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INA181: IC overheating

Part Number: INA181

Hello,

for a bidirectional current sensing application I'm using an INA181.
I'm sensing the current in- and outflow of a leadacid-battery, so approx. 12V. As a µC I use a CC2640R2F.
I applied 3.3V to VS, ~12V to IN+ and IN- and gave it as a reference the VDDR pin of the CC2640, which is ~1.7V, the output goes directly to an analog pin of the µC.
There are two problems occuring:

1. When my board is completely switched off, but the 12V on IN+ and IN- still applied, there is a Voltage on VDDS present, which I don't want. If switched off, the board is required to carry no voltage at all  (but the 12V) .

2. After a while the INA181 heats up (you can't touch it with the finger), and eventually it broke due to overheating.

So my question is, is there anything I'm missing out on? I sticked to the datasheet as close as possible, but the INA181 shows the stated behaviour. 
Maybe using the VDDR pin of the CC2640 as a reference was a mistake, would that explain it? Any suggestions on how to make it work on the prototype?

Any help is highly appreciated, thank you. Alex

  • In reply to Alexander Gemmer:

    Hey Alex,

    That is correct, when there is only voltage at IN+ and IN- then OUT, VS, and REF should all be 0V. I look forward to hearing how those tests go.

    Best Regards,

    Mitch M, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    TI makes no warranties and assumes no liability for applications assistance or customer product design. You are fully responsible for all design decisions and engineering with regard to your products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning your designs.

  • In reply to Mitch M:

    Hi Mitch,

    today I received the order of some more INA181 devices from TI-Store. I soldered a device on my PCB with no parts on it at all, but the INA181 and shunt. I then applied 12V to IN+ and IN- and measured the outputs. 

    VS = 1.468V
    OUT = 1.468V
    REF = 11.19V

    None of the pins have a physical connection to each other on the board in this setup.
    I then measured the Pin-to-Pin resistance.

    Soldered on the board:

    OUT - IN+/IN- --> ~500 kOhm
    REF - IN+/IN- --> ~500 kOhm
    VS - IN+/IN- --> ~9.6 MOhm

    The IC just itself NOT soldered on the board:

    OUT - IN+ --> ~500 kOhm
    OUT - IN- --> ~500 kOhm
    REF - IN+ --> ~500 kOhm
    REF - IN- --> ~500 kOhm
    VS - IN+ --> ~9 MOhm
    VS - IN- --> ~9 MOhm
    IN+ - IN- --> ~2.5 kOhm

    As you stated above the Pins VS, REF and OUT are supposed to show 0V, which is not the case right out of the box.
    The functional block diagramm in Figure 39 in Section 8.2 of the datasheet shows probably the resistance I measured.

    I'm confused, maybe this information tells you something.
    Regards, Alex

  • In reply to Alexander Gemmer:

    Hey Alex,

    I'll look over this information. Meanwhile, could you also try applying the 12V to IN+ and IN- on a new part not connected to your board and measure the voltage on VS, OUT, and REF? This way we will know for sure there is no short on your board.

    Best Regards,

    Mitch M, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    TI makes no warranties and assumes no liability for applications assistance or customer product design. You are fully responsible for all design decisions and engineering with regard to your products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning your designs.

  • In reply to Mitch M:

    Hi Mitch,

    I checked all the connections on the board and there was no short. But to be 100% sure I took a brandnew INA181 and applied 12V directly to the pins and measured the OUT-Pin. In order to achieve this I soldered small wires to the IC, which I then connected to my DC power supply and voltmeter.

    The result is exactly 1.468V on OUT at 12V on common mode. So it can be definately said that my board is not involved since the upper results could be reproduced without it. I hope that helps.

    Regards Alex
  • In reply to Alexander Gemmer:

    Hey Alex,

    Looking into it a little bit farther, it turns out that a small parasitic voltage on these pins is possible as explained in this forum thread:
    e2e.ti.com/.../1785831

    However, this still does not explain the overheating. You mentioned that the LDO was overheating as well, but does the LDO still overheat when the INA181 has been removed? Also, could you send me some waveforms of the voltage at each and every pin of the INA181 while it is in the operating condition that leads to increased temperature?

    Best Regards,

    Mitch M, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    TI makes no warranties and assumes no liability for applications assistance or customer product design. You are fully responsible for all design decisions and engineering with regard to your products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning your designs.

  • In reply to Mitch M:

    Hi Mitch,

    it's good to know, that the parasitic voltage is normal. I now approached the intended configuration step by step by soldering a INA181 to a fully functional board. I first applied 12V to common mode, which was fine, then I switchen on the whole board, that worked out as well. I then applied 3.3V from the Launchpad via JTAG, that was fine too.

    The measured voltage on OUT fits perfectly to the applied current to the shunt.
    So my guess is, that probably either the first IC that I used was broken, some ESD issues occured or I messed up something when soldering it.

    It's a bit unsatisfying, because I can't recreate the overheating and thus never know what actually happened. But on the other hand two of my boards now work as intended for some hours without any problems.
    I'd like to thank you for your help, that is much appreciated. And after all, it didn't work in the beginning and now it does and you were a great support on the way!
    Thank you!

    Regards Alex

  • In reply to Alexander Gemmer:

    Hey Alex,

    I'm glad that you were able to get it working!

    Best Regards,

    Mitch M, TI Sensing Products Applications Support

    Getting Started with Current Sensing Video Training Series

    TI makes no warranties and assumes no liability for applications assistance or customer product design. You are fully responsible for all design decisions and engineering with regard to your products, including decisions relating to application of TI products. By providing technical information, TI does not intend to offer or provide engineering services or advice concerning your designs.

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