# Question about a differential mount board pressure sensor

## Main Question or Discussion Point

Hello
I have a query regarding how this diffrential pressure sensor works
http://www.mouser.com/ds/2/302/MP3V5010-783555.pdf

I have used the formula they suggested and I am getting a negative pressure output. I am unable to interpret the reason to such an output, It would be great if someone can shed some light on it.

I am trying to utilize this to measure the variation in atmospheric pressure and vacuum but i have done this test only under normal pressure (basic working test) but since i am not able to understand this result , i haven't ventured testing it under those conditions.

Thank you.

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berkeman
Mentor
Hello
I have a query regarding how this diffrential pressure sensor works
http://www.mouser.com/ds/2/302/MP3V5010-783555.pdf

I have used the formula they suggested and I am getting a negative pressure output. I am unable to interpret the reason to such an output, It would be great if someone can shed some light on it.

I am trying to utilize this to measure the variation in atmospheric pressure and vacuum but i have done this test only under normal pressure (basic working test) but since i am not able to understand this result , i haven't ventured testing it under those conditions.

Thank you.
Welcome to the PF.

How are you testing it? It probably has an offset that needs to be calibrated out, but I could be wrong about that.

If you hook up a flexible tube to its input and blow and suck, do you get more normal changes in output voltage?

Svein
What output are you getting?

jim hardy
Gold Member
2019 Award
Dearly Missed
A question well stated is half answered.
You didnt give us much to go on.

What output are you getting?
I'll add, for sake of completeness you could have included the following:

Which one do you have ? From datasheet you linked there are four varieties

as to its operation, again from datasheet

Again from datasheet

Vout = VSx (0.09 x P + 0.08)

so
If VS = 3 as recommended
(sanity check : 0 pressure gives +0.24 volts)
Vout/3 = .09P +.08
P= (Vout/3 - .08) /.09
P = Vout X 3.704 - .8889

i do not see how any Vout more positive than 0.24 volts could be interpreted as negative pressure.

What voltage did you get ?

SO

If VS is 3 volts and is decoupled per fig 3
and you're getting less output than +0.24 volts(caveat - see offset specification below)
i'd guess either
pressure is applied to wrong port
or you're trying to read vacuum with a pressure gage
or you've hooked the voltmeter up backward.

Caveat - here's their offset spec from datasheet
that 0.24 volts is only guaranteed to be within this range:

my two cents and overpriced at that.

Let us know what you find ?

old jim

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Averagesupernova
Gold Member
If you hook up a flexible tube to its input and blow and suck....
Something I never thought I would see on physics forums.

Thank you all so much.
I apologize for the lack of information provided.

So first to answer your query regarding the output i am obtaining a pressure of -0.79.

Then, yes i tried varying the pressure by blowing and sucking air and it works fine showing me variation of pressure.

I am obtaining a positive voltage of more than 0.24. But i am getting a standard negative pressure (eg : Pressure = -0.79 Voltage= 0.280645).

Thank you Sir Jim Hardy, I shall check where i have made an error again and discuss the results i get with all of you.

jim hardy
Gold Member
2019 Award
Dearly Missed
Voltage= 0.280645
Again from datasheet

Vout = VSx (0.09 x P + 0.08)

so
If VS = 3 as recommended
(sanity check : 0 pressure gives +0.24 volts)
Vout/3 = .09P +.08
P= (Vout/3 - .08) /.09
P = Vout X 3.704 - .8889
Voltage= 0.280645 (What the heck kind of meter do you have ? Gives me digit envy !)
+0.280645 X 3.704 -0.8889 = + 0.1506
now datasheet didn't say what are units
but i'd think from their graph it's kilopascals

That's .15 kpa out of 10 kpa range? That's error of only about 1.5% of span. Not bad for a "fresh out of the box" untouched ±5% gizmo.

so next question is at what condition did you measure .280+++ volts? Both ports open to air? No water inside the sensor?
Horizontal or vertical?
You'll probably find a small difference when the sensor is held in vertical vs horizontal plane. That's because Gravity pulls on its "differential sensing element" .

So
open both ports to air, place it in the position you intend to use it and measure Vout. That is the number you'll use in place of of 0.24 in the transfer function. Notice that your 0.28++++ falls within the specified range of 0.1 to 0.38 volts.

That's a neat gizmo, thank you for making me aware of it.

old jim

Thank you Sir Jim. I tried that it, but now we had to make another choice because we need Absolute pressure measurement in vaccum. Thus we have planned on using MPX2100AP but we are facing an issue with it as well, the data sheet doesnt provide the formula for calculation of pressure , its confusing as this has two pins for Vout with one positive and one negative , i dont seem to place the idea and reason behind it . Could you plz help me out with this.

Data sheet : http://www.mouser.com/ds/2/302/MPX2100-783865.pdf

jim hardy
Gold Member
2019 Award
Dearly Missed
You should study the datasheets carefully asking yourself "What did they mean by this?" for every entry.

That sensor is different than the first one you tried. It does not have an amplifier to give output refernced to common
it's more of a Wheatstone bridge
you must measure the voltage between those two pins
either with a differential A/D converter
or by two measurements

If you supply the device with 10 volts dc,
voltage between the pins will be nominal 40mv at 14.5 psia
and nominal 0 mv at 0 psia

y = m x pressure + b
where m is a fraction of applied input voltage, that's what "ratiometric" means
0.04 X pressure in kPa from that chart
Output (in millivolts) = kPaabsolute at P1 X 0.04 X Vsupply + offset,
so with 10 volt supply Output (in millivolts) = kPaabsolute at P1 X 0.4 + offset,

offset being error as given here(for 10V supply)

and Output being the voltage difference between the two pins.

You can surely solve that for pressure as f(volts) . At whatever supply voltage you use.

check my arithmetic - it's dinnertime here and Fair Anne has it ready !

old jim

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