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MEMS Acceleromter

lhagva99
Associate

Hello,

I am looking for MEMS accelerometer for seismic application.

We need low noise, up to 100Hz bandwidth, full scale +-2g range.

I consider IIS3DHHC as a candidate, which has features as below:

  • 3-axis, ±2.5 g full-scale
  • Ultra-low noise performance: 45 µg/√Hz
  • 16-bit data output

I would like to clarify what is the effective number of bits for 16bit output?

How does the Sensitivity 0.076mg/LSB relates to the noise performance?

Is there any other products with lower noise?

Best regards,

Lhagva

3 REPLIES 3
Federica Bossi
ST Employee

Hi @lhagva99 ,

The sensitivity of the IIS3DHHC accelerometer is 0.076 mg/LSB. This means that each least significant bit (LSB) of the digital output corresponds to a change of 0.076 mg in acceleration.

You could also look at LIS2DW12

In order to give better visibility on the answered topics, please click on 'Accept as Solution' on the reply which solved your issue or answered your question.

Hello Federica,

Thank you for your reply.

Yes I know (LSB) of the digital output corresponds to a change of 0.076 mg (76ug) in acceleration.

Data sheet specifies noise density as 45μg/√(Hz).

So for bandwidth of 100Hz it will make noise of 450ug.

450ug/76ug=5.92bits, I think almost 6 lower bits occupied by noise.

I think effective number of bits will be 12 bits.

Am I right? Please advice me. 

Is there any other MEMS accelerometer with lower noise density?

Best regards,

Lhagva  

 

Federica Bossi
ST Employee

Hi @lhagva99 ,

To get the noise density value reported in DS I suggest you do a PSD.

Alternatively you can calculate the noise RMS (std on 10s test) and divide the obtained value by the square root of the BW, which in this case is 440Hz.

This is our MEMS accelerometer with lower noise density, you can consider LIS2DW12 if you want a good trade off between Noise and Current Consumption.

In order to give better visibility on the answered topics, please click on 'Accept as Solution' on the reply which solved your issue or answered your question.