User Manual

Voltage to acceleration example:
”The X output reads 2.14V. What acceleration does this correspond to?”
The 0g point is approximately Vcc / 2 = 3.3 / 2 = 1.65V
2.14V – 1.65V = +0.49V with respect to the 0g point
If sensitivity is 660mV/g, 0.49 / 0.660 = 0.742g
Therefore the acceleration in the X direction is +0.742g
Acceleration to voltage example:
”What voltage will correspond to an acceleration of -0.5g?”
The 0g point is approximately 1.65V
If sensitivity is 660mV/g, -0.5 * 0.660 = -0.33V with respect to the 0g point.
1.65V – 0.33V = 1.32V
Therefore you can expect a voltage of approximately 1.32V 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 1.650V. When my robot goes
uphill, Yout increases to 1.779V. What is the slope of the hill?”
1.779V – 1.650 = +0.129V with respect to the 0g point.
With a sensitivity of 660mV/g, 0.129 / 0.660 = 0.195g
Sin
-1
(0.195) = 11.2º
The slope of the hill is 11.2º 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 30º with respect to ground in
any direction. I have measured the 0g bias point to be 1.682V, and I want to know what voltage to trigger the
alarm at.”
Sin(30º) = 0.5 so acceleration with a tilt of 30º will be 0.5g
0.5g * 0.660V/g = 0.330V with respect to the 0g point
1.682 + 0.330 = 2.012V
1.682 – 0.330 = 1.352V
Sound the alarm when the voltage reaches more than 2.012V or less than 1.352V.