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TPS7a4700 heating up and automatic shutdown.

Tejas s j
Intellectual 395 points
Other Parts Discussed in Thread: TPS7A47, TIPD168, TIPD116

Hi,


I am using TPS7a4700 in my design to give 5V output. I have followed the data sheet   but I cannot figure what is wrong.
The IC gets heated up very quickly and goes into thermal shutdown. There is no shorting in the PCB.

  • I have shorted 'Sense' pin to 'output' pin.
  • I have shorted thermal pad to ground.
  • I have used 1uF for 'NR' pin
  • Also the output is 5.043V.

  • 0
    Intellectual 395 points
    The capacitance at output is 20uF (10uF in parallel).
  • 0
    TI__Expert 3700 points

    Hi Tejas,

    The Thermal Pad is internally connected to GND, there is no damage in connecting it again, however, you should also connect the Thermal Pad to a a "large-area ground plane", as specified by the datasheet.

    Now, even when the Thermal Pad is connected your device might go into shutdown if the Power Dissipated on your device is too much, please refer to this answer from a previous post in order to help you calculate your thermal limitations.

    If you have any other question, please let us know the Input Voltage, as well as Output Current and whether or not your Thermal Pad is also connected to a large ground area.

    Regards,
    Victor

  • 0 in reply to Victor Salomon
    Intellectual 395 points
    Thank you for your reply Victor.

    The input voltage is 15v and current consumed by my circuit is around 210mA.
    The Thermal Pad was not connected to large ground area. I am trying another pcb with thermal pad connected to ground pouring.
    But main doubt is the regulator is rated at 1A, even with 210-250 mA how can thermal shutdown happen?

    regards,
    Tejas
  • 0 in reply to Tejas s j
    TI__Expert 3700 points

    Tejas,

    Even though the regulator is rated for 1 A, you must also take into account the power dissipated (PD), which depends not only in the Output Current (IOUT), but also the Input and Output Voltages.

    Your PD = (VIN - VOUT) * IOUT = (12.0 - 5.0) * .250  = 1.75 W

    Then your thermal resistance RθJA ≤ (125 - 25) / 1.75 = 57°C/W, which would be met given two assumptions:

    1. Ambient Temperature is 25°C
    2. The area connected to your Thermal Pad is 9 in2

    See the following screenshot from the TPS7A47 datasheet:

    So from Figure 27, found in the datasheet in page 20, we can see that if the copper area is less than 1 in2 you are already not meeting your minimum RθJA, so your device will go into thermal shutdown.

    To avoid this issue you will need to have the Thermal Pad connected to a ground area that will need to be greater than 1 in2, the minimum area needed is going to depend on other parameters such as the expected ambient temperature and power dissipation as I have mentioned before.

    Regards,
    Victor

  • 0 in reply to Victor Salomon
    Intellectual 395 points
    I made a new PCB with a larger ground area. The two observations made are :

    1.Even though the thermal shutdown is not happening , the regulator is still heating up. With the amount of power dissipation, there is a concern about the efficiency of the regulator.
    2. Regarding the ripple rejection:
    This regulator was used to test an EEG amplifier. There was a significant amount of 50hz noise observed. I am not sure if this degradation was due to the overheating of the regulator.
    Has there been any testing done in bio-potential measurment application?

    Regards,
    TEJAS S J
  • 0 in reply to Tejas s j
    TI__Expert 3700 points

    Sorry for the delay in response, we are in holidays in the US.

    1. Because of the drop between input and output being relatively high (relative to the dropout voltage), it is expected that there will be a certain amount of power dissipation, in your case 1.75 W, so it is normal that your regulator heats up.

      LDOs are known to have the best efficiency, so if you are looking for better efficiency, but are still concerned about the noise added by the device you might want to cascade a DC/DC converter with an LDO, where the DC/DC converter steps down the voltage and the LDO serves as some kind of a "filter".

    2. Depending on your configuration, you should be expecting somewhere in between 70 and 80 dB of attenuation at 50 Hz, so your 50 Hz noise is going to depend on the source, in other words, whatever you are using to power your LDO.

      Over heating of the LDO is not going to increase the noise at a specific spot, but rather the broad band or white noise, so if the noise you are noticing were due to heating up, you would probably see noise from all frequencies.

    So, what are you using to power your LDO? is there any other power sources connected to your set up? 

    Perhaps you could check these TI reference designs: TIPD116 and TIPD168