MPX5050GP ( reading kPa )

[tbalon]

I'm using an MXP505GP pressure sensor to measure dry air pressure. I have the sensor connected to a regulated 5V supply bypassed with a 1uF ceramic at the sensor input. The output pin is connected via a 750 ohm and 0.33uf

filter as per AN1646.  This is connected to my HP meter to test. The power supply is well regulated 5V 1.5A output. 

The sensor outputs a 0.2V with no pressure applied as expected from the datasheet.  

I applied various pressure using a water column and measured the voltage on the output of the MXP5050GP. 

I'm using the transfer function from the data sheet Vout = 5 .13( P x 0.018 + 0.04 ) (  Vcc = 5.13V )

When I plot the values, they are indeed linear, but the pressure (kPa) using the formula seems way off. 

If I apply 6 kPa , I get 0.583 volts output, which using the formula indicates about 4 kPa ! 

Pressure   |   MPX5050 Reading (using formula)

1 kPa           0.58 kPa   0.259V on Meter

2 kPa           1.32 kPa   0.327V

3 kPa           1.93 kPa   0.383V

3.74 kPa      2.43 kPa   0.430V

5.98 kPa      4.09 kPa   0.583V

12.95 kPa    8.88 kPa   1.02 V

To generate the pressure, I am using a column of water. The MXP5050 is connected to a 12 foot section of

1/4" hose which is then submerged into the column of water. To generate a pressure of 2 kPa, I submerge

the end of the tube into the column so that it extends 8 inches below the surface.  

Using the formula, I should get 2 kpa, but instead I read only 1.32. At 24 inches ( a pressure of ~ 6 kPa ) it

only reads 4 kPa and is off by 2 kPa !  

I must be missing something here ... 

Tom

[darioarias]

Hello Tom,

The correct procedure is to calibrate the device to a known pressure, not a theoretical pressure, we know that often this two pressures are far from each other because there are other variables involved in the process. If there is a difference between the real pressure and theoretical pressure, this difference will increase as the pressure increase making the error bigger for higher pressures. Regarding the MPX5050, the output of the device is ratiometric, so there is no chance for the error increasing proportionally to the pressure, if there is an offset, it would remain constant for any pressure. I suggest to compare the readings of our sensor, with the readings of another well calibrated device.

Best regards,

Darío

[tbalon]

I've double checked my work.  I ran a simple test. 

I applied 6 kPa pressure to my MPX5050GP by submerging a head tube 24" below the water surface 

in my water column.  A 24" water column applies 6 kPa pressure so the MXP5050 should read something

close to 6 kPa.

When I read the voltage off the sensor I get  0.583 Vdc.  I am using the following detail from the data sheet

to calculate the pressure P. 

Vout = Vs ( P x 0.018 + 0.04 )   +/- ( Pressure Error x Temp Factor x 0.018 x Vs )

0.583 = 5.0 ( P x 0.018 + 0.04 ) +/- ( 1.25 (max) x 1.0 x 0.018 x 5 )

0.583 = 0.09P + 0.2  +/- ( 0.1125 )

P = 4.26 kPa  +/- 0.1125 kPa

4.1431 kPa   <= P  <= 4.37 kPa

Since we know the actual pressure is 6 kPa , the sensor is off by  approx 2 kPa ! 

The more pressure I apply the larger the error. At 12.95 kPa the result is only 8.8 for an error of 4 kPa ! 

I can compensate for this in the software but the sensor itself seems way off. 

[darioarias]

Hi Tom,

You are right, for the part you are using the supply voltage is 5.0 V. I've noticed that you misinterpreted the formula a bit, the units for the Error = (1.25 x 1.0 x 0.018 x 5) is not kPa but Volts.The formula gives you the output voltage where the independent variable is the pressure Vout(P). It wouldn't be right is the formula gives you, let's say any example, 3.5 V +/- 0.1125 kPa; these units aren't compatible. If you want to use a formula where the independent variable is the output voltage you can use this one, (obtained by isolating P):

P(Vout) = (Vout / (Vs x 0.018)    -     0.04/0.018) +/- 1.25kPa

Using this one I get that: 3.01 kPa <= P <= 5.51 kPa

Since the actual pressure is 6 kPa, the sensor is off by 0.5 kPa, to make the sensor to be in range the  "Implementing Auto-Zero for integrated pressure sensors" technique can still help you correct the offset of 0.5 kPa.

For a pressure of 12.95 kPa you should have a voltage of 1.253 <= Vout <= 1.478. How much voltage do you have when applying 12.95 kPa?

Regards,

Darío

[tbalon]

The pressure error term in the formula to compute the sensor error is quoted in kPa, that's what made

me think the error term was in kPa, overall you show the error as +/- 1.25 kPa so I will use that. 

