I am trying to calculate the resolution of my simple force sensor system in Newton [N].

The sensor system consists of a simple voltage divider, where the voltage across the FSR is measured by an 8-bit ADC.

I have calibrated it using a force push-pull gauge, in increments of 1 N in a range from 1-10N, noting down the ADC output, see table.

According to this link, the resolution would be calculated as follows:

Resolution = (V_ref * FS_input) / (V_fso * ADC_resolution)

Looking at the graph, ignoring the fact that my transfer function is non-linear and thus the sensitivity degrades with higher forces and thus the resolution with it, the theoretical resolution of my system assuming a linear voltage change to force input, would be:

FS_input = 10N-1N = 9N ---> because i measured from 1-10 Newtons [N]

V_fso = 2.462V - 1.702V = 0.76V ---> this is the output voltage [V] range for the input force range of 1-10 Newtons [N]

V_ref = 3.3V ---> this is the ADC reference voltage and the voltage across the whole voltage divider

ADC_resolution = 2^n = 2^8 = 256

Resolution = (3.3V * 9N) / (0.76V * 256) = 0.153 N

Meaning the smallest detectable change in force [N] with my 8-bit ADC would be 0.153N.

Is this the correct way to calculate the resolution?

Voltage divider




1 Answer 1


Your answer is right, but the way you got there is very complicated, and I don't really follow it. From the highest measurement to the lowest measurement is 191 - 132 = 59 counts, and that covers a range of
10 - 1 = 9 Newtons, so the average resolution over that range is 9 / 59 ~ 0.153 N / count.

The curve-fit you did also looks fine, though again, going via voltage instead of directly relating force to ADC count only serves to complicate things.

  • \$\begingroup\$ The transfer function is very non-linear so it's better not to take the average. Divide the steps in the force column (all 1s) by the steps in the count column. This gives a resolution of 0.07 N between 1 and 2 N vs. a resolution of 0.5 N between 9 and 10 N. \$\endgroup\$
    – DamienD
    Sep 4, 2021 at 7:02
  • \$\begingroup\$ @DamienD I entirely agree. In fact I had that in my answer, but I decided to eliminate clutter. \$\endgroup\$
    – hobbs
    Sep 4, 2021 at 8:06
  • \$\begingroup\$ IMHO it's not clutter and would improve your answer a lot -- there's a factor of seven difference between the ends of the scale. \$\endgroup\$
    – DamienD
    Sep 4, 2021 at 8:49

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