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I have a weighing scale project using a 200kg load cell with a rated output of 1mV/V. I used 5V as excitation voltage and an Ad620 in-amp for my amplifier. I only need atleast the 0-100kg so I only want to amplify the 0 - 2.5mV part of my output which I believe corresponds to the 0 - 100kg load.

On choosing the gain, I have read the AD620 datasheet: http://users.ece.utexas.edu/~valvano/Datasheets/AD620.pdf which says that the output swing at overtemp(worst case) is -Vs + 1.6 to +Vs - 1.5 which is equal to 1.6V to 3.5V at my case because i'm only using 5V single supply. So i have targeted my output on that part and settled with a gain of 381 or Rg = 130 ohms. So using 2.5V as my reference voltage, I will have an expected output swing of 2.5V to 3.4525V which is within the safe range I assumed.

But the problem is, I don't get an output of 2.5V at zero load even my reference voltage is at half the supply or 2.5V. I only get about 2.47V which is only near it by 30mV or something. According to the book I have read and according to this simulation by analog devices: http://designtools.analog.com/dt/inamp/inamp.html?inamp=AD620, the output at no load or zero differential voltage must be (V+ - (V-)) + Vref. So the V+ and V- will be cancelled out leaving the Vref as my output voltage which must be 2.5V at my case.

This is my preliminary circuit: enter image description here

What could be the problem? Any help would be appreciated. thanks!

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In an ideal in amp the equation you have is true. In a real in amp we have to deal with something called offset voltage. This will be the real difference between the inverting and non-inverting inputs, as opposed to the 0 found in an ideal in amp.

Looking at the datasheet we see that it has a worst case value of 125uV. Multiplying this by the gain of 381 gives us about 48mV. Your output is therefore completely within spec.

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  • \$\begingroup\$ ok now I understand, it is in the input offset right? thanks! \$\endgroup\$
    – Rhonald Rei Pahayac
    Dec 27, 2014 at 16:38
  • \$\begingroup\$ so considering the offset voltages, am I right by assuming that 0 - 2.5mV of my output would cover the 0 - 100kg load on my load cell? \$\endgroup\$
    – Rhonald Rei Pahayac
    Dec 27, 2014 at 16:43
  • \$\begingroup\$ Are you sure the output unit is correct? \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Dec 27, 2014 at 16:44
  • \$\begingroup\$ yes that is before the amplification of the AD620. the rated output of the load cell is 1mV/V, so with 5V excitation, the full load output of 200kg will be 5mV. and I assumed that dividing that into half(2.5mV) would be my output at 100kg since the load cell's output is linear \$\endgroup\$
    – Rhonald Rei Pahayac
    Dec 27, 2014 at 16:47
  • \$\begingroup\$ Is the load cell ideal? In a real load cell, it could be offset as well. If you short Sig+ to Sig- do you get VOUT=VREF? \$\endgroup\$
    – Dave X
    Nov 9, 2015 at 21:26
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Of course there's an offset voltage. Also, you're less than 2% off, so it can be because of resistor mismatch of your 10Ks, even for 1% resistors. In short, there's nothing wrong. If you need better, you need to trim it with a pot.

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