If I use your above calculations and plug in the voltage I get off the sensor at 6 kPa I get

the following

P(Vout) = (Vout / (Vs x 0.018)    -     0.04/0.018) +/- 1.25kPa

P(Vout) = (0.583/(5 x 0.018) -     0.04/0.018) +/- 1.25kPa

P(Vout) = (0.583/0.09)  -    2.22 +/- 1.25kPa   

P(Vout) = 6.48  -    2.22 +/- 1.25kPa  

P(Vout) = 4.26 kPa   +/- 1.25kPa  

I would be great if we always knew the actual pressure (6 kPa) but then we wouldn't need the sensor

to measure it. In this case the sensor reading indicates 4.26 kPa  and its off by 1.74 kPa, not 0.5 kPa. 

Regarding your question ( voltage at 12.95 kPa)

-----------------------------------------------------------------

Applying 12.95 kPa I read a voltage of 1.02 Volts. 

Using the same method I and a reading of 1.02 Volts

P(Vout) = (1.02/0.09)  -    2.22 +/- 1.25kPa   

P(Vout) = 11.33  -    2.22 +/- 1.25kPa   

P(Vout) = 9.12 +/- 1.25kPa   

Since in this case we know the pressure ahead of time is 12.95 and the sensor is reads 9.12 +/- 1.25 kPa

it is off by 3.83 kPa which is much larger than the error term of 1.25 kPa. 

Even if I could somehow know to add the error so that  the sensor reads 9.12 kPa + 1.25 = 10.37 kPa,

it's still off by 2.6 kPa. 

[darioarias]

Hi Tom,

This is very very strange, considering that you have 0.2V when the pressure is zero, meaning that you don't have any offset affecting the device.

In response to what you said: "It would be great if we always knew the actual pressure (6 kPa) but then we wouldn't need the sensor to measure it."

I absolutely agree with you, however if you want to properly calibrate the device, you need be certain on which is the pressure being applied to the MPX5050GP. So, this led me to the following questions: besides doing the calculation based on the diameter and length of the hose,  do you have any other device to accurately measure and compare the pressure like a manometer? have you followed the advice of implementing auto zero technique?

Regards

Darío

[tbalon]

Dario,

I was hoping that I could use a simple water column and the well established formula to 

convert inches of water to kPa.   6 kPa ~ 24 " of water  to calibrate the sensor.

I wasn't expecting this to be exceedingly accurate, I was expecting that if I connected the sensor to a tube

and then submerged the tube into 24" of water, the sensor would give me a value that was approx 6 kPa. 

What I noticed is that it was off by quite a bit, and got worse as the pressure increased. 

I'm using a very simple set-up. The length of the tube is approx 15 ft. and the diameter is on the MPX5050GP

spec sheet. This is static pressure so there is no volume of air moving in the tube, just static pressure so I

don't see how the diameter of the tube should matter. 

I have implemented an auto-zero and I am using a linear regression to correct the errors. Unfortunately I

do not have access to a manometer at the moment. 

Thanks for your help 

[darioarias]

Hello Tom,

In the datasheet, in section 2.2 Operating Characteristics, specifies a supply voltage of 3 V and a supply current of 7 mA. When you are using the device with 5 V and 1.5 A, it cannot be guaranteed that you'll have the accuracy of +/- 1.25 kPa, mainly because the accuracy of the device is given as the +/-2.5 %Vfss, Full-Scale Span Voltage, defined as the algebraic difference between the output voltage at full-rated pressure and the output voltage at the minimum rated pressure, typically of 2.7 V. With 5 V all the formulas and transfer function fail to reproduce accurately the relation between pressure and voltage.

Also when using the transfer function you have to consider the error, later in the datasheet described as (Pressure Error x Temp. Factor x 0.018 x Vs) = (1.25 kPa x 1 x 0.018 x Vs).

Finally you can try using the Auto zero referencing technique to help prevent this errors in the output, the following application note "Implementing Auto-Zero for integrated pressure sensors" would be a very helpful guide to do so.

Regards,

Darío

[tbalon]

Here is what I have on my MPX5050GP data sheet. It shows Vs = 5 Vdc not 3.0

Also from section 2 of the data sheet, supply voltage is listed as 5 Vdc not 3. 

[tbalon]

From the data sheet :

Pressure Error (Max)
0 to 50 (kPa) ±1.25 (kPa)

Which still doesn't explain the values I get using the formula. If I apply 6 kPa and measure the voltage

and plug that into the formula I get 4 kPa ( ~ 2kPa error ) , for a 13 kPa pressure its off by 4 kPa !