Ratiometric Help

Discussion in 'General Electronics Chat' started by Philbilly, Feb 11, 2010.

  1. Philbilly

    Thread Starter New Member

    Aug 18, 2009
    7
    0
    Greetings,
    I have a question regarding ratiometric conversion of LVDT sensors. I have been performing testing on a engine control system. One input to this system is LVDT sensor which is excited at 3Vrms (nominally),and the secondary output from this device is 0 to 1.5 Vrms representing -10 to 95%.

    During testing I monitored the control unit using a GUI program designed by the Mfg, additionally I recorded the sensor excitation and feedback using a high speed data aqusition device on test points at the input of the engine contol. The high speed data was recorded in Vrms.

    The control system calculates the position of this sensor by monitoring its own excitation voltage provided to the LVDT and the secondary feedback signal. Because the power supply has an output that can vary between 2-4Vrms, a ratiometric conversion is required. The problem is, I subjected this device to numerous types of faults and the Excel equation I created for the high speed data doesn't plot the same curve that GUI is reporting.

    My equation is: ((Vin/(Vexc/3))/3)/0.007)-10

    Where as the Vin/(Vexc/3) corrects the secondary output for back to a 3Vrms nominal excitation. This corrected Vin is then divided by 3 to obtain the the ratiometric value and then divided by 0.007 (0.007 volts per percent) to obtain the percentage. Finally, I subtract 10 because the scale is between -10% and 95%.

    To further define the problem, when a direct short form the Vexc hi to lo was applied, the high speed data, using the calculation above, never change (reamined at 60%). This is because Vexc and Vin simultaneouslyfalling at the same rate.

    Is my calculation sound? The other possibility of why the control system may being seeing a percentage decrease and my calculation is not; the control sytem samples Vexc at 96ms and Vin at 48ms.

    Sorry for the long post, but any ideas/comments would be appreciated.

    Thanks!
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    That was very confusing. Are you trying to correct for the variability of Vexc? What are you using to perform the arithmetic?
     
  3. Philbilly

    Thread Starter New Member

    Aug 18, 2009
    7
    0
    The high speed data is exported to .csv file. I use an excel macro with the equation listed previously. If the power supply was stable (3V ± 5%)the equation would be Vin/Vexc. Because the power supply significaltly drifts (± 1Vrms), Vin has to to be corrected back to a 3Vrms supply.

    Thanks for your assistance.
     
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Well, I still don't understand why Vin/Vexc will not suffice to normalize the data.
    Maybe it does.
    Vin/(Vexc/3)/3=Vin/Vexc.
     
  5. Philbilly

    Thread Starter New Member

    Aug 18, 2009
    7
    0
    Vin/Vexc is the correct ratio for determining displacement of the LVDT core. I will try to better explain my problem.

    With an excitation voltage (vexc) of 3Vrms and Vin at 1Vrms the ratio is 0.3333. When I apply a short circuit to the excitation path, both Vexc and Vin fall at the same rate. My recorded data at Xms later indicates Vexc at 1.5Vrms and Vin at 0.5Vrms which is still equivelent to 0.3333. So my plot of the recorded data appears as though there was never a problem in this circuit.

    If I use a fixed value of 3Vrms for Vexc instead of using the recorded value the plots shows the short circuit, but at the sacrifice of LVDT displacement accuracy prior to fault insertion.

    Is there a better equation or is this just one of life trade-offs? My gut tells me I'm overlooking something.
     
Loading...