: 28VDC. 
Output voltage : 5VDC.
load current : 100mA.

I have tested this regulator in the Breadboard, Please find the attached circuit diagram.
where R1 is 220ohm, R2 is 660ohm. and input voltage is +28VDC.
with this combination I will get a 5VDC O/P from a regulator.
50ohm load resistor connected to the output side of a regulator.
Whenever the circuit is powered up, the initial load current will be  110mA. 
Within 30 seconds after powering ON, regulator temperature reaches to 65°C and output current decreases gradually.

As per the datasheet it should provide output current up to 1.5A and the input voltage can be given from 3-V to 40-V.

Why in my case regulator output current even cannot handle the 100mA?
Kindly clarify on this subject.

Thanks,
Sachin N

  • 0
    TI__Genius 12395 points

    Hi Sachin, 

    When the load current decreases, does it settle to a specific value, or does it completely come to 0? 

    Is it possible to share some scope shots showing this behavior? 

    Also, how is temperature being measured? 

    Since the device is being tested on a bread board, I assume there is no heat sink on the LM317, therefore, after 30 seconds, it could be heating up enough were the device is shutting off. 

    You are dissipating 2.3W

    Pdo=((Vin-Vout)*Iout)= (28V-5V)*100mA= 2.3W

    Looking at the Junction temperature: 

    Tj=Ta+(Rja*Pdo)

    Where Ta is the Ambient Temperature, Rja is the Junction-to-ambient thermal resistance of the device and Pdo the dissipated power

    Tj=65+(23.3*2.3)=118.6 C

    Although this is less than maximum recommended Tj of 125C, after 30 seconds, this could possible increase if there is no way to dissipate the heat. 

    Is it possible to place some sort of heat sink? 

    Best, 

    Edgar Acosta

  • 0 in reply to Edgar Acosta1
    Prodigy 10 points

    Hi Edgar,

    When the load current decreases, the current drop from 110mA to 65mA in the mean of 5 minutes. 

    The temperature measured using temperature measurement Gun.

    Attaching the test image for the reference.



    What is the maximum power dissipation of regulator without the Heat sink?


  • +1 in reply to sachin shet
    TI__Genius 12395 points

    Hi Sachin, 

    The maximum power dissipation is a function of TJ (max), Rja, and Ta. And the maximum allowable power dissipation at any allowable ambient temperature is Pdo = (Tj (max) – Ta) / Rja.

    Therefore, Pdo=(150-27)/23.3, assuming 27C as ambient, 5.28W

    However, operating at the absolute maximum TJ of 150°C can affect reliability and is not recommended. 

    When you measured the current at 30 seconds, what was the value then? Also, after 5 min, was the temp gun still reading 65C? 

    Best, 

    Edgar Acosta