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Hi all,
Im trying to read analog values from ADXL330 3 axis accelerometer but first i have to power it with 3V while Arduino is 5v, can i step voltage down using a simple voltage divider? or i would need a regulator to do the job?
Help will be greatly apreciated!
Thanks

[Reply #1 on:]

Hi VictorH,

You can use a voltage divider, but a low-dropout regulator like the L4931 series is better - you'll need to add two capacitors as described in the datasheet.  Also, if you connect the 3V to your ADC reference, you can get full resolution on the 3V analog output.

D.

[Reply #2 on:]

This might be a stupid question, but why couldn't you just put a 6.8 kOhm resisitor in between the 5V source and the ADXLs VDD pin?  At 3V the ADXL draws roughly 0.3 mA, so a resistance of R = U/I = (5-3)/0.0003 = 6.67 kOhm should bring the 5V down to 3V.  The chosen resistor then needs to be able to deal with a power of P = U * I = (5-3) * 0.0003 = 0.3 mW.  Am I missing something here or is it indeed easiest and safe to just use a 6.8 kOhm resistor in series in front of the ADXL?

Thanks!
-Jonas

[Reply #3 on:]

Newer arduinos have a small 3.3V supply available.

If that isn't available on your arduino, I'd get a small LDO regulator and feed it from the 5V line.

The problem with an inline resistor is that the voltage drop depends on the actual current draw, which can (and probably does) vary.  Likewise for the divider network.  This is particularly bad because a) you can end up with too much or too little voltage for correct and safe operation, and b) this is an analog output device whose output depends on the input voltage (as well as the sensor).

You need a regulator (or tap the 3.3V line on a newer arduino).

-j

[Reply #4 on:]

Thanks, will get a LDO regulator then.

[Reply #5 on:]

Sorry to exhume an old topic, but I had the same issue and instead of using an LDO regulator, I just used a red LED in series since it drops 1.7 volts.  On the breadboard it works.  Anyone see any problem with this solution?

[Reply #6 on:]

If your device takes 40mA, for example, this is the current going through the LED as well. It won't last very long.

[Reply #7 on:]

Anyone see any problem with this solution?

As well as the above.
Also LEDs are very noisy, so you will need extra decoupling.
Look at the data sheet for the LED and see the spread of turn on voltage you have.
Another point is that the forward voltage drop of an LED is not so stable with temperature so you could be over or under voltaging your device under temperature extremes.