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INA186-Q1: INA186 output error

LIN XUE
Prodigy 10 points

Part Number: INA186-Q1
Other Parts Discussed in Thread: INA186, INA240

Hello, when we were using the INA186A2QDCKRQ1 with a gain of 50, we found that there was a significant deviation between its input and output results.

VPWR is 13.51V, the coil resistance is 5.1 ohms, driving the low-side MOS at a frequency of 16 kHz, and we tested waveforms with duty cycles of 9%, 18%, and 27%, but the input-output error is very large.

9%: Theoretical current value = 13.51 / 5.1 * 9% = 0.238 A; INA186 output = 0.173 A; Actual measured current via clamp = 0.137 A

18%: Theoretical current value = 13.51 / 5.1 * 18% = 0.477 A; INA186 output = 0.555 A; Actual measured current via clamp = 0.394 A

27%: Theoretical current value = 13.51 / 5.1 * 27% = 0.715 A; INA186 output = 0.911 A; Actual measured current via clamp = 0.635 A

We also checked the PCB layout and used Kelvin routing, so please help investigate why the output error of the INA186 is so large.

 

  • 0 in reply to LIN XUE
    Prodigy 10 points

    18% duty

  • 0 in reply to LIN XUE
    TI__Genius 10140 points

    Hi Lin,

    From my preliminary assessment of your schematic, I believe the output resistor (511k) is too large, please try a value between 1k and 10k and check what the output is then averaged to.

    For more information on selecting output filter resistors, please see Section 8.1.3 of the datasheet.


    Best,
    Mallika Senthil

  • 0 in reply to Mallika Senthil
    Prodigy 10 points

    Hi Mallika：

    We changed the resistor to 1K，the result as ：CH1 OP output;CH2 Actual measured current;CH3 LSD;CH4 High side voltage，Why is the output signal of the op-amp inconsistent with the VPWR signal?

    BR

  • 0 in reply to Mallika Senthil
    Prodigy 10 points

    Hi Mallika：

    Another question, what is the value of Rdiff in the INA186?

    BR

  • 0 in reply to LIN XUE
    TI__Genius 10140 points

    Hi Lin,

    1. Could you please measure the differential voltage across shunt at the pin of the device as shown below. This can help us debug whether the INA186 is amplifying the input correctly. After finding the voltage, check if the output matches what you'd expect given the input shunt voltage.

    Note: Measure the voltage at pins 4 and 5.

    2. You can find Rdiff value from the graph below. For a gain version of A2 at 25C, Rdiff ~ 4.5MOhms.

    Best,
    Mallika Senthil

  • 0 in reply to Mallika Senthil
    Prodigy 10 points

    I use a DC load test to verify the relationship between input and output, which is accurate in this case.But when using 16K PWM drive, the OP output deviates significantly. I will measure the data and send it later. Why is this?

  • 0 in reply to Mallika Senthil
    Prodigy 10 points

    Hi Mallika：

    Test frequency 16 kHz, duty cycle 18%，F1 is IN+-IN-,but this is not tested with a differential probe, F1 is the result obtained using an oscilloscope probe to measure the difference of VIN+ and VIN-. C3 is VPWR,C4 is LSD,C5 is OP out，C6 is current;The average value is 26.09 mV, while the average output of the operational amplifier is 744.4 mV, which deviates somewhat from the theoretical value calculated from the input (26.09*50=1304.5 mV). Even multiplying the differential input value by the gain results in a significant deviation.Using the maximum supply voltage to calculate I=V/R*DUTY=13.6V/4.01Ω*0.191=647.8 mA.

    BR 

  • 0 in reply to LIN XUE
    TI__Genius 10140 points

    Hi Lin,

    Thank you for providing all the requested information. 

    I believe the issue here is the PWM input. 
    As frequency increases, the CMRR decreases, hence more common mode error couples into your input signal, hence the increase in output voltage.  

    The INA186 isn't designed to handle PWM signals, I'd recommend looking into the INA240 / INA241 instead.

    Best,
    Mallika Senthil