## What is load cell sensitivity?

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

## 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 the sensitivity for each individual cell. This value can be used to convert the load cell’s output signal to usable units of weight or force. To calculate the raw output voltage of the cell relative to the full capacity of the cell, use the following equation:

$V_{out, max}=S V_e$
where

$V_{out, max}=\text{output voltage from load cell when loaded to 100\% of rated capacity (mV)}$ $S = \text{sensitivity (mV/V)}$ $V_e = \text{excitation voltage (V)}$
In other words, the above result will be the voltage output from the cell when fully loaded relative to the full rated capacity.

Typically, load cells are used with amplifiers to transform the small output voltage to a 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

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

If the load cell is not factory calibrated, it must be field calibrated to determine the sensitivity. To do so, record the no-load output value, then load the cell with a known load and record the output. Subtract the zero output 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

$S = \text{load cell sensitivity (mV/V)}$ $L_{tot} = \text{the total rated capacity of the cell (Kg, Lb, N, etc.)}$ $V_l = \text{loaded output voltage (mV)}$ $V_0 = \text{unloaded output voltage (mV)}$ $V_e = \text{excitation voltage (V)}$ $L_{cal} = \text{known load (Kg, Lb, N, etc.)}$ $A = \text{amplifier gain (V/V)}$ Note: The units for the known load and total rated capacity should match and thus cancel out.

## Does load cell excitation voltage matter?

Load cells require an excitation voltage to produce an output signal. This is directly related to the fundamental workings of the internal Wheatstone bridge. In sort, 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 also good from a data quality standpoint.

However, 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 circuits.

Load cell excitation vales are often listed as recommend and maximum excitation voltage. As long as your excitation voltage is not higher than the maximum, you will be fine. The recommended value from the cell manufacturer is a good rule of thumb but there is no harm in a lower excitation voltage. eg. 5v is a very common (modern) excitation voltage. Modern instrumentation amplifiers are much better than old designs and do not suffer from the lower excitation voltage. It is perfectly fine to use a 5V excitation amplifier with a 10V (recommended) excitation load cell. Just not the other way round.

Finally, be sure your amplifier or other signal conditioning electronics can handle the common mode voltage produced by the cell’s output relative to the excitation voltage. This value is 50% of the excitation voltage. eg. 10V excitation produces a 5V common mode voltage. Most commonly, amplifiers will provide the excitation voltage to the cell. When this is the case, you should never provide your own excitation voltage. Or at the very least never exceed the excitation voltage provided by the amplifier.