# Arduino Pro Mini Reliability, and related questions

Discussion in 'Microcontrollers, Programming and IoT' started by 4x4_e30, Apr 12, 2016.

1. ### 4x4_e30

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Apr 17, 2015
I've been looking into using an Arduino Pro Mini to modify an analog signal. I'm replacing an old wiper style potentiometer with a hall effect sensor. The old setup output starts at 0.3vdc and goes up to 4.5vdc at 90 degrees (linear), whereas the hall sensor I'm looking at using has an output that starts at 0.5vdc and ends up at 4.5vdc (also linear). Since the potentiometer is acting as a sensor, the voltage output is crucial and I'm trying to stay away from reprogramming the computer. Unless there is another way I can modify the output of the hall sensor without using a microcontroller??

The whole point is to make something rock solid reliable, as in function continuously for hours on end on a daily basis. I already have regulated 5vdc going to the old potentiometer setup, and all of my research has told me the only reliability problem with the Pro Mini is trying to supply it with 12v and relying on the built in voltage regulator.

Also if the computer needs more current than the arduino can supply, does anyone know where I can look at how to amplify the current of a PWM output? I'm assuming I'll have to add a capacitor to smooth the PWM output to true analog as well...

Thanks in advance for the help. I'm trying not to be one of those newbs that just wants an easy answer, so if you guys have any material I can look up, that would be awesome. I'd rather have a fundamental understanding of how everything works than just copying what other people tell me.

2. ### Harald KappModeratorModerator

11,653
2,694
Nov 17, 2011
Welcome to EP.

To get from 0.5V...4.5V to 0.3V...4.5V the transfer function is y=1.05*x-0.225.

You can achive this by an analog circuit using an amplifier with gain 1.05 an d subtracting a constant value of 0.225V.

You will need a rail-to-rail opamp for this to work.

3. ### 4x4_e30

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Apr 17, 2015
Oh my.. Thank you, thank you, and thank you again. I don't understand a single part of what you just posted, but now I know what to research. Once I have a better understanding, I'll be back.

Edit: For anyone else trying to figure this out, here is where I'm starting:

http://www.analog.com/library/analogDialogue/archives/39-05/Web_Ch1_final_R.pdf

It's starting to look like I'll need a bachelors degree to get this...

Last edited: Apr 12, 2016
4. ### Harald KappModeratorModerator

11,653
2,694
Nov 17, 2011
That's a good link you have there.

The op-amp together with R1,R2 has a gain of 1.05 to realize the term x*1.05.
The constant voltage source 4.5V provides the required offset. By connecting it to the resistive divider in the feedback loop of the op-amp this constant voltage shifts the output by the required amount to realize the term -0.225V

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Apr 17, 2015
6. ### 4x4_e30

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Apr 17, 2015
Last edited: Apr 13, 2016