simulate this circuit – Schematic created using CircuitLab
Vo = V1* R2 /(R1+R2)
Total power Po = V1*V1/(R1+R2)
Adc unit (5V reference) = 5V/1023 ~= 5mV
Adc unit (1.1 internal reference) = 1.1/1023 ~= 1mV
No pressure R1 = 470 ohm:
- Vo = 5*50/520 = 480mV ~= 96(5V) or 480(1.1V) ADC units
- Po = 25/520 = 48mW
No pressure, R1 = 10 kohm:
- Vo = 5*50/10050 = 25mV ~= 5(5V) or 25(1.1V) ADC units.
- Po = 25/10050 = 2.4mW
No pressure, R1 = 10 ohm:
- Vo = 5*50/60 = 4.16V ~= 803(5V) ADC units.
- Po = 25/60 = 0.41W
Same calculation for pressure applied until the sensor has 10 ohms:
R1 = 10 ohm:
- Vo = 5*10/20 = 2.5V ~= 500(5V) ADC units.
- Po = 25/20 = 1.25W, that's why the resistor is heating
It's better to put the sensor on the ground side because you can use 0 to no pressure voltage instead no pressure voltage to 5V then subtract the no pressure voltage. You also can use a lower reference for better resolution which is not possible with the sensor on the upper side since V0 will be higher than the reference voltage.
By using the internal reference you will lose the ratiometric output but the gain in resolution will compensate this.