Questions about voltage ranges, and Amp saturation

Hi,

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!

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?

So, that's a gain of around 200, right?

It shouldn't saturate until you get to about 1.5V of the power supply... With a 10V power supply that's around 8.5V. It will probably only go down to 1.5V. You'll have to check the datasheet. A "rail-to-rail" op-amp will get very-close to the negative & positive power supplies. (Or very close to ground with a single supply.)

If you are only getting 1.8V and you need more, you need more gain (or a bigger input signal). If you are not familiar with op-amps, you change the gain by changing resistor ratios (less feedback = more gain... That is, less negative feedback = more gain or more negative feedback = less gain).

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)?

It sounds like you're basically on the right track there.

With the op-amp using a 5V virtual ground, it's output will be 5V when the (differential) input is zero. A divide-by-two voltage divider will put that at 2.5V.

With zero-volts into the Arduino's ADC, it will read ~512 (assuming the default 5V reference).

When the op-amp's input is net-postive, the ADC input will be above 2.5V and you'll read more than 512. When the op-amp's input is net-negative, the ADC input will be below 2.5V and it will read less than 512.

8.6mv to 1.8V

If the op-amp gain produces 1.8V, but you have a 5V bias (a 5V virtual ground), the result is 6.8V (relative to "true ground"). If you run that through your voltage divider, it's 3.4V.

If the 8.6mV input polarity is reversed, that's (5V - 1.8V)/2 = 1.6V

yes 200

"It shouldn't saturate until you get to about 1.5V of the power supply... With a 10V power supply that's around 8.5V. It will probably only go down to 1.5V. You'll have to check the datasheet. A "rail-to-rail" op-amp will get very-close to the negative & positive power supplies. (Or very close to ground with a single supply.) " I didnot understand this.

If i increase the gain, it saturates sooner. or values like 2mv are amplified less.

"If the op-amp gain produces 1.8V, but you have a 5V bias (a 5V virtual ground), the result is 6.8V (relative to "true ground"). If you run that through your voltage divider, it's 3.4V.

If the 8.6mV input polarity is reversed, that's (5V - 1.8V)/2 = 1.6V "

I did not understand this. I measured these values using a multimeter.