User Manual

Voltage to acceleration example:
”The X output reads 2.96V. What acceleration does this correspond to?”
The 0g point is approximately Vcc / 2 = 3.33 / 2 = 1.66V
2.96V - 1.66V = +1.20V with respect to the 0g point
Sensitivity is 222mV/g, 1.2 / 0.312 = 5.41g
Therefore the acceleration in the X direction is 5.41g
Acceleration to voltage example:
”What voltage will correspond to an acceleration of -0.5g?”
The 0g point is approximately 1.66V
Sensitivity is 222mV/g, -0.5 * 0.222 = -0.111V with respect to the 0g point.
1.66V – 0.111V = 1.55V
Therefore you can expect a voltage of approximately 1.55V when experiencing an acceleration of -0.5g.
Voltage to tilt example:
“With the accelerometer oriented flat and parallel to ground in my robot, Yout is 1.66V. When my robot goes
uphill, Yout increases to 1.765V. What is the slope of the hill?”
1.765V – 1.66 = +0.105V with respect to the 0g point.
With a sensitivity of 222mV/g, 0.105 / 0.222 = 0.471g
Sin
-1
(0.471) = 28.1º
The slope of the hill is 28.1º in the Y axis
Tilt to voltage example:
“I am making an antitheft device that will sound an alarm if it is tilted more than 10º with respect to ground in
any direction. I have measured the 0g bias point to be 1.663V, and I want to know what voltage to trigger the
alarm at.”
Sin(10º) = 0.1736 so acceleration with a tilt of 10º will be 0.1736g
0.1736g * 0.222V/g = 0.0385V with respect to the 0g point
1.663 + 0.0385 = 1.7015V
1.663 – 0.0385 = 1.6245V
Sound the alarm when the voltage reaches more than 1.7015V or less than 1.6245V.