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mpred
Contributor I

Hello!

I'm studying MPX2200GP and I have a problem.

Without pressure, the differential voltage is output by about 20 mv.

The manual says 0mV. Why is that?

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TomasVaverka
NXP TechSupport
NXP TechSupport

Hi  Yenhaeng,

MPX2200AP means that you are using absolute type and not gauge (GP) as you wrote earlier.

 

When using MPX2200AP, it is expected to have about 20mV on the output as the ambient local pressure is around 100kPa absolute. Then considering the sensitivity of 0.2mV/kPa you will get 20mV.

 

Only for D/DP/GP variants the output voltage should be between -1mV and +1mV when no pressure is applied.

Best regards,

Tomas

PS: If my answer helps to solve your question, please mark it as "Correct" or “Helpful”. Thank you.

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mpred
Contributor I

It was my mistake.

I didn't know about absolute pressure and gauge pressure.

I'm gonna need to replace it with a GP type.

Thank you for your kind response.

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TomasVaverka
NXP TechSupport
NXP TechSupport

Hi  Yenhaeng,

It is hard to say where the problem might be.

 

How many sensors do you have in total and on how many of them you are observing the offset being out of specification?

Can you please share your complete schematic to make sure there is not anything influencing the sensor’s output?

What is the complete marking found on the top side of the sensor?

Best regards,

Tomas

 

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mpred
Contributor I

Hi!  Tomas.

Thank you for your reply.

The voltage used is 12v and the voltage is only applied without any connection.

The markings for the sensor are MPX2200AP, KGL1716A.

I'll wait for your opinion.

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368 Views
TomasVaverka
NXP TechSupport
NXP TechSupport

Hi  Yenhaeng,

MPX2200AP means that you are using absolute type and not gauge (GP) as you wrote earlier.

 

When using MPX2200AP, it is expected to have about 20mV on the output as the ambient local pressure is around 100kPa absolute. Then considering the sensitivity of 0.2mV/kPa you will get 20mV.

 

Only for D/DP/GP variants the output voltage should be between -1mV and +1mV when no pressure is applied.

Best regards,

Tomas

PS: If my answer helps to solve your question, please mark it as "Correct" or “Helpful”. Thank you.

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