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INA233: Asking for the design on INA233 power register read back value to get actual output power

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Replies: 20

Views: 129

Part Number: INA233

Hi Team,

Customer is designing INA233 on power calculation.

Would you please provide comment for the following calculation is correct or not?

 

Register setting

0xD4 => 0x2800

0x4A => 0x2710

0xD0 => 0x4827

0xD2 => 0x00FB

 

Customer read back 0x96=22327(10)

 

So the power is calculated as below

B=0

R=0

m=20000

Y=22327

22327= (20000*X+0)*10^0

then X=22327/20000 = 1.11635W

 

 

BR,

SHH

 

  • Hello SHH,

    Thanks for using our forums.  From the information provided it looks like you are calculating the real-world value correctly.  Is your customer's value different from what they are expecting?  If so, it would be helpful for us if you provide the expected current and bus value as well as the current and bus values the device reads.  The shunt value would also be helpful. 

    Best Regards,

    Patrick Simmons, 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 Patrick Simmons:

    Hi Patrick,

    Would you please review the design and calculation with comment and suggestions?

    please refer to the actual measurement data. it seems the error is very big.

    V1=18.9V, I = 1.2A

    P =  22.68w

    this is the schematic.

    BR,

    SHH

  • In reply to SHH:

    Hello SHH,

    I am looking into your issue.  Unfortunately our IT has blocked installation of the INA233EVM software for the moment, while I wait to fix that issue.  Perhaps I can get a few more details from you and your customer to determine whether the part is broken, improperly programmed, or have a communication issue.  To rule out whether its broken, If your customer supplies the part with an external supply, does it show the supply sourcing well above the typical quiescent current (mAs)?  If not, its likely a programming issue.  If your customer reads register 0xD4 after writing, do they actually read 0x2800?

    Best Regards,

    Patrick Simmons, 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 Patrick Simmons:

    Hi Patrick,

    Please refer to customer's read back register value and let us know your comment.

    0xD4 => 0x2800

    0x4A => 0x2710

    0xD0 => 0x4827

    0xD2 => 0x00FB

    BR,

    SHH

  • In reply to SHH:

    Hello SHH,

    I think I found part of the issue. Your slope coefficient, m, for power should be calculated as 1/(25*current_LSB) = 800.  You can then shift that left 1 place and use R=-1.  While that should be considerably closer to what you are wanting, from my calculations that would be 28W.  With an 18.9V bus and 12mV Vsense I get 22.8525W. I think it would be worthwhile to look at both the Read_VIN and Read_IIN registers to find where the discrepancy is with your measurements.

    Best Regards,

    Patrick Simmons, 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 Patrick Simmons:

    Hi Patrick,

    can you please confirm the calculation is correct or not?

    b=0

    R=0

    m=800

    Y=22327

     

    X=(Y*10^(-R) -b)/m=(22327*10^0-0)/800=27.9W

    for Read_VIN and Read_IIN, I will check with customer.

    BR,

    SHH

  • In reply to SHH:

    Hello SHH,

    You are on the right track, but there is one additional step the datasheet recommends to avoid possible rounding errors. That is maximizing m.   Below, I will summarize all the steps so that it is clear for you and any person who looks at this post in the future.

    Calculate current_LSB: (typically calculated from maximum expected current, however is calculated from CAL in this case)

    CAL = 0x2800 = 10240

    Rshunt = 0.01ohm

    Current_LSB = 0.00512/(CAL*RShunt) = 50uA

    Calculate m for power:

    m=1/(25*Current_LSB) = 800

    Shift (maximize) m and get R:

    if 1000*m>32767>100*m, R=-2, new m = m*100

    else if 100*m>32767>10*m, R=-1, new m = m*10

    else if 10*m>32767>m, R = 0

    10*m = 10*800 = 8000<32767:  m =8000, R=-1

    Calculate the real-world value, x:

    Y=22327
    x=1/m(Y*10^(-R)-b) = 1/(50uA)*(22327*10^(-1)-0)=27.90875W

    Best Regards,

    Patrick Simmons, 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 Patrick Simmons:

    Hi Patrick

    thanks for reply. Would you please provide comment for the following questions?

    1. Why the Iin and read back value (convert to real word value) is different?
    2. Why case 1 and case 2 output power are not match the multi-meter measurement data?
    3. Please provide comment for further debug.

     

    Here are the two different project with INA233 read back 0x88, 0x89 and 0x96 and actual voltage and current data.

    project 1

     

    project 2

     

    BR,

    SHH

  • In reply to SHH:

    Hello SHH,

    When I convert the values for case 1, I get pretty close to your multimeter measurements.  As for case 2, the current value does deviate quite a bit.  Is this a different device?  If so, you need to check the calibration register and make sure the correct value is there.  If the correct value is there, then either there is a problem with the device or there is some parasitic path siphoning off current between the multimeter and the INA233.  There are a few ways you can go about testing this. The first would be to swap out the part. If the reading gets better than the device may have been damaged.  The other thing you can do is put your multimeter on the low-side of the shunt (assuming your previous multimeter readings were on the high side).  If a low side multimeter reading shows lower current as well, then you need to start to look at components around the shunt that may be damaged and are pulling more current than expected. 

    To further assist you with your debug, I am attaching an excel file with my calculations based on the information you provided above.

    /cfs-file/__key/communityserver-discussions-components-files/14/INA233Analysis.xlsx

    Best Regards,

    Patrick Simmons, 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 SHH:

    Hello SHH,

    When I convert the values for case 1, I get pretty close to your multimeter measurements.  As for case 2, the current value does deviate quite a bit.  Is this a different device?  If so, you need to check the calibration register and make sure the correct value is there.  If the correct value is there, then either there is a problem with the device or there is some parasitic path siphoning off current between the multimeter and the INA233.  There are a few ways you can go about testing this. The first would be to swap out the part. If the reading gets better than the device may have been damaged.  The other thing you can do is put your multimeter on the low-side of the shunt (assuming your previous multimeter readings were on the high side).  If a low side multimeter reading shows lower current as well, then you need to start to look at components around the shunt that may be damaged and are pulling more current than expected. 

    To further assist you with your debug, I am attaching an excel file with my calculations based on the information you provided above.

    /cfs-file/__key/communityserver-discussions-components-files/14/1172.INA233Analysis.xlsx

    Best Regards,

    Patrick Simmons, 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.