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Im using a px26 001GV differential pressure sensor for a project with arduino, here is the data sheet. (http://www.omega.com/manuals/manualpdf/M1608.pdf) Now, I have few questions regarding this.

1) I'm using this around a valve to know the differential pressure, lets say side A and side B Now if side A>B i get a positive voltage in milli volts, what if side B>A would it give a negative voltage? so would I get a range of voltages for its full functionality from negative to positive like say -8mv to +8mv?

2)I'm using an op amp to amplify the reading so I could amplify 8.6mv to 1.8V and then the opamp saturates, is there any way around this?

3) the opamp AD620AN which is a instrumental amp to be precise , needs voltage ranges -5 t0 +5 so i used a 10 volts dc, made a voltage divider, created a virtual ground. So what if I connect this system to arduino (0to5V range) would the divider cut off the system to (-2.5 to +2.5)?If so, should I check my circuit by using a 5V Dc and using the divider and check how the opamp responds?

PS: I'm new to this , please try to avoid all the electrical lingo, I wouldn't understand it. Thank you for your effort!

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  • \$\begingroup\$ Note that the "virtual ground" made with the divider MUT be the ARduino true ground if interfacing the opamp/instamp directly to the Arduino without level shifting. \$\endgroup\$ – Russell McMahon Mar 12 '15 at 22:39
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What you have to remember about the AD620 (and other instumentation amps and most opamps) is that their inputs only work correctly when the voltages applied to them are within a limited range. This limited range is governed by the power rails.

For instance, the AD620 data sheet says: -

Input Voltage Range = -Vs+1.9 volts to +Vs-1.2 volts

This means, that for a +/-5V supply, the valid input voltage range is: -

-3.1 volts to +3.8 volts. 

For a 0 to 10V supply it is: -

+1.9 volts to 8.8 volts.

Whatever you do, make sure you don't get too close to these limits or you'll get disappointing results. Next is the output voltage swing. The data sheet says this: -

Output voltage range = -Vs+1.1 volts to +Vs-1.2 volts

For a 0 to 10V supply this means the output can fall as low as +1.1 volts and rise as high as 3.8 volts.

If you can get your sensor to input the right voltage range that's a start. Next, use the REF pin to bias the output so that it remains within the output limits imposed above and finally, try using a potential divider to reduce the output range that that suitable for your arduino.

One last bit of advice - read the data sheet and keep asking questions if there is something you don't understand in the data sheet. A lot of the more experienced guys on this site will tell you that the data sheet is all you have - no belief system involved except to believe the manufacturers have told the truth - invariably they have of course.

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  • \$\begingroup\$ I have tried to read the data sheet but, i dont really understand many terms in it, so as you said about the input ranges im using millivolts, that would do right falling within ranges? what i dont understand is why is it saturating when im within the range? im new to all this so bear with my understanding ability. \$\endgroup\$ – Krishna Chaitanya Mar 12 '15 at 22:03
  • \$\begingroup\$ I think a circuit diagram would help - if you post one somewhere someone will undoubtedy paste it into your question \$\endgroup\$ – Andy aka Mar 12 '15 at 22:39
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First, let's get something straight - if you come to an electrical engineering forum, and ask for advice about an electrical circuit, you are just going to have to expect "electrical lingo".

With a 10 volt excitation, according to the data sheet you should expect a +/- 16.8 mV on your sensor outputs, centered around 5 volts.

Because you have created a virtual ground, your outputs and inputs have to be in terms of the virtual ground, not the circuit ground. Your sensor will provide the input levels you need, but I believe you've gotten yourself in trouble with the reference.

If you look at the data sheet http://www.analog.com/media/en/technical-documentation/data-sheets/AD620.pdf bottom of page 4, you will see that the reference input range is -Vs + 1.6, and I bet you've tied it to circuit ground. Right? So you're operating the AD620 outside its range.

What you need to do is provide an offset on the AD620 output.

schematic

simulate this circuit – Schematic created using CircuitLab

The two op amps should be general-purpose types with rail-to-rail outputs. The AD620 gain should be set to give +/- 2.5 volts. The output of op amp 2 will be 0 to 5 volts.

And make sure you use 1% resistors.

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  • \$\begingroup\$ hi, i donot know how to ad a picture here, so i posted an answer, please check it. thanks \$\endgroup\$ – Krishna Chaitanya Mar 13 '15 at 0:06
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SOrry, I'm nt from an electrical background so it was hard for me to understand those words.

I'm attaching a picture of my circuit, i think i connected it to the circuit ground. And also i have to use this sensor with arduino which works with 0 to 5, so then would my voltage divider make -2.5 to +2.5? would my circuit be valid then? enter image description here

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