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TPS92520-Q1: how to improve LED current accuracy

Iris Shen
Expert 4002 points
Part Number: TPS92520-Q1

Hi Francis,

I see in datasheet "The internal offset of the device can be measured and compensated using the lower LSBs of the 10-bit CHxIADJDAC." I'd like to know how to measure the offset of one device and how to calculate the compensation.

I ask this because my customer found one specific 92520 IC has lower accuracy.

They set CHxIADJ to output 187mA. One 92520 output current is 164mA while the others output current is around 190mA. They did ABA test (put this 92520 to another PCB) and found the this follows specific IC.

Thanks!

  • 0
    TI__Mastermind 23256 points

    Hello Iris,

    I shared the equations with you in the .pdf (page 17) I sent to you previously.  It includes VOS.  

    You simply have to back calculate what the VOS would have to be for it to output that current given the equation.  You know all the other variables so you solve for VOS.  

    Do you know the RCS they used?  I want to see where they are operating at when at 164-187mA.  If they are supporting the full scale current of 1.6A and are using something close to 0.1 ohms for RCS then it isn't surprising that they are seeing this difference at light loads.  That equates to a decimal value close to 100.  When compared to the specs it will have a tolerance between +/-15.94% (decimal value of 192) to +/-48% (decimal value of 63).  What CHxIADJ are they programming it to?

    from the ATE test data above we have a typical value of 12.5mV at a decimal value of 63.  If this were ideal we would have 10.6mV assuming RCS doesn't have any tolerance. If we assume that most of the error at light load is VOS then we can say that VOS is 12.5-10.6mV = 1.9mV.  You can now use that value to recalculate the desired output current using the equation from above.  

    Does this make sense.  Let me know if you want to setup a meeting to go over it.  Not sure why you didn't put this on the internal forum.  

    -Francis Houde

  • 0 in reply to fhoude
    TI__Expert 4002 points

    Hi Francis,

    Great thanks for your explanation. I sent you a meeting invitation to quickly go over it.

    To reply your question:

    1. They are using 100mohm current sense resistor.

    2. They are programming CHxIADJ to 108. 

    I have several questions:

    1. What does the 10.6mV come from? From my understanding, my customer calculation is VCSP-VCSN=2.48/14*108/1024=18.68mV. The real current measurement is 164mA, so VCSP-VCSN=100ohm*164mA=16.4mV. Vos=18.68mV-16.4mV=2.28mV. (Assume VCSN<17V). For each 92520 device, customer need to test the output current to calculate the Vos. Am I understanding correctly?

    fhoude said:
    from the ATE test data above we have a typical value of 12.5mV at a decimal value of 63.  If this were ideal we would have 10.6mV assuming RCS doesn't have any tolerance. If we assume that most of the error at light load is VOS then we can say that VOS is 12.5-10.6mV = 1.9mV. 

    You are using 108 decimal value, I am using the 63 decimal value from the datasheet.  If you put 63 in the decimal value in the equation you get current or the Vcsp-csn;

    They idealy want to test it at the end of the production line and then store those values in the micro controller.  If they want to do the statistical method based on the datasheet then they can see what the average Vos is by the datasheet.  The datsheet shows 12.5mV, but the calculated is 10.77mV, then the Vos is 12.5-10.77 = 1.73 mV.

    2. I see the VDACX(FS) also has variation. Do we need to take this into consideration when calculate Vos?

    This error is factored in the Vcsp_csn that is measured in ATE.  When we test the entire loop then all the factors are included except the Rcs tolerances.

    3. For +-4% accuracy, we can only guarantee when VDAC= full scale 1023 right? If customer want to output 187mA, they should increase the current sense value to increase the CHxIADJ for smaller threshold error?

    Yes, the guarantee is at full scale.  They can take Vos and then include that into the equation.  

    Let me know if this isn't clear. -fhoude

    Thanks!