2
\$\begingroup\$

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!

| improve this question | |
\$\endgroup\$
4
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ ok now I understand, it is in the input offset right? thanks! \$\endgroup\$ – Rhonald Rei Pahayac Dec 27 '14 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 '14 at 16:43
  • \$\begingroup\$ Are you sure the output unit is correct? \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 27 '14 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 '14 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 '15 at 21:26
3
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.