Load cell measurement on i2c bus - amplification problem...

Discussion in 'General Electronics Chat' started by diddl14, Nov 14, 2013.

  1. diddl14

    Thread Starter New Member

    Nov 14, 2013
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    I'm working on a relatively small, low power i2c-enabled device for measuring relative force on a load cell.

    As a quick test I've got one of those cheap kitchen scales and hooked up the load cell via an ADS1100 to a Raspberry Pi. Vdd is 3.3v.

    [​IMG]

    It works but I'm only getting about 3 bits of resolution instead of something in the range of 10+ bit.

    I'm kind of stuck here as it seems to be an analogue amplification problem which unfortunately is beyond my rather basic electronics skills...

    Can anyone give me a hint what kind of circuitry could be added to make this work?

    ADS1100: http://www.ti.com/lit/ds/symlink/ads1100.pdf

    Load-cell: http://www.manyyear.com/Weighing/Weighing_scale_sensor/handle scale load cell MLC611N.html
     
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Load-cell, it says 1-25 kg. Am I missing the output voltages that corresponds to the mass I am measuring? If I measure 1 kg, what the output voltage? If I measure 5 kg, what the output voltage?
     
  3. MikeML

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    Oct 2, 2009
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    What are you setting the gain (PGA) of the ADS1100 to? (1, 2, 4, or 8?)

    The differential output of the load cell is likely to be mV, while to get the full 16bit span out of the ADS1100, the differential input must be 3.3V/PGA, where PGA is 1, 2, 4, or 8.
    You will almost certainly need a differential amplifier between the load cell and the ADS1100.
     
  4. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    Your cell is rated .03mv/V. At 10volts excitation, your full scale is 3mV.
     
  5. diddl14

    Thread Starter New Member

    Nov 14, 2013
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    Thanks for all the quick answers. Really appreciate.

    My current understanding (thx to 'GetDeviceInfo') is that the output voltage is a linear fraction of the output rating, multiplied by the excitation voltage and divided by the full load. So (.03mv/V * 3.3V ) / 25kg would give approx. 1/25 mV/Kg.

    Gain is at 8 with 8 sps (=16 bit resolution).
    The ADS1100 spec states

    "the ADS1100 would typically be operated at a gain of 8. The input range in this case is ±0.75 volts".

    So to get 16bit resolution, the 0-1mv would need to be amplified to 0-750mv?

    Is that what the "differential amplifier" would do that MikeML refers to?
    There are tons of references to that but I honestly have no clue how to pick the right components for this specific case. Does this come close?
    [​IMG]

    How would that sit between the load-cell and the ads1100?
     
  6. shteii01

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    The default gain is 1. Please see Table VII of the datasheet.
     
  7. diddl14

    Thread Starter New Member

    Nov 14, 2013
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    Thx, I'm aware of that and set it to 8 before reading. With the default gain I only got 1 bit resolution due to the low excitation voltage and output rating of the load cell. For the Wheatstone bridge example, the spec mentions that a "external instrumentation amplifier" might be required to bring the signal to full scale.

    Is that the same as the differential amplifier mentioned by MikeML?
     
  8. shteii01

    AAC Fanatic!

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    It looks like.
    http://en.wikipedia.org/wiki/Instrumentation_amplifier
    "An instrumentation (or instrumentational) amplifier is a type of differential amplifier"
     
  9. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    In reviewing the specs, I see the cell rating is 1mv/V (the .03mv/V is error).

    Let's do some math to flush that out first. Your converter's full scale is the power rails, so at an internal gain of 1, you'll be looking for a 0 - 3.3v swing. your cell, at an excitation of 3.3v, full scale, will output 3.3mv, so a total gain of 1000 is required. Higher resolution for partial scale will increase that accordingly.

    Your converter can be operated as single input, so you may want to go with the amp as depicted in this post, capable of rail to rail performance.
     
  10. MikeML

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    Oct 2, 2009
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    So why not take advantage of the internal Gain of 8, thereby reducing the external gain to 125?

    The OP needs to purchase an IC chip instrumentation amplifier rather than trying to build one. It is a bitch to match resistors to a sufficient tolerance to minimize offset trying to make one from scratch.
     
  11. shteii01

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    I was wondering about that. I saw +/- 0.03 mV/V, but everyone was ignoring 1 in front of it for some reason.
     
  12. GetDeviceInfo

    Senior Member

    Jun 7, 2009
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    Sure, why not. One reason might be to programmatically change gain, providing a reduced scale 'zoom'. Another may be to actively filter the signal.
     
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