Wiring 3.3V Sensors to 5V Analog Breakout...

[TD912]

Because we somehow managed to fry the KoP accelerometer, we decided to purchase another one (this time analog) from SparkFun electronics - the MMA7260Q (http://www.sparkfun.com/commerce/pro...roducts_id=252 )
We're trying to ensure that we don't short out this one, so we read the directions and other documenation before attempting to wire it to the analog breakout. We Just noticed the analog breakout outputs 5V of current, but the sensor calls for 3.3V...

Will it with with 5V? Did we get the wrong sensor? Is there a way to lessen the voltage?

[Al Skierkiewicz]

Chris,
You cannot use a 3.3 volt device with a 5 volt power supply. The good news is there are three terminal regulators that will take the 5 volt power and regulate down to 3.3 volts. You are looking for one that is defined as a "low dropout regulator". There is some in the Digikey catalog but I bet Mouser and others have some to.

[billbo911]

Originally Posted by TD912

Because we somehow managed to fry the KoP accelerometer, we decided to purchase another one (this time analog) from SparkFun electronics - the MMA7260Q (http://www.sparkfun.com/commerce/pro...roducts_id=252 )
We're trying to ensure that we don't short out this one, so we read the directions and other documenation before attempting to wire it to the analog breakout. We Just noticed the analog breakout outputs 5V of current, but the sensor calls for 3.3V...

Will it with with 5V? Did we get the wrong sensor? Is there a way to lessen the voltage?

Since you already have a working relationship with Sparkfun, one of my favorite sites, try using this regulator from them. It will more than handle your needs.
You could also use this one, but it will require a small bit of support circuitry.

[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).

[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.)

[EricVanWyk]

Originally Posted by RyanN

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

As Bill said, this idea only works in theory. In practice, it is likely to end in tears. I've been down that road many times, each time thinking "this is the time I can get away with it!". That time has never come.

It is bad for two reasons:
1) Inaccurate voltage.
just as Bill said, your output voltage is dependent on your load. Your load is unknown AND variable. Since your output is ratiometric ( the signal is expressed as a fraction of the input voltage ), this is very very very bad.
2) High output impedence.
Output impedence is a measure of the effective resistance of a power rail. Usually it is (much) less than an ohm. This technique would place it in the thousands of ohms. It is an important measure for many reasons, but the most relevant reason here is the ability to deal with switching. (( Bypass capacitors are used to lower it further (for high frequencies))).

You may think that an analog gyro is an analog device, but internally they are chopping in the high tens of kilohertz. A poor output impedence will interfere with this chopping and do "funky things". I'm not positive what, but I'd imagine that you'd end up with a weird offset error.


Bottom Line: Use a regulator. They are cheap and good.

[JCharlton]

Um, can you send it back and get one that's 5V?