[billbo911]

Originally Posted by RyanN

Any reason not to use a resistor connected to the 5V Analog Breakout?

Let's do the math.

V=IR
R=V/I

Voltage drop needed across the resistor will be 5V - 3.3V = 1.7V

From the data sheet, it uses 500μA typical. I'm not sure how the sleep current will work out.

So..
R = V/I
R = 1.7V/500μA
R = 1.7/500x10^-6
R = 1.7/0.0005
R = 3400Ω = 3.4kΩ

Now, the accelerometer has a bit of flexibility. You should be good with any resistor between:

V = 5V-2.2V = 2.8V
R = 2.8V/500μA = 5.6kΩ

and

V= 5V-3.6V = 1.4V
R = 1.4V=500μA = 2.8kΩ

I would stay away from either extreme, preferably lower since I'm not sure what will happen with the sleep current.

Before doing this though, let some of the more experience engineers take a look at what I provided before going with it. You could get a 3.3V voltage regulator, and it will work fine. Resistors seemed simpler and easier to acquire (@ RadioShack or in your shop).

Ryan, your logic is sound, but there may be other things to consider. The one that pops to mind is, how is the output of this device affected by varying input voltage. If the device is stable across the input range, then your process will work. Now, if the output varies much at all with varying supply voltage, then I would go for a clean requlator.


Here is a quote from the spec sheet for this device

Within the supply range of 2.2 and 3.6 V, the device operates as a fully calibrated linear accelerometer. Beyond these supply limits the device
may operate as a linear device but is not guaranteed to be in calibration.

So, I would say it should work.

( Honestly, you can pick up a 3.3v regulator at Radio Shack for a couple bucks and just be more confident.)