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INA250: 3x INA250 in parallel for 0-40A current monitor

SteeVeeDee
Intellectual 700 points
Part Number: INA250

I've read TIDUA90 as well as the applications section of the data sheet regarding paralleling 2 INA250 for measuring > 15A.

I would like to put 3 of them in parallel to measure 0-40A. 

Rather than daisy chain the outputs as shown in the aforementioned documentation, I would prefer to use a resistor based voltage summing circuit to to get the sum of the outputs.

I imagine that summing the outputs is probably not an issue.

However, I am wondering what the true tolerance of the internal shunts may be- when we start hooking many in parallel, a device having a lower shunt value will "hog" more of the current?

Has anyone analyzed the limit of how many of these things can be hooked in parallel?

Thanks, Best, Steve

  • 0
    TI__Mastermind 21720 points
    Hello SteeVeeDee,

    Thanks for considering to use Texas Instruments in your design. The package resistance has a tolerance of 15%. So in the worst case scenario 2 INA250s have a resistance of 5.175mohm and one INA250 will have a resistance of 3.825mohm, thereby yielding a max current of 16.14A through the 3.825mohm device. This would violate the absolute max of the part, so to ensure that none of your units fail, you should probably use 4 devices. As for the limit you could use in this configuration, I would suspect board space and price would be your limiting factor. If you are using the daisy chain configuration, 4 would be the max. I am sure you are aware, but just in case, your board traces will also determine how much current passes through the INA250 devices. So be sure to match the trace impedances as close as possible.
  • 0 in reply to Patrick Simmons
    Intellectual 700 points

    Please clarify:

    The shunt inside the INA250 is "2 mOhm". 

    You mention that one INA250 will have a resistance of 3.825 mOhm.  Is the package resistance then 1.825 mOhm?

    How to you get 5.175 Ohms?  If these are connected in parallel for a high current application, wouldn't the overall resistance get divided by two, i.e. 3.825 Ohm/2 = 1.91 Ohm?

  • 0 in reply to SteeVeeDee
    Intellectual 700 points
    OK, so I've taken a closer look at the data sheet.

    Page 5 states that the IN+ to IN- package resistance is typically 4.5 mOhm, which includes the lead frame and the internal shunt. You're suggesting that the tolerance on this is +/-15%, so 3.825 mOhm to 5.175 Ohm., thank you for this additional information that is not in the data sheet.

    To follow up with your "worst case" analysis of two devices, where one device has value (1-t)*R,typ and a parallel one has (1+t)*R,typ.

    Let t = 15% = 0.15.

    In this special worst case, the current flowing through the parallel combination of the two devices will divide as:

    I1 = I*(1 + t)/2 and I2 = I*(1 - t )/2, and for t = 0.15, a 57.5%/42.5% split.

    Note how even though the tolerance of the resistor is 15%, the current splitting is half that, or 7.5%.

    The next special case would be 2 devices at the high extreme, and the third at the low resistance extreme, i.e. R1 = R2 = (1+t)*R, and R3 = (1 - t)*R

    Rather than slug this through analytically, I just did a quick SPICE simulation, and in this case the low resistance R3 caries 40.35% of the current.

    At 40A, this is where you get the 16.14A number.

    HOWEVER, it is VIRTUALLY IMPOSSIBLE for this combination to occur.

    In order for this to happen, you would have to have all three devices all built at their extremes, which I am assuming is +/-3 sigma from the mean. The probability of this is ~(650 ppm)^3 = not happening.

    This is why I was asking if anyone at TI had ever done a real statistical analysis of this. As you use 3 or more devices, the current will start approaching a mean value per device. In my SPICE simulation, adding a fourth device indeed reflects this.

    I can do a Monte-Carlo of this arrangement with 3 devices, but I suggest perhaps doing a more thorough analysis of the statistics to come up with a better analysis than "worst case".