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UA78: IC Top High Temperature 125C Question

Barton Huang
Prodigy 20 points
Part Number: UA78
Other Parts Discussed in Thread: LM2674

Hi Team,

My circuit is 15Vin and 5V output with output current 0.3A. The ambient temperature is 25C. There is no hot sources around UA7805.

The UA7805 measured top temperature is about 125C. Is it normal for this IC? According the the datasheet, should I use theta JA or other thermal parameter to calculate  

the estimated junction temperature?

Regards,

Barton

  • 0
    Guru 140200 points
    Hi Barton,

    the regulator has to dissipate a heat of (15V - 5V) x 0.3A = 3W. A TO220 package shows a junction to ambient temperature resistance of about 20K/W. So, the regulator will internally heat up 3W x 20K = 60K over the ambient temperature, which results in a chip temperature of 85°C in your application.

    This should be sufficient cooling. But if there are situations when the ventilating get worse, the regulator could heat up to much higher chip temperatures which can decrease the liftetime of regulator.

    If you measure a temperature of 125°C at the regulator then the regulator is probably already damaged. You should improve the cooling by a mounting the regulator onto a heatsink. This will tremendeously improve the cooling.

    Kai
  • 0 in reply to kai klaas69
    TI__Mastermind 23361 points
    HI Barton,

    The recommended operating temperature for this device is 125C. As Kai mentioned, the device is burning 3W of power. Depending on your layout, you could get or worse thermal performance than the thermal numbers listed on the datasheet.

    Could you post your layout especially the size of the copper that is used for heat sink?

    Regards,
    Jason Song
  • 0 in reply to Jason Song
    Prodigy 20 points

    Hi teams

     

    The regulators operating environment is no air cooling and no heat sink,

     

    J3 is the regulators,

     

    Regards,

    Barton

     

  • 0 in reply to Barton Huang
    Guru 140200 points
    Hi Barton,

    with this small piece of copper it's impossible to make the regulator dissipate 3W. You could increase the copper area or add a heatsink or decrease the input voltage of UA7805.

    If you insert a 22R resistor between the 15V line and the input of UA7805, you can move (0.3A)^2 x 22R = 2W heat away from the regulator. For proper cooling the 22R resistor would need to have a power rating of let's say 5W. Another method is to use a zener diode instead of 22R resistor. A 6V8/5W zener diode could be used.

    But these remedies all have one disadvantage: The overall heat stays unchanged. And if your application cannot move away the 3W, the temperature can still rise to dangerous levels. There is a method which allows to reduce the overall heat: Insert a DC/DC-switcher in front of the UA7805 which makes 8.5V from the 15V. The LM2674 could be used, for instance. Or you could even generate the 5V from the 15V totally with the LM2674.

    Kai