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KrisBlueNZ
Sadly passed away in 2015
Do you mean "potentiometer"?
Does your potentiometer have 11.1V on one end, and 0V on the other end? Is that what you're trying to say? It needs to be connected between some voltage and 0V, and its 0V connection needs to connect to the micro's 0V rail.
Then you use a voltage divider to reduce the wiper voltage range (which is presumably 0~11.1V) to the range you want at the micro's input.
See http://en.wikipedia.org/wiki/Voltage_divider for information about voltage dividers. In this application, the ratio of the top resistor to the bottom resistor needs to be 6.1:5.0, or 1.22:1.
For best linearity, your potentiometer should have a relatively low resistance compared to the resistors in the voltage divider. For example, a 1k potentiometer and voltage divider resistors of 120k (top) and 100k (bottom). If you don't do this, the current drawn by the voltage divider will interact with the resistance of the potentiometer and this will affect the linearity of the potentiometer.
You should also connect a small capacitor (e.g. 10 nF) from the micro's input to ground, to provide a low AC impedance into the ADC in the microcontroller. Otherwise the ADC conversions will be inaccurate and may jump around.
Your request is quite vague. Perhaps you could upload a circuit diagram of what you have, and a more detailed description of what you want to do.
Does your potentiometer have 11.1V on one end, and 0V on the other end? Is that what you're trying to say? It needs to be connected between some voltage and 0V, and its 0V connection needs to connect to the micro's 0V rail.
Then you use a voltage divider to reduce the wiper voltage range (which is presumably 0~11.1V) to the range you want at the micro's input.
See http://en.wikipedia.org/wiki/Voltage_divider for information about voltage dividers. In this application, the ratio of the top resistor to the bottom resistor needs to be 6.1:5.0, or 1.22:1.
For best linearity, your potentiometer should have a relatively low resistance compared to the resistors in the voltage divider. For example, a 1k potentiometer and voltage divider resistors of 120k (top) and 100k (bottom). If you don't do this, the current drawn by the voltage divider will interact with the resistance of the potentiometer and this will affect the linearity of the potentiometer.
You should also connect a small capacitor (e.g. 10 nF) from the micro's input to ground, to provide a low AC impedance into the ADC in the microcontroller. Otherwise the ADC conversions will be inaccurate and may jump around.
Your request is quite vague. Perhaps you could upload a circuit diagram of what you have, and a more detailed description of what you want to do.
If the 11.1V is just a supply for the pot, you clould use a 6v2 zener to drop the voltage dovn to 5V and then use the pot over 5V.
If th input voltage is from 0-11.1V and you want to have this represented as 0 - 5V on the input, you'll need to convert the voltage either with a voltage divider or an opamp.
How you do it depends on the output impedance of the voltage driver.
If th input voltage is from 0-11.1V and you want to have this represented as 0 - 5V on the input, you'll need to convert the voltage either with a voltage divider or an opamp.
How you do it depends on the output impedance of the voltage driver.
