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Original question:

LP2985: Behavior when output voltage is short-circuited

LP2985: junction temperature when the output is short-circuited

Tsukui Yusuke
Guru 21930 points
Part Number: LP2985

Hi,

Please tell me how to calculate the junction temperature when the output is short-circuited.

Customer conducted an output short circuit test under the following conditions.
However, the device did not undergo a thermal shutdown.

・Vin=13.5V
・Vout=0V
・TA=25℃
・Short-circuit current＝104mA

I understand that the loss calculation of LDO is the following formula.
・Pd＝Vin×Ishort＝1.4W
・θJA＝206℃/W
・Junction temperature rise ≒ 289℃

However, the thermal protection and  protections did not destroy the device.
Could you tell me how to calculate QJ when output short circuit condition?
And In this short-circuit test, what is the state inside the LP2985 to protect it, and what is the junction temperature at that time?

Best regards,
Yusuke


  • 0
    TI__Genius 13085 points

    Hi Yusuke-san,

    Your understanding of thermals and power dissipation are correct. With that much power dissipation we expect ~289C rise above the ambient temperature and we'd expect it to start cycling in and out of thermal shutdown to protect the device. So I have some questions to help verify what is happening. 

    How was the current measured? was this measurement made using a DMM or an oscilloscope? Thermal shutdown will cause the device to turn on and off as the junction temperature heats and cools. Unfortunately multimeters will show the average current so to see the thermal shutdown we need to evaluate this using an oscilloscope. The datasheet reports that the short circuit current should be ~400mA, so the fact that you are seeing ~100mA leads me to believe this measurement was mad using a DMM and it is showing you the average current but please verify. 

  • 0 in reply to Kyle Van Renterghem
    Guru 21930 points

    Hi kyle-san

    Thank you for your kind response.
    As you pointed out, the average current was 104mA.

    A rectangular wavy current value is observed, and the output voltage is 0V.
    The peak current of the square wave was about 500mA.
    However, it is observed that thermal protection is not working and only overcurrent protection is working.

    Is my understanding correct?
    Also, is there a way to calculate the period of the square wave and the heat generation status?

    I want to send you a private message to share the waveform.

    Best regards,
    Yusuke

  • 0 in reply to Tsukui Yusuke
    TI__Genius 13085 points

    Hi Yusuke-san,

    The description of the square wave current seems to indicate that the device is going into over current protection first (limiting the current into the short to GND) and then goes into thermal protection and which shuts down when the temperature gets too high (this is what causes the square wave).

    I've accepted your friend request so you can send the waveform privately. 

  • 0 in reply to Kyle Van Renterghem
    Guru 21930 points

    Hi Kyle-san,

    Thank you for your kind support.
    I used a private message to share the waveform.
    Could you tell me a comment on the waveform and a Pd and Junction temperature calculation method that takes in short circuit ?

    Best regards,
    Yusuke