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AD2S1210 DOS questions

Hello

  1. How to know the DOS MISMATCH from resolver waveform?  
  2. What’s the optimum value about DOS MISMATCH THRESHOLD REGISTER?

BR

Patrick

        

  • Patrick,

    The monitor signal (Page 16, equation 4) is used to determine, internal to the RDC, the difference in the peak, differential magnitudes at the sine (A1) and cosine (A2) inputs.   A mismatch in A1 and A2 magnitudes as described in the datasheet results in ripple in the monitor output.  Over the window period the maximum and minimum values of the monitor signal are stored internally and the difference compared against the DOS threshold.  If the difference exceeds the programmed threshold the flag is set.

    For your purposes a simple comparison of the differential magnitudes of the input envelopes (Sin-SINlo) - (COS-COSLO) should be sufficient to determine what amplitude mismatch you have under normal conditions from your resolver. 

    The optimum value for the DOS mismatch threshold will be determined by your specific resolver selection and system performance needs.  Note that while some mismatch is expected due to mechanical tolerances in the system, excessive mismatch will cause positional errors in the output.   By default the DOS Mismatch register value is programmed to set the threshold at 380mV.

    Sean

  • Hello Sean

       F1 waveform is (Sin-SINLO) - (COS-COSLO). Customer measure value is 3.2V. If consider derating, they will set DOS MISMATCH THRESHOLD REGISTER as 0x5E  (3.572V). Does the threshold value has any problem? Thank you. 

    BR

    Patrick 

            

  • Patrick,

    The methodology they are using is incorrect.  Instead of subtracting COS from SIN they should be using the arithmetic sum and then looking at the peak to peak ripple.  If the SIN and COS channels have equivalent magnitude  and are truly orthogonal they should get a DC value.   Again I would refer you to the monitor signal description.

    Sean  

  • Hello Sean

                  Per your reply, while A1=A2, the sum of SIN & COS will get a DC value. 

    1. SIN=3.2Vpp, COS=3.42Vpp, why customer can not get a DC value by SIN+COS as waveform? 

    2. If the DC value >DOS mismatch threshold , it will happen DOS fault, right? 

    3. How to calculate by SIN+COS waveform to get a reasonable DOS mismatch threshold?  

    4. Per below table describe on datasheet page 22, 1.71V will be over 0.38V DOS mismatch threshold, so mismatch threshold needs to set 1.71V? Thank you.  

    BR

    Patrick

  • Patrick,

    Consider by substitution that A1 = 3.2Vpp and A2 = 3.42Vpp as you have entered and that the loop is in steady state such that we can assume for small errors that PHI = THETA and then we can essentially build a model in excel that proves the monitor signal will be A1 sin^2(theta) + A2 cos^2 (theta) and we'd get the following.

    As illustrated for one rotation the monitor magnitude changes at twice the rate of rotation at a magnitude of 220mV pk-pk.   Thus under this condition the MISMATCH flag should NOT be indicated.

    The DOS RESET thresholds should not be used to set the MISMATCH.  These are values set to allow for the monitor comparison to start at some reasonable level outside of the selected LOS and DOS Thresholds to avoid automatically tripping indications erroneously.  

    If the solution is producing the correct angular output I would leave the default 380 mV setting for now as the customers inputs appear to be in line with expectations.

    Sean