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INA105: How to convert differential to single ended (load cell amplifier)

John Lintern
Prodigy 100 points
Part Number: INA105
Other Parts Discussed in Thread: INA2132, INA132

Tool/software:

I am using a 6-axis load cell which has a built in amplifier.  However, the output signal is differential and the analogue inputs I am using are single ended. 

Therefore I need to convert the differential signal to single ended.

I was wondering if I could use an INA105 Precision Unity Gain DIFFERENTIAL AMPLIFIER to convert the differential signal to single ended and if so how ?

Details about the load cell are below (M3815C) :

The supplier provided the following example of the differential signal:  "+signal" is 3.5V, "-signal" is 1.5V, thus the common mode is (3.5+1.5) / 2 = 2.5V and the input voltage is (3.5-1.5) = 2V.

Alternatively are there existing off the shelf devices that will convert differential to single ended ?

  • 0
    TI__Mastermind 33011 points

    Hey John,

    A difference amplifier is exactly what you are looking for. What supply voltage do you want to use? I'm guessing this will be fed to a low voltage ADC, there may be an additional amplifier needed to buffer the output of the difference amp in front  of the ADC, depending on what ADC is used.

    What is your desired accuracy and what is the range of input voltages that need to be converted?

    Best,
    Jerry

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 100 points

    Thanks Jerry

    I am not sure what the desired accuracy should be.  I have a +/-15V supply from a DCDC, I was then going to use +/-12V linear regulators to get a smoother supplier.

    The datasheet shows that the full scale for each channel is 5V...

    I have assumed that: 

    • +SIG can swing from -2.5V to +2.5V (with respect to ground)
    • -SIG can swing from -2.5V to +2.5V (with respect to ground)

    One things that is not clear to me is whether the output of the load cell will always be in a condition where +SIG is greater than -SIG ?

     i.e. is it possible that -SIG could be greater than +SIG ?  In these conditions, VDIFF would be a negative value.

    See powerpoint attachment for different combinations of +SIG and -SIG...

    Differential2SingleEndedTI.pptx

  • 0 in reply to John Lintern
    TI__Genius 15410 points

    Hi John, 

    Just to clarify, is the +/-12V what the supply voltage into the differential amplifier will be? Also is there any package preferences? 

    It does not seem the system needs a high common-mode range based on the powerpoint, but are there any requirements for input offset or CMRR? 

    Regards,
    Ashley

  • 0 in reply to Ashley Kang
    Prodigy 100 points

    Hi Ashley, I plan to power the +/-15V for the load cell (M3815C) using a DCDC converter (DTJ1524D15).

    I then plan to use the +/-15V DCDC that powers the load cell to also power the differential amplifiers, but with a +12V and -12V linear regulator to provide a smoother supply.

    The schematic so far is shown below (in this case using AD8426 differential amplifiers)...

    There are no package preferences (I will need to design a PCB anyway).

    I have no idea about input offset or CMRR ?

    Is it ok to share the 5V reference (AD4550) between all 6 amplifiers ?

    And is the current rating of 100mA ok for the +12V and -12V regulators ?

  • +1 in reply to John Lintern
    TI__Mastermind 33011 points

    Hey John,

    INA132 may work for this application, which luckily is also available in a dual package (INA2132), which can save you some board space. 100mA should be fine, and the reference should be fine to be common across the amplifiers.

    Negative values should be acceptable as long as the ref value biases the output to a level above ground. I don't believe that the output resistor divider should be needed.

    Best,
    Jerry