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How should we select a load cell?

The common procedure is as follows:

1.       Decide on the optimal means to conduct the measurement and then determine the loading conditions of the load cells. For example tension, compression, bending, etc.

2.       The determination of the rated capacity is done by the following calculations:

The capacity of the load cells≥ ((Impact coefficient*Load+ initial load)*Load eccentricity coefficient * Load imbalance coefficient) / Number of load cells

Impact coefficient:  (1.1~1.5) Normally 1.2

Load eccentricity coefficient:  (1.1~1.3) Normally 1.2

Load imbalance coefficient:   (If 1 or 3 points:1, If 4 points:1.2)

Load:                                Maximum load of the measured object

Initial load:                       Own weight of the instrument

Number of the load cells:  Number of the load cells to be used

When the load is 3kg, the capacity of the load cell should be approx. 3.6kg. To measure 3kg, a load cell whose capacity is more than 3.6kg is required.

3.       See whether the desired resolution can be obtained

Please note: if the capacity is too large, it will be difficult to conduct minute measurements. But the strength of the load cell is also important. You also have to consider whether it can withstand sudden shocks. It is also necessary to select the appropriate indicator in accordance with the degree of precision of the load cell.

Below is an example of how to connect an indicator and a load cell.

Suppose we prepared a load cell whose capacity is larger than 3.6kg to measure 3kg. We would like to display the results using an indicator, but what kind of indicator should we use? The specification of the load cell and the indicator are as follows:


FAQ1

First, look at the rated output of the load cell, which is 1m V/V. This means that it will output 1mV when 1V is input. As the recommended excitation voltage is 12V, if we input 12V to the load cell, it will output up to 12Mv.

Let’s presume that the load cell with output approx. 12mV when 3kg is loaded.

                    Load 3kg----à12mV Output voltage

Suppose the indicator displays the weight in units of 0.1g between 0 and 3kg. This 0.1g is called the minimum scale division. As 3kg is divided by 30000 divisions (0.1g), the resolution is said to be 1/30000. Per 0.1g, the output will be:

12mV /3000=0.0004mV =0.4 uV

The input sensitivity of the indicator is 0.33uV; the indicator cannot display the weight in units of 0.1g.

The following graph shows the relationship between the load and the output voltage. This is an ideal result.

 FAQ2