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XTR106: Design a 4-20mA transmitter for a 700 Ohm input impedance Load Cell

Pablo Fonovich
Intellectual 765 points
Part Number: XTR106

Hi:

I have to design a 4-20 transmitter for a H9D-N3-2.5K-6B Zemic Load Cell. Datasheet here:

www.zemiceurope.com/.../H9D_Datasheet.pdf

In datasheet it states an Input Impedance of 700 Ohm, a Sensitivity of 3+0.008 [mv/V] (15mV Vfs for 5V excitation?) and combined Error of (+-0.023%). I'm trying to calculate the resistors Values. 

1) Using equation 4 in figure 3 of XTR106 datasheet, i get a RLin of near 6Ohm. I think this is a bit low.. could it be right?

2) I use equations in figure 3 of XTR106 datasheet for calculating Rg, but it dependes of Vfs. However, i must add series resistance to the bridge cause it's input impedance is lower than 1K. In page 13 it says that this changes the bridges output... So Vfs changes? Should i calculate Rg with new Vfs modified by this series resistances? how to do that?

3)For calculating bridge balance, i need Vtrim... how does this value relates to my Load Cell?

Thanks!

  • 0
    Guru 140200 points
    Hi Pablo,

    are you sure that your sensor is specified of having a well known parabolic nonlinearity?

    Kai
  • 0 in reply to kai klaas69
    Intellectual 765 points
    Hi, Thanks very much for your answer.
    No i'm not sure, and i really don't know how to measure it, as it's the first time i work with a load cell with real need of accuracy. In previous works i just ignored that (No linearization). I thought it was something to do with the combined error.
    And what happens to the Vfs when i add series resistances because of the low impedance of the sensor?
  • 0 in reply to Pablo Fonovich
    Guru 140200 points
    Hi Pablo,

    if you choose a bridge current of 1mA and a Vref of 2.5V you must insert two 910R resistors in series to the bridge, one above and the other below the bridge, like shown in datasheet. 1mA causes a voltage drop of 1mA x 700R = 0.7V across the bridge. To get the output signal of bridge you must multiply the FS output sensitivity of sensor with the bridge voltage, resulting in 3mV/V x 0.7V = 2.1mV. This is the FS output voltage your sensor is producing in this configuration.

    If you don't know whether your sensor actually shows a parabolic nonlinearity, which could be linearized by the linearization of XTR106, better turn-off the linearization, like shown in datasheet.

    Kai
  • 0 in reply to kai klaas69
    Intellectual 765 points
    Thanks a lot! I'll turn off linearization.
    So, my sensor needs 5V excitation... for 1mA i need two 2.2k resistors in series, which would give me the 2.1mV Vfs... Am i right? i think i could also leave the 910R resistors and get arround 2mA in the bridge, that would give me 1.4V in the bridge, and 4.2mV Vfs?

    And what about the Vtrim for balance? i was asked to include a digital potentiometer for controlling the balance.
    Thanks!
  • 0 in reply to Pablo Fonovich
    Guru 140200 points
    Hi Pablo,

    you can also choose Vref=5V and 2mA bridge current. In the datasheet of your sensor an excitation voltage (= bridge voltage) of 5...12V is recommended. Keep in mind that in the XTR106 configuration with Vref=5V and 910R resistors the actual bridge voltage will only be 1.4V and not 5V. But I don't think that this is a problem.

    Kai
  • 0 in reply to kai klaas69
    Intellectual 765 points
    Thanks a lot! you've been very helpful and now i think i understand better. The only thing i need to figure out now is how to set VTrim.
    Thanks!
  • 0 in reply to Pablo Fonovich
    Guru 140200 points
    Hi Pablo,

    VTrim is optional. It helps to eliminate offset errors. In your apllication you could take a 50k cermet trimmer and a 100k resistor between wiper and bridge.

    Kai
  • 0 in reply to kai klaas69
    Intellectual 765 points

    Hi,

    Thanks very much for everything! you helped me a lot.