Load Cell FAQ

What is load cell sensitivity?

Typical unamplified analog load cells have a sensitivity rating with a unit of mV/V. This load cell sensitivity rating is specified on a load cell data sheet often in the 1 mV/V – 3 mV/V range. Most quality load cells will come with a factory calibration sheet which specifies the actual sensitivity rating for that individual load cell. This load cell sensitivity value can be used to convert the load cell’s output to scientific units of weight or force.

What is excitation voltage?

Load cells require an excitation voltage to produce an output signal. This is directly related to the fundamental workings of the internal Wheatstone bridge. Load cell excitation values are often listed by a manufacturer as “recommended” and “maximum” excitation voltage. As their names imply, the recommended value is that which the manufacturer recommends for best output results, and the maximum value is that which should not be exceeded. See “Does load cell excitation voltage matter?” below for further pointers on selecting excitation voltage.

How to convert load cell output voltage to pounds or kilograms or Newtons

As mentioned above, each load cell should come with a calibration certificate denoting its sensitivity. This value can be used to convert the load cell’s output signal to units of weight or force. To calculate the raw output voltage of the cell relative to its rated full capacity, use the following equation:

    \[V_{out, max}=S V_e\]


where

V_{out, max}=\text{output voltage from load cell when loaded to 100\% of rated capacity (mV)} S = \text{sensitivity (mV/V)} V_e = \text{excitation voltage (V)}
Typically, load cells are used with amplifiers to transform the small output voltage to an easily measurable voltage (while conditioning the signal with filters, etc.). If we add an amplifier to the mix, we have:

    \[  V_{out, max}=S V_e A$ \]


where

V_{out, max}=\text{output voltage from load cell when loaded to 100\% of rated capacity (mV)} S = \text{sensitivity (mV/V)} V_e = \text{excitation voltage (V)} A = \text{amplifier gain (V/V)}
If we were to convert this output voltage to force or weight, we would use the ratio of the actual output voltage to the maximum output voltage which is equal to the ratio of the actual load to the maximum rated load:

    \[\displaystyle \frac{V_{out}}{V_{out, max}}=\frac{L}{L_{tot}}$ \]


or

    \[\displaystyle \frac{V_{out}}{S V_e A}=\frac{L}{L_{tot}}$ \]


Rearranging, we can solve for L:

    \[\displaystyle L=\frac{L_{tot}V_{out}}{S V_e A}$ \]


where

L = \text{load on cell (Kg, Lb, N, etc.)} L_{tot} = \text{total rated capacity of the load cell (Kg, Lb, N, etc.)} V_{out}=\text{output voltage from load cell when loaded (mV)} S = \text{sensitivity (mV/V)} V_e = \text{excitation voltage (V)} A = \text{amplifier gain (V/V)}

How to calculate the load cell sensitivity if a factory calibrated value is not given

If the load cell is not factory calibrated, it must be field calibrated to determine the sensitivity. This is done through the following steps:

  1. Record the no-load output voltage of the load cell,
  2. Load the cell with a known load,
  3. Record the output voltage with the known load.
  4. Subtract the zero output voltage from the known load voltage and use the resulting voltage as the calibration factor:

    \[\displaystyle S = \frac{L_{tot}(V_l-V_0)}{V_e L_{cal}A}$ \]


where

S = \text{load cell sensitivity (mV/V)} L_{tot} = \text{the total rated capacity of the cell (Kg, Lb, N, etc.)} V_l = \text{known load output voltage (mV)} V_0 = \text{no-load output voltage (mV)} V_e = \text{excitation voltage (V)} L_{cal} = \text{known load (Kg, Lb, N, etc.)} A = \text{amplifier gain (V/V)}
The units for the known load and total rated capacity should match and thus cancel out.

Does load cell excitation voltage matter?

As stated previously, load cells require an excitation voltage to produce an output signal. A higher excitation voltage will produce a higher output voltage swing when a load is applied to the cell. So, bigger is better, right?

Yes, to a degree. Larger signals are easier to measure and digitize. Additionally, assuming the noise is constant, the ratio of signal to noise (SNR) increases. This is good from a data quality standpoint.

However, a high excitation voltage has drawbacks. Higher voltages through the resistive strain gauges (which comprise the Wheatstone bridge) will cause more current to flow and heat the strain gauges. The cell body acts as a heat sink to keep the gauges cool. If the maximum rated excitation voltage is exceeded, the heating will cause signal perturbation or gauge failure. Additionally, in battery operated devices, high excitation voltage (and thus, current) will cause the battery to be depleted much faster than lower excitation voltages through the circuit.

Excitation voltages lower than the manufacturer’s stated maximum are acceptable for the given load cell. The manufacturer’s recommended value is obviously best, but there is no harm in a lower excitation voltage. For example, 5V is a very common excitation voltage for modern load cells. Modern instrumentation amplifiers are much better than old designs and their output signals are not compromised by the lower excitation voltage. It is perfectly fine to use a 5V excitation amplifier with a 10V recommended excitation load cell. (However, the converse is not true!)

Finally, when choosing the excitation voltage, consider the common mode voltage produced by the cell’s output. The amplifier or other signal conditioning electronics must be able to handle this common mode voltage, whose value is 50% of the excitation voltage. For example, a 10V excitation produces a 5V common mode voltage. Typically these amplifiers provide the excitation voltage to the cell; when this is the case, an external excitation voltage is not recommended